10 JEE MAIN CHAPTERWISE EXPLORER
10. (d) : Method I : Let z1 = cosq 1 + i sinq 1, Þ w6 ( –8) + (8) = 0 Þ 0 = 0
z 2 = cosq 2 + isinq 2 \ – 1, 1 – 2w, 1 – 2w2 are roots of
(x – 1)3 + 8 = 0 but on the other hand the other roots does
\ z1 + z2 = (cosq 1 + cosq 2) + i(sinq 1 + sinq 2) not satisfies the equation (x – 1)3 + 8 = 0.
Now |z 1 + z2 | = |z 1| + |z2 |
13 (a) : z + iw = 0
Þ (cosq1 + cos q2 )2 + (sin q1 + sin q2 )2 = 1+1 Þ z = -iw Þ z = iw
Þ 2(1 + cos(q 1 – q 2) ) = 4 (by squaring) Þ w = –iz \ arg (–iz2 ) = p
Þ arg (–i) + 2arg (z)= p
Þ cos(q 1 – q 2) = 1 Þ q 1 – q 2 = 0 (Q cos0° = 1)
Þ 2arg(z) = p + p/2 = 3p/2
Þ Arg z1 – Arg z2 = 0.
arg(z) = 3p/4
3 z 3 z
11. (d) : Given w = 3 z - i \ w = 3 z - i 14. (d) : z1 /3 = p + iq
Þ x – iy = (p + iq) 3
Þ 3z - i = 3 z Þ x – iy = p3 – 3pq2 + i(3p2 q – q 3)
Þ x = p 3 – 3pq 2 and y = –(3p 2q – q3 )
Þ 3(x) + i(3 y - 1) = 3(x + iy) ( z = x + iy) (*)
x = p2 – 3q 2 and y = –(3p 2 – q 2 )
Þ (3x) 2 + (3y – 1)2 = 9(x 2 + y 2) Þ 6y – 1 = 0 which is straight p q
line.
Adding the equations of (*) we get
12. (d) : Method (1) : (By making the equation from the given
roots) x + y = –2(p2 + q 2)
Let us consider x = –1, –1, –1 pq
\ Required equation from given roots is
(x + 1)(x + 1)(x + 1) = 0 15. (b) : |z 2 – 1| = |z|2 + 1 x = 0 Imaginary axis
(x + 1) 3 = 0 which does not match with the given equation
(x – 1) 3 + 8 = 0 so x = –1, –1, –1 cannot be the proper choice. Þ Let z = x + iy For real axis y = 0
Again consider x = –1, –1 + 2w, –1 –2w2 Þ (x – 1) 2 + y2
\ Required equation from given roots is
Þ (x + 1)(x + 1 – 2w)(x + 1 + 2w2 ) = 0 =(x2 + y 2) + 1
Þ (x + 1)[(x + 1)2 + (x + 1)(2w2 – 2w) – 4w3 ] = 0
Þ (x + 1)[(x + 1) 2 + 2(x + 1)(w 2 – w) – 4 ] = 0 Þ 2x = 0
Þ (x + 1) 3 + 2(x + 1)2 ( w 2 – w) – 4(x + 1) = 0 x = 0
which cannot be expressed in the form of given equation (x –
1) 3 + 8 = 0. Now consider the roots Þ z lies on imaginary axis.
x i = –1, 1 – 2w, 1 – 2w2 (i = 1, 2, 3)
and the equation with these roots is given by 16. (d) : Given æ 1 + i ö x = 1
x3 – (sum of the roots)x2 + x(Product of roots taken two at a time) çè 1 - i ÷ø
– Product of roots taken all at a time = 0
Now sum of roots x1 + x2 + x3 æ 2i öx
= –1 + 1 – 2w + 1 – 2w2 = 3 Þ çè 2 ÷ø = 1
Product of roots taken two at a time
= –1 + 2w – 1 + 2w2 + 1 + 2(w 2 + w) + 4w3 = 3 Þ ix = 1 Þ ix = (i) 4n
Product of roots taken all at a time
= (–1)[(1 – 2w)(1 – 2w2 ) ] = –7 Þ x = 4n, n Î I+
\ Required equation is x 3 – 3x2 + 3x + 7 = 0
Þ x 3 – 3x 2 + 3x – 1 + 8 = 0 Þ (x – 1)3 + 8 = 0 which matched 17. (c) : | zw | = 1 Þ | z || w | = 1 So | z | = | 1 ... (1)
with given equation. w |
Method 2 (by taking cross checking) Again Arg (z) – Arg (w) = p
As (x – 1) 3 + 8 = 0 ...(*) 2
and x = –1 satisfies (x – 1)3 + 8 = 0
i.e. (–2) 3 + 8 = 0 Þ 0 = 0 \ z = z i = z 2 i from (1)
Similarly for 1 – 2w we have (x – 1)3 + 8 = 0 w w
Þ (1 – 2w – 1)3 + 8 = 0
Þ (–2w)3 + 8 = 0 Þ –8 + 8 = 0 and for 1 – 2w2 \ z = zzi Þ z w = 1 = - i .
we have (1 – 2w2 – 1)3 + 8 = 0 w i
18. (b) : As z1 , z2 are roots of z 2 + az + b = 0
\ z 1 + z2 = –a, z1 z 2 = b
Again 0, z 1 , z2 are vertices of an equilateral triangle
0
z1 z2
\ 02 + z 12 + z2 2 = 0z 1 + z 1 z 2 + z2 0 = 0
z 1 2 + z2 2 = z1 z 2
Þ (z 1 + z2 ) 2 = 3z1 z2
a 2 = 3b
Complex Numbers 11
19. (b) : Let |z| = |w| = r 21. (b) : r z 3
\ z = re ia and w = re ib
where a + b = p (given) a
Now Z- = re ia = re i(p – b) z 1 b
= re ip × e –ib
= – re– ib z 2
= – w
z1 z 3 – z3 z 2 = (a + r) – (b + r)
20. (c) : |z – 4| < |z – 2|
or |a – 4 + ib| < |(a – 2) + ib| by taking z = a+ib = a - b = a constant, which represent a hyperbola
Þ (a – 4) 2 + b2 < (a – 2)2 + b2 Since, A hyperbola is the locus of a point which moves in such
Þ –8a + 4a < –16 + 4 a way that the difference of its distances from two fixed points
Þ 4a > 12 Þ a > 3 (foci) is always constant.
Þ Re(z) > 3
12 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER MATRICES
AND DETERMINANTS
3
é1 a 3 ù 7. Consider the system of linear equations
1. If P = êê1 3 3 úú is the adjoint of a 3 × 3 matrix A and x1 + 2x2 + x 3 = 3
êë2 4 4ûú 2x1 + 3x2 + x 3 = 3
|A| = 4, then a is equal to 3x1 + 5x 2 + 2x 3 = 1
(a) 11 (b) 5 (c) 0 (d) 4 (2013) The system has
2. The number of values of k, for which the system of equations (a) infinite number of solutions
(k + 1)x + 8y = 4k, kx + (k + 3)y = 3k – 1, has no solution, is (b) exactly 3 solutions
(a) 1 (b) 2 (c) 3 (d) infinite (2013) (c) a unique solution (2010)
(d) no solution
æ1 0 0 ö 8. The number of 3 × 3 non‑singular matrices, with four entries
as 1 and all other entries as 0, is
3. Let A = ç 2 1 0 ÷÷ . If u 1 and u 2 are column matrices such
ç
è 3 2 1 ø
(a) less than 4 (b) 5
æ1ö æ0 ö (c) 6 (d) at least 7 (2010)
that Au1 = ç 0÷÷ and Au 2 = ç 1 ÷÷ , then u 1 + u2 is equal to 9. Let A be a 2 × 2 matrix with non‑zero entries and let
ç ç A2 = I, where I is 2 × 2 identity matrix. Define Tr(A) = sum
è0ø è0 ø of diagonal elements of A and |A| = determinant of
matrix A.
æ -1ö æ 1 ö æ -1ö æ -1ö
Statement‑1 : Tr(A) = 0.
(a) ç -1÷÷ (b) ç -1÷÷ (c) ç 1 ÷÷ (d) ç 1 ÷÷ ( 2 0 1 2 )
ç ç ç ç
è 0 ø è -1ø è 0 ø è -1ø
4. Let P and Q be 3 × 3 matrices with P ¹ Q. If P 3 = Q 3 and Statement‑2 : |A| = 1.
P2 Q = Q 2P, then determinant of (P 2 + Q2 ) is equal to (a) Statement1 is true, Statement2 is true; Statement2 is a
(a) 0 (b) – 1 (c) – 2 (d) 1 (2012) correct explanation for Statement1.
5. Let A and B be two symmetric matrices of order 3. (b) Statement1 is true, Statement2 is true; Statement2 is
Statement‑1 : A(BA) and (AB)A are symmetric matrices. not a correct explanation for Statement1
Statement 2 : AB is symmetric matrix if matrix multiplication
of A with B is commutative. (c) Statement1 is true, Statement2 is false.
(a) Statement1 is true, Statement2 is false. (d) Statement1 is false, Statement2 is true.
(2010)
(b) Statement1 is false, Statement2 is true. 10. Let A be a 2 × 2 matrix
(c) Statement1 is true, Statement2 is true; Statement2 is Statement1: adj (adj A) = A
a correct explanation for Statement1. Statem ent2 : |adj A| = |A|
(a) Statement1 is true, Statement2 is true; Statement2 is
(d) Statement1 is true, Statement2 is true; Statement2 is
not a correct explanation for Statement1. (2011) not a correct explanation for Statement1
6. The number of values of k for which the linear equations (b) Statement1 is true, Statement2 is false.
4x + ky + 2z = 0 (c) Statement1 is false, Statement2 is true.
kx + 4y + z = 0 (d) Statement1 is true, Statement2 is true; Statement2 is a
2x + 2y + z = 0 correct explanation for Statement1.
possess a non‑zero solution is (2009)
(a) 1 (b) zero (c) 3 (d) 2 (2011) 11. Let a, b, c be such that b(a + c) ¹ 0. If
Matrices and Determinants 13
a a +1 a -1 a +1 b +1 c -1 18. Let æ1 2ö æ a b0 ÷öø , a, b Î N. Then
A = ç ÷ and B = ç
-b b +1 b - 1 + a -1 b -1 c +1 = 0, è 3 4 ø è 0
c c - 1 c +1 (-1)n+2 a (-1)n+1b (- 1) n c (a) there cannot exist any B such that AB = BA
then the value of n is (b) there exist more than one but finite number B’s such that
(a) any even integer (b) any odd integer AB = BA
(c) any integer (d) zero (2009) (c) there exists exactly one B such that AB = BA
(d) there exist infinitely many B’s such that AB = BA.
12. Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 (2006)
identity matrix. Denote by tr(A), the sum of diagonal entries
of A. Assume that A 2 = I. 19. If A2 – A + I = 0, then the inverse of A is
Statement1 : If A ¹ I and A ¹ –I,then det A =–1.
Statement2 : If A ¹ I and A ¹ – I, then tr(A) ¹ 0. (a) A (b) A + I (c) I – A (d) A – I. (2005)
(a) Statement1 is true, Statement2 is false
(b) Statemen1 is false, Statement2 is true 20. If A = é1 0 ù é1 0 ù
(c) Statement1 is true, Statement2 is true; Statement2 is êë1 1 úû and I = êë0 1 úû , then which one of the
a correct explanation for Statement1
(d) Statement1 is true, Statement2 is true; Statement2 is following holds for all n ³ 1, by the principle of mathematical
not a correct explanation for Statement1 (2008) induction
(a) A n = 2 n – 1 A – (n – 1)I
(b) An = nA – (n – 1)I
(c) A n = 2 n – 1A + (n – 1)I
13. Let a, b, c be any real numbers. Suppose that there are real (d) An = nA + (n – 1)I. (2005)
numbers x, y, z not all zero such that x = cy + bz, y = az + 21. If a 2 + b 2 + c 2 = – 2 and
1 + a2 x (1 + b2 ) x
cx and z = bx + ay. Then a 2 + b 2 + c 2 + 2abc is equal to (1 + c2 ) x
f ( x) = (1+ a2 ) x 1 + b2 x (1 + c2 ) x
(a) 1 (b) 2 (c) – 1 (d) 0 (2008) (1+ a2 ) x (1 + b2 ) x 1 + c2 x
14. Let A be a square matrix all of whose entries are integers. then f (x) is a polynomial of degree
Then which one of the following is true? (a) 0 (b) 1 (c) 2 (d) 3. (2005)
(a) If det A = ± 1, then A –1 need not exist
(b) If det A = ± 1, then A –1 exists but all its entries are not 22. The system of equations ax + y + z = a – 1,
necessarily integers x + ay + z = a – 1, x + y + az = a – 1 has no solutions,
(c) If det A ¹ ± 1, then A –1 exists and all its entries are non
if a is
integers
(d) If det A = ± 1, then A –1 exists and all its entries are integers (a) either –2 or 1 (b) –2
(2008) (c) 1 (d) not –2. (2005)
1 1 1 æ 0 0 -1 ö
15. If D = 1 1 + x 1 for x ¹ 0, y ¹ 0 then D is ç ÷÷÷ø .
23. Let A = ççè 0 -1 0 The only correct statement about the
1 1 1 + y -1 0 0
(a) divisible by x but not y matrix A is
(b) divisible by y but not x
(c) divisible by neither x nor y (a) A– 1 does not exist
(d) divisible by both x and y.
(b) A = (–1)I, where I is a unit matrix
(2007) (c) A is a zero matrix (d) A 2 = I. (2004)
é5 5a a ù æ 1 -1 1 ö æ 4 2 2 ö
16. Let A = êê0 a 5a ú . If | A 2 | = 25, then |a| eq uals 24. Let A = ç 2 1 -3÷÷ and 10(B ) = ç -5 0 a÷ . If B is the
ú ç ç ÷
è 1 -2 3ø
ëê0 0 5 úû è 1 1 1 ø
inverse of matrix A, then a is
(a) 1/5 (b) 5 (c) 5 2 (d) 1. (2007) (a) 2 (b) –1 (c) –2 (d) 5. (2004)
17. If A and B are square matrices of size n × n such that determinant
25. If a1 , a 2, a3 , ...., a n , ... are G.P., then the value of the
A2 – B2 = (A – B)(A + B), then which of the following will
log an log an+1 log a n +2
be always true? log an+3 log an+4 log a n + 5 , is
(a) A = B (b) AB = BA
(c) either A or B is a zero matrix log an + 6 log an + 7 log an + 8
(d) either A or B is an identity matrix. (2006)
(a) 2 (b) 1 (c) 0 (d) –2. (2004)
14 JEE MAIN CHAPTERWISE EXPLORER
26. If 1, w, w2 are the cube roots of unity, 28. If a > 0 and discriminant of ax 2 + 2bx + c is –ve, then
1 wn w2 n a b ax + b
then D = wn w2 n 1 is equal to b c bx + c is
ax + b bx + c 0
w2 n 1 w n
(a) 1 (b) w (c) w2 (d) 0. (2003) (a) +ve (b) (ac – b 2) (ax2 + 2bx + c)
(c) –ve
(d) 0. (2002)
27. If A = éa b ù and A2 = éa b ûúù , then 29. If l, m, n are the p th, q th and r th term of a GP, all positive, then
êëb a ûú êë b a
log l p 1
(a) a = a 2 + b 2, b = 2ab
log m q 1 equals
(b) a = a 2 + b 2 , b = a 2 – b 2 log n r 1
(c) a = 2ab, b = a 2 + b 2 (a) –1 (b) 2
(d) a = a 2 + b 2 , b = ab. (2003) (c) 1 (d) 0. (2002)
Answer Key
1. (a) 2. (a) 3. (b) 4. (a) 5. (d) 6. (d)
7. (d) 8. (d) 9. (c) 10. (a) 11. (b) 12. (a)
16. (a) 17. (b) 18. (d)
13. (a) 14. (d) 15. (d) 22. (b) 23. (d) 24. (d)
19. (c) 20. (b) 21. (c) 28. (c) 29. (d)
25. (c)
26. (d) 27. (a)
Matrices and Determinants 15
é1 a 3 ù 4. (a) : P3 = Q 3, P 2Q = Q 2 P, PQ 2 = P 2 Q
1. (a) : P = êê1 3 3 úú Þ P(P2 + Q 2 ) = (Q 2 + P2 )Q
Þ P(P2 + Q 2) = (P2 + Q 2 )Q
êë2 4 4 úû P ¹ Q Þ P 2 + Q2 is singular.
Let P = 1(12 – 12) – a(4 – 6) + 3(4 – 6) = 2a – 6 Hence, |P2 + Q 2| = 0
Also, det(adj A) = (det A) 2
Þ 2a – 6 = 16 Þ 2a = 22. \ a = 11 5. (d) : Let A(BA) = P
Remark : det(adj A) = (det A) n – 1, where A is a matrix of Then PT = (ABA) T = A T B T A T (Transversal rule)
order n. = ABA = P
ék + 1 8 ù éxù é 4 k ù Thus P is symmetric.
2. (a) : The equation is ê k + 3ûú ëêyûú = êë3k - 1ûú Again, A(BA) = (AB)A by as sociativity.
ë k Also (AB) T = B T A T = BA = AB
For no solution of AX = B a necessary condition is
det A = 0.
k +1 8 (Q A and B are commutative)
Þ = 0 Þ AB is also symmetric.
k k +3
Þ (k + 1)(k + 3) – 8k = 0 Þ k 2 + 4k + 3 – 8k = 0 6. (d) : For the system to possess nonzero solution,
Þ k2 – 4k + 3 = 0 Þ (k – 1)(k – 3) = 0 \ k = 1, 3
For k = 1, the equation becomes 4 k 2
2x + 8y = 4, x + 4y = 2 we have k 4 2 = 0
which is just a single equation in two variables.
x + 4y = 2 It has infinite solutions. 2 2 1
For k = 3, the equation becomes which on expansion gives k 2 – 6k + 8 = 0
4x + 8y = 12, 3x + 6y = 8 Þ (k – 2)(k – 4) = 0. \ k = 2, 4
which are parallel lines. So no solution in this case. 7. (d) : x 1 + 2x 2 + x 3 = 3
é1 0 0ù é1ù é0 ù 2x 1 + 3x2 + x3 = 3
3x 1 + 5x 2 + 2x3 = 1
3. (b) : A = êê2 1 0úú , Au1 = êê0úú , Au2 = êê1 úú A quick observation tells us that the sum of first two equations
ëê3 2 1úû ëê0úû ëê0 úû
yields
éa ù
Let u1 = êêb úú (x 1 + 2x 2 + x 3 ) + (2x 1 + 3x 2 + x 3 ) = 3 + 3
Þ 3x 1 + 5x 2 + 2x3 = 6
ëê c úû But this contradicts the third equation, i.e.,
Au1 = é1 ù Þ a = 1, 2a + b = 0 3x 1 + 5x 2 + 2x3 = 1
êê0 úú Þ b = -2, 3a + 2b + c = 0 Þ c = 1 As such the system is inconsistent and hence it has no solution.
êë0 úû éa1 a2 a3 ù
é p ù 8. (d) : A= ê b1 b2 b3 úú
ê q úú ê
Let u2 = ê êëc1 c2 c3 úû
ëê r ûú Let A = (a 1 b 2 c 3 + a2 c1 b3 + a 3 b 1 c 2 ) – (a1 c 2 b 3 + a 2 b 1 c 3 + a 3 c 1b 2 )
If any of the terms be nonzero, then det A will be nonzero
Au 2 = é0ù Þ p = 0, 2 p + q = 1 Þ q = 1, and all the elements of that term will be 1 each.
êê1 úú 3 p + 2q + r = 0 Þ r = -2 Number of nonsingular matrices = 6 C 1 × 6 C 1 = 36
êë0ûú We can also exhibit more than 6 matrices to pick the right
choice.
é 1ù é 0ù é 1ù
êê-2úú ê 1úú = êê-1úú
u1 + u2 = + ê
ëê 1úû ëê-2ûú ëê-1ûú
16 JEE MAIN CHAPTERWISE EXPLORER
9. (c) : Let A = éa bù 12. (a) : Let A = éa bdûúù . We have
êë g dûú êë g
A2 = é a2 + bg b(a + d) ù = é1 0 ù A2 = é a 2 + bg b(a + d) ù
ê ú êë0 1 úû ê ú
ëêg(a + d) d + bg ûú ëêg (a + d) d2 + bg ûú
Which gives a + d = 0 and a 2 + bg = 1 é a2 + bg b(a + d)ù é1 0 ù
ê ëê0 1 ûú
So we have Tr(A) = 0 A2 = I = d2 + ú=
ûú
det A = ad – bg = –a 2 – bg = – (a 2 + bg) = –1 ëêg (a + d) bg
Thus statement1 is true but statement2 is false. giving a2 + bg = 1 = d 2 + bg
10. (a) : We have adj(adj A) = |A| n–2A and g(a + d) = b(a + d) = 0
Here n = 2, which gives adj(adj A) = A As A ¹ I, A ¹ – I, we have a = – d
The statement1 is true. 1 - bg b
Again |adj A| = |A| n – 1 detA = g - 1 - bg = -1+ bg - bg = -1
Here n = 2, which gives |adj A| = |A|
Thus statement2 is also true. But statement2 doesn’t explain Statement1 is therefore true.
tr (A) = a + d = 0 {a = – d}
statement1. Statement2 is false because tr (A) = 0
a a +1 a -1 a +1 b +1 c - 1 13. (a) : System of equations
b +1 b -1 + a -1 b -1 c - 1 = 0 x – cy – bz = 0
11. (b) : -b c -1 c + 1 (-1)n+2 a (-1)n+1 b (- 1) n c cx – y + az = 0
c bx + ay – z = 0
has non trivial solution if the determinant of coefficient matrix
a a +1 a -1 a + 1 b + 1 c - 1 is zero
Þ -b b + 1 b - 1 + (-1)n a - 1 b - 1 c - 1 = 0
c c - 1 c + 1 a - b c
a + 1 a - 1 a 1 -c -b
Þ D + (-1)n b + 1 b - 1 -b = 0 Þ c -1 a = 0
c + 1 c - 1 c b a -1
(Changing rows to columns)
Þ 1(1 – a2 ) + c(– c – ab) – b(ca + b) = 0
a a + 1 a - 1 Þ a2 + b 2 + c 2 + 2abc = 1
Þ D + (-1)n -b b + 1 b - 1 = 0
14. (d) : Each entry of A is an integer, so the cofactor of every
c c + 1 c - 1 entry is an integer. And then each entry of adjoint is integer.
