MATHS COMBIND

Jpacademy FUNCTIONS

PREVIOUS EAMCET BITS

1. If f: [2, 3] → IR is defined by f (x ) = x3 + 3x − 2 , then the range f(x) is contained in the interval :

1) [1,12] 2) [12,34] 3) [35,50] [EAMCET 2009]

4) [−12,12]

Ans: 2

Sol. f (2) = 12 and f (3) = 34

∴ Range = [12, 34]

2. ⎧⎨x ∈ IR : x3 2x −1 3x ∈ IR ⎫ = [EAMCET 2009]
⎩ + 4x2 + ⎬
⎭

1) IR −{0} 2) IR −{0,1,3} 3) IR −{0, −1, −3} 4) IR − ⎧⎨0, −1, −3, + 1⎫
⎩ ⎬
2 ⎭

Ans: 3

Sol. x ( 2x −1 + 3) = x (x 2x −1 + 3)
x2 + 4x
+ 1) ( x

is not defined if x ( x +1)( x + 3) = 0 ⇒ x = −3, −1, 0

3. Using mathematical induction, the numbers an ‘s are defined by a0 = 1, an+1 = 3n2 + n + an (n ≥ 0)

, then an = [EAMCET 2009]

1) n3 + n2 +1 2) n3 − n2 +1 3) n3 − n2 4) n3 + n2

Ans: 2

Sol. a0 = 1, a1 = 1, a2 = 3 +1+ a1 = 5 and so on. Verify (2) is correct

4. The number of subsets of {1, 2, 3, …….9} containing at least one odd number is[EAMCET 2009]

1) 324 2) 396 3) 496 4) 512

Ans: 3

Sol. Number of subsets = 29 − 24 = 512 −16 = 46

4. If → C is defined by f (x ) = e2ix for x ∈ R , then f is (where C denotes the set of all complex

numbers) [EAMCET 2008]
3) one-one and onto 4) neither one-one nor onto
1) one-one 2) onto

Ans: 4

Sol. f ( x ) = e2ix = cos 2x + i sin 2x

f (0) = f (π) = 1 ⇒ f is not one one

There exists not x ∈ R ∋ f (x) = 2 ⇒ f is not onto.

5. If f : R → R and g : R → R are defined by f ( x) = x and g ( x) = [x − 3]for x ∈ R , then

⎧⎨g (f ( x ) ) : − 8 < x < 8 ⎫⎬ = [EAMCET 2008]
⎩ 5 5 ⎭ 4) {2, 3}

1) [0, 1] 2) [1, 2] 3) {–3, –2}
1

Jpacademy Functions

Ans: 3

Sol. − 8 < x < 8 ⇒ 0 ≤ x < 8 ⇒ −3 ≤ x − 3 < 8 − 3
55 5 5

⇒ −3 ≤ x −3< −7 ⇒ ⎡⎣ x − 3⎤⎦ = −3or − 2
5

⇒ ⎨⎧g ( f ( x )) : − 8 < x < 8 ⎫ = {−3, −2}
⎩ 5 5 ⎬
⎭

6. If f :[−6, 6] → R defined by f (x) = x2 − 3 for x ∈ R then

(fofof )(−1) + (fofof )(0) + (fofof )(1) = [EAMCET 2008]

(1) f 4 2 ) (2) f 3 2 ) (3) f 2 2 ) 4) f ( 2)

Ans: 1

Sol. (fofof )(−1) + (fofof )(0) + (fofof )(1) = −2 + 33 − 2 = 29

( )f 4 2 = 32 − 3 = 29

7. If Q denotes the set of all rational numbers and f ⎛ p ⎞ = p2 − q2 for any p ∈ Q, then observe
⎜ q ⎟ q
⎝ ⎠

the following statements [EAMCET 2007]

I) f ⎛ p ⎞ is real for each p ∈Q
⎜ q ⎟ q
⎝ ⎠

II) f ⎛p⎞ is complex number for each p ∈Q
⎜ ⎟ q
⎝ q ⎠

Which of the following is correct ? 2) I is true, II is false
1) Both I and II are true