(Changing columns in cyclic order doesn’t change the deter Also det A = ± 1 and we know that
minant)
Þ D + (–1) n D = 0 Þ {1 + (–1) n }D = 0 A-1 = 1 (adj A )
det A
a a + 1 a - 1
This means all entries in A –1 are integers.
Now D = -b b + 1 b - 1 = 0
c c - 1 c + 1 1 1 1
a 2 a - 1 15. (d) : D = 1 1 + x 1
= -b 2 b - 1 C 2 ® C 2 – C 3
c -2 c + 1 1 1 1 + y
(Apply C2 ® C 2 – C 1, C3 ® C 3 – C1 )
a + c 0 a + c
= -b + c 0 b + c R 1 ® R 1 + R 3 , R 2 ® R 2 + R 3 1 0 0
= 1 x 0 = 1( xy - 0) = xy
c -2 c + 1
Expanding along 2nd column 1 0 y
D = 2{(a + c)(b + c) – (a + c) (c – b)}
Hence D is divisible by both x and y.
= 2(a + c)2b é5 5a a ù é5 5a a ù
= 4b(a + c) ¹ 0 (By hypothesis) 16. (a) : A2 = êê0 a 5aúú êê0 a 5 aúú
Now {1 + (–1) n }D = 0 Þ 1 + (–1) n = 0
Which mean n = odd integer. ëê0 0 5 úû ëê0 0 5 ûú
Matrices and Determinants 17
é25 25a + 5a2 5a + 5a + 25 a2 ù 22. (b) : For no solution | A | = 0 and (adj A)(B) ¹ 0
ê ú
A2 = ê 0 a2 25a + 5 a2 ú a 1 1
Now | A | = 0 Þ 1 a 1 = 0
ê 0 0 25 ú
êë úû 1 1a
Þ a3 – 3a + 2 = 0 Þ (a – 1) 2 (a + 2) = 0
Given |A 2 | = 25, 625a 2 = 25 Þ | a |= 15 . Þ a = 1, –2.
But for a = 1, | A | = 0 and (adj A)(B) = 0
17. (b) : Give A 2 – B2 = (A + B)(A – B) Þ for a = 1 there exist infintiely many solution.
Þ 0 = BA – AB Also the each equation becomes
Þ BA = AB x + y + z = 0 again for a = –2
|A| = 0 but (adj A)(B) ¹ 0 Þ $ no solution.
18. (d) : æ1 2ö æ a 0ö
A = èç 3 4÷ø and B = çè 0 b ø÷
23. (d) : (i) |A| = 1 \ A– 1 does not exist is wrong statement
Now AB = æ 1 2ö æ a 0ö = æa 2b ö ... (i) æ -1 0 0 ö
çè 3 4÷ø çè 0 b÷ø èç 3a 4bø÷ ... (ii)
(ii) (–1) I = ç 0 -1 0 ÷ ¹ A Þ (b) is false
ç 0 0 ÷
è -1 ø
a 0ö æ1 2ö æa 2a ö
and BA = æ 0 b÷ø çè 3 4ø÷ = çè 3b 4b ø÷ (iii) A is clearly a non zero matrix \ (c) is false
çè
We left with (d) only.
As AB = BA Þ 2a = 2b Þ a = b 24. (d) : Given A –1 = B = 10 A– 1 = 10 B
\ B = æ a 0 ö = aI 2 Þ $ infinite value of a = b Î N æ 4 2 2 ö
çè 0 aø÷ ç a÷
Þ ç -5 0 ÷ = 10 A –1.
è 1 -2 ø
3
19. (c) : A2 – A + I = 0 Þ I = A – A × A æ 4 2 2 ö
IA –1 = AA –1 – A(AA –1) , A –1 = I – A. ç a÷
Þ ç -5 0 ÷ (A) = 10I
è 1 -2 ø
3
20. (b) : A = æ1 0 ö æ 4 2 2 ö æ 1 -1 1 ö
çè1 1 ÷ø
Þ ç -5 0 a ÷ ç 2 1 -3 ÷
ç ÷ ç ÷
æ1 0 ÷ö , æ 1 0ö æ 1 0 ö è 1 -2 3 ø è 1 1 1 ø
1 ø ç 3 ç n 1 ÷ø
\ A2 = ç 2 A3 = è 1 ÷ so A n = è æ 10 0 0 ö
è ø
ç 0 10 0 ÷
= ç 0 ÷ ...(*)
è ø
æ n 0 ö æ n - 1 0 ö 0 10
ç n n ÷ ç 0 n -1 ÷ø
and nA – (n – 1)I = è ø - è
= æ1 0 ö = A n . Þ –5 + a = 0
ç 1 ø÷
è (equating A 21 entry both sides of (*))
Þ a = 5
n 25. (c) : a2 = a3 = ... a n = r
21. (c) : Applying C 2 ® C2 + C3 + C 1 a1 a2 an - 1
f ( x) = 1 + 2 x + x(a2 + b2 + c2 )
which means an , an + 1 , an + 2 Î G.P.
1 + a2 x 1 (1 + c2 ) x 2
(1 + a2 ) x 1 (1 + c2 ) x Þ a n + 1 = a n a n + 2
(1 + a2 ) x 1 1 + c2 x
Þ 2 log a n + 1 – log an – log a n + 2 = 0...(i)
Applying R1 ® R1 – R 2, R2 ® R2 – R3 and
using a 2 + b 2 + c2 = – 2 we have Similarly 2 log an + 4 – log an + 3 – log an + 5 = 0 ...(ii)
and 2 log an + 7 – log an + 6 – log a n + 8 = 0 ...(iii)
Using C1 ® C 1 + C3 – 2C 2
we get D = 0
26. (d) : As w is cube root of unity \ w3 = w 3n = 1
1 - x 0 0 1 wn w2 n
w2 n 1
(1 + 2 x - 2 x) 0 0 x -1 = (1 – x)2 \ wn
w2 n 1 w n
(1 + a2 ) x 1 1 + c2 x
= x 2 – 2x + 1 \ degree of f (x) is 2. = (w 3n – 1) – wn ( w 2n – w 2n ) + w 2n( wn – w n ) = 0
18 JEE MAIN CHAPTERWISE EXPLORER
27. (a) : A2 = AA = æa böæa b ö \ l = tp = ARp – 1
Þ log l = log A + (p – 1) log R
ç b a ÷ ç b a ÷
è ø è ø
Similarly, log m = log A + (q – 1) log R
æ a2 + b2 2 ab ö æ a b ö and log n = log A + (r – 1) log R
= ççè 2 ab b2 ø÷÷ =
a2 + ç b a ÷ log l p 1
è ø \ log m q 1
28. (c) : C 1 ® xC 1 + C2 – C3 log n r 1
= 1 0 b ax + b log A + ( p - 1) log R p 1
0 c bx + c
x ax2 + 2bx + c bx + c 0 = log A + (q - 1) log R q 1
log A + (r - 1) log R r 1
(ax2 + 2bx + c ) log A - log R p 1 p log R p 1
= [b2 x + bc – acx – bc] = log A - log R q 1 + q log R q 1
x log A - log R r 1 r log R r 1
= (b 2 – ac) (ax2 + 2bx + c)
= (+ve) (–ve) < 0 c 1 µ c 3 c 1 µ c 2
= 0 + 0
29. (d) : Let A be the first term and R be the common ratio of = 0
G. P.
Quadratic Equations 19
CHAPTER QUADRATIC EQUATIONS
4
1. The real number k for which the equation 9. If the roots of the quadratic equation x 2 + px + q = 0 are tan30°
2x 3 + 3x + k = 0 has two distinct real roots in [0, 1] and tan15°, respectively then the value of 2 + q – p is
(a) lies between 2 and 3 (a) 2 (b) 3 (c) 0 (d) 1. (2006)
(b) lies between –1 and 0 10. All the values of m for which both roots of the equation x2
(c) does not exist
(d) lies between 1 and 2 (2013) – 2mx + m 2 – 1 = 0 are greater than –2 but less than 4, lie
in the interval
2. If the equations x 2 + 2x + 3 = 0 and ax 2 + bx + c = 0, a, b, (a) –2 < m < 0 (b) m > 3
c Î R have a common root, then a : b : c is (c) –1 < m < 3 (d) 1 < m < 4. (2006)
(a) 3 : 2 : 1 (b) 1 : 3 : 2 11. The value of a for which the sum of the squares of the roots
(c) 3 : 1 : 2 (d) 1 : 2 : 3 (2013) of the equation x2 – (a – 2)x – a – 1 = 0 assume the least value
3. The equation e sinx – e– sinx – 4 = 0 has is
(a) exactly one real root.
(b) exactly four real roots. (a) 0 (b) 1 (c) 2 (d) 3. (2005)
(c) infinite number of real roots.
(d) no real roots. 12. If the roots of the equation x 2 – bx + c = 0 be two consecutive
integers, then b 2 – 4c equals
(2012) (a) 3 (b) –2 (c) 1 (d) 2. (2005)
4. Let a, b be real and z be a complex number. 13. If both the roots of the quadratic equation
If z 2 + az + b = 0 has two distinct roots on the line x2 – 2kx + k2 + k – 5 = 0
Re z = 1, then it is necessary that are less than 5, then k lies in the interval
(a) |b| = 1 (b) b Î (1, ¥) (a) (6, ¥) (b) (5, 6]
(c) b Î (0, 1) (d) b Î (–1, 0) (2011) (c) [4, 5] (d) (– ¥, 4). (2005)
5. If a and b are the roots of the equation x 2 – x + 1 = 0, then 14. If the equation a n x n + a n – 1 x n – 1 + ... + a 1 x = 0,
a 2009 + b 2009 = a 1 ¹ 0, n ³ 2, has a positive root x = a, then the equation
(a) –2 (b) –1 (c) 1 (d) 2 (2010) nan x n – 1 + (n – 1)an – 1 x n – 2 + ... + a 1 = 0 has a positive root,
6. If the roots of the equation bx 2 + cx + a = 0 be imaginary, which is (b) greater than a
then for all real values of x. The expression (a) smaller than a
3b 2 x 2 + 6bcx + 2c 2 is (c) equal to a (d) greater than or equal to a.
(a) less than 4ab (b) greater than – 4ab (2005)
(c) less than – 4ab (d) greater than 4ab (2009) 15. Let two numbers have arithmetic mean 9 and geometric
7. The quadratic equations x2 – 6x + a = 0 and x 2 – cx + 6 = 0 mean 4. Then these numbers are the roots of the quadratic
have one root in common. The other roots of the first and second equation
equations are integers in the ratio 4 : 3. Then the common root (a) x 2 + 18x – 16 = 0 (b) x 2 – 18x + 16 = 0
is (c) x2 + 18x + 16 = 0 (d) x 2 – 18x – 16 = 0. (2004)
(a) 2 (b) 1 (c) 4 (d) 3 (2008) 16. If (1 – p) is a root of quadratic equation x2 + px + (1 – p) = 0
then its roots are
8. If the difference between the roots of the equation (a) 0, –1 (b) –1, 1 (c) 0, 1 (d) –1, 2. (2004)
x 2 + ax + 1 = 0 is less than 5 , then the set of possible 17. If one root of the equation x 2 + px + 12 = 0 is 4, while the
values of a is equation x 2 + px + q = 0 has equal roots, then the value of
(a) (3, ¥) (b) (– ¥, –3) q is
(c) (–3, 3) (d) (–3, ¥). (2007) (a) 3 (b) 12 (c) 49/4 (d) 4. (2004)
20 JEE MAIN CHAPTERWISE EXPLORER
18. The value of a for which one root of the quadratic equation (a) 3x 2 – 25x + 3 = 0 (b) x 2 + 5x – 3 = 0
(c) x 2 – 5x + 3 = 0 (d) 3x 2 – 19x + 3 = 0. (2002)
(a 2 – 5a + 3)x 2 + (3a – 1)x + 2 = 0 is twice as large as the
other is 22. Difference between the corresponding roots of
(a) –2/3 (b) 1/3 (c) –1/3 (d) 2/3. (2003) x2 + ax + b = 0 and x2 + bx + a = 0 is same and a ¹ b, then
19. If the sum of the roots of the quadratic equation (a) a + b + 4 = 0 (b) a + b – 4 = 0
ax 2 + bx + c = 0 is equal to the sum of the squares of their (c) a – b – 4 = 0 (d) a – b + 4 = 0. (2002)
reciprocals, then a , b and c are in 23. Product of real roots of the equation x 2 + | x | + 9 = 0
c a b
(a) is always positive (b) is always negative
(a) geometric progression
(b) harmonic progression (c) does not exist (d) none of these. (2002)
(c) arithmeticgeometric progression 24. If p and q are the roots of the equation x2 + px + q = 0, then
(d) arithmetic progression. (2003) (a) p = 1, q = –2 (b) p = 0, q = 1
20. The number of real solutions of the equation (c) p = –2, q = 0 (d) p = –2, q = 1. (2002)
x 2 – 3|x| + 2 = 0 is 25. If a, b, c are distinct +ve real numbers and a 2 + b 2 + c2 = 1
(a) 4 (b) 1 (c) 3 (d) 2. (2003) then ab + bc + ca is
21. If a ¹ b but a2 = 5a – 3 and b 2 = 5b – 3 then the equation (a) less than 1 (b) equal to 1
whose roots are a/b and b/a is
(c) greater than 1 (d) any real no. (2002)
Answer Key
1. (c) 2. (d) 3. (d) 4. (b) 5. (b) 6. (b)
8. (c) 9. (b) 10. (c) 11. (b) 12. (c)
7. (a) 14. (a) 15. (b) 16. (a) 17. (c) 18. (d)
13. (d) 20. (a) 21. (d) 22. (a) 23. (c) 24 (a)
19. (b)
25. (a)
Quadratic Equations 21
1. (c) : Let f (x) = 2x 3 + 3x + k , f ¢(x) = 6x 2 + 3 > 0 7. (a) : Let a and 4b be the root of
Thus f is strictly increasing. Hence it has atmost one real root. x 2 – 6x + a = 0
But a polynomial equation of odd degree has atleast one root.
Thus the equation has exactly one root. Then the two distinct and a and 3b be those of the equation x 2 – cx + 6 = 0
roots, in any interval whatsoever is an impossibility. No such From the relation between roots and coefficients
(c) exists. a + 4b = 6 and 4ab = a
a + 3b = c and 3ab = 6
2. (d) : In the equation x2 + 2x + 3 = 0, both the roots are imaginary. we obtain ab = 2 giving a = 8
The first equation is x2 – 6x + 8 = 0 Þ x = 2, 4
Since a, b, c Î R, we have a = b = c For a = 2, 4b = 4 Þ 3b = 3
1 2 3 For a = 4, 4b = 2 Þ 3b = 3/2 (not an integer)
So the common root is a = 2.
Hence a : b : c : : 1 : 2 : 3
8. (c) : x 2 + ax + 1 = 0
3. (d) :es inx – e –sinx – 4 = 0 Let roots be a and b, then a + b =–a and ab = 1
Þ (e sinx ) 2 – 4e sinx – 1 = 0 Þ t2 – 4t – 1 = 0
| a - b | = (a + b)2 - 4ab , | a - b | = a2 - 4
Þ t = 4 ± 16 + 4 = 2 ± 5
2 Since, | a - b |< 5 Þ a2 - 4 < 5
Þ a2 – 4 < 5 Þ a2 < 9 Þ – 3 < a < 3.
i.e., es in x = 2 + 5 or 1244-2 4453 (neglected)
9. (b) : a = tan 30°, b = tan 15° are roots of the equation
- ve x2 + px + q = 0
\ tan a + tan b = – p and tan a ∙ tan b = q
sin x = ln (2 + 5) > 1 \ No real roots. using tan a + tan b = tan (a + b)
(1 – tan a tan b)
4. (b) : Let roots be 1 + ai, 1 + bi, then we have, (a Î R) Þ – p = 1 – q Þ q – p = 1 Þ 2 + q – p = 3
(1 + ai) + (1 + bi) = – a Þ 2 + (a + b)i = – a 10. (c) : Let a, b are roots of the equation
(1 + ai)(1 + bi) = b (x 2 – 2mx + m 2) = 1
Comparing we have, a = – 2 and a = – b
Now (1 + ai)(1 – ai) = b
Þ 1 + a 2 = b Þ b = 1 + a 2
As a 2 ³ 0 we have b Î (1, ¥)
5. (b) : We have x 2 – x + 1 = 0 giving x = 1 ± i 3 .
2
Identifying these roots as w and w2 ,
we have a = w, b = w 2. We can also take the other way round –3 –1 3 5
that would not affect the result.
Now a 2009 + b 2009 = w 2009 + w 4018 Þ x = m ± 1 = m + 1, m – 1
= w 3k + 2 + w3 m + 1 (k, m Î N) Now –2 < m + 1 < 4 ...... (i)
= w 2 + w = –1. (. .. w 3k = 1)
and –2 < m – 1 < 4 ....... (ii)
6. (b) : The roots of bx 2 + cx + a = 0 are imaginary means ì Þ -3 < m < 3 ......(A)
c 2 – 4ab < 0 Þ c 2 < 4ab îíand -1 < m < 5 .......(B)
Again the coeff. of x2 in By (A) & (B) we get –1 < m < 3 as shown by the number
3b 2 x 2 + 6bcx + 2c 2 is +ve, so the minimum value of the line.
expression 11. (b) : Let f (a) = a2 + b 2 = (a + b) 2 – 2ab
= (a – 2) 2 + 2(a + 1)
= - 36b2c2 - 4(3b2)(2c2) = 12b 2c 2 = - c 2 \ f ¢(a) = 2(a – 2) + 2
(3b2 ) 12b 2 For Maxima | Minima f ¢(a) = 0
Þ 2[a – 2 + 1] = 0 Þ a = 1
As c 2 < 4ab we have –c2 > – 4ab Again f ¢¢(a) = 2,
f ¢¢(1) = 2 > 0 Þ at a = 1, f (a) will be least.
Thus the minimum value is – 4ab.
22 JEE MAIN CHAPTERWISE EXPLORER
12. (c) : Let a, a + 1 are consecutive integer By (i) and (ii) we have
\ (x + a)(x + a + 1) = x2 – bx + c
Comparing both sides we get Þ –b = 2a + 1 9a 2 (1 - 3a)2 a2 - 5a + 3
c = a 2 + a 2a 2 = (a2 - 5a + 3) 2 ´
\ b2 – 4c = (2a + 1)2 – 4(a 2 + a) = 1. 2
Þ 9(a2 – 5a + 3) = (1 – 3a) 2
13. (d) : Given x2 – 2kx + k 2 + k – 5 = 0 2
Þ a = 3
Roots are less than 5 Þ D ³ 0
Þ (–2k)2 ³ 4(k2 + k – 5) Þ k £ 5 ...(A) 19. (b) : Given a + b
1 1 (a + b)2 - 2ab
Again f (5) > 0 = a2 + b2 =
a 2b2
Þ 25 – 10k + k 2 + k – 5 > 0
Þ 2a2 c = bc2 + ab2
Þ k 2 – 9k + 20 > 0 Þ (k – 4)(k – 5) > 0
2a c b c , a , b Î A.P
Þ k < 4 È k > 5 ...(B) Þ b = a + c Þ abc
Also sum of roots < 5 Þ k < 5 ...(C) Þ reciprocals are in H.P
2
20. (a) : Given x2 – 3|x| + 2 = 0
from (A), (B), (C) we have
k Î (–¥, 4) as the choice gives number k < 5 is (d). If x ³ 0 i.e. |x| = x
14. (a) : If possible say \ The given equation can be written as
f ( x) = a0 xn + a1 xn -1 + ... + an x \ f (0) = 0
x 2 – 3x + 2 = 0
Now f (a) = 0 (\ x = a is root of given equation) Þ (x – 1) (x – 2) = 0
Þ x = 1, 2
\ f ¢( x) = nan xn-1 + (n -1)an -1xn -2 + ... + a1 = 0 has at least Similarly for x < 0, x 2 – 3|x| + 2 = 0
one root in ]0, a [ Þ x2 + 3x + 2 = 0
Þ na n x n–1 + (n – 1) an –1 x n–2 + .... + a1 = 0
has a +ve root smaller than a. Þ x = –1, –2
15. (b) : Let the two number be a, b Hence 1, –1, 2, –2 are four solutions of the given equation.
a + b
ab
\ 2 = 9 and ab = 4 21. (d) : We need the equation whose roots are b and a which
\ Required equation
are reciprocal of each other, which means product of roots is
x2 – 2(Average value of a, b)x +
x2 – 2(9)x + 16 = 0 2 ab
= 1. In our choice (a) and (d) have product of roots 1,
G.M = 0
b a
so choices (b) and (d) are out of court. In the problem choice,
16. (a) : As 1 – p is root of x2 + px + 1 – p = 0 None of these is not given. If out of four choices only one
Þ (1 – p) 2 + p(1 – p) + (1 – p) = 0 choice satisfies that product of root is 1 then you select that
(1 – p) [1 – p + p + 1] = 0 choice for correct answer. Now for proper choice we proceed
Þ p = 1
\ Given equation becomes x2 + x = 0 as,
Þ x = 0, – 1 a ¹ b, but a 2 = 5a – 3 and b 2 = 5b – 3,
Changing a, b by x
\ a, b are roots of x2 – 5x + 3 = 0
17. (c) : As x 2 + px + q = 0 has equal roots \ p2 = 4q Þ a + b = 5, ab = 3
and one root of x 2 + px + 12 = 0 is 4. now, S = a + b a2 + b2 = 19 and product a . b = 1
b a = ab 3 b a
\ 16 + 4p + 12 = 0 \ p = –7
\ Required equation,
49
\ p2 = 4q Þ q = 4 x 2 – (sum of roots) x + product of roots = 0
18. (d) : Let a, 2a are roots of the given equation Þ x 2 – 19 1 = 0
x +
3
\ sum of the roots
Þ 3x 2 – 19x + 3 = 0 is correct answer.
1 - 3a ...(i) 22. (a) : Let a, b are roots of x 2 + bx + a = 0
a + 2a = 3a = a2 - 5a + 3 ...(ii) \ a + b = –b and ab = a
and product of roots again let g, d are roots of x 2 + ax + b = 0
\ g + d = –a and gd = b
a(2a) = 2a2 = 2 + 3
a2 - 5a Now given
Quadratic Equations 23
a – b = g – d Now if q = 0 then p = 0 Þ p = q
Þ (a – b) 2 = (g – d) 2 If p = 1, then p + q = – p
Þ (a + b) 2 – 4ab = (g + d)2 – 4gd q = – 2p
Þ b2 – 4a = a 2 – 4b q = – 2(1)
Þ b2 – a 2 = –4(b – a) q = – 2
Þ p = 1 and q = – 2
Þ (b – a) (b + a + 4) = 0
25. (a) : In such type of problem if sum of the squares of number
Þ b + a + 4 = 0 as (a ¹ b) is known and we needed product of numbers taken two at a
time or needed range of the product of numbers taken two at
23. (c) : x2 + |x| + 9 = 0 a time. We start square of the sum of the numbers like
(a + b + c) 2 = a2 + b 2 + c2 + 2(ab + bc + ca)
Þ |x|2 + |x| + 9 = 0 Þ 2(ab + bc + ca) = (a + b + c) 2 – (a 2 + b2 + c 2 )
(a + b + c)2 - 1
Þ \ no real roots (³ D < 0) Þ ab + bc + ca = 2 < 1
24 (a) : Given S = p + q = – p and product pq = q
Þ q(p – 1) = 0
Þ q = 0, p = 1
24 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER PERMUTATIONS
AND COMBINATIONS
5
1. Let Tn be the number of all possible triangles formed by joining 6. From 6 different novels and 3 different dictionaries, 4 novels
vertices of an nsided regular polygon. and 1 dictionary are to be selected and arranged in a row on
If T n + 1 – T n = 10, then the value of n is a shelf so that the dictionary is always in the middle. Then
(a) 5 (b) 10 (c) 8 (d) 7 the number of such arrangements is
(2013) (a) at least 500 but less than 750
2. Assuming the balls to be identical except for difference in (b) at least 750 but less than 1000
colours, the number of ways in which one or more balls can (c) at least 1000
be selected from 10 white, 9 green and 7 black balls is (d) less than 500 (2009)
(a) 630 (b) 879 (c) 880 (d) 629
(2012) 7. In a shop there are five types of icecreams available. A child
3. Statement1 : The number of ways of distributing buys six icecreams.
10 identical balls in 4 distinct boxes such that no box is empty Statement1 : The number of different ways the child can
is 9 C3 . buy the six icecreams is 10 C 5 .
Statement2 : The number of ways of choosing any 3 places Statement2 : The number of different ways the child can
from 9 different places is 9 C3 . buy the six icecreams is equal to the number of different
(a) Statement1 is true, Statement2 is false. ways of arranging 6 A’s and 4 B’s in a row.
(b) Statement1 is false, Statement2 is true. (a) Statement1 is true, Statement2 is false
(c) Statement1 is true, Statement2 is true; Statement2 is (b) Statemen1 is false, Statement2 is true
a correct explanation for Statement1. (c) Statement1 is true, Statement2 is true; Statement2 is
(d) Statement1 is true, Statement2 is true; Statement2 is a correct explanation for Statement1
not a correct explanation for Statement1. (2011) (d) Statement1 is true, Statement2 is true; Statement2 is
not a correct explanation for Statement1
10 10 (2008)
å å 4. Let S1 = j( j - 1) 10C j , S2 = j 10C j
j=1 j =1 8. How many different words can be formed by jumbling the
10 letters in the word MISSISSIPPI in which no two S are adjacent?
å and S3 = j2 10C j . (a) 7 ∙ 6 C 4 ∙ 8 C 4 (b) 8 ∙ 6 C 4 ∙ 7 C 4
j =1 (c) 6 ∙ 7 ∙ 8C 4 (d) 6 ∙ 8 ∙ 7 C 4 (2008)
Statement1 : S 3 = 55 × 2 9 .
Statement2 : S 1 = 90 × 2 8 and S 2 = 10 × 2 8 . 9. The sum of the series
(a) Statement1 is true, statement2 is true; statement2 is a 20 C 0 – 20 C 1 + 20 C 2 – 20 C 3 + ..... – ...... + 20 C 10 is
correct explanation of statement1. (a) 0 (b) 20 C 10 (c) – 20 C 10 (d) 1 20 C 10 .
(b) Statement1 is true, statement2 is true; statement2 is not 2
a correct explanation for statement1. (2007)
(c) Statement1 is true, statement2 is false.
(d) Statement1 is false, statement2 is true. (2010) 10. At an election, a voter may vote for any number of candidates,
5. There are two urns. Urn A has 3 distinct red balls and urn B not greater than the number to be elected. There are 10
has 9 distinct blue balls. From each urn two balls are taken candidates and 4 are to be elected. If a voter votes for at least
out at random and then transferred to the other. The number one candidate, then the number of ways in which he can vote
of ways in which this can be done is is
(a) 3 (b) 36 (c) 66 (d) 108 (a) 5040 (b) 6210 (c) 385 (d) 1110.
(2010) (2006)
Permutations and Combinations 25
11. If the letters of the word SACHIN are arranged in all possible (a) 196 (b) 280 (c) 346 (d) 140.
(2003)
ways and these words are written out as in dictionary, then
the word SACHIN appears at serial number
(a) 602 (b) 603 (c) 600 (d) 601. 17. If n C r denotes the number of combinations of n things taken
r at a time, then the expression nC r + 1 + n Cr – 1 + 2 × n C r equals
(2005) (a) n + 2C r + 1 (b) n + 1C r (c) n + 1 Cr + 1 (d) n + 2 C r .
12. The value of 50 C4 + 6 56 -r C3 is (2003)
å
r =1 18. Number greater than 1000 but less than 4000 is formed using
(a) 56C 4 (b) 56 C 3 (c) 55C 3 (d) 55 C 4. the digits 0, 2, 3, 4 repetition allowed is
(2005)
13. How many ways are there to arrange the letters in the word (a) 125 (b) 105 (c) 128 (d) 625.
GARDEN with the vowels in alphabetical order? (2002)
(a) 360 (b) 240 (c) 120 (d) 480. 19. Five digit number divisible by 3 is formed using 0, 1, 2, 3,
(2004) 4, 6 and 7 without repetition. Total number of such numbers
14. Then number of ways of distributing 8 identical balls in 3 are
distinct boxes so that none of the boxes is empty is (a) 312 (b) 3125 (c) 120 (d) 216.
(a) 3 8 (b) 21 (c) 5 (d) 8C 3 . (2002)
(2004)
20. The sum of integers from 1 to 100 that are divisible by 2 or
15. The number of ways in which 6 men and 5 women can dine
at a round table if no two women are to sit together is given 5 is
by
(a) 3000 (b) 3050 (c) 3600 (d) 3250.
(a) 30 (b) 5 !´ 4 ! (c) 7 ! ´ 5 ! (d) 6 ! ´ 5!. (2002)
(2003) 21. Total number of four digit odd numbers that can be formed
16. A student is to answer 10 out of 13 questions in an examination using 0, 1, 2, 3, 5, 7 are
such that he must choose at least 4 from the first five questions.
The number of choices available to him is (a) 216 (b) 375 (c) 400 (d) 720.
(2002)
Answer Key
1. (a) 2. (b) 3. (c) 4. (c) 5. (d) 6. (c)
7. (b) 8. (a) 9. (d) 10. (a) 11. (d) 12. (a)
13. (a) 14. (b) 15. (d) 16. (a) 17. (a) 18. (c)
19. (d) 20. (b) 21. (d)
26 JEE MAIN CHAPTERWISE EXPLORER
1. (a) : 1 st solution : n + 1C 3 – nC 3 = 10 Since the dictionary is fixed in the middle, we only have to
Þ (n + 1)n(n - 1) - n(n - 1)(n - 2) = 10 arrange 4 novels which can be done in 4! ways.
6 6 Then the number of ways = 6C 4 ∙ 3 C1 ∙ 4!
Þ 3n(n – 1) = 60 Þ n(n – 1) = 20 Þ n 2 – n – 20 = = 6∙5 ∙ 3∙ 24 = 1080 = 1080
0 0
2
Þ (n – 5)(n + 4) = 0 \ n = 5 7. (b) : We have to find the number of integral solutions
2 nd solution : n + 1 C3 – n C3 = 10 if x 1 + x 2 + x 3 + x 4 + x 5 = 6
and that equals 5+6–1 C5 –1 = 10 C 4
Þ n C2 = 10 Þ n(n - 1) = 10 Thus Statement1 is false.
2
Number of different ways of arranging 6A’s and 4B’s in a
Þ n2 – n – 20 = 0. \ n = 5
row
Here we have used n Cr + nC r + 1 = n + 1 C r + 1
2. (b) : Number of ways in which one or more balls can be = 10 10 C4 = Number of different ways the child can buy
selected from 10 white, 9 green, 7 black balls is =
6 ´ 4
= (10 + 1) (9 + 1) (7 + 1) – 1 the six icecreams.
= 880 – 1 = 879 ways \ Statement2 is true
3. (c) : x 1 + x 2 + x 3 + x 4 = 10 So, Statement1 is false, Statement2 is true.
The number of positive integral solution is 6 + 4 – 1 C4 – 1 8. (a) : Leaving S, we have 7 letters M, I, I, I, P, P, I.
= 9 C 3 7
It is the same as the number of ways of choosing any 3 balls way of arranging them = 2 4 = 7 × 5× 3
from 9 different places.
å 4. (c) : S1 = j( j -1) 10C j And four S can be put in 8 places in 8C 4 ways.
å = j( j -1) × 10j((1j0 --11)) × 8C j -2 The required number of ways = 7 ∙ 5 ∙ 3 ∙ 8 C 4 = 7 ∙ 6 C4 ∙ 8 C4 .
9. (d) : Q 20C0 +20 C1x + ....... +20 C10 x10 +
....... +20 C20 x20 = (1 + x) 20
After putting x = –1, we get
10 20 C0 -20 C1 +20 C2 -20 C3 + ......
+20 C10 -20 C11 -20 C12 + ..... +20 C20 = 0
å = 9 ´10 8C j -2 = 90 ´ 28
j =2
10 10 2(20 C0 -20 C1 +20 C2 -20 C3 + ...... -20 C9 ) +20 C10 = 0
20 C0 -20 C1 +20 C2 -20 C3 + ..... -20 C9 +20 C10 = 12 20 C10
å å S2 = j × 10C j = 10 9C j -1 = 10 ´ 29
j=1 j =1
10 10 10. (a) : A voter can vote one candidate or two or three or four
candidates
å å S3 = j2 × 10C j = ( j( j -1) + j) × 10C j \ Required number of ways
j=1 j =1 = 10c 1 + 10 c 2 + 10c 3 + 10c 4 = 385
10 10 F ix e d
å å = j( j - 1) 10C j + j × 10 C j
j=1 j =1
= 90 ∙ 2 8 + 10 ∙ 2 9 = (45 + 10)2 9 = 55 ∙ 2 9 . 11. (d) : S A C H I N
Then statement1 is true and statement2 is false. No. of word start with A = 5!
No. of word start with C = 5!
5. (d) : The number of ways = ( 3C 2 ) × ( 9 C2 ) = 3´ 9 ´ 8 = 108 No. of word start with H = 5!
2 No. of word start with I = 5!
No. of word start with N = 5!
6. (c) : Out of 6 novels, 4 novels can be selected in 6C 4 ways. Total words = 5! + 5! + 5! + 5! + 5! = 5(5 !) = 600
Also out of 3 dictionaries, 1 dictionary can be selected in 3C 1
ways.
Permutations and Combinations 27
Now add the rank of SACHIN so required rank of SACHIN 16. (a) : Case (i) :
= 600 + 1 = 601.
Required ways for first case = 5C 4 × 8 C 6 = 140
6 Case (ii):
12. (a) 50 C4 + å 56 -r C3 No. of question 5 No. of question 8
r = 1
Putting r = 6, 5, 4, 3, 2, 1 we get No. of question 4 No. of question 6
No. of question 5 No. of question 8
50 C4 + 50C3 + 51C3 + 52C3 + 53C3 + 54C3 + 55 C3
5C 5 8C 5
( ) Q nCr + nCr +1 = n + 1C r +1
\ Required ways for case (ii) = 5C 5 × 8C 5
= 51C 4 + 51C 3 + 52C 3 + 53C 3 + 54C 3 + 55C 3 8 ´ 7 ´ 6
= 52C 4 + 52C 3 + 53C 3 + 54C 3 + 55C 3
= 53C 4 + 53C 3 + 54C 3 + 55C 3 = 54C 4 + 54C 3 + 55C 3 = 3 ´ 2 ´ 1 = 56
= 55C 4 + 55C 3 = 56C 4 Total number of ways = 140 + 56
13. (a) : Number of letters = 6 = 196
Number of vowels = 2 namely A.E these alphabets can be 17. (a) : Consider n Cr – 1 + 2 n Cr + nC r + 1
arrange themselves by 2! ways = ( nC r – 1 + nC r ) + (n C r + nC r + 1 )
= n + 1 Cr + n + 1 Cr + 1
6! = n + 2 Cr + 1
\ Number of words = 2! = 360
18. (c) : Let number of digits formed x.
14. (b) : (i) Each box must contain at least one ball since no
box remains empty so we have the following cases \ 1000 < x < 4000, which means left extreme digit will be
Box Number of balls
I 1, 1, 1, 2, 2, either 2 or 3.
I 1, 2, 3, 3, 2, \ Required numbers = 2 C1 × H T U
where H = Hundred place
III 6, 5, 4, 3, 4, = 2 C 1 × 4 × 4 × 4 T = Ten’s place
= 128 U = Unit place
\ Number of ways 19. (d)
1 ´ 3! 20. (b) : Set of numbers divisible by 2 are 2, 4, 6, ....100
3 × + 3! × 2
Set of numbers divisible by 5 are 5, 10, 15, ....100
2!
= 9 + 6 × 2 = 21 Set of numbers divisible by 10 are 10, 20, 30, ....100
Now sum of numbers divisible by 2 is given by
As 1, 1, 6 2, 3, 4 2, 2, 4 have case ways and
1, 2, 5 1, 3, 4 have equal number of ways of arranging the 50 n
balls in the different boxes. S5 0 = 2 [2 + 100] using S n = [a + l]
2
(ii): Let the number of balls in the boxes are x, y, z respectively
S5 0 = 25[102]
then x + y + z = 8 and no box is empty so each x, y, z ³ 1
20
Þ l + m + n + 3 = 8 where l = x – 1, Similarly, S2 0 = [5 + 100] = 10 × 105 = 1050
2
m = y – 1, n = z – 1
10
i.e. (l + 1) + (m + 1) + (n + 1) = 8 are non negative integers and S1 0 = 2 [10 + 100] = 5 × 110
\ Required number of ways = n + r – 1C r \ Required sum = 25 × 102 + 1050 – 550
= 3 + 5 – 1C 5 = 7C 5 = 7C 2 M = 25[102 + 42 – 22]
15. (d) : Number of women = 5 M M
= 25 × 122
Number of men = 6
= 3050
Number of ways of 6 men at a M M 21. (d) : Odd numbers are 1, 3, 5, 7
We have to fill up four places like TH H T U
round Table is n – 1! = (6 – 1)!= 5! M (Case: If repetition is allow)
5C 1 62 4C 1 = 5 × 6 2 × 4
Now we left with six places between the men and there are = 5 × 36 × 4
5 women, these 5 women can be arranged themselves by 6P 5
way. = 720
\ Required number of ways = 5! × 6P 5 = 5! × 6!
28 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER MATHEMATICAL INDUCTION
6 AND ITS APPLICATION
1. Statement1 : For every natural number n ³ 2, 2. Let S(k) = 1 + 3 + 5 + ... + (2k – 1) = 3 + k 2. Then which
1 + 1 + ... + 1 > n . of the following is true? (b) S(k) Þ S(k + 1)
1 2 n (a) S(k) Þ S(k – 1)
(c) S(1) is correct
Statement2 : For every natural number n ³ 2,
n(n + 1) < n + 1. (d) principle of mathematical induction can be used to prove
(a) Statement1 is true, Statement2 is false
the formula. (2004)
(b) Statemen1 is false, Statement2 is true 3. If an = 7+ 7+ 7 + .... having n radical signs then by
(c) Statement1 is true, Statement2 is true; Statement2 is a
methods of mathematical induction which is true
correct explanation for Statement1
(a) a n > 7, " n ³ 1 (b) a n > 3, " n ³ 1
(d) Statement1 is true, Statement2 is true; Statement2 is
not a correct explanation for Statement1 (c) a n < 4, " n ³ 1 (d) a n < 3, " n ³ 1.
(2008) (2002)
1. (d) 2. (b) Answer Key
3. (b)
Mathematical Induction and Its Application 29
1. (d) : Statement1 (k + 1) - 1 1
Let P(n) : 1 + 1 + ... + 1 > n Þ k> Þ k > k + 1 -
k + 1 k + 1
1 2 n
Þ k + 1 > k + 1
1+ 1 > k + 1 ...(iv)
1 2
Step 1 : For n = 2, P(2) : 2 is true
Step 2 : Assume P(n) is true for n = k, i.e. From (iii) & (iv)
11 1 1 + 1 + 1 + ... + 1 + 1 > k + 1
+ + ... + > k ...(i) 123 k k + 1
1 2 k . . . ( i i )
hence (ii) is true for n = k + 1
Step 3 : For n = k + 1, we have to show that hence P(n) is true for n ³ 2
1 + 1 + 1 + ... + 1 + 1 > k + 1 So, Statement1 and Statement2 are correct but
123 k k + 1 Statement2 is not explanation of Statement1
By Assumption step, we get 2. (b) : S(k) = 1 + 3 + .... + (2k – 1) = 3 + k 2 ...(i)
When k = 1, L.H.S of S(k) ¹ R.H.S of S(k)
1 + 1 + ... + 1 > k So S(1) is not true.
1 2 k
Now S(k + 1); 1 + 3 + 5 + .... + (2k – 1) + (2k +1)
1 = 3 + (k + 1) 2 ...(ii)
Adding k + 1 on both sides, we get Let S(k) is true \ 1 + 3 + 5 + .... + (2k – 1) = k2 + 3
Þ 1 + 3 + 5 + ... + (2k – 1) + (2k + 1)
1 + 1 + ... + 1 + 1 > k + 1 ...(iii) = 3 + k2 + 2k + 1 = (k + 1) 2 + 3
12 k k +1 k + 1 Þ S(k + 1) true \ S(k) Þ S(k + 1)
Statement2 3. (b) : an = 7 + an
For n = k Þ an 2 – a n – 7 = 0
k (k + 1) < k + 1
Þ k k + 1 < k + 1 k + 1 Þ k < k + 1
Q k + 1 > k For k ³ 2
Þ 1 > k Þ k > k \ an = 1 ± 1 + 28
k + 1 k + 1 2
Multiplying by k
1 ± 29
= 2 > 3
30 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER BINOMIAL THEOREM
7
1. The term independent of x in expansion of bn - a n an - b n
(a) (b)
æ x2/3 x +1 - x x -1 ö10 (2006)
çè - x1/3 + 1 - x1/2 ø÷ is b - a b - a
an+1 - b n +1 bn+1 - a n +1
(a) 120 (b) 210 (c) 310 (d) 4 (c) (d) .
b - a b - a
(2013) 8. For natural numbers m, n if
(1 – y) m (1 + y)n = 1 + a 1y + a 2 y2 + ...,
2. If n is a positive integer, then ( 3 +1)2n - ( 3 - 1) 2 n is
and a 1 = a 2 = 10, then (m, n) is
(a) an even positive integer. (a) (20, 45) (b) (35, 20)
(b) a rational number other than positive integers. (c) (45, 35) (d) (35, 45). (2006)
(c) an irrational number. úùû1 1 equals the coefficient
( ) 3. é 1
(d) an odd positive integer. (2012) 9. If the coefficient of x7 in ëê ax 2 + bx
The coefficient of x 7 in the expansion of (1 – x – x 2 + x3 ) 6
ø÷öùúû1 1 ,
is of x –7 in éêax - æ 1 then a and b satisfy the relation
(a) –144 (b) 132 ë çè bx2
(c) 144 (d) –132
(a) a + b = 1 (b) a – b = 1
(2011)
a = 1.
4. The remainder left out when 82 n – (62)2 n +1 is divided by 9 is (c) ab = 1 (d) b (2005)
(a) 2 (b) 7 (c) 8 (d) 0 10. If x is so small that x 3 and higher powers of x may be neglected,
(2009) ( ) (1+ x)3/ 2 - 1 + 1 x 3
n then 2 may be approximated as
(1- x)1 / 2
å 5. Statement1 : (r + 1) nCr = (n + 2)2n -1
r = 0 (a) 3 x + 3 x2 (b) 1 - 3 x2
n 8 8
å Statement2 : (r + 1) nCr xr = (1 + x)n + nx(1 + x) n -1
r = 0 (c) x - 3 x2 (d) - 3 x2 . (2005)
2 8 8
(a) Statement1 is true, Statement2 is false
(b) Statemen1 is false, Statement2 is true
(c) Statement1 is true, Statement2 is true; Statement2 is a 11. The coefficient of the middle term in the binomial expansion
correct explanation for Statement1 in powers of x of (1 + ax)4 and of (1 – ax)6 is the same if
(d) Statement1 is true, Statement2 is true; Statement2 is a equals
not a correct explanation for Statement1 (a) –3/10 (b) 10/3 (c) –5/3 (d) 3/5.