3) I is false, II is true 4) Both I and II are false

Ans: 3

Sol. f ⎛ 1 ⎞ = 1−4 = −3 is an imaginary ⇒ I is false
⎜⎝ 2 ⎟⎠

f ⎛ p ⎞ = p2 − q2 it is a complex number ⇒ II is true
⎜ q ⎟
⎝ ⎠

8. If f : R → is defined by f (x) = 1 for each x ∈ R , then the range of f is

2 − cos 3x
[EAMCET 2007]

1) ⎛ 1 ,1⎞⎟⎠ 2) ⎡ 1 ,1⎤⎦⎥ 3) (1, 2) 4) [1, 2]
⎜⎝ 3 ⎣⎢ 3

Ans: 2

Sol. Max. and Min. values of 2 – cos3x are 3 and 1

∴ Range = ⎡ 1 ,1⎤⎦⎥
⎢⎣ 3

9. If f : R → and g : → are defined by f(x) = x – [x] and g(x) = [x] for x ∈ , where [x] is

The greatest integer not exceeding x, then for every x ∈ , f (g ( x)) = [EAMCET 2007]

2

Jpacademy Functions

1) x 2) 0 3) f(x) 4) g(x)

Ans: 2

Sol. f (g (x))

= g (x) − ⎣⎡g (x)⎤⎦
=[x]−[x] = 0

10. If f = → is defined by f (x) = x − [x] − 1 for x ∈ , where [x] is the greatest integer not

2

exceeding x, then ⎧⎨x ∈ : f ( x) = 1 ⎫ =…… [EAMCET 2006]
⎩ 2 ⎬
⎭

1) Z, the set of all integers 2) IN, the set of all natural number

3) φ, the empty set 4)

Ans: 3

Sol. f (x) = x −[x] − 1 , x ∈

2

f (x) = 1

2

⇒ x −[x]− 1 = 1

22

⇒ x −[x] =1

⇒ {x} = 1 which is not possible, where {x} denotes the fractional part

11. If f = → is defined by f (x) = [2x] − 2[x]for x ∈ . where [x] is the greatest integer not

exceeding x, then the range of f is [EAMCET 2006]

1) {x ∈ : 0 ≤ x ≤ 1} 2) {0, 1}

3) {x ∈ : x > 0} 4) {x ∈ : x ≤ 0}

Ans: 2

Sol. f (x) = [2x] − 2[x], x ∈ = 0

= ∀x ∈ where x = a + f

∋ 0 < f < 0.5
= 1,∀x ∈

x = a + f where 0.5 ≤ a < 1

∴ Range = {0, 1}

12. If f : → ⎧ x + 4 for x < −4

is defined by f (x ) = ⎨⎪3x + 2 for −4 ≤ x < 4 then the correct matching of List I

⎪⎩ x − 4 for x ≥ 4

from List II is [EAMCET 2006]

List – I List – II

A) f (−5) + f (−4) i) 14

B) f ( f (−8) ) ii) 4

C) f (f (−7) + f (3)) iii) – 11

3

Jpacademy Functions

D) f (f (f (f (0)))) + 1 iv) – 1

v) 1
vi) 0

A BC D AB CD
ii v
1) iii vi ii v 2) iii iv v ii

3) iv iii ii i 4) iii vi

Ans: 1

Sol. (A)f (−5) + f (−4) = (−5 + 4) + 3(−4) + 2 = −11

(B)f (−8 + 4) = f (−4) = 3 ⇒ f (4) = 0

(C) f ⎣⎡(−3) +11⎦⎤ = f (8) = 4

(D)f (f (f (2))) = f (f (8)) +1= f (4) +1= 0 +1=1

{ }13. x ∈ : ⎡⎣x − x ⎤⎦ = 5 = [EAMCET 2005]