(2004)
(2008)
6. In the binomial expansion of (a – b) n, n ³ 5, the sum of 5 th 12. The coefficient of x n in expansion of (1 + x)(1 – x) n is
(a) (–1) n – 1( n – 1)2 (b) (–1)n ( 1 – n)
and 6 th terms is zero, then a/b equals
(c) (n – 1) (d) (–1)n – 1 n. (2004)
n - 5 n - 4 5 (d) n -6 5 .
(a) 6 (b) 5 (c) n - 4 (2007) 13. If s n = n 1 and t n = n r , then t n is equal to
n C r n C r s n
å å
r =0 r =0
1 (a) n – 1 (b) 12 n - 1 (c) 12 n (d) 2n2- 1 .
7. If the expansion in powers of x of the function (1 - ax)(1 - bx) (2004)
is a 0 + a 1x + a2 x 2 + a3 x3 + ..., then an is
Binomial Theorem 31
14. If x is positive, the first negative term in the expansion of (1 equal, then n equals
+ x) 27/5 is
(a) 5t h term (b) 8 th term (c) 6t h term (d) 7t h term. (a) 3r (b) 3r + 1 (c) 2r (d) 2r + 1.
(2003) (2002)
18. The coefficients of x p and x q in the expansion of
15. The number of integral terms in the expansion of ( 3 + 8 5)256 (1 + x) p + q are
is (a) equal
(a) 33 (b) 34 (c) 35 (d) 32. (b) equal with opposite signs
(2003) (c) reciprocals of each other
16. The positive integer just greater than (1 + .0001)1 000 is (d) none of these. (2002)
(a) 4 (b) 5 (c) 2 (d) 3. 19. If the sum of the coefficients in the expansion of (a + b)n
(2002) is 4096, then the greatest coefficient in the expansion is
17. r and n are positive integers r > 1, n > 2 and coefficient of (a) 1594 (b) 792 (c) 924 (d) 2924.
(r + 2) th term and 3r th term in the expansion of (1 + x)2 n are
(2002)
Answer Key
1. (b) 2. (c) 3. (a) 4. (a) 5. (c) 6. (b)
7. (d) 8. (d) 9. (c) 10. (d) 11. (a) 12. (b)
13. (c) 14. (d) 15. (a) 16. (c) 17. (c) 18. (a)
19. (c)
32 JEE MAIN CHAPTERWISE EXPLORER
x -1 ö10 ( ) 6. a n -5 4 .
1. (b) : æ x +1 - - x1/ 2 ø÷ (b) : n C4 an-4 (-b)4 = - n C5 an -5 (- b)5 Þ b =
èç - x1/3 + 1
x2/3 x 7. (d) : From given
= ìï (x1/3 + 1)(x2/3 - x1/3 +1) - ( x +1)( x -1) üï10 1 = (1 - ax )-1(1 - bx ) -1
í x2/3 - x1/3 + 1 ý (1 - ax)(1 - bx)
îï x( x - 1) þï
= (a 0 + a1 x + ..... + an x n + .....) (1 – bx) –1
20 - 5 r = (1 + ax + a 2x 2 + ....... + an –1 x n–1 + a n x n + ......)
= (x1/3 - x-1/2 )1 0 \ Tr +1 = (- 1)r 10C r x 6 (1 + bx + b 2x 2 ....... + b n x n + .......)
Þ (a 0 + a 1 x + ....... + an x n + .......)
Thus 20 - 5r = 0 Þ r = 4 \ Term = 10 C 4 = 210.
6 = 1 + x (a + b) + x 2 (a 2 + ab + b 2 ) + x 3
(a 3 + a 2 b + ab 2 + b 3 ) + .... + .... + .... + x n
2. (c) : ( 3 +1)2n - ( 3 - 1) 2 n (an + an –1 b + a n–2 b 2 + .... + ab n–1 + bn ) + ....
= 2 é 2n C1∙ ( 3)2n-1 + 2nC3 ∙( 3)2n -3 + ....ùû ,
ë
On comparing the coefficient of x n both sides we have
an irrational number. an = an + an –1 b + a n–2 b 2 + ...... + a bn –1 + b n
3. (a) : (1 – x – x 2 + x 3 ) 6 = ((1 – x)(1 – x 2) ) 6 = (a n + an-1b + a n-2b2 + ..... + abn-1 + bn )(b - a )
= (1 – x)6 (1 – x 2 ) 6 b - a
= (1 – 6 C1 x + 6C 2x 2 – 6 C 3x 3 + 6 C 4 x 4 – 6 C5 x 5 + 6 C 6 x6 ) (Multiplying and dividing by b – a)
(1 – 6 C1 x 2 + 6C 2 x 4 – 6C 3x 6 + 6C 4x 8 – 6 C5 x1 0 + 6 C 6 x 12 )
= b n +1 - a n +1 .
b - a
Coeff. of x7 = (– 6 C1 ) (–6 C 3 ) + (– 6C 3) ( 6 C2 ) + (– 6C 5 )(– 6 C1 ) 8. (d) : (1 – y) m (1 + y) n
= 6 ∙ 20 – 20 ∙ 15 + 6 ∙ 6 = 120 – 300 + 36 = –144 = 1 + a 1 y + a2 y 2 + a 3y 3 + ...... + ...... (*)
4. (a) : Using Modulo Arithmetic Differentiating w.r.t. y both sides of (*) we have
8 = –1 (modulo 9) Also 62 = –1 (modulo 9) –m(1 – y)m –1( 1 + y) n + (1 – y)m n(1 + y) n–1
Þ 8 2n – (62) 2n + 1 = [(–1) 2n – (–1) 2n + 1 ]mod 9
= (1 + 1)mod 9 = 2 mod 9 Þ Remainder = 2 = a 1 + 2a 2 y + 3a 3 y 2 + 4a4 y 3 + .......
Þ n(1 + y) n–1( 1 – y)m – m(1– y) m–1 (1 + y)n
= a1 + 2a 2 y + 3a 2 y2 + 4a 4y 3 + ....... .....(**)
n n n Again differentiating (**) with respect to y we have
5. (c) : å (r + 1) nCr = å r × nCr å + n Cr [n(n – 1)(1 + y) n–2 (1 – y)m + n(1 + y)n –1( –m) (1 – y) m–1 ]
r =0 r =0 r =0
n n × n –[m(1 + y)n (m – 1)(1 – y) m–2 (1–y)m –1n (1 + y)+n–1]
r
n -1 Cr n Cr
å å = r × +
-1 = 2a 2 + 6a3 y + ....... ....... (***)
r =0 r = 0 Now putting y = 0 in (**) and (***) we get
= n × 2n-1 + 2n = 2n -1 (n + 2) n – m = a1 = 10 (A)
and m 2 + n2 –(m + n) –2 mn = 2a2 = 20 ....(B)
Thus Statement1 is true.
Solving (A) and (B)
n n
n = 45, m = 35
Again å (r + 1) nCr xr = å r × nCr xr + å nCr xr
\ (m, n) = (35, 45)
r =0 r = 0
11 11 - r
n n ( ) ( ) 9. 1 1
(c) : Tr + 1 of ax2 + bx = 11Cr (ax 2 ) r bx
å å = n n-1C r -1 xr + nCr xr
ö11 ö11 - r
r =0 r =0 ÷ ÷
ø ø
= nx(1+ x)n-1 + (1 + x) n
Substitute x = 1 in the above identity to get Tr + 1 of æ ax - 1 = 11C r ( ax ) r æ - 1
å(r + 1) nCr = n × 2n-1 + 2 n èç bx2 èç bx2
Statement2 is also true & explains Statement1 also. ax2 + 1 11 a 6
bx b5
( ) \ Coeff. of x7 in = 11C 5
Binomial Theorem 33
and coefficient of x– 7 in æ ax - 1 ö11 = 11C 6 a 5 16. (c) : Let R = æ 1 + 1 ö1000
çè bx2 ø÷ b6 ç ÷
è 104 ø
Now 11 C5 a6 = 11 C6 a 5 \ ab = 1. = 1 + 1000 èæç 1 ö1 999 æ 1 ö 2 æ 1 ö 103
b5 b6 104 ø÷ 1000 çè 104 ø÷ + .... + çè 104 ø÷
+
2
(1+ x)3/ 2 - 1 + 1 x 3
( ) 10. (d) : 2 < 1 + 1 + 1 1 10
10 102 + 103 + ...¥ =
(1 - x)1 / 2
9
( ) ( ) = 2 2 2 2! 10
1+ 3 x + 3 1 1 x2 + ... - 1 + 3 × 1 x + 3 × 2 × 1 x 2 + .... \ R <
2 2! 4
9 10
(1 - x)1 / 2 \ The positive integer just greater than is 2.
= - 3 x2 (1- x) -1/ 2 = - 3 x2 éëê1 + 1 x + 1 × 3 × 1 x2 + ... úùû 9
8 8 2 2 2 2! 17. (c) : Given r > 1, n > 2 and
= - 3 x2 + higher powers of x2 . Coefficient of
8 Tr + 2 = Coefficient of T 3r in (1 + x) 2n
Þ 2nC r + 1 = 2n C 3r – 1
11. (a) : Coefficient of middle term in (1 + ax)4 = coefficient of ì2n - 3r + 1 = r + 1
middle term in (1 – ax)6 Þ 3r – 1 = r + 1 and ï Þ 2n = 4 r
í
\ 4 C 2a2 = 6C 3( – a)3 Þ a = - 3 2r = 2 ïî n = 2 r
10
12. (b): (1 + x) (1 – x) n = (1 – x)n + x(1 – x) n ìQ nCx = n C y
ï
\ Coefficient of x n is = (–1) n + (–1)n – 1 nC 1 r = 1 í Þ x + y = n
= (–1) n [1 – n]
ï or x = y
î
å n r 18. (a) : In the expansion of (1 + x) p + q
13. (c): t n = r = 0 n Cr Tr + 1 = p + qC r x r
\ Coefficient of x p = p + qC p
n n - (n - r ) n n
å å å tn = Þ tn = n =0 1 - = 0 n - r ( p + q)! = ( p + q )!
r = 0 nCr n Cn - r p!( p + q - p)! p!q!
n Cn - r r r = ...(i)
å å tn = n 1 - n r replacing n – r by r Also coefficient of x q in (1 + x) p + q is
nCr = 0 n Cr = p + q Cq
r
( p + q )!
tn = ns n – tn = q!( p + q - q)!
t n
\ sn n
=
2
27 ( p + q )!
=
14. (d) : General term in the expansion of (1 + x) 5 ....(ii)
q! p!
n(n - 1) ..... (n - r + 1) x r
Tr + 1 = r ! \ By (i) and (ii)
Coefficient of x p in (1 + x) p + q = Coefficient of x q in
27 32 (1 + x)p + q
\ n – r + 1 < 0 Þ + 1 < r Þ r >
19. (c) : Consider (a + b)n = C 0 a n + C1 a n – 1b
5 5 + C2 an – 2 b2 + .... + C nb n
Þ r > 6 Putting a = b = 1
\ 2n = C 0 + C1 + C 2 + .... + Cn
15. (a) : (3 1/2 + 51 /8 )2 56
2 n = 4096 = 2 12
r
Þ n = 12 (even)
Tr + 1 = 256C r ( 3)256 - r 58 Now (a + b)n = (a + b) 12
For integral terms 256 - r , r are both positive integer as n = 12 is even so coefficient of greatest term is
2 8
n C n = 12C12 = 12 C6
\ r = 0, 8, 16, ...256
2 2
\ 256 = 0 + (n – 1)8 using tn = a + (n – 1)d
256 256 = 12 ´ 11 . 10 9 . 8 . 7 = 11 ´ 9 .8. 7 3×4 × 7 = 924
6 54 3 2 1 3.2.1 = 11 ×
\ = n – 1 \ n = + 1
8 8
n = 32 + 1 Þ n = 33
34 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER SEQUENCES AND SERIES
8
1. The sum of first 20 terms of the sequence 0.7, 0.77, 6. The sum to infinity of the series 1 + 32 + 6 10 1344 + ...... is
0.777, ..., is 32 33
+ +
(a) 79 (99 - 10-20 ) (b) 7 (179 + 10-20 ) (a) 3 (b) 4 (c) 6 (d) 2
81 (2009)
(c) 97 (99 + 10-20 ) 7 - 10-20 ) 7. The first two terms of a geometric progression add up to 12.
81
(d) (179 (2013) The sum of the third and the fourth terms is 48. If the terms
2. Statement 1 : The sum of the series 1 + (1 + 2 + 4) + of the geometric progression are alternately positive and
(4 + 6 + 9) + (9 + 12 + 16) + ... + (361 + 380 + 400) is 8000. negative, then the first term is
n (a) 4 (b) – 4 (c) – 12 (d) 12
Statement 2 : å (k3 - (k - 1)3) = n 3 for any natural number n. (2008)
k =1 8. The sum of the series 1- 1+ 1 - .... upto infinity is
2! 3! 4!
(a) Statement 1 is true, Statement 2 is true; Statement 2 is - 1 + 1
not a correct explanation for Statement 1. (a) e 2 (b) e 2 (c) e –2 (d) e –1 .
(b) Statement 1 is true, Statement 2 is false. (2007)
(c) Statement 1 is false, Statement 2 is true.
(d) Statement 1 is true, Statement 2 is true; Statement 2 is a 9. In a geometric progression consisting of positive terms, each
term equals the sum of the next two terms. Then the common
correct explanation for Statement 1. (2012)
ratio of this prog ression is equals
3. If 100 times the 100 th term of an A.P. with nonzero common (a) 5 (b) 12 ( 5 - 1)
difference equals the 50 times its 50 th term, then the 150 th
term of this A.P. is (b) zero (c) (1 1 - 5 ) (d) 1 5. (2007)
(a) 150 2
2
(c) – 150 (d) 150 times its 50 th term
(2012) 10. Let a 1, a2 , a 3, ... be terms of an A.P. If
4. A man saves ` 200 in each of the first three months of his a1 + a2 + ... + a p = p 2 p ¹ q , then a6 equals
service. In each of the subsequent months his saving increases , a2 1
a1 + a2 + ... + aq q2
by ` 40 more than the saving of immediately previous month. (a) 41/11 (b) 7/2 (c) 2/7 (d) 11/41.
His total saving from the start of service will be ` 11040 (2006)
after 11. If a1 , a 2 , ..., an are in H.P., then the expression a 1 a 2 + a 2a 3
(a) 20 months (b) 21 months + ... + an – 1a n is equal to
(c) 18 months (d) 19 months (2011) (a) n(a1 – an ) (b) (n – 1)(a1 – a n)
5. A person is to count 4500 currency notes. Let an denote the (c) na1 a n (d) (n – 1)a 1a n . (2006)
number of notes he counts in the n th minute. 12. If the coefficients of r th, (r + 1) th and (r + 2) th terms in the
If a 1 = a 2 = ... = a1 0 = 150 and a 10 , a 11 , .... are in an A.P. with
binomial expansion of (1 + y) m are in A.P., then m and r satisfy
common difference –2, then the time taken by him to count
the equation
all notes is (a) m 2 – m(4r – 1) + 4r 2 + 2 = 0
(a) 24 minutes (b) 34 minutes (b) m 2 – m(4r + 1) + 4r2 – 2 = 0
(c) 125 minutes (d) 135 minutes (c) m2 – m(4r + 1) + 4r2 + 2 = 0
(2010) (d) m2 – m(4r – 1) + 4r2 – 2 = 0. (2005)
Sequences and Series 35
13. If x = ¥ an , y= ¥ bn , z = ¥ cn where (a) are in G.P. (b) are in H.P.
å å å a, b, c are in A.P. (c) satisfy a + 2b + 3c = 0 (d) are in A.P. (2003)
n=0 n=0 n = 0
and |a| < 1, |b| < 1, |c| < 1 then x, y, z are in 20. Let f (x) be a polynomial function of second degree. If f (1)
(a) H.P. = f (–1) and a, b, c are in A.P., then f ¢(a), f ¢(b) and f ¢(c)
(b) ArithmeticGeometric progression are in
(c) A.P. (a) G.P. (b) H.P.
(d) G.P. (2005) (c) ArithmeticGeometric Progression
14. If a 1, a2 , a3 , ..., an , ... are in G.P., then the determinant (d) A.P. (2003)
log an log an +1 log a n + 2
21. The sum of the series 1 - 1 + 1 - ... upto ¥ is equal to
1× 2 2×3 3 × 4
D = log an + 3 log an + 4 log an + 5 , is equal to
(a) log e 2 – 1 (b) log e 2
log an + 6 log an + 7 log an + 8
(c) log e (4/e) (d) 2log e 2. (2003)
(a) 0 (b) 1 (c) 2 (d) 4. 22. If x 1, x2 , x 3 and y 1, y 2 , y3 are both in G.P. with the same common
ratio, then the points (x1 , y 1 ), (x2 , y2 ) and (x3 , y3 )
(2005) (a) lie on an ellipse
15. The sum of the series
1 + 1+ 1 + 1 + .... ¥ is (b) lie on a circle
× 2! 16 × 4! 64 × 6!
4 (c) are vertices of a triangle
(a) e + 1 (b) e - 1 (c) e + 1 (d) e - 1 . (d) lie on a straight line. (2003)
ee 2 e 2 e
(2005) 23. Let R 1 and R 2 respectively be the maximum ranges up and
down on an inclined plane and R be the maximum range on
16. Let Tr be the r th term of an A.P. whose first term is a and the horizontal plane. Then, R 1, R, R2 are in
common difference is d. If for some positive integers m, n, (a) A.P.
(b) G.P.
m ¹ n, T m = n1 , and T n = m1 , then a – d equals (c) H.P.
(d) ArithmeticGeometric Progression (A.G.P.). (2003)
(a) 1/mn (b) 1 (c) 0 (d) 1 + n1 .
m
(2004) 24. If 1, log9 (31 – x + 2), log 3[ 4 × 3x – 1] are in A.P. then x equals
17. The sum of first n terms of the series (a) log3 4 (b) 1 – log3 4
12 + 2 × 22 + 32 + 2 × 4 2 + 52 + 2 × 62 + .... is n(n + 1) 2 when (c) 1 – log4 3 (d) log4 3. (2002)
2
25. 13 – 2 3 + 3 3 – 4 3 + ... + 9 3 =
n is even. When n is odd, the sum is
(a) 425 (b) – 425 (c) 475 (d) – 475.
(a) n(n + 1) 2 (b) n 2 (n + 1) (2002)
4 2
26. Sum of infinite number of terms in GP is 20 and sum of their
3n(n + 1) é n(n + 1) ûùú 2 .
(c) 2 (d) ëê 2 (2004) square is 100. The common ratio of GP is
1 1 1 (a) 5 (b) 3/5 (c) 8/5 (d) 1/5.
2! 4! 6!
18. The sum of series + + + .... is (2002)
(a) (e - 1) 2 (b) (e 2 - 1) 27. The value of 21/4 × 41/8 × 81/6 ... ¥ is
2 e 2e
(a) 1 (b) 2 (c) 3/2 (d) 4.
(2002)
(c) (e2 - 1) (d) (e 2 e- 2) . (2004)
2
28. Fifth term of a GP is 2, then the product of its 9 terms is
19. If the system of linear equations x + 2ay + az = 0, (a) 256 (b) 512 (c) 1024 (d) none.
x + 3by + bz = 0, x + 4cy + cz = 0 has a nonzero solution,
then a, b, c (2002)
Answer Key
1. (b) 2. (d) 3. (b) 4. (b) 5. (b) 6. (a)
7. (c) 8. (d) 9. (b) 10. (d) 11. (d) 12. (b)
13. (a) 14. (a) 15. (c) 16. (c) 17. (b) 18. (a)
19. (b) 20. (d) 21. (c) 22. (d) 23. (c) 24. (c)
25. (a) 26. (b) 27. (b) 28. (b)
36 JEE MAIN CHAPTERWISE EXPLORER
1. (b) : tr = 014.74474727474..4.4.73 Hence, n = 34 is the only answer.
r terms 2+ 6 10 1344 + .....
3 32 33
7 7 7 7 10- 6. (a) : Let S =1+ + + ...(1)
10 102 10 r 9 (1 ...(2)
= + + .... + = - r )
...(1)
20 æ 20 10- r ÷ö 1S = 1 + 2 + 6 + 10 + ..... ...(2)
ç 20 r =1 ø 3 3 31 31 3
å å S20 = tr 7 è 7 íîì20 1 10-20 ) üýþ
= 9 - = 9 - 9 (1 - Subtracting (2) from (1), we get
r =1
2 1+ 4 4 4
= 7 (179 + 10-20 ) S =1+ 3 32 + 33 + 34 + .....
81
3
2. (d) :Statement 1 : = 4 éëê1 + 1 + ......to ¥ùûú = 4 ∙ 1 = 4 ∙ 1 = 2
3 3 3 1 - 1 3 2 / 3
1 + (1 + 2 + 4) + (4 + 6 + 9) + .... + (361 + 380 + 400) is 8000
3
n Þ 2 = 2 \ S = 3
S
å Statement 2 : (k3 - (k - 1)3) = n3 3
k = 1
7. (c) : Let the G.P. be a, ar, ar 2, ar 3, ...
Statement 1 : T 1 = 1, T 2 = 7 = 8 – 1, we have a + ar = 12
T 3 = 19 = 27 – 8 Þ T n = n 3 – (n – 1) 3 ar2 + ar 3 = 48
\ Statement 2 is a correct explanation of statement 1.
on division we have
3. (b) :100 (a + 99d) = 50 (a + 49d) ar2 (1 + r ) 48 r 2 = 4
Þ a + 149d = 0 i.e., T 150 = 0 =Þ
a(1+ r) 12
4. (b) : Let it happens after n months.
\ r = ± 2
But the terms are alternately positive and negative,
3 ´ 200 + n -2 3 {2 ´ 240 + (n - 4)40} = 11040 \ r = – 2
æ n -2 3 öø÷ (480 Now a = 12 = 12 = 1-21 = -12 From (1)
çè 1+ r 1- 2
Þ + 40n - 160) = 11040 - 600 = 10440
Þ n 2 + 5n – 546 = 0 Þ (n + 26)(n – 21) = 0 8. (d) : Q e- x = 1- x + x2 - x3 + x 4 - ........