1) , the set of all real numbers 2) φ, the empty set

3) {x ∈ : x < 0} 4) {x ∈ : x > 0}

Ans: 2

Sol. x− | x |= 2x,∀x < 0

= 0,∀x ≥ 0

∴ x - |x| ≠ 5

14. The function f : c → c defined by f ( x) = ax + b for x ∈ c where bd ≠ 0 reduces to a constant

cx + d

function if [EAMCET 2005]

1) a = c 2) b = d 3) ad = bc 4) ab = cd

Ans: 3

Sol. f (x) = ax + b

cx + d

cx + d) ax + b (a / c

ax + ad / c

bc − ad

c

f (x) = a + bc − ad = constant bc = ad
c
c(cx + d)

2004

15. For any integer n ≥ 1, the number of positive divisors of n is denoted by d(n). Then for a prime P,

d(d(d(P7 )))= [EAMCET 2004]

1) 1 2) 2 3) 3 4) P

Ans: 3

( )Sol. d d(d (p7 )) = d (d (8)) = d (d (23 )) = d (4)

= d(22 ) = 2 +1 = 3

4

Jpacademy Functions

⎧ 2 if n = 3k, k ∈ Z

16. If f : N → Z is defined by f (n) = ⎨⎪10 if n = 3k +1, k ∈ Z then {n ∈ N : f (n) > 2}=

⎪⎩ 0 if n = 3k + 2, k ∈ Z

1) {3, 6, 4} 2) {1, 4, 7} 3) {4, 7} [EAMCET 2004]
4) {7}

Ans: 2

Sol. f (n ) > 2 ⇒ n = 3k +1

⇒ n = 1; n = 4; n = 7

17. The function f : → is defined by f (x ) = 3−x . Observe the following statements of it :

I. f is one-one II) f is onto III) f is a decreasing function [EAMCET 2004]

Out of these, true statements are

1) only I, II 2) only II, III 3) only I, III 4) I, II, III

Ans:

Sol. f : R → R;f ( x) = 3−x

∴ f(x) is one-one and it is decreasing function

⎧ [x] if −3 < x ≤ −1
⎪
18. If f (x) = ⎨ x if 1 < x < 1 , then (x : f (x) ≥ 0) = [EAMCET 2004]

⎪⎩⎣⎡[x]⎦⎤ if 1 ≤ x < 3

1) (–1, 3) 2) [–1, 3) 3) (–1, 3] 4) [–1, 3]

Ans: 1

Sol. Verification
19. I f : → and g : → are definite by f(x) = 2x + 3 and g(x) = x2 + 7 then the values of x such

that g(f(x)) = 8 are [EAMCET 2003]
4) 1, –2
1) 1, 2 2) –1, 2 3) –1, –2

Ans: 3

Sol. g (f (x)) = 4x2 +12x +16

⇒ 4x2 +12x +16 = 8

⇒ (x +1)(x + 2) = 0 ⇒ x = −1, −2

20. Suppose f :[−2, 2] → is defined f (x) = ⎧ −1 for − 2 ≤ x ≤ 0
⎨⎩x −1 ,

for 0 ≤ x ≤ 2

{ }then x ∈[−2, 2]: x ≤ 0 and f ( x ) = x = .... [EAMCET 2003]

1) {–1} 2) {0} 3) ⎧⎨− 1⎫ 4) φ
⎩ ⎬
2 ⎭

Ans: 3

Sol. Now take x = − 1
2

∴f ⎛ − 1 ⎞ = f ⎛1⎞ = 1 −1 = − 1
⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎠⎟ 2 2

Hence f(|x|) = x

5

Jpacademy Functions

∴ Domain of f (x) = ⎨⎧− 1 ⎬⎫
⎩ 2 ⎭

21. If f : → and g : → are given f(x) = |x| and g(x) = [x] for each

{x ∈ : g (f (x)) ≤ f (g (x))} [EAMCET 2003]

1) z ∪ (−∞, 0) 2) (−∞,0) 3) z 4)

Ans: 4

Sol. f (x) = x ;g (x) = [x]

g(f (x)) ≤ f (g(x))
g (f (x)) = g ( x ) = ⎡⎣ x ⎤⎦ = [x]
f (g(x)) = f [x] = [x]