2! 3! 4!
\ n = 21. upto infinity
5. (b) : We have a 1 + a 2 + ... + a n = 4500 Then put x = 1, we get
Þ a1 1 + a1 2 + ... + a n = 4500 – 10 × 150 = 3000
Þ 148 + 146 + .... = 3000 e -1 = 1- 1 + 1 - 1 + 41! - ........ upto infinity.
1! 2! 3!
Þ n -21 0 × (2 ´ 148 + (n - 10 - 1)(-21)) = 3000
9. (b) : Given, a = ar + ar 2 Þ r 2 + r – 1 = 0
Þ r = -1 + 5 .
2
Let n – 10 = m 10. (d) : Given a1 , a2 , a 3, ... be terms of A.P.
Þ m × 148 –m(m – 1) = 3000 a1 + a2 + ....a p = p 2
Þ m2 – 149m + 3000 = 0 a1 + a2 + ... aq q2
Þ (m – 24)(m – 125) = 0
\ m = 24, 125, p [ 2 a1 + ( p - 1) d ] p 2
giving n = 34, 135 2
But for n = 135, we have =
a 135 = 148 + (135 – 1)(–2) = 148 – 268 < 0 Þ 2q [ 2 a1 + ( q - 1) d ] q 2
But a 34 is positive.
Þ 2 a1 + ( p - 1) d = p
2 a1 + ( q - 1) d q
Sequences and Series 37
Þ [2a 1 + (p – 1)d]q = p[2a1 + (q – 1)d] as a, b, c Î AP \ 2b = a + c
Þ 2a 1 (q – p) = d[(q – 1)p – (p – 1)q] 2 èæç y - 1 ö = x -1 + z -1 Þ 2 = 1 + 1z
y ø÷ x z y x
Þ 2a1 ( q – p) = d(q – p) Þ 2a 1 = d
Þ a6 = a1 + 5d = a1 + 10 a1 Þ x, y, z Î H.P.
a21 a1 + 20d a1 + 40 a1
14. (a) : Let tr denote the r th term of G.P. with first term b and
Þ a6 = 11 common ratio R
a21 41
\ tr = b Rr - 1 . \ log r = logb + (r - 1)log R
11. (d) : Given a 1 , a 2 , ... an are in H.P. Now from given determinant we have
Þ 1 , 1 ..... 1 Î A.P.
a1 a2 a n
log b + (r -1)log R logb + r log R logb + (r +1)log R
Þ 1 - 1 = d log b + (r + 2)log R log b + (r + 3)log R log b + (r + 4)log R
a2 a1 log b + (r + 5)log R logb +(r + 6)log R log b + (r + 7)log R
Þ a1a 2 = a1 - a2 = a1 - a 2 ... (i) Þ using (applying C2 ® 2C 2 – (C 1 + C 3) )
d dd ... (ii)
a2 a3 = a2 - a3 ... (n) log b + (r -1) log R 0 log b + log R(r +1)
dd = 1 log b + (r + 2) log R 0 log b + (r + 4) log R
M 2
log b + (r + 5) log R 0 logb + (r + 7) log R
an -1 - an
an-1 an = dd = 1 ´ 0 = 0.
2
Adding (i), (ii) ............ (n) equations we get 15. (c) : 1 + 1 + 1 + 1 + .... ¥
4 ( 2!) 16 ( 4!) 64(6!)
a1 - an
a 1a 2 + a2 a 3 + a3 a 4 + .... an – 1 a n = d d = 1 + 1 + 1 + 1 + .... ¥
22 2! 24 (4!) 26 (6!)
Also 1 = 1 + (n -1) d
an a1 12 êéêë2ççèæ1 + 1 1 1 öù
22 2! 24 (4!) 26 (6!) .... ¥ ÷÷øûúú
a1 - a n = + + +
d
Þ = ( n - 1) a1 a n
\ a 1a 2 + a2 a 3 + ... an – 1 a n = (n – 1)a1 a n . = 12 éêë2çèæ1 + x2 + x4 + x6 + ...¥ öù
2! 4! 6! ÷øúû
12. (b) : Tr + 1 = m C r y r . \ m C r – 1 + m Cr + 1 = 2 × m Cr = 12 éë ex + e- x ùû
Þ m! + m ! = 2 r m ! where x = 1/2
1)!(m - 1)!(m - m - r
(r - r + 1)! ( r + r - 1)! !( )!
Þ r(r + 1) + (m - r +1)(m - r ) = 1 [e1/ 2 + e -1/ 2 ] = e + 1
(r +1)!(m - r +1)! (r +1)!(m - r + 1)!
2 2 e
1 ...(i)
= 2(r +1)(m - r +1) 16. (c) : Tm = a + (m – 1) d = n ...(ii)
(r +1)!(m - r +1)!
1
Þ r (r + 1) + (m – r + 1)(m – r) = 2(r + 1)(m – r + 1) Tn = a + (n – 1) d = m
Þ r (r + 1) + (m – r)2 + m – r – 2(r + 1) (m – (r – 1))=0 1 1
Now Tm – Tn = - = (m – n) d
n m
Þ r (r + 1) + m2 + r2 – 2mr + m – r + 2 (r2 – 1) 1 1
– 2m(r + 1) = 0 Þ d = and a =
mn mn
Þ m 2 – m (4r + 1) + 4r2 – 2 = 0. \ a – d = 0
13. (a) : Given | a | < 1, | b | < 1, | c | < 1, a, b, c Î A.P. 17. (b): As Sn is needed for n is odd let n = 2k + 1
and ¥ an = 1, ¥ bn = 1 1 b , ¥ c n = 1 \ Sn = S2 k + 1
1- a - 1 - c = Sum up to 2k terms + (2k + 1)t h term
å å å
2k (2k + 1) 2
n=0 n=0 n = 0 = + last term
\ x = 1 1 a , y = 1 1 b , z = 1 2
- - 1 - c
(n - 1)n2 n2 (n + 1)
Þ a = x -1, b = y -1 , c = z -1 = 2 + n 2 as n = 2k + 1 = 2
x yz
38 JEE MAIN CHAPTERWISE EXPLORER
é x2 x4 ù = –[log e 2 – 1]
18. (a) : e x + e– x = 2 ê1 + + + ... ¥ ú = 1 – log e 2
ëê 2! 4! úû Now s = s 1 – s 2 = (A) – (B) ...(B)
= log e2 – 1 + log e 2 = log(4/e)
e + e-1 1 + 1 + ... ¥
– 1 = 22. (d) : Let x1 = a \ x2 = ar, x 3 = ar2
2 2! 4! and y 1 = b \ y2 = br, y3 = br2
Now A(a, b), B(ar, br), C(ar 2, br 2 )
(e - 1) 2 = 1 + 1 + 1 + ... ¥
2e 2! 4! 6!
19. (b) : For non trivial solution the determinant of the coefficient b(1 - r) b
of various term vanish Now slope of AB = = and
1 2 a a a(1 - r) a
i.e. 1 3b b = 0
br(1 - r) b
1 4 c c slope of BC = =
Þ (3bc – 4bc) – 2a(c – b) + a(4c – 3b) = 0 ar(1 - r) a
as slope of AB = slope of BC
2ac \ AB || BC, but point B is common so
Þ a + c = b
Þ a, b, c Î H.P. A, B, C are collinear.
23. (c) : Let q be the angle of inclination of plane to horizontal
20. (d) : Let the polynomial be f (x) = ax2 + bx + c and u be the velocity of projection of the projectile
given f (1) = f (–1) Þ b = 0 u 2 u 2
\ f (x) = ax 2 + c
\ R1 = g (1 + sin q) , R2 = g (1 - sin q)
now f ¢ (x) = 2ax
\ f ¢ (a) = 2a 2 , f ¢ (b) = 2ab, f ¢ (c) = 2ac \ 1 + 1 = 2 g = 2
as a, b, c Î A.P. R1 R2 u 2 R
Þ a 2 , ab, ac Î A.P. Þ 2a2 , 2ab, 2ac Î A.P.
Þ f ¢ (a), f ¢ (b), f ¢ (c) Î A.P. Þ R 1 , R, R2 Î H.P.
1
24. (c) : As 1, 2 log 3 (3 1 – x + 2), log 3( 4∙3x – 1) Î A.P.
Þ log3 ( 31 – x + 2) = log 3( 4∙3x – 1) + 1
1 1 1
21. (c): s = - + – ....¥ Þ 31 – x + 2 = (4∙3 x – 1) × 3 Q log3 3 = 1.
1× 2 2 × 3 3 × 4 Þ 31 – x + 2 = 12∙3 x – 3
1 1 1
Let s 1 = 1× 2 + 3 × 4 + 5 × 6 + ....¥ Þ 3x [(31 – x) + 2] = 12∙3 2x – 3∙3 x
1 1 1 (multiplying 3 x both side)
\ tn = (2n - 1)(2n) = 2n - 1 - 2n
Þ 12t 2 – 5t – 3 = 0 where t = 3 x
Þ (3t + 1) (4t – 3) = 0
å \ s n = åtn = æ 1 1 - 1 ö Þ t = –1/3, t = 3/4
çè 2n - 2 n ÷ø Þ 3 x = –1/3 which is not possible
1 1 1 1 ...(A) 3 3
= 1 - + - + .....¥ and t = Þ 3 x =
2345 44
= loge 2
Þ x log3 3 = log3 3 – log3 4
1 1 1 (By taking logarithm at the base 3 both sides)
Again s 2 = 2 × 3 + 4 × 5 + 6 × 7 + ....¥
1 Þ x = 1 – log 34
t¢ n = (2n) (2n + 1)
25. (a) : (13 + 33 + 53 + ..... + 93 ) – (23 + 43 + 63 + 83 )
s 2 = å tn ¢ =
= (1 3 + 3 3 + 5 3 + .... + 93 ) – 23 ( 1 3 + 2 3 + 33 + 43 )
å å 1 æ 1 - 1 ö
= çè 2n 2n + 1 ÷ø = [1 3 + 3 3 + ..... + (2n – 1)3 ] n = odd = 5
(2n)(2n + 1) – 23 [ 13 + 2 3 + .... + n 3] n = even = 4
= [2n(n + 1) (n + 2) (n + 3) – 12n(n + 1) (n + 2) + 13n(n
= æ 1 - 1 ö + æ 1 - 14 ÷öø + ....¥ é 2 1) 2 ù
çè 2 3 ø÷ çè 3 ê ú
+ 1) – n]n = 5 (odd) – 2 3 n (n +
é 1 1 1 1 + ù êë 4 úû n = 4
= - ëê - 2 + 3 - 4 + 5 ...¥ úû (even)
(Remember this result)
Sequences and Series 39
= [2 × 5 × 6 × 7 × 8 – 12 × 5 × 6 × 7 + 13 27. (b) : S¥ = 1 + 2 +136 + 342 + ... ¥ = 2l (say) ...(*)
8
24
2 3 çæè 16 ´ 25 ö
× 5 × 6 – 5] – 4 ÷ø Where l = 1 + 2 + 3 + 4 + ....¥ ... (A)
4 8 16 32 ... (B)
= [3750 – 5(505)] – 2 × 16 × 25
l = 0 + 1 + 2 + 3 + 4 + ...¥
= 1225 – 800 = 425 2 8 16 32 64
26. (b) : Let terms of G.P. are a, ar, ar 2, .... Now (B) – (A) Þ l = 1 + 1 + 1 + 1 + 1 + ...
2 4 8 16 32 64
a
\ S¥ = 1 - r where a = first term, r = common ratio
l a = 1´ 2
S¥ = 20 2 = 1- r 4 1 \ l = 1
According to question a = 20 so S¥ = 2 1
1 - r
Þ a = 20(1 – r) ...(i) 28. (b) : Let first term of a G.P is a and common ratio r
a 2 \ t 5 = ar 4 = 2
Also 1 - r 2 = 100
9
Þ 1 a r . a = 100
- 1 + r Õ \ ai = a × ar ar2 .....ar8
Þ a = 5(1 + r) ...(ii) i = 1
Solving (i) and (ii) we have r = 3/5 8 ´ 9
= a9 r 2
= a9 r 36
= (ar 4 ) 9
= 2 9 = 512
40 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER DIFFERENTIAL CALCULUS
9
1. lim (1 - cos2x)(3 + cosx) is equal to Statem ent 2 : a = 1 and b = -1
x tan4x 2 4
x ®0
(a) Statement 1 is true, Statement 2 is true; Statement 2 is
(a) 1/2 (b) 1 (c) 2 (d) –1/4
(2013) not a correct explanation for Statement 1.
(b) Statement 1 is true, Statement 2 is false.
2. At present, a firm is manufacturing 2000 items. It is estimated (c) Statement 1 is false, Statement 2 is true.
that the rate of change of production P w.r.t. additional number (d) Statement 1 is true, Statement 2 is true; Statement 2 is
of workers x is given by dP = 100 - 12 x . If the firm employs a correct explanation for Statement 1. (2012)
dx 7. A spherical balloon is filled with 4500 p cubic metres of helium
25 more workers, then the new level of production of items is
(a) 3000 (b) 3500 (c) 4500 (d) 2500 gas. If a leak in the balloon causes the gas to escape at the
(2013) rate of 72p cubic metres per minute, then the rate (in metres
3. If y = sec(tan –1x), then dy x = 1 is equal to per minute) at which the radius of the balloon decreases 49
dx at minutes after the leakage began is
(a) 1 (b) 1 (c) 2 1 (a) 2/9 (b) 9/2 (c) 9/7 (d) 7/9
2 (d) 2 (2012)
(2013)
4. Consider the function, f(x) = |x – 2| + |x – 5|, x Î R 8. d 2 x equals to
Statement 1 : f ¢(4) = 0 dy 2
Statement 2 : f is continuous in [2, 5], differentiable in æ d2 y ÷øö÷ çæè dy ö -2 æ d 2 y öø÷÷ çèæ dy ö-3
(2, 5) and f(2) = f(5). ççè dx2 dx ÷ø èçç dx 2 dx ÷
(a) Statement 1 is true, Statement 2 is true; Statement 2 is (a) (b) - ø
not a correct explanation for Statement 1. (c) æ d 2 y ö-1 (d) - æ d 2 y ø÷ö÷ -1 çèæ dy ö-3 (2011)
(b) Statement 1 is true, Statement 2 is false. ççè dx2 ÷÷ø ççè dx2 dx ÷ø
(c) Statement 1 is false, Statement 2 is true.
(d) Statement 1 is true, Statement 2 is true; Statement 2 æ 2)} ÷÷øö
xli®m2 ççè
is a correct explanation for Statement 1. (2012) 9. 1 - cos {2(x -
x -2
5. If f : R ® R is a function defined by
f(x) = [x] cos æçè 2x - 1 ö÷ø p , where [x] denotes the greatest integer (a) equals - 2 (b) equals 1
2 (c) does not exist (d) 2
function, then f is equals 2
(a) discontinuous only at nonzero integral values of x. (2011)
(b) continuous only at x = 0. 10. The values of p and q for which the function
(c) continuous for every real x. ì sin(p + 1)x + sin x , x < 0
(d) discontinuous only at x = 0. x
(2012) ï
6. Let a, b Î R be such that the function f given by q , x = 0
f (x) = ïï
í
f(x) = ln |x| + bx 2 + ax, x ¹ 0 has extreme values at ï x + x2 - x , x>0
x = – 1 and x = 2. ï
ïî x3/2
Statement 1 : f has local maximum at x = – 1 and at x = 2. is continuous for all x in R, are
Differential Calculus 41
(a) p = - 3 , q = 1 (b) p = 1 , q = 3 17. Let y be an implicit function of x defined by x 2x – 2x x cot y –
22 22 1= 0. Then y¢(1) equals
(c) p = 1 , q = - 3 (d) p = 5 , q = 1 (2011) (a) 1 (b) log 2 (c) –log 2 (d) –1
22 22 (2009)
11. Let f : (–1, 1) ® R be a differentiable function with
f (0) = –1 and f ¢(0) = 1, g(x) = [f (2f (x) + 2)] 2. Then g¢(0) 18. Suppose the cubic x 3 – px + q has three distinct real roots
where p > 0 and q > 0. Then which one of the following
= holds?
p p
(a) 4 (b) – 4 (c) 0 (d) –2 (a) The cubic has maxima at both 3 and – 3
p p
(2010) (b) The cubic has minima at 3 and maxima at – 3
12. Let f : R ® R be a positive increasing function p
(c) The cubic has minima at – 3 and maxima
with lim f (3x) = 1. Then lim f (2x) = p
x®¥ f (x) x®¥ f (x) at 3
p p
(a) 1 (b) 2/3 (c) 3/2 (d) 3 (2010) (d) The cubic has minima at both 3 and – 3
(2008)
13. Let f : R ® R be defined by
ìk - 2x, if x £ -1
f (x) = íî 2x + 3, if x > -1
If f has a local minimum at x = –1, then a possible value of
k is
(a) 1 (b) 0 (c) –1/2 (d) –1 ïì( 1
í x - 1
(2010) 19. Let f ( x ) = x - 1) sin if x ¹ 1
14. Let f : R ® R be a continuous function defined by îï 0 if x = 1
1 Then which one of the following is true?
+ 2 e - x
f (x) = ex . (a) f is differentiable at x = 1 but not at x = 0
(b) f is neither differentiable at x = 0 nor at x = 1
Statement1 : f (c) = 1/3, for some c Î R. (c) f is differentiable at x = 0 and at x = 1
Statement2 : 0 < f (x) £ 1 , for all x Î R. (d) f is differentiable at x = 0 but not at x = 1
22
(2008)
(a) Statement1 is true, Statement2 is true; Statement2 is a
correct explanation of Statement1. 20. How many real solutions does the equation
(b) Statement1 is true, Statement2 is true; Statement2 is x 7 + 14x 5 + 16x 3 + 30x – 560 = 0 have?
not a correct explanation for Statement1. (a) 5 (b) 7
(c) Statement1 is true, Statement2 is false. (c) 1 (d) 3 (2008)
(d) Statement1 is false, Statement2 is true. (2010)
21. If p and q are positive real numbers such that p 2 + q 2 = 1, then
15. Let f (x) = x|x| and g(x) = sin x. the maximum value of (p + q) is
Statement1 : gof is differentiable at x = 0 and its 1 1
2 (b) 2
derivative is continuous at that point. (a) (d) 2.
Statement2 : gof is twice differentiable at x = 0. (c) 2 (2007)
(a) Statement1 is true, Statement2 is true; Statement2 is 22. The function f : R – {0} ® R given by
not a correct explanation for Statement1.
(b) Statement1 is true, Statement2 is false. f ( x ) = 1x - 2
e2 x - 1
(c) Statement1 is false, Statement2 is true.
(d) Statement1 is true; Statement2 is true; Statement2 is a can be made continuous at x = 0 by defining f (0) as
correct explanation for Statement1. (2009) (a) 0 (b) 1
16. Given P(x) = x 4 + ax 3 + bx 2 + cx + d such that x = 0 is the only (c) 2 (d) –1. (2007)
real root of P¢(x) = 0. If P(–1) < P(1), then in the interval [–1, 1] : 23. Let f : R ® R be a function defined by
(a) P(–1) is not minimum but P(1) is the maximum of P
(b) P(–1) is the minimum but P(1) is not the maximum of P f (x) = min {x + 1 , |x| + 1}. Then which of the following is
(c) neither P(–1) is the minimum nor P(1) is the maximum
true ?
of P
(d) P(–1) is the minimum and P(1) is the maximum of P (a) f (x) is differentiable everywhere
(b) f (x) is not differentiable at x = 0
(c) f (x) ³ 1 for all x Î R
(2009) (d) f (x) is not differentiable at x = 1. (2007)
42 JEE MAIN CHAPTERWISE EXPLORER
24. The function f (x) = tan –1 (sin x + cos x) is an increasing 33. Let a and b be the distinct roots of ax 2 + bx + c = 0, then
function in 1- cos(ax2 + bx + c )
( x - a ) 2
çæè 0, p ÷öø çèæ - p , p ÷øö lim is equal to
2 2 2
(a) (b) x ®a
çæè p , p ÷öø çèæ - p , p ÷öø . (a) 0 (b) a2 (a - b ) 2
4 2 2 4 2
(c) (d) (2007)
-a2 (a - b )2 .
25. A value of c for which conclusion of Mean Value Theorem (c) 1 (a - b ) 2 (d) 2 (2005)
2
holds for the function f (x) = log e x on the interval [1, 3] is 34. The normal to the curve x = a(cosq + q sinq),
(a) log 3 e (b) log e 3 y = a(sinq – qcosq) at any point q is such that
(c) 2 log 3 e (d) 12 loge 3. (2007) (a) it makes angle p + q with xaxis
2
26 If x m ∙ y n = (x + y)m + n , then dy/dx is (b) it passes through the origin
x + y (c) it is at a constant distance from the origin
(b) xy
y ( ) (d) it passes through
(a) x a p , - a . (2005)
(d) y . 2
x
(c) xy (2006) 35. If f is a realvalued differentiable function satisfying
| f ( x) - f ( y) | £ ( x - y)2 , x, y Î R and f (0) = 0, then f (1)
27. A triangular park is enclosed on two sides by a fence and on equals
the third side by a straight river bank. The two sides having
fence are of same length x. The maximum area enclosed by (a) 1 (b) 2 (c) 0 (d) – 1.
the park is
(2005)
36. Let f be the differentiable for " x. If f (1) = –2 and f ¢(x) ³ 2
(a) 3 x 2 x 3 for [1, 6], then
2 (b)
(a) f (6) < 8 (b) f (6) ³ 8
(c) 1 x 2 8
2 (d) px2 . (2006) (c) f (6) = 5 (d) f (6) < 5. (2005)
28. The set of points where f (x ) = x is differentiable, is 37. Suppose f (x) is differentiable at x = 1 and
1+ | x |
lim 1 f (1 + h ) = 5, then f ¢(1) equals
(a) (– ¥, 0) È (0, ¥) (b) (– ¥, –1) È (–1, ¥) h ®0 h
(a) 4 (b) 3 (c) 6 (d) 5. (2005)
(c) (– ¥, ¥) (d) (0, ¥). (2006) 38. Area of the greatest rectangle that can be inscribed in the ellipse
29. Angle between the tangents to the curve y = x 2 – 5x + 6 at x 2 + y 2 = 1 is
a2 b2
the points (2, 0) and (3, 0) is
(a) p/2 (b) p/3 (c) p/6 (c) p/4. (a) ab (b) 2ab (c) a/b (d) ab
(2006) (2005)
30. The function g(x ) = x + 2 has a local minimum at 39. If 2a + 3b + 6c = 0, then at least one root of the equation
2 x ax2 + bx + c = 0 lies in the interval
(a) (2, 3) (b) (1, 2) (c) (0, 1) (d) (1, 3).