[x]≤ [x]

∴x∈ 3) 1 [EAMCET 2002]
4) – 1
22. If f (x) = ax + a2 , then f (a ) =

ax
1) a 2) 0
Ans: 2

Sol. f ( x) = ax + a2

ax

f′(x) = 2 1 .a + a2 ⎣⎡⎢− 1 (ax )−3/2 a ⎤
ax ⎦⎥
2

f ′(a ) = a − a3.a−3 = 0

2a 2

23. If f (x) = cos2 x + sin4 x for x∈R , then f(2002)= [EAMCET 2002]
sin2 x + cos4 x 4) 4

1) 1 2) 2 3) 3

Ans: 1

Sol. f (x ) = cos2 x + sin4 x
sin 2 x + cos4 x

1− 1 sin2 2x
=4
1− 1 sin2x =1

4

⇒ f (2002) = 1

24. The function f : R → R is defined by f (x) = cos2 x + sin4 x for x ∈ R .Then f(R) =

[EAMCET 2002]

1) ⎛ 3 ,1⎦⎤⎥ 2) ⎡3 ,1⎠⎞⎟ 3) ⎡ 3 ,1⎤⎥⎦ 4) ⎛ 3 ,1⎠⎟⎞
⎜⎝ 4 ⎣⎢ 4 ⎢⎣ 4 ⎝⎜ 4

Ans: 3

6

Jpacademy Functions

Sol. f ( x) = cos2 x + sin4 x

( )= cos2 x + sin2 x 1− cos2 x

= 1− 1 sin2 2x
4

sin2 2x ∈[0,1]

∴ Maximum of f(x) = 1− 1 (0) = 1

4

Minimum of f(x) = 1− 1 (1) = 3

44

∴ Range of f(x) = ⎡3 ,1⎤⎥⎦
⎢⎣ 4

25. If the functions f and g are defined by f (x) = 3x − 4, g ( x) = 2 + 3x for x ∈ respectively, then

( )g−1 f −1 (5) = [EAMCET 2002]

1) 1 2) 1/2 3) 1/3 4) 1/4

Ans: 3 [EAMCET 2001]
4) 2−1
Sol. f −1 ( x ) = x + 4 , g−1 ( x ) = x − 2

33

( )f −1 (5) = 3 g−1 f −1 (5) = g−1 (3) = 1
3

If f ( x ) = 25 − x4 1/4 for 0 < x < ⎣⎢⎡f ⎛ 1 ⎞⎤
( )26. 5 then f ⎜⎝ 2 ⎟⎠⎥⎦ =

1) 2−4 2) 2−3 3) 2−2

Ans: 4

( )Sol. f (x ) = 25 − x4 1/4

( )⇒ f (f ( x)) = ⎡⎣25 − 25 − x4 ⎤⎦1/4 = x

∴ f (f (1/ 2)) = 1 = 2−1

2

27. Let z denote the set of all integers Define f : z →z by f (x) = ⎪⎧x / 2 (x is even)
⎨ ( x is odd) . Then f is =
⎪⎩ 0

[EAMCET 2001]

1) On to but not one-one 2) One –one but not onto

3) One-one and onto 4) Neither one-one nor onto

Ans: 1

Sol. ---

⎧x + 2 (x ≤ −1)
⎪
28. Let f :R →R be defined by f ( x ) = ⎨ x2 (−1 ≤ x ≤ 1) . Then the value of f(–1.75)+f(0.5) +

⎪⎩2 − x (x ≥ 1)

f(1.5) is [EAMCET 2001]

1) 0 2) 2 3) 1 4) – 1

Ans: 3

7

Jpacademy Functions

Sol. f (−1.75) + f (0.5) + f (1.5)

= (−1.75 + 2) + (0.5)2 + 2 −1.5 = 1

29. The functions f : → , g : → are defined as follows: [EAMCET 2001]

f ( x ) = ⎧⎪0 (x rational) ; g(x) = ⎧⎪−1 (x rational)
⎨⎪⎩1 (x irrational) ⎨ ( x irrational) . The (f0g) (π) + (gof) (e) =
⎩⎪ 0