(a) x = 2 (b) x = –2 (2004)
(c) x = 0 (d) x = 1. (2006)
3x2 + 9x +17 40. A function y = f (x) has a second order derivative f ¢¢(x) = 6(x
31. If x is real, the maximum value of 3x2 + 9x + 7 is
– 1). If its graph passes through the point (2, 1) and at that
(a) 1/4 (b) 41 (c) 1 (d) 17/7. point the tangent to the graph is y = 3x – 5, then the function
(2006) is
(a) (x + 1)3 (b) (x – 1)3
32. A spherical iron ball 10 cm in radius is coated with a layer (c) (x – 1)2 (d) (x + 1) 2. (2004)
of ice of uniform thickness that melts at a rate of 50 cm3 / min. 41. Let f (x ) = 14- xt-anp x , x ¹ p4 , x Î ëêé0, p2 ùúû .
When the thickness of ice is 5 cm, then the rate at which the
thickness of ice decreases, is ( )f p p
2 4
(a) 1 cm/min (b) 1 cm/min (x ) is continuous in êëé0, úùû , then f is
18p 36p
(c) 5 cm/min (d) 1 cm/min. (2005) (a) - 12 (b) 1 (c) 1 (d) –1.
6p 54p 2 (2004)
Differential Calculus 43
lim çèæ1 + a + b ö÷ø 2 x = e 2 , 51. If 2a + 3b + 6c = 0 (a, b, c Î R) then the quadratic equation
x x 2
42. If x ®¥ then the values of a and b, are ax 2 + bx + c = 0 has
(a) a Î R, b = 2 (b) a = 1, b Î R (a) At least one in (0, 1)
(c) a Î R, b Î R(d) a = 1 and b = 2. (b) At least one root in [2, 3]
(2004) (c) At least one root in [4, 5]
43. Let f (a) = g(a) = k and their n th derivatives (d) none of these (2002)
f n (a), gn (a) exist and are not equal for some n. Further if
52. Let f (2) = 4 and f ¢(2) = 4 then lim x f (2) - 2 f ( x ) equals
lim f (a)g(x) - f (a) - g(a) f (x) + g(a ) = 4, x ® 2 x - 2
g(x) - f (x)
x®a (a) 2 (b) – 2
then the value of k is (c) – 4 (d) 3. (2002)
(a) 2 (b) 1 1
(c) 0 (d) 4. (2003) 53. lim æ x2 + 5x + 3 ö x
x ®¥ çè x2 + x + 3 ø÷
44. lim [1 - tan(x / 2)][1 - sin x ] is (a) e 4 (b) e2
x ®p/ 2 [1 + tan( x / 2)] [p - 2x]3 (d) 1.
(c) e3 (2002)
(a) 0 (b) 1/32 (c) ¥ (d) 1/8. 54. If f (x + y) = f (x) × f (y) " x, y and f (5) = 2, f ¢(0) = 3, then
(2003)
f ¢(5) is
45. The value of (a) 0 (b) 1
(c) 6 (d) 2. (2002)
lim 1 + 24 + 34 + .... + n4 - lim 1 + 23 + 33 + .... + n3 is 1 - cos 2 x
n5 n5
n®¥ n ® ¥ 55. lim is
(a) zero (b) 1/4 (c) 1/5 (d) 1/30. x ® 0 2 x
(2003) (a) 1 (b) –1
(d) does not exist.
46. The real number x when added to its inverse gives the minimum (c) 0 (2002)
value of the sum at x equal to 56. The maximum distance from origin of a point on the curve
(a) 1 (b) –1 (c) –2 (d) 2. sin èæ at ö
b ø
(2003) x = a sint – b
47. If f (x ) = ïì -çæ |1x| +1x ø÷ö , x ¹ 0 , then f (x) is y = a cost - b cos æ abt øö , both a, b > 0 is
í è
xe è
îï 0 , x = 0 (a) a – b (b) a + b
(a) continuous for all x, but not differentiable at x = 0
(b) neither differentiable not continuous at x = 0 (c) a2 + b2 (d) a2 - b2 . (2002)
(c) discontinuous everywhere f ( x ) -1
(d) continuous as well as differentiable for all x. 57. If f (1) = 1, f ¢( 1) = 2, then Lt is
x ®1 x - 1
(2003)
(a) 2 (b) 4
48. If the function f (x) = 2x 3 – 9ax 2 + 12a 2 x + 1, where a > 0, (c) 1 (d) 1/2.
attains its maximum and minumum at p and q respectively (2002)
such that p 2 = q, then a equals 58. f (x) and g (x) are two differentiable function on [0, 2] such
(a) 1 (b) 2 (c) 1/2 (d) 3. t h a t f ¢¢(x) - g¢¢ (x ) = 0, f ¢(1) = 2g¢(1) = 4 , f (2) = 3g(2) = 9
(2003) then f (x)– g(x) at x = 3/2 is
49. If f (x) = x n, then the value of (a) 0 (b) 2
.... + (-1)nn! f n (1) (c) 10 (d) 5.
f (1) - f ¢(1) + f ¢¢(1) - f ¢¢¢(1) + is (2002)
1! 2! 3!
59. f is defined in [–5, 5] as
(a) 2 n – 1 (b) 0 (d) 2 n .
(c) 1 ìx, if x is rational and
f (x ) = îí -x, if x is rational. Then
(2003) (a) f(x) is continuous at every x, except x = 0
50. If lim log(3 + x) - log(3 - x ) = k , the value of k is
x
x ®0 (b) f(x) is discontinuous at every x, except x = 0
(a) –1/3 (b) 2/3 (c) –2/3 (d) 0. (c) f(x) is continuous everywhere
(2003) (d) f(x) is discontinuous everywhere. (2002)
44 JEE MAIN CHAPTERWISE EXPLORER
60. lim log xn - [ x ] , n Î N , ([x] denotes greatest integer less than 61. If y = ( x + 1 + x 2 )n , then (1 + x 2 ) d 2 y + x ddyx is
d x 2
x ®0 [ x] (a) n2 y
(c) –y
or equal to x) (b) –n2 y
(d) 2x 2y .
(a) has value – 1 (b) has value 0 (2002)
(c) has value 1 (d) does not exist (2002)
Answer Key
1. (c) 2. (b) 3. (d) 4. (a) 5. (c) 6. (a)
7. (a) 8. (b) 9. (c) 10. (a) 11. (b) 12. (a)
13. (d) 14. (a) 15. (b) 16. (a) 17. (d) 18. (b)
19. (b) 20. (c) 21. (c) 22. (b) 23. (a) 24. (d)
25. (c) 26. (a) 27. (c) 28. (c) 29. (a) 30. (a)
31. (b) 32. (a) 33. (b) 34. (a), (c) 35. (c) 36. (b)
37. (d) 38. (b) 39. (c) 40. (b) 41. (a) 42. (b)
44. (b)
43. (d) 50. (b) 45. (c) 46. (a) 47. (a) 48. (b)
49. (b) 56. (a) 51. (a) 52. (c) 53. (d) 54. (c)
55. (a) 57. (a) 58. (d) 59. (b) 60. (d)
61. (a)
Differential Calculus 45
1. (c) : lim (1 - c os2x)(3 + c os x) Þ f ¢ (x) = 1 + 2bx + a
x( t a n 4 x) x
x ®0 [Given]
f ¢(– 1) = 0 and f¢(2) = 0
= lim 1 - cos2 x (3 + cosx)
x ® 0 x2 æ tan4x ö Þ - 1 - 2b + a = 0 Þ b = - 1
× èç x ÷ø 4
= æ 2sin 2 x ö æxö (3 + c o s x ) and 1 + 4b + a = 0 Þ a = 1
xli®m0 çè x2 ø÷ × èç tan4x ÷ø 22
= 2 ´ 1 ´ 4 = 2 f ¢¢(x) = - 1 + 2b = - 1 - 1 = - æ 1 + 21 ø÷ö < 0
4 x2 x2 2 çè x2
for all x Î R – {0}
2. (b) : dP = 100 - 12 x Þ f has a local maximum at x = – 1, x = 2
dx
\ Statement 1 : f has local maxima at x = –1, x = 2
Integrating, we have, dP = (100 - 12 x )dx \ Statem ent 2 : a = 1 , b = - 1
2 4
P = 100x - 12 × 2 × x3/ 2 + l
3 7. (a) : dv = - 72p m3 / min, v0 = 4500p
dt
P = 100x - 8x3 / 2 + l
P(0) = 2000 = l. \ l = 2000 v = 4 pr3 \ dv = 4 p ´ 3r 2 ´ dr
P(25) =100 × 25 – 8 × 25 3/2 + 2000 = 3500. 3 dt 3 dt
3. (d) : y = sec (tan –1 x) + 49 ∙ dv
dt
dy = sec(tan-1x) × t a n ( t a n -1 x ) × 1 After 49 min, v = v 0 = 4500p – 49 × 72p
dx + x2
1 = 4500p – 3528p = 972p
dy = 2 × 1 × 1 = 1 Þ 972 p = 4 pr3 Þ r 3 = 243 × 3 = 3 6 Þ r = 9
dx x =1 2 2 3
4. (a) : f(x) = |x – 2| + |x – 5| \ - 72p = 4p ´ 81 ´ dr Þ dr = - 18 = - 2
dt dt 81 9
ì7 - 2x, x < 2
ï 2
Þ f (x) = í 3, 2 £ x £ 5 Thus, radius decreases at a rate of 9 m/min
ïî 2x - 7, x > 5
Statement1 : f ¢(4) = 0. True d 2 x d æ dx ö d íìîïïèæç dy ö -1 ïü
dy2 dy èç dy ø÷ dy dx ø÷ ý
8. (b) : = = ïþ
Statement2 : f is continuous in [2, 5], differentiable in ìíïïîèçæ ö -1 üï ö-2 d 2 y ö -1
(2, 5) and f(2) = f(5). True ÷ø ý× ÷ø dx 2 ÷ø
þï
But Statement 2 is not a correct explanation for statement 1. = d dy dx = - æ dy × æ dy
dx dx dy çè dx çè dx
5. (c) : f : R ® R, f (x) = [x] cos èçæ 2 x- 1 ÷öø p ö-3 d2 y
2 ÷ø dx2
= - æ dy
èç dx
= [x] cos æçè px - p ö px 9. (c) : Let x = 2 + h
2 ø÷ = [x] sin
Let n be an integer. lim 1 - cos 2h = lim |sin h|
h®0 h h®0 h
lim f (x) = 0, lim f (x) = 0 RHL = 1, LHL = –1. Thus limit doesn’t exist.
x®n+ x®n-
\ f(n) = 0 ì sin(p + 1)x + sin x , x < 0
ï x = 0
Þ f(x) is continuous for every real x. ïï x
í q, x>0
10. (a) : f (x) =
6. (a) : f(x) = ln|x| + bx 2 + ax, x ¹ 0 has extreme values at ï x + x2 - x,
x = – 1, x = 2. ï x 3/ 2
îï
46 JEE MAIN CHAPTERWISE EXPLORER
lim f ( x ) = 1 Then H (x ) = ì- sin x2 , x < 0
2 îí sin x2 , x ³ 0
x ® 0 +
Again, lim f (x) = sin(p + 1) + sin x = p + 2 H (0 - h) - H (0)
x ® 0 - x LH ¢(0) = lim
h ®0 - - h
Now, p + 2 = q = 1/2
\ p = –3/2, q = 1/2. = lim - sin h2 = lim × sin h 2 × h = 1× 0 = 0
h ®0 - h ® 0-
11. (b) : g(x) = {f (2f (x) + 2} 2 - h h2
We have on differentiation with respect to x, RH ¢ (0) = lim H (0 + h) - H (0)
g¢(x) = 2f (2f (x) + 2) ∙ f ¢(2f (x) + 2) ∙ 2f ¢(x) h ®0 +
h
Let x = 0 = lim sin h 2 - 0 = lim æ sin h 2 ö
g¢(0) = 2f (2f (0) + 2) ∙ f ¢(2f (0) + 2) ∙ 2f ¢(0) h ®0 + h çè h2 ÷ø h
h ® 0 +
= 2f (0) ∙ f ¢(0) ∙ 2f ¢(0) = (–2)(1)(2) = – 4. = 1 ∙ 0 = 0
12. (a) : As f is a positive increasing function, we have Thus H(x) is differentiable at x = 0
f (x) < f (2x) < f (3x) ì-2x cos x2 , x < 0
, x = 0
Dividing by f (x) leads to 1 < f (2x) < f (3x) Also H ¢(x) = ï 0 , x > 0
f (x) f (x) í
As lim f (3x) = 1, we have by Squeez theorem ïî 2x cos x2
x®¥ f (x)
H¢(x) is continuous at x = 0 for
H¢(0) = LH¢(0) = RH¢(0)
or Sandwich theorem, lim f (2x) = 1. Again H ¢¢(x ) = ì -2 cos x2 + 4x2 sin x2 , x < 0
x®¥ f (x) í
î 2 cos x2 - 4x2 sin x2 , x ³ 0
13. (d) : lim f (x) = 1
LH¢¢(0) = –2 and RH¢¢(0) = 2
x ®1+
Thus H(x) is NOT twice differentiable at x = 0
As f (–1) = k + 2
As f has a local miminum at x = –1 16. (a) : P(x) = x 4 + ax 3 + bx 2 + cx + d
P¢(x) = 4x 3 + 3ax 2 + 2bx + c
f (–1 +) ³ f (–1) ³ f (–1 – ) Þ 1 ³ k + 2 P¢(0) = 0 Þ c = 0
Also P¢(x) = x(4x 2 + 3ax + 2b)
Þ k + 2 £ 1. \ k £ –1 As P¢(x) = 0 has no real roots except
Thus k = –1 is a possible value.
14. (a) : Using A.M.G.M. inequality,
ex + 2 e-x ³ ex × 2 e-x . Thus, ex + 2e-x ³ 2 2 x = 0, we have
2 D of 4x 2 + 3ax + 2b is less than zero
i.e., (3a)2 – 4 ∙ 4 ∙ 2b < 0
Then ex 1 £ 1 then 4x 2 + 3ax + 2b > 0 " x Î R
+ 2 e - x 2 2 (If a > 0, b2 – 4ac < 0 then ax2 + bx + c > 0 " x Î R)
So P¢(x) < 0 if x Î [–1, 0) i.e., decreasing
1 and P¢(x) > 0 if x Î (0, 1] i.e., increasing
As ex + 2e-x is always positive, we have
0 < ex 1 £ 1
+ 2 e - x 2 2
Max. of P(x) = P(1)
Observe that f (0) = 1/3. Thus such that But minimum of P(x) doesn’t occur at x = –1, i.e., P(–1) is
not the minimum.
f (c) = 1/3.
Using extremevalue theorem, we can say that as f is continuous, 17. (d) : x2 x – 2x x cot y – 1 = 0 .....(i)
f will attain a value 1/3 at some point. Here we are able to
identify the point as well. At x = 1 we have
1 – 2 cot y – 1 = 0
15. (b) : gof (x) = g(f (x)) = sin(x|x|) Þ cot y = 0 \ y = p/2
ì- sin x2 , x < 0 Differentiating (i) w.r.t. x, we have
= íî sin x2 , x ³ 0
Let the composite function gof (x) be denoted by H(x). 2x2x (1 + ln x) - 2[x x (- cosec2 y) dy + cot y × x x (1 + ln x )] = 0
dx
At P(1, p/2) we have
Differential Calculus 47
2(1 + ln1) - 2[1(-1) ( ) dy + 0] = 0 By using A.M ³ G.M., p2 + q 2 ³ pq Þ pq £ 1
dx 2 2
P
( ) dy (p + q)2 = p 2 + q2 + 2pq Þ ( p + q) £ 2.
Þ 2+ 2 dx = 0 f (0) = xli ®m0 êéë 1x - 2 ù
e2 x - 1 úû
P 22. (b) :
( ) \ dy = - 1 = xli ®m0 ex2 (x e-2 1x - 2x çèæ 00 form ÷øö
dx - 1)
P
18. (b) : Denote x 3 – px + q by f (x)
i.e. f (x) = x3 – px + q By using L’ Hospital rule
Now for expression, f ¢(x) = 0, i.e. 3x 2 – p = 0 f (0) = lim 2e 2 x - 2 çæè 0 form ÷öø
x ®0 ( - 1) + 2 xe2 x 0
p p e2 x
x = - ,
Again use L’ Hospital rule
3 3
f ¢¢(x) = 6x 4 e 2 x
4e2 x + 4 xe2 x
æ p ö æ p ö f (0) = lim = 1.
èç 3 ÷ø èç 3 ø÷
f ¢¢ - < 0 Þ f ¢¢ > 0 x ®0
23. (a) : f (x) = min {x + 1, |x| + 1}
p 3p . Þ f (x) = x + 1, x Î R
Thus maxima at - 3 and minima at
Hence f (x) is differentiable for all x Î R.
19. (b) : By definition 1
x + cos
f (1 + h) - f (1) 24. (d) : f ¢( x) = 1 + (sin x ) 2 .(cos x - sin x )
f ¢ (1) = lim h , if the limit exists.
h ®0 cos x - sin x
2 + sin 2 x
f (1+ h) - f (1) f ¢( x ) =
\ lim h If f ¢(x) > 0 then f (x) is increasing function
h ®0
(1 + h -1) sin 1 - 0 p p4 ,cos
2
(1+ h -1) 1 For - < x < x > sin x
= lim = lim sin
h®0 h h ®0 h
çæè - p , p ÷öø .
As the limit dosen’t exist, Hence y = f (x) is increasing in 2 4
\it is not diffentiable at x = 1 25. (c) : By LMVT,
Again f ¢ (0) = lim f (h) - f (0) , if the limit exists f (b) - f (a) f (3) - f (1)
b - a 3 - 1
h ®0 h f ¢(c ) = =
( h - 1) sin 1 1 - sin1 f ¢(c) = loge 3 - loge 1 = 21 loge 3
lim 2
\ lim f (h) - f ( 0) = h -
h®0 h h ®0 h 1 1 1
c 2 2 log 3 e
But this limit dosen’t exis t. Hence it is not differentiable at Þ = loge 3 = \ c = 2 log3 e.
x = 0.
26. (a) : x m × y n = (x + y)m + n
20. (c) : Let f (x) = x7 + 14x5 + 16x3 + 30x – 560 Taking log both sides we get
\ f ¢(x) = 7x 6 + 70x4 + 48x2 + 30 m log x + n log y = (m + n) log(x + y)
Þ f ¢(x) > 0 " x Î R Differentiating w.r.t. x we get
i.e. f (x) is an strictly increasing function. m + n dy = m+ n çèæ1 + dy ö
x y dx x+ y dx ÷ø
so it can have at the most one solution. It can be shown that
it has exactly one solution. Þ dy æ n - m+ n ö = m + n - m
21. (c) : Let p = cos q, q = sin q dx çè y x+ y ø÷ x + y x
0 £ q £ p/2
p + q = cos q + sin q Þ dy æ nx + ny - my - ny ö = mx + nx - mx - my
Þ maximum value of (p + q) = 2 dx èç y(x + y) ø÷ x( x + y)
Second method
Þ dy = æ nx - my ö y = y Þ dy = y .
dx çè nx - my ø÷ x x dx x
48 JEE MAIN CHAPTERWISE EXPLORER
27. (c) : AT = x sin a K = b2 – 4ac = –123
BT = x cos a A i.e., solve –3y 2 + 126 + y – 123 ³ 0
T
Area of triangle Þ 3y2 – 126y + 123 £ 0
ABC = 1 base × height Þ y2 – 42y + 41 £ 0
2
Þ (y – 1)(y – 42) £ 0
= 12 (2BT )( AT ) Þ 1 £ y £ 42
= 1 (2 x2 cos a sin a ) a Þ maximum value of y is 42
2 B
C 32. (a) : v = 4 p( y + 10)3 where y is thickness of ice
3
1 x2 1 x 2 as
= 2 sin 2a £ 2 - 1 £ sin 2a £ 1
\ Maximum are of DABC = 1 x2 10
2
28. (c) : Given f ( x ) = x
1+ | x |
10 + y
Þ f ( x ) = ïì1 -x x , x < 0 Þ dv = 4p(10 + y ) 2 dy
í x ³ 0 dt dt
ï x ,
î 1 + x çèæ dy ö 50 ( ) as dv = 50 cm3 / min.
dt ø÷at 4p(15) 2 dt
ì 1 =
ïï - x ) 2
, x < 0 y = 5
x ³ 0
Þ f ¢( x ) = í (1 = 1 cm/min.