1) –1 2) 0 3) 1 4) 2

Ans: 1

Sol. O

= f (0) + g (1) ( ∵ π and e are irrationals)

=0–1=-1

30. If f : R → R is defined by f(x) = 2x + |x|, then f(2x) + f(–x) – f(x) = [EAMCET 2000]

1) 2x 2) 2|x| 3) –2x 4) –2|x|

Ans: 2

Sol. f (x) = 2x + x

∴ f (2x) + f (−x) − f (x)

= 2(2x) + 2x + 2(−x) + −x − (2x + x ) = 2 x

31. If f : R → R and g : R → R are defined by f(x) = 2x + 3 and g(x) = x2+ 7, then the value of x for

which f(g(x)) = 25 are [EAMCET 2000]

1) ±1 2) ±2 3) ±3 4) ±4

Ans: 2

Sol. f (g (x)) = 25 ⇒ f (x2 + 7) = 25

⇒ 2(x2 + 7) + 3 = 25

∴ x = ±2 [EAMCET 2000]

{ }32. x ∈ R : x − 2 = x2 =

1) {–1, 2} 2) {1, 2} 3) {–1, –2} 4) {1, –2}
Ans: 4

Sol. {1, −2} satisfies

™™™

8

Jpacademy MATHEMATICAL INDUCTION

PREVIOUS EAMCET BITS

1. Using mathematical induction, the numbers an ‘s are defined by a0 = 1, an+1 = 3n2 + n + an (n ≥ 0)

, then an = [EAMCET 2009]

1) n3 + n2 +1 2) n3 − n2 +1 3) n3 − n2 4) n3 + n2

Ans: 2

Sol. a0 = 1, a1 = 1, a2 = 3 +1+ a1 = 5 and so on. Verify (2) is correct

n [EAMCET 2008]

2. For any integer n ≥ 1, then sum ∑ k (k + 2) is equal to
k =1
n (n +1)(2n + 7) n (n +1)(2n + 9)
n (n +1)(n + 2) n (n +1)(2n +1)
3) 4)
1) 2) 6 6
66

Ans: 3

Sol. n k (k + 2) = n (k2 + 2k ) = n k2 n k = n (n + 1) ( 2n +1) + 2n (n + 1)

∑ ∑ ∑ +2∑ 6 2
k=1 k=1 k=1 k=1

n (n +1) [2n +1+ 6] = n (n +1)(2n + 7)

66

3. If Sn = 13 + 23 + ...... + n3 and Tn = 1+ 2 + ..... + n then [EAMCET 2007]

1) Sn = Tn3 2) Sn = Tn2 3) Sn = Tn5 4) Sn = Tn4

Ans: 2

n3 = n2 (n +1)2 n 2 = Tn2

4
∑ (∑ )Sol. Sn =

4. For all integers n ≥ 1, which of the following is divisible by 9 [EAMCET 2006]

1) 8n +1 2) 4n − 3n −1 3) 32n + 3n +1 4) 10n +1

Ans: 2

Sol. by verification n = 2 [EAMCET 2005]

5. {n (n +1)(2n +1) : n ∈ Z} ⊂

1) {6k : k ∈ Z} 2) {12k : k ∈ Z} 3) {18k : k ∈ Z} 4) {24k : k ∈ Z}

Ans: 1

Sol. n (n +1)(2n +1)

= 6. n (n +1)(2n +1) = 6k, k ∈ Z ⎡⎣∵∑ n2 is an int eger⎤⎦

6

∑6. 5 13 + 23 + ...... + k3 [EAMCET 2004]
4) 32.5
k=1 1+ 3 + 5 + .... + (2k −1)

1) 22.5 2) 24.5 3) 28.5

Ans: 1

∑5 k2 (k +1)2

Sol.
k=1 4k 2

1

Jpacademy Mathematical Induction

= 22 + 32 + 42 + 52 + 62 = 22.5
4

7. If t1 = 1 ( n + 2) (n + 3) for n = 1, 2, …….then 1+1 +1 + .... + 1 = [EAMCET 2003]
4 t1 t2 t3 t 2003

1) 4006 2) 4003 3) 4006 4) 4006
3006 3007 3008 3009

Ans: 4

Sol. 1 = 4
tn
(n + 2)(n + 3)

= 4 ⎡ n 1 2 − n 1 ⎤
⎢⎣ + + 3 ⎥⎦

∴ 1 + 1 + .... + 1
t1 t2 tn

4 ⎡ 1 − 1 + 1 − 1 + .... + n 1 2 − n 1 3 ⎤
⎣⎢ 3 4 4 5 + + ⎥⎦

= 4n 3) = 4 ( 2003) = 4006
3 ( 2006 ) 3009
3(n +

8. In the sequence {1}, {2, 3}, {4, 5, 6}, {7, 8, 9, 10}…….. of sets the sum of elements in the 50th

set is [EAMCET 2002]

1) 62525 2) 65225 3) 56255 4) 55625

Ans: 1
Sol. The sum of elements in the nth set is

n (n2 +1)

Sn = 2

50(502 +1)

∴ S50 = 2 = 62525

∑9. = 1 ⎛ n ⎞2 =
If ak k ( k+ 1) , for k = 1, 2,3,...n then ⎜⎝ ⎠⎟ [EAMCET 2000]
ak n6

k =1 4) (n +1)6

1) n n2 n4
n +1
2) (n +1)2 3) (n +1)4

Ans: 1

Sol. Given ak = 1 = 1 − 1
k k +1
k (k +1)

= 1 + 1 + 1 + .... ⎡ 1 + 1 + ......⎦⎥⎤
2 3 ⎢⎣ 2 3

1 =1− 1 = n +1−1 = n
k +1 n +1 n +1 n +1

™™™

2

Jpacademy ADDITION OF VECTORS

PREVIOUS EAMCET BITS

1. IJnJJGa quaJdJJrGilatJeJrJaGl ABJJCJGD, thJeJJGpoint P divides DC in the ratio 1 : 2 and Q is the midpoint of AC. If
AB + 2AD + BC − 2DC = kPQ , then k =
[EAMCET 2009]

1) – 6 2) – 4 3) 6 4) 4

Ans: 1

Sol. A = a, B = b, C = c, D = d

∴ P = c + 2d ,Q = a + c
JJJG J3JJG JJJG 2JJJG JJJG

∴ AB + 2AD + BC − 2DC = KPQ

⇒ k = −6 HJJG

2. THJhJeG posHiJtJiGon vectors of P and Q are respectively a and b. If R is a point on PQ such that
P R = 5P Q , then the position vector of R is
[EAMCET 2008]

1) 5b – 4a 2) 5b + 4a 3) 4b – 5a 4) 4b + 5a

( )AJJnJGs: 1 JJJG JJJG JJJG JJJG JJJG JJJG JJJG JJJG GG G

Sol. PR = 5PQ ⇒ OR − OP = 5 OQ − OP ⇒ OR = 5OQ − 4OP = 5b − 4a
GGGGG G
3. If the points whose position vectors are 2i + j + k, 6i − j + 2k and 14i − 5 j + pk are collinear, then

the value of p is [EAMCET 2007]

1) 2 2) 4 3) 6 4) 8

Ans: 2

Sol. ( x1, y1, z1 ) = (2,1,1);

( x2, y2, z2 ) = (6, −1, 2);

( x3, y3, z3 ) = (14, −5, P)

x1 − x2 = z1 − z2 ⇒ P = 4
x2 − x3 z2 − z3

4. TGheGposGition vGecGtor oGf a point lying on the line joining the points whose position vectors are
i + j − k and i − j + k is [EAMCET 2006]

GG G G
1) j 2) i 3) k 4) 0

Ans: 2

Sol. Vector which is collinear with given two vector by verification answer is i.

5. I : Two non-zero, non-collinear vectors are linearly dependent. [EAMCET 2005]

II: Any three coplanar vectors are linearly dependent.

Which of the above statements is true?

1) Only I 2) Only II 3) Both I and II 4) Neither I nor II

Ans: 3

Sol. By conceptual

6. Observe the following statements : [EAMCET 2005]

1

Jpacademy Addition of Vectors

A : Three vectors are coplanar if one of them is expressible as a linear combination of the other
two.
R : Any three coplanar vectors are linearly dependent.
The which of the following is true?
1) Both A and R are true and R is the correct reason for A
2) Both A and R are true but R is not the correct reason for A
3) A is true, R is false
4) A is false, R is true
Ans: 2
Sol. FroGm thGe defGinitGion AG anGd R are true but R is not correct explanation of A
7. If i + 2 j + 3k,3i + 2 j + k are sides of a parallelogram, then a unit vector parallel to one of the

diagGonGals oGf the parallelGogrGam Gis GGG G G[EAGMCET 2004]
1) i + j + k 2) i − j + k 3) i + j − k 4) −i + j − k
33
3 3

Ans: 1 GGG

Sol. diagonal = 4i + 4 j + 4k GGG

∴ unit vector parallel to diagonal = i + j + k
JJ3JG JJJG JJJG
8. If G JiJsJGthe centroid of theJΔJJAG BC, then GA + BGG + GC =
1) 2GB 2) 2GA 3) 0 JJJG [EAMCET 2004]
4) 2BG

AJJnJGs: 4JJJG JJJG
Sol. GAJJ+JGGBJ+JJGGCJ=JJG0 JJJG
⇒ GA + BG + GC = 2BG
JJJG JJJG
9. If D, E and F are respectively the midpoints of AB, AC and BC in ΔABC, then BE + AF = ....

JJJG JJJG JJJG JJJG [EAMCET 2003]
1) DC BF 3) 2BF BF
2) 1 4) 3

2 2

Sol. Ans:JJ1JG G JJJG G JJJG = G A
Let OA = a, OB = bGb,+OcGC c DF
G G
JJJG = a + c , JJJG =
OF OE
JJJG JJJG2 JJJG JJJG2 JJJG JJJG
BE + AF = OE − OB + OF − OA
G JJJG B EC
( )=G 1 G b = DC
c − a +
2
10. If GGG are three non-coplanar vectors, then the vector equation G = (1 − p − q ) G + G + G
a, b, c r a pb qc

represents is [EAMCET 2003]

1) Straight line 2) Plane

3) Plane passing through the origin 4) Sphere

Sol. Ans: 2 p − q ) G + G + G is a plane passing through a, b and c where p and q are scalars.
a Pb qc
r = (1−

2

Jpacademy Addition of Vectors

11. If three points A, B and C having position vector is (1, x, 3) (3, 4, 7) and (y, -2, -5) respectively

and if they are collinear, then (x, y) = [EAMCET 2002]

1) (2, –3) 2) (–2, 3) 3) (–2, –3) 4) (2, 3)

Ans: 1

Sol. AB = t AC ⇒ (2, 4 − 4, 4)

= t ( y −1, −2 − x, −8)

2 = 4 − x = 4 ⇒ 2 = −1 ⇒ y = −3
y −1 −2 − x −8 y −1 2

4 − x = −1 ⇒ x = 2
−2 − x 2
G G GG G G GGG
12 If the position vectors of the vertices of a triangle are 2i − j + k, i − 3 j − 5k and 3i − 4 j − 4k then

it is a …….triangle [EAMCET 2002]

1) Equilateral 2) Isosceles 3) Right angled isosceles 4) Right-angled

Ans: 4

Sol. Let A = (2, −1,1), B = (1, −3, −5), C = (3, −4, −4) are the vertical of ΔABC

AB2 = 1+ 4 + 36 = 41

BC2 = 4 +1+1 = 6 ; AC2 = 1+ 9 + 25 = 35

AB2 = AC2 + BC2

∴ GΔAGBC iGs rGight GangleGdGtrianGgle. G
13. If a=i G+ 4 j, b = 2i −3 jaGc+=25bGi +9 j , then C G[EAMCET 2001]
1) G b 2) G G G b
5a + 3) a + 3b 4) 3a +

Ans:G4 G G
Sol. Let C = ta + b

⇒ 5i + 9 j = t (i + 4 j) + (2i − 3j) G
G G b
t=3 ∴ C = 3a + JJJG JJJG

14. ABCJDJJGis a parallelogramJ,JwJG ith AC, BD as diagoJnJaJGls. Then AC − BD =JJJG [EAMCET 2001]

1) 4AB 2) 3AB 3) 2AB 4) AB

AJJnJGs: 3JJJG JJJG JJJG JJJG JJJG

( )Sol. AC − BD = AB + BC − BA + AD
( )JJJG JJJG JJJG JJJG JJJG
JJJG G JJJG G JJJG
= AB + BC − −AB + BC = 2AB

15. If OACB is a parallelogram with OC = a and AB = b then OA [EAMCET 2000]

G G G G 1 GG 1 GG
a b a b b−a a−b
1) + 2) − ( )3) ( )4)
2 2

Ans: 4 JJJG JJJG B
O
Sol. MG id JpJoJGint oJJfJGOC = Mid point of AB C
a= OA + OB A

22 JJJG JJJG
G JJJG JJJG JJJG 2OA G OA GG
G = + b ⇒ = 1 a−b
( )a = 2OA + OB − OA ; a
2
””””

3

Jpacademy SCALAR (DOT) PRODUCT OF TWO VECTORS

PREVIOUS EAMCET BITS

1. If m1, m2 , m3 and m4 are respectively the magnitudes of the vectors [EAMCET 2009]

a1 = 2i − j + k, a2 = 3i − 4 j − 4k , a3 = i − j + k and a4 = − i + 3 j + k

Then the correct order of m1, m2, m3, m4 is

1) m3 < m1 < m4 < m2 2) m3 < m1 < m2 < m4

3) m3 < m4 < m1 < m2 4) m3 < m4 < m2 < m1

Ans: 1

Sol. m1 = 6, m2 = 41, m3 = 3, m4 = 11
∴ m3 < m1 < m4 < m2

2. Suppose a = λ i − 7 j + 3k, b = λ i + j + 2λk . If the angle between a and b is greater than 90°, then

λ satisfies the inequality : [EAMCET 2009]

1) −7 < λ < 1 2) λ > 1 3) 1 < λ < 7 4) −5 < λ < 1

Ans: 1

Sol. a.b < 0

⇒ λ2 + 6λ − 7 < 0

(λ −1)(λ + 7) < 0

−7 < λ < 1

3. If the position vectors of A, B and C are respectively 2i − j + k,i − 3j − 5k and 3i − 4 j − 4k , then

cos2A = [EAMCET 2008]

1) 0 2) 6/41 3) 35/41 4) 1

Ans: 3

Sol. AB = OB − OA = (i − 3j − 5k) − (2i − j + k ) = −i − 2 j − 6k

AC = OC − OA = (3i − 4 j − 4k) − (2i − j + k) = i − 3j − 5k

cos A = AB.AC = (−i − 2 j − 6k)(i − 3j − 5k)

AB AC −i − 2 j − 6k i − 3j − 5k

1

Jpacademy Scalar (dot) Product of Vectors

= −1+ 6 + 30 35 = 35
1+ 4 + 36 1+ 9 + 25 35 41 41

∴ cos2 A = 35
41

( )4. If a = i − j − k and b = λ i − 3 j + k and the orthogonal projection of b on a is 4 i − j − k ,
3

then λ = [EAMCET 2007]

1) 0 2) 2 3) 12 4) –1

Ans: 2

Sol. Orthogonal projection of b on a = (b.a) a

a2

⎛ λ +3 − 1 ⎞ ( i − J − k) = 4 (i − J − k) ⇒ λ = 2
⎝⎜ 3 ⎠⎟ 3

5. If a + b + c = 