ï 18p
1 ,
ïî (1 + x) 2 33. (b) : As a is root of ax2 + bx + c = 0
\ aa 2 + ba + c = 0. Now
f ¢ (x) is finite quantity " x Î R
\ f ¢ (x) is differentiable " x Î (-¥,¥ ) lim 1 - cos(ax2 + bx + c )
( x - a ) 2
x ® a
29. (a) : Given equation y = x2 – 5x + 6, given points (2, 0), (3, 2 sin 2 çèæ ax2 + bx + c ÷øö
0) 2
\ dy = 2x - 5 = lim ( x - a ) 2
dx
x ® a
2sin 2 êéë a (x - a )( x - b) ù a2 b ) 2
2 úû
say m1 = æ dy ö = 4 - 5 = -1 = lim ´ ( x -
èç dx ø÷ x =2
x ®a a2 é ( x - a )2 ( x - b ) 2 ù 4
ê 4 ú
y = 0 êë ûú
and m2 = æ ddxy øö÷ at x =3 = 6 - 5 = 1 é sin æçè a ( x - a)( x - b ) ö ù 2
çè ê 2 ÷ø ú
y = 0 ê ú
a2 ( x - b) 2
since m 1m 2 = –1 = lim ´
Þ tangents are at right angle i.e. p x ® a ê a( x - a)( x - b) ú 2
2
êë 2 úû
30. (a) : Let g( x ) = x + 2 = 1 ´ a2 (a - b )2 .
2 x 2
\ g¢( x ) = 1 - 2 34. (a), (c) : dy = dy × d q = tan q = slope of tangent
2 x2 dx dq dx
for maxima and minima g¢(x) = 0 Þ x = ± 2 \ Slope of normal to the curve = – cotq
g¢¢( x) = 4 > 0 for x = 2 = tan (90 + q).
x3
Again Now equation of normal to the curve
< 0 for x = –2 \ x = 2 is point of minima [ y - a(sin q - qcosq ) ]
31. (b) : For the range of the expression = - cos q ( x - a(cosq + asin q))
sin q
3x2 + 9x + 17 ax2 + bx + c
3x2 + 9x + 7 = y = px2 + qx + r , Þ xcos q + y sinq = a(1)
Now distance from (0, 0) to x cosq + y sinq = a is
[find the solution of the inequality (0 + 0 - a)
distance (d) = 1
A y 2 + B y + K ³ 0] \ distance is constant = a.
where A = q 2 – 4pr = –3, B = 4ar + 4PC – 2bq = 126
Differential Calculus 49
35. (c) : Given f ( x) - f ( y) £ ( x - y) 2 2a + 3b + 6 c
= 6
lim f ( x) - f ( y ) £ lim x - y = 0 given 2a + 3b + 6c = 0
x®y x - y
x® y \ x = 0 and x = 1 are roots of
Þ f ¢(x) £ 0 , f (x) = 0 ( f ¢(x) < 0, not possible ) f (x) = ax3 + bx 2 + cx = 0
3 2
Þ f ( x) = k (by integration)
\ at least one root of the equation ax2 + bx + c = 0 lies in
Þ f (x) = 0 Q f (0) = 0 (0, 1)
40. (b) : Given f ² (x) = 6(x – 1)
Þ f ( x) (" x Î R) = 0 \ f (1) = 0. 6(x - 1) 2
36. (b) : Let if possible f ¢(x) = 2 for Þ f ¢ (x) = 2 + c
Þ f (x) = 2x + c (Integrating both side w.r.t. x)
\ f (1) = 2 + c, –2 = 2 + c Þ 3 = 3 + c éQ f (x) = y = 3x + 5 ù
Þ c = – 4 \ f (x) = 2x – 4
\ f (6) = 2 × 6 – 4 = 8 \ f (6) ³ 8. Þ c = 0 ê f ¢(x) = 3 " x ÎRúû
so f ¢ (x) = 3(x – 1)2 ë
37. (d) : As f (x) is differenatiable at x = 1 Þ f (x) = (x – 1) 3 + c 1 as curve passes through (2, 1)
Þ 1 = (2 – 1) 3 + c 1 Þ c1 = 0
\ f (x) = (x – 1)3
5 = lim f (1 + h ) assumes 0/0 form 41. (a) : Lt 1 - tan x putting 4x – p = t
h x ® p/4 4 x - p
h ®0
5 = lim f ¢(1) \ f ¢(1) = 5. \ Lt (1 - tan x) ´ (1 + tan x )
h ®0 1
38. (b) : Any point on the ellipse x ® p/4 + tan x) éëê -4 çæè p- x ö ù
4 ÷ø ûú
x 2 y 2 2x (1
a2 b2 2y
+ = 1
is (acosq, bsinq) so the area of tan æ p- x ö ´ (1 + tan x )
èç 4 ø÷
rectangle inscribed in the ellipse is Lt - = –1/2
given by x ® p/4 4 èçæ p - x ö
4 ÷ø
A = (2acosq) (2bsinq)
\ A = 2absin 2q Þ dA = 4ab cos 2 q 42. (b) : e 2 = çæè1 + a+ b ö2 x (1¥ form)
d q x x2 ÷ø
Now for maximum area
dA = 0Þ q = p and æ d 2 A ö = -8ab sin 2 q Lt éê1+ a+ xb 2 -1ùúû ( 2 x )
d q 4 ç ÷ ë x
è d q2 ø q e 2 = e x ® ¥
= p / 4
as d 2 A < 0. \ Area is maximum for q = p/4. e 2 = e 2a
dq2
Þ 2a = 2 \ a = 1 and b Î R
\ sides of rectangle are 2a , 2 b
2 2 Lt f (a) g(x) - f (a) - g(a) f (x) + g (a )
g(x) - f (x)
43. (d) : x ® a = 4
Required area = 2ab.
ax3 bx 2 Þ Lt f (a)[g(x) - f (x )] Þ Lt f (a) = 4
39. (c) : Let f (x) = + + cx = 4
x ® a g(x) - f (x) x ® a
3 2
Note : In such type of problems we always consider f (x) as k = 4
the integration of L.H.S of the given equation without constant.
Here integration of ax2 + bx + c is ax3 bx 2 called tan æ p - x ö (1 - sin x )
+ + cx çè 4 2 ÷ø
3 2
it by f (x). Now use the intervals in f (x) if f (x) satisfies the 44. (b) : Lt æ p -42 x ÷öø (p
èç
x ®p/2 4. - 2 x) 2
given condition then at least one root of the equation ax 2 +
bx + c = 0 must lies in that interval. æ p x ö cos çæè p- x ÷øö
çè 4 2 ÷ø 2
ax3 bx 2 tan - 1 -
Now f (x) = + + cx
Lt 4. æçè p x ö (p - 2x) 2
3 2 4 2 ÷ø
x ®p/2 -
f (0) = 0 and f (1) = a + b + c
3 2
50 JEE MAIN CHAPTERWISE EXPLORER
tan æ p - x ö 2 sin 2 çèæ p - x ö 1 ´ 2 1 Þ a2 = 2a Þ a 2 – 2a = 0
çè 4 2 ÷ø 4 2 ÷ø 4 16 = 32 Þ a(a – 2) = 0 Þ a = 0, a = 2
Lt
4. èçæ p x ö ö2 = 49. (b) : f (x) = x n \ f (1) = 1 = n C 0
x ® p/2 4 2 ÷ø 4 2 çèæ p - 2 x ÷ø
- 4
\ f ¢ (x) = nx n – 1 so –f ¢ (1) = – n = – nC 1
14 + 24 + 34 + ... + n 4 - Lt 13 + 23 + ... + n3 f ² (x) = n(n – 1)x n – 2 so f ¢¢(1) = n(n -1) = nC 2
2! 2!
45. (c) : Lt n ® ¥ n5
n5 f n (1)(- 1) n
n ® ¥ f n (x) = n(n – 1) ...1 \ n! = (–1)n n C n
= Lt n(n + 1)(2n + 1)(3n2 + 3n - 1) - Lt n2 (n + 1) 2
30 × n5 4 ´ n5
n ® ¥ n ® ¥ f ¢(1) + f ¢¢(1) - f ¢¢¢(1) (- 1)n f n (1)
+ ... +
\ f (1) – 1! 2! 3! n! =
- 0 éëê Using 14 + 24 + .... + n4 nC 0 – nC 1 + n C 2 ....+ (–1)n C n ...(i)
Now (1 + x) n = C0 + C1 x + C2 x 2 + ... + Cn x n
n(n + 1)(2n + 1)(3n2 + 3n - 1)
= Putting x = –1 in both side of (i) we get
30 0 = C0 – C 1 + C2 – C3 + ...
= n(n + 1)(6n3 + 9n2 + n - 1) ù log(3 + x) - log(3 - x )
ú
30 úû 50. (b) : Lt x = k
6 x ® 0
= - 0
log çèæ1 + x ö - log çèæ1 - x ö
30 3 ÷ø 3 ÷ø
1 \ k = Lt
=
x ® 0 x
5
46. (a) : f (x) = x + 1/x log çèæ1 + x ö log çèæ1 - x ö
3 ÷ø 3 ÷ø
f ¢ (x) = 1 – 1/x2 and f ² (x) = 2 k = Lt + Lt
x 3 x ´3 x ® 0 - x ´ 3
x®0
now f ¢ (x) = 0 3 3
Þ x = ± 1 \ f ²(1) > 0 k = 1 + 1 = 2
Þ x = 1 is point of minima. 33 3
ì - æ | 1 | + 1 ö 51. (a) : Let us consider f (x) = ax3 + bx2 + cx
ïí x èç x 3 2
47. (a) : Given f (x) = e x ÷ø , x ¹ 0
x = 0
ïî 0 , \ f (0) = 0 and f (1) = a + b + c
3 2
Lt f (x) = Lt xe-2/ x = 0
...(A) 2a + 3b + 6 c
x ® 0+ x ® 0 + ...(B) = = 0 given.
é - 1 + 1 ù 6
x x úû As f (0) = f (1) = 0 and f (x) is continuous and differentiable
and Lt f (x) = Lt eëê = 0
x ® 0- x ® 0 -
also in [0, 1].
As LHL = RHL \ f (x) is continuous at x = 0 \ By Rolle’s theorem f ¢ (x) = 0
x ® 0 x ® 0 Þ ax2 + bx + c = 0 has at least one root in the interval (0, 1).
Again RHD at x = 0 is
- æ 1 + 1 ö 52. (c) : Lt xf (2) - 2 f (x) + 2 f (2) - 2 f (2)
èç h h ÷ø x - 2
(0 + h)e - 0 x ® 2
Lt h = 0 Lt (x - 2) f (2) - 2[ f (x) - f (2)]
x ® 0 +
x ® 2 x - 2
also we have L.H.D at x = 0
h)e -èæç 1 - 1 ö = lim [f (2) – 2 f ¢(x)]
h h ÷ø
(0 - - 0 x® 2
is = 1 = 4 – 2 × 4 = – 4
- h
x2 + 5x + 3 ö1/ x
so L.H.D ¹ R.H.D at x = 0 53. (d) : We have xl i®m¥ çèæ x2 + x + 3 ÷ø
\ f (x) is non differentiable at x = 0
48. (b) : For maximum and minima f ¢ (x) = 0 æ1 + 5 + 3 1 / x
Þ 6x2 – 18ax + 12a2 = 0 and f ² (x) = 12x – 18a
f ¢ (x) = 0 ç ö
x 2 ÷
Þ x = a, 2a and f ² (a) < 0 and f ² (2a) > 0 = lim ç x ÷ = 10 = 1
Now p = a and q = 2a and p2 = q ç 1+ 3 ÷ø
x ®¥ 1 + x x 2
è
Differential Calculus 51
54. (c) : Given f (x + y) = f (x) f (y) 1 ´ 2 x
f (x) 1
\ f (0 + 0) = (f (0))2 Lt × f ¢ (x)
x ® 1 2
Þ f (0) = 0 or f (0) = 1 but f (0) ¹ 0
Lt f (x + h) - f ( x ) Lt f (x) f (h) - f (x ) 2 ´1´ 2
h h = 2 = 2
Now f ¢(x) = h ® 0 = h ® 0
f ¢(x) = f (x) Lt f (h ) - 1 58. (d) : As f ¢¢ (x) – g¢¢ (x) = 0
h ® 0 h Þ f ¢ (x) – g¢(x) = k
\ f ¢(0) = f (0) Lt f (h ) - 1 f ¢ (1) – g¢(1) = k \ k = 2
h
h ® 0 So f ¢ (x) – g¢(x) = 2
3 = Lt f (h ) - 1 (³ f (0) = 1) Þ f (x) – g(x) = 2x + k1
h f (2) – g(2) = 4 + k1
h ® 0 k1 = 2
Now f ¢(x) = f (x) Lt f (h ) - 1 So f (x) – g(x) = 2x + 2
h ® 0 h
\ [ f (x) - g ( x)] 3 = 2 ´ 3 + 2 = 5
\ f ¢(5) = f (5) × 3 = 2 × 3 = 6 59. (b) 2
x =
2
55. (a) : Lt 2 sin2 x = Lt sin x = 1.
x®0 x 2 x ®0 x
56. (a) : ì a sin t - at x 60. (d) : log xn - [x ] = Lt n log x - 1
Let ï - b sin = y Lt [ x ]
x ® 0 [x] x ® 0
b
A(0,0), B(x, y) = í b cos at =
ïa cos t which does not exist as Lt log x does not exist
îb x ® 0 [x]
a2 (sin2 t + cos2 t) + b 2 æ sin 2 çèæ at ö 61. (a) : y1 = n ëêé x + 1 + x 2 ûùún -1 ëêêé1 + x ù
ç b ÷ø ú
è
1 + x2 ûú
\ x2 + y2 = AB = + cos2 çæè at ö ö - 2ab cos çæè t - at ö é x 2 ù n 1
b ÷ø ÷ b ÷ø êë úû 1 + x 2
ø y1 = n x + 1 + .
= a2 + b2 - 2ab cos a (since |cos a| £ 1) y1 = ny æ y1 = dy ö
1 + x 2 èç dx ÷ø
£ a2 + b2 - 2 ab = a – b.
57. (a) : Lt f (x ) - 1 (0/0 form) Þ y 12 ( 1 + x 2) = n2 y 2
Þ y1 2 ( 2x) + (1 + x 2 ) (2y 1 y 2) = 2yy1 n 2
x ®1 x - 1 Þ y 2( 1 + x 2 ) + xy1 = n 2y
52 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER INTEGRAL CALCULUS
10
1. If ò f (x)dx = y(x) then ò x5 f (x3 ) dx is equal to x
6. If g(x) = ò cos 4t dt , then g(x + p) equals
1 x3y(x3 ) - 3 x3y(x3 ) dx + C 0
3
ò (a) (a) g(x) – g(p) (b) g(x)∙g(p)
1 x3y(x3) - x2y(x3 ) dx + C g( x ) (d) g(x) + g(p) (2012)
3 (c) g(p )
ò (b)
ò (c) 31 éëx3y(x3) - x3y(x3 ) dxûù + C 7. The area bounded between the parabolas x 2 = y and
4
x2 = 9y and the straight line y = 2 is
(d) ò 31 ëéx3y(x3) - x2y(x3 )d xùû + C (2013) (a) 20 2 (b) 10 2
3
2. The intercepts on xaxis made by tangents to the curve, (c) 20 2 (d) 10 2 (2012)
3
x
y = ò| t | dt, x Î R , which are parallel to the line y = 2x, are
0 8. The population p(t) at time t of a certain mouse species satisfies
equal to
(a) ± 2 (b) ± 3 (2013) the differential equation dp(t) = 0.5p(t) - 450 . If p(0) = 850,
(c) ± 4 (d) ± 1 dt
then the time at which the population becomes zero is
3. The area (in sq. units) bounded by the curves y = x, (a) 1 ln 18 (b) ln 18
2y – x + 3 = 0, xaxis and lying in the first quadrant is 2
(a) 36 (b) 18 (c) 27 (d) 9 (c) 2 ln 18 (d) ln 9 (2012)
4
9. Let I be the purchase value of an equipment and V(t) be the
(2013) value after it has been used for t years. The value V(t)
4. StatementI : The value of the integral depreciates at a rate given by differential equation
p/3 dx is equal to p/6. dV(t) = - k(T - t); where k > 0 is a constant and T is the total
tanx dt
pò/ 6 1 +
life in years of the equipment. Then the scrap value V(T) of
bb
the equipment is
StatementII : ò f (x)dx =ò f (a + b - x)dx. (a) I - k(T - t) 2
aa
2 (b) e –kT
(a) StatementI is true, StatementII is true, StatementII is
not a correct explanation for StatementI. (c) T 2 - I (d) I - kT 2 (2011)
(b) Statement1 is true, StatementII is false. k 2
(c) StatementI is false, StatementII is true. 10. If dy = y + 3 > 0 and y(0) = 2, then y(ln 2) is equal to
(d) StatementI is true StatementII is true, StatementII is a dx
correct explanation for StatementI. (2013) (a) 13 (b) – 2
5. If the integral ò 5 tan x dx = x + a ln | sin x - 2 cos x| + k , then (c) 7 (d) 5 (2011)
tan x - 2
1 8 log (1 + x) dx is
a is equal to 11. The value of 1 + x2 (b) log2
ò
(a) 1 (b) 2 (c) – 1 (d) – 2 (a) p log 2
2 0
(2012)
Integral Calculus 53
(c) p log2 (d) p log 2 (2011) ò ò 20. Let I = 1 sin x dx and J = 1 cos x dx .
8
0x 0 x
x
12. For x Î æ 0, 5p ÷øö , define f (x) = t sin t dt Then f has Then which one of the following is true?
çè 2 ò
2 2
0 I > 3 and J < 2 I > 3 and J > 2
(a) local minimum at p and local maximum at 2p. (a) (b)
(b) local maximum at p and local minimum at 2p. 2 2
3 3
(c) local maximum at p and 2p. (c) I < and J < 2 (d) I < and J > 2 (2008)
(d) local minimum at p and 2p. (2011) 21. The area enclosed between the curves y 2 = x and y = |x| is
13. The area of the region enclosed by the curves y = x, x = e, (a) 1/6 (b) 1/3 (c) 2/3 (d) 1. (2007)
y = 1/x and the positive xaxis is ò cos dx equals
+3
(a) 3/2 square units (b) 5/2 square units 22. x sin x
(c) 1/2 square units (d) 1 square units (2011) (a) log tan çèæ 2x + 1p2 ÷öø + C (b) log tan æèç 2x - 1p2 øö÷ + C
14. Let p(x) be a function defined on R such that
p¢(x) = p¢(1 – x), for all x Î [0, 1], p(0) = 1 and p(1) = 41. (c) 12 log tan çæè 2x + 1p2 ÷øö + C (d) 1 log tan èçæ x - 1p2 ÷øö + C.
2 2
1
(2007)
Then ò p(x) dx equals
0 (c) 41 (d) 42 (2010) 23. The solution for x of the equation x dt = p
t 2 - 1 2 is
(a) 41 (b) 21 ò
2 t
15. The area bounded by the curves y = cosx and y = sinx between (a) 3 (b) 2 2 (c) 2 (d) p.
2
the ordinates x = 0 and x = 3 p is
2 (2007)
(a) 4 2 - 2 (b) 4 2 + 2 24. Let F ( x) = f ( x) + f çæè 1x ÷øö , where f ( x) = 1xò 1lo +g tt dt ,
(c) 4 2 - 1 (d) 4 2 + 1 (2010) Then F(e) equals
16. The area of the region bounded by the parabola (a) 1 (b) 2 (c) 1/2 (d) 0.
(2007)
(y – 2) 2 = x – 1, the tangent to the parabola at the point a
(2, 3) and the xaxis is 25. The value of ò [x] f ¢(x) dx, a > 1, where [x] denotes the greatest
(a) 6 (b) 9 1
integer not exceeding x is
(c) 12 (d) 3 (2009) (a) a f (a) – {f (1) + f (2) + ... + f ([a])}
p (b) [a] f (a) – {f (1) + f (2) + ... + f ([a])}
17. ò [cot x]dx, where [.] denotes the greatest integer function, is (c) [a] f ([a]) – {f (1) + f (2) + ... + f (a)}
0 (d) a f ([a]) – {f (1) + f (2) + ... + f (a)}. (2006)
equal to
(a) 1 (b) –1
(c) – p/2 (d) p/2 (2009) - p / 2
18. The value of is 2 ò sins ièænx x-dxp4 øö 26. ò [(x + p)3 + cos2 (x + 3p)] dx is equal to
-3p / 2
p 4 p4 p p (d) p - 1 .
(a) (b) + (c) 2
32 2 2 (2006)
32
(a) x - log cos æè x - pö + c (b) x + log cos èæ x - pö + c p
4 ø 4 ø
27. ò x f (sin x) dx is equal to
0
sin æè pö sin èæ pö p p
4 ø 4 ø
(c) x - log x - + c (d) x + log x - + c (a) p ò f (cos x) dx (b) p ò f (sin x) dx
0 0
(2008) (c) p p/2 f (sin x) dx p/2 (2006)
ò
19. The area of the plane region bounded by the curves 2 (d) p ò f (cos x) dx .
0
0
x + 2y 2 = 0 and x + 3y 2 = 1 is equal to 6 x dx is
4 5 1 2 28. The value of the integral, ò
(a) 3 (b) 3 (c) 3 (d) 3 3 9 - x + x
(2008) (a) 1/2 (b) 3/2 (c) 2 (d) 1.
(2006)
54 JEE MAIN CHAPTERWISE EXPLORER
29. lim é 1 sec2 1 + 2 sec2 4 + .... + n1 sec2 1ûùú equals 38. If f (x ) = e x x , I1 = f (a ) and
êë n2 n2 n2 n2 1 + e
n ® ¥ ò xg{x(1 - x)} dx
f (-a )
(a) 1 cosec1 (b) 1 sec1 I 2 = f (a ) then the value of I 2 is
2 2 I 1
ò g{x(1 - x)}dx,
1 tan1
2 f (-a )
(c) (d) tan1. (2005) (a) –1 (b) –3 (c) 2 (d) 1.
p cos 2 x (2004)
1 + a x
30. The value of ò dx , a > 0, is p p/ 2
-p 39. If ò x f (sin x)dx = A ò f (sin x)dx, then A is
(a) p/2 (b) ap (c) 2p (d) p/a 0 0
(2005)
(a) p/4 (b) p (c) 0 (d) 2p.
31. The parabolas y2 = 4x and x 2 = 4y divide the square region (2004)
bounded by the lines x = 4, y = 4 and the coordinate axes. If 40. The value of I = p/2 (sin x + cos x ) 2 dx is
S1 , S 2 , S 3 are respectively the areas of these parts numbered ò
0 1 + sin 2 x
from top to bottom; then S1 : S2 : S 3 is (a) 2 (b) 1 (c) 0 (d) 3.
(a) 1 : 2 : 3 (b) 1 : 2 : 1 (2004)
(d) 1/3.
(c) 1 : 1 : 1 (d) 2 : 1 : 2. (2005) 3 2
32. The area enclosed between the curve y = log e( x + e) and the 41. The value of ò | 1 - x | dx is
coordinate axes is
-2
(a) 7/3 (b) 14/3 (c) 28/3
(a) 2 (b) 1 (c) 4 (d) 3. (2004)
(2005) dx
ò cos x - sin x
1 2x2 dx, 1 x3 I3 = 2ò 2 x 2 dx 2 2 x 3 42. is equal to
33. If I1 = ò I2 = ò 2 dx, and I4 = ò dx
( ) (a)
then 0 0 1 1 1 log tan 2x - 3p +C
2 8
(a) I1 > I 2 (b) I2 > I 1 (c) I 3 > I 4 (d) I3 = I 4 .
(2005) 1 log cot x + C
2 2
34. Let f (x ) be a nonnegative continuous function such that ( ) (b)
the area bounded by the curve y = f (x), xaxis and the ordinates ( ) (c)
1 log tan 2x - p8 +C
( ) x= p p 2
4 a n d x = b > 4 is b sin b + p cosb + 2b . Then
4
( ) (d)
( )f p 1 log tan 2x + 3p + C. (2004)
2 2 8
is
( ) (a) ( ) (b) 43. If ò sins(ixn- x a ) dx = Ax + B logsin(x - a) + C , then value of
p - 2 + 1 p + 2 - 1 (A, B) is
4 4
( ) (c) ( ) (d)
1 - p + 2 1 - p - 2 . (2005) (a) (–sina, cosa) (b) (cosa, sina)
4 4
(c) (sina, cosa) (d) (–cosa, sina) (2004)
35. Let F : R ® R be a differentiable function having lim n 1 e r / n
n
( ) f (2) = 6, 1 f ( x ) 4 t 3 dt 44. n ®¥ å is
48 x - 2
f ¢(2) = . Then lim ò equals r =1
x ®2 6 (a) 1 – e (b) e – 1 (c) e (d) e + 1.
(a) 36 (b) 24 (c) 18 (d) 12. (2004)
(2005) 45. Let d F (x) = æ e sxin x ö÷ø , x > 0.
dx çè
ò ïíîìï (log x - 1) ïü 2
36. 1 + (log x)2 þïý is equal to 4 3 esin x 3 dx
dx x
ò
If = F (k) - F (1), then one of the possible values
1
x + C log x + C
(a) x2 + 1 (b) (log x)2 + 1 of k is
x + C xe x + C . (a) 16 (b) 63
(log x)2 + 1 1 + x2
(c) (d) (2005) (c) 64 (d) 15. (2003)
(d) 3 (2003)
37. The area of the region bounded by the curves x 2
y = |x – 2|, x = 1, x = 3 and the xaxis is
ò sec 2 tdt
46. The value of lim 0 is
(a) 3 (b) 2 (c) 1 (d) 4. x ®0 x sin x
(a) 2 (b) 1 (c) 0
(2004)
Integral Calculus 55
1 52. lim 1p + 2 p + 3p + ... + n p is
47. The value of the integral I = ò x(1 - x) n dx is n ®¥ n p + 1
0
(a) 1 (b) n 1 - n 1 1 1
n + 2 +1 + 2 (a) p + 1 (b) 1 - p
(c) n 1 + n 1 (d) n 1+ 1. (2003) (c) 1 - 1 (d) p 1+ 2 . (2002)
+1 + 2 p p - 1
b 53. The area bounded by the curves y = lnx, y = ln |x|, y = |ln x| and
48. If f (a + b – x) = f (x), then ò x f (x) dx is equal to
a y = |ln|x|| is
(a) a + b b f ( x ) dx (b) b - a b f ( x) dx (a) 4 sq. units (b) 6 sq. units
2 2
ò ò (c) 10 sq. units (d) none of these (2002)
a a
(c) a + b b f (a + b - x) dx p
2 -p
ò ò 54. 2x (1+ sin x ) is
1 + cos 2 x dx
a
(d) a + b b f (b - x)dx.
ò (2003)
2 (a) p 2 /4 (b) p 2 (c) 0 (d) p/2.
a (2002)
49. If f (y) = e y , g(y) = y ; y > 0 and 55. If y = f (x) makes +ve intercept of 2 and 0 unit x and y and
t
F(t) = ò f (t - y)g(y)dy , then encloses an area of 3/4 square unit with the axes then
0
2
(a) F(t) = et - (1 + t) (b) F (t) = t e t ò x f ¢( x) dx is
(c) F (t) = te –t (d) F (t) = 1 – e t (1 + t). (0a ) 3/2 (b) 1 (c) 5/4 (d) –3/4.
(2003) (2002)
10 p
50. Let f (x) be a function satisfying f ¢(x) = f (x) with f (0) = 1 56. ò sin x dx is
and g(x) be a function that satisfies f (x) + g(x) = x 2 . Then
p
1 (b) 8 (c) 10 (d) 18.
(a) 20
the value of the integral ò f (x)g(x) dx is
p/4 (2002)
0 tann x dx, then lim n In + In -2 equals
n ®¥
(a) e + e 2 - 32 (b) e - e 2 - 32 ò [ ] 57. In =
2 2
0
(a) 1/2 (b) 1 (c) ¥ (d) 0.
(c) e + e 2 + 52 (d) e - e 2 - 52 . (2003) (2002)
2 2
2
51. The area of the region bounded by the curves y = |x – 1| and y
58. ò [ x2 ] dx is
= 3 – |x| is 0
(a) 3 sq. units (b) 4 sq. units (a) 2 - 2 (b) 2 + 2
(c) 6 sq. units (d) 2 sq. units. (2003) (c) 2 - 1 (d) 2 - 2. (2002)
Answer Key
1. (b) 2. (d) 3. (d) 4. (c) 5. (b) 6. (a, d)
7. (a) 8. (c) 9. (d) 10. (c) 11. (b) 12. (b)
13. (a) 14. (b) 15. (a) 16. (b) 17. (c) 18. (d)
19. (a) 20. (c) 21. (a) 22. (c) 23. (*) 24. (c)
25. (b) 26. (c) 27. (d) 28. (b) 29. (c) 30. (a)
31. (c) 32. (b) 33. (a) 34. (c) 35. (c) 36. (c)
37. (c) 38. (c) 39. (b) 40. (a) 41. (c) 42. (d)
43. (b) 44. (b) 45. (c) 46. (b) 47. (b) 48. (a, c)
49. (a) 50. (b) 51. (b) 52. (a) 53. (a) 54. (b)
55. (d) 56. (d) 57. (b) 58. (c)
56 JEE MAIN CHAPTERWISE EXPLORER
1. (b) : Let x3 = u, then 3x 2d x = du 5. (b) : ò ta5ntaxn- x2 dx = x + aln|sin x - 2 cos x| + k
Also suppose ò f (x)dx = y(x) Differentiating both sides, we get
Now òx5 f (x3 )dx = 31 ò u f (u) du
= 31 ëéuò f (u)du - ò (ò f (u)) duùû 5 tan x = 1 + a (cos x + 2 sin x)
tan x - 2 sin x - 2 cos x
ò = 1 x3y(x3) - x2y(x3 ) dx + C
3 Þ 5sin x = sin x (1 + 2a) + cos x (a - 2)
sin x - 2cos x sin x - 2 cos x
dy Þ a = 2
(d) : dx
2. = | x |= 2. \ x = ± 2 x
6. (a, d) : g(x) = ò cos 4t dt
0
We can solve for y to get é sin ù x sin 4 x
4t 4
t2 2 = Þ g ( x) = êë 4 úû 0 =
2 2 2 0
| t | dt tdt
0 0
ò ò y1 = = = 2
-2 -2 Þ g(x + p) = sin 4(x + p) = sin 4 x
44
and y2 = ò| t | dt = - ò tdt = -2
0 0 Þ g(p) = 0 Þ g(x + p) = g(x) + g(p) or g(x) – g(p).
Tangents are y – 2 = 2(x – 2) and y + 2 = 2(x + 2) 7. (a) : x2 = y , x 2 = 9y
4
Then the x intercepts are obtained by putting y = 0.
Area bounded by the parabolas and y = 2
We then get x = ±1 2 æ y ö 2
3. (d) : Solving y = x with 2y – x + 3 = 0, we have = 2 ´ ò çè3 y- ø÷ dy = 5 ò ydy
0 2 0
2 x - x + 3 = 0 Þ ( x - 3)( x + 1) = 0 = 5 ´ (y)3/2 = 10 ´ 2 2 = 20 2
\ x = 1, 9 3/2 3 3
8. (c) : d(p(t)) = 0.5p(t) - 450
dt
p p 2 dp = t Þ 2ln p - 900 = t
- 900 -5 0
ò ò dt
850 0
3 - y3 3 Þ p = 900 - 50∙et/2
3 0
ò Area = [(2y + 3) - y2 ] dy = y2 + 3 y If p = 0, then 900 = et/2 Þ t = 2 ln 18
50
0
= 9 + 9 – 9 = 9
p/3 dx p / 3 1 9. (d) : dV = -k(T - t)
tanx cotx dt
p/61 + =
ò ò 4. = p / 6 1 +
(c) : I
k(T - t) 2
On integration, V = 2 + a
p/ 3æ 1 + 1 + tan x ö dx At t = 0, V(t) = I Þ I = kT 2 + a
tanx tanx ø÷ 2
Adding, 2 I = pò/ 6 çè1 +
ò =p/3 = æp - pö = p \ I = p \ a = I - kT 2
çè 3 6 ø÷ 6 12 2
1 × dx
kT 2
p/ 6 2
Again StatementII is true. As t = T, we have V(T) = a = I -
Integral Calculus 57
dy = y + 3 Þ dy = dx 15. (a) :
10. (c) : dx y + 3
As y(0) = 2, we have ln5 = C
Now ln(y + 3) = x + ln5
As x = ln2 we have
ln(y + 3) = ln2 + ln5 = ln10
Þ y + 3 = 10 Þ y = 7. The desired area =
11. (b) : I = ò1 8 ln +(1x+2 x) dx p/ 4 5p/4 3p/2
1
0 ò (cos x - sin x)dx + ò (sin x - cos x)dx + ò (cos x - sin x)dx
0 p/4 5p/4
ò Let J = 1 ln (1 + x) = 2[sin x + cos x ]p / 4 + [- cos x - sin x]5 pp/4/ 4
0 1+ x2
dx 0
(As the first and third integrals are equal in magnitude)
p/4 æ 1+ 1 - 1 ÷øö æ 1+ 1+ 1+ 1 ö
çè 2 2 èç 2 2 2 2 ÷ø
Let x = tanq Þ J = ò ln(1 + tan q)dq = 2 +
0
ò Now J = p/4 + tan æ p - q) ø÷ö dq = 8 - 2 = 4 2 - 2
èç 4 2
ln(1
0
ò Adding 2J p/4 + tan q) + ln æ + tan èçæ p - q ø÷ö ö dq 16. (b) : (y – 2) 2 = x –1
èç1 4 ø÷
= ln(1 Differentiating w.r.t. x, we have 2(y –2)y¢ = 1
p/4 0
= ò ln íì(1 + tan q) æèç1 + tan æèç p - q ÷öø öü dq Þ y ¢ = 1 y¢ = 1/2
î 4 ø÷þý 2( y - 2) at (2, 3),
0
p/4 p p The equation of the tangent to the parabola at (2, 3) is
4 4
2J = ò (ln 2)dq = ln 2 Þ 8J = 4 ln 2 1
y – 3 = 2 (x –2) Þ x – 2y + 4 = 0
0
Þ I = 8J = p ln 2.
x
12. (b) : f (x) = ò t sin tdt
0
f ¢ (x) = x sin x
f ¢¢( x) = x cos x + 12 x-1/ 2 sin x
f ¢¢ (p) = - p < 0 ; f ¢¢ (2p) = 2p > 0 The area of the bounded region
Thus at p maximum and at 2p minimum.
3
13. (a) : Area = 1 + e dx = 1 + ln x e= 3
2 x 2 2 = ò [( y - 2)2 + 1 - (2 y - 4)] dy
ò 1
0
1 3 3
= ò ( y2 - 6 y + 9) dy = ò ( y - 3) 2 dy
0 0
14. (b) : p¢(x) = p¢(1 – x) (Let 3 – y = t)
On integration, = 3 (3 - y)2 dy = 3 = é t 3 ù3 = 33 = 9
êë 3 úû 0 3
p(x) = –p(1 – x) + k, ò ò t 2 dt
0 0
k being the constant of integration. p
Set x = 0 to obtain p(0) = –p(1) + k 17. (c) : I = ò [cot x] dx
Þ 1 = – 41 + k. \ k = 42 0 p
p
I = ò [cot(p - x)]dx = ò [- cot x]d x
1 1 0 0
Now, I = ò p(x)dx = ò p(1 - x)dx Adding we have
0 0
p
On adding we get
2I = ò {[cot x] + [- cot x]}d x
1 1 1
0 \ I = –p/2
2I = ò p(x) + p(1 - x) dx = ò kdx = ò 42 dx = 42. p
0 0 0
2I = ò (-1) dx = -p
Thus I = 21.
0
Note that [x] + [–x] = 0, x Î Z = –1, x Ï Z.
58 JEE MAIN CHAPTERWISE EXPLORER
18. (d) : 2 ò sin èæsixn -x p4 öø dx 23. (*) : ëésec -1 t ù x = p
û 2 2
sin æè x - p pö sec-1 x - sec-1 2 = p Þ sec -1 x = p + p = 3 p
sin èæ x 4 4 ø 2 2 4 4
+
- x = - 2. There is no correct option.
= 2 ò pö dx
4 ø
24. (c) : F ( x) = x ln t dt + 11/ò x 1l n+ tt dt
2 ò éëêcos p æ pö sin p4 ûùú 1+ t
= 4 + cot è x - 4 ø dx ò
1
2× 1 x+ 1 ln sin èæ pö F ( x) = x æ ln t + ln t ö dt = 1òx lnt t dt = 12 (ln x ) 2
2 2 4 ø çè 1+ t (1 + t) t ÷ø
= 2× x - + c ò
1
F(e) = 1/2.
= x + ln sin èæ x - pö + c a
4 ø
25. (b) : ò [ x] f ¢( x) dx , say [a] = K such that a > 1
c being a constant of integration.
2
19. (a) : Solution x + 2y2 = 0 and x + 3y 2 = 1 we have 2 3
1 – 3y 2 = – 2y 2 Þ y 2 = 1 \ y = ± 1 = ò1 f ¢( x)dx + ò 2 f ¢( x)dx + ...... +
K a
1 2
ò ( K -1) f ¢( x)dx + ò Kf ¢( x) dx
y = – 1 Þ x = – 2
K -1 K
y = 1 Þ x = – 2
The bounded region is as under = f (2) – f (1) + 2[ f (3) – f (2)] + 3[ f (4) – f (3)] + ...
(K – 1) [ f (K) – f (K –1)] + K[ f (a) – f (K)]
y
(– 2, 1) = – [ f (1) + f (2) + ...... + f (K)] + K f (a)
= [a] f (a) – [ f (1) + f (2) + ....... + f ([a])]
A
O
x¢ (1, 0) x
B
(– 2, – 1) x + 3y2 = 1 -p
x + 2y2 = 0 y¢ 26. (c) : Let I = 2 éë(x + p)3 + cos2 (x + 3p) ùû dx
ò
1 -3 p
The desired area = 2ò [(1 - 3 y2 ) - ( -2 y2 )] dy 2
Putting x + p = z
0 -p p -3p -p
2 2 2 2
ò 1 é y 3 ù1 also x = Þ z = and x = Þ z =
= 2 êë y - 3 úû 0
= 2 (1 - y2 ) dy \ dx = dz
0 and x + 3p = z + 2p
= 2 ´ 2 = 4 sq. units p pp
33 2
2 2
20. (c) : In the interval of integration sin x < x \ l= ò [z3 + cos2 (2p + z)] dz =
ò z3dz + ò cos 2 z dz
1 x 1 -p -p -p
dx =
x 0
ò ò ò 1 sin x é 2 x 3 / 2 ù1 2 2 2 2
êë 3 úû0 3
I = dx < xdx = = p
0x 0 = 0 (an odd function) +2 2ò cos 2 z dz
2 0
\ I < 3
= 0 + 2 12 ´ 2p
ò òAlso J = 1 cos x dx < 1 1 dx = [2
0x 0 x x ]10 = 2 ìp
ïíusing fact 2ò sin n x dx
\ J < 2 Y ï 0
î
21. (a) : Required area y = |x |
x = y 2
1 ì n - 1 × n - 3 ...... 12 ´ p if n = 2m ü
(1, 1) ï n n - 2 2 ï
= ò ( x - x)d x = í 1 n - ý
X ï n- 3 2
0 î n- 2 ...... 2 if n = 2m + 1ï
þ
é 2 x3/ 2 x 2 ù1 2 1 1 (0, 0) n
ê 3 3 2 6.
= ë - 2 ú = - = = p
û 0 2
(c) : 1 dx = 21 log tan èçæ 2x + 1p2 öø÷ + c . p ..... (i)
( ) 22. 2 ò x+ p
6 27. (d) : Let I = ò xf (sin x) dx
sin
0
Integral Calculus 59
p 0
I = ò (p - x) f (sin x) dx ..... (ii) 32. (b) : Required area = ò loge ( x + e) dx Y
y = loge( x + e )
0 1 - e
(0, 1)
a a e
O X
using ò f ( x)dx = ò f (a - x) dx = ò logz dz
0 0 1 1 – e
By (i) & (ii) on adding = ëéz (loge z -1)ùû 1e = 1.
p
\ I = pp f (sin x)dx = 2 p 2 f (sin x) dx
ò 2 0ò
2 33. (a) : For 0 < x < 1, x 2 > x 3 \ 2x2 > 2 x3
0
2 a a and for 1 < x < 2, x 3 > x2 \ 2x3 > 2 x2
i.e. 2x2 < 2 x 3 Þ I3 < I4
[using ò f ( x)dx = 2ò f ( x) dx if f (2a – x) = f (x)]
0 0
p p
= 2 f æ sin( p2 - x)øö dx = 2 (cos x) dx
è
pò p ò f
0 0 as 2x2 > 2 x3
28. (b) : Using fact
b f (a f ( x ) f (x) = b ( x) dx = b - a \ 1 2x2 dx > 1ò 2 x 3 dx
+b+ x)+ 2
ò òf ò
a a 0 0
\ 6 x dx = 6 - 3 = 3 \ I 1 > I 2.
a - x + x2 2
ò 34. (c) : According to question
3
lim 1 1 2 æ 4 ÷öø + 1 sec2 1 B ( > p/ 4)
n ®¥ n2 n2 n2 çè n2 n
ò
p/ 4
sec2 sec2 ( ) B
ò
p/4
29. (c) : + ... + f ( x)dx = Bsin B + p cos B + B 2
4
= lim 1 sec2 1 + 2 sec 2 çèæ 4 ÷øö + ... + n sec 2 çæ n 2 ö f (b) = sin B + Bcos B - p sin B + 2
n2 n2 n2 n2 n2 è n2 ÷ 4
n ®¥ ø
( ) \ f
( ) ( ) ( ) =r = n 2 r = n 1 r 2 p = 1- p + 2.
n n 2 4
å æ r ö sec 2 r = lim å sec2 r
lim ç n2 ÷ø n n
r =1 è n ®¥ r = 0 f ( x ) 4t 3 dt
n ®¥ x - 2
lim ò (0/0) form,
35. (c) :
x ® 2 6
1
= 1 tan1. f ¢( x) ´ 4( f ( x)) 3
= ò x sec2 ( x2 ) dx 2
= lim
0 1
x ® 2
30. (a) : Let f (x) = p cos 2 x dx ( a > 0) ...(1) = 4 f ¢(2) ´ ( f (2))3 = 1 ´ 4 ´ 6 ´ 6 ´ 6 = 18.
1 + a x 48
ò
-p
\ f ( x) = p cos 2 x dx \ b b 36. (c) : Method by cross check
1 + a- x
ò ò f ( x)dx = ò f (a + b - x) Consider f ( x ) = x
-p a a
\ f ( x) = p ax cos 2 x dx ... (2) (log x)2 + 1
1 + a x
ò 1 + (log x ) 2 - 2 x log x
-p
pp \ f ¢( x ) = x
2 f ( x) = ò cos2 x dx = 2 ò cos 2 x dx (1 + (log x )2 ) 2
-p 0
= p/2 dx , 2 f (x) = 4 ´ 1 ´ p \ f ¢( x ) = 1 + (log x )2 - 2 log x = æ (1 - log x ) ö 2
2 2 (1 + log 2 x )2 ççè (1 + log x ) 2 ÷÷ø
2 ´ 2 ò cos 2 x
0
é By using p/2 sin n x dx ù æ (1 - log x ) 2 ö
ê ú \ ò çç ÷÷ dx = ò f ¢( x ) dx = f ( x )
ò è ) 2 ø
ê 0 ú 1 + (log x
êëê= n -1 × n - 3 ..... 21 ´ p if n is even úûú ö÷2 dx
n n- 2 2 ø
\ ò æ 1 - log x = x
ç + (log x)2 1 + (log x) 2
f ( x) = p è 1
2
Hence (c) is correct answer and we can check the other choices
31. (c) : Total area = 4 × 4 = 16 sq. units C by the similar argument.
y = 4
Area of S3 = 4 x 2 dx = 16 = S1 ìx - 2 if x > 2
4 3 O A ï x = 0
ò 37. (c) : y = í 0 if x < 2
n = 4
0
\ S2 = 16 - 16 ´ 2 = 16 . îï 2 - x if
3 3
\ S 1 : S2 : S 3 is 1 : 1 : 1. Required area = Area of DLAB + Area of DMBC
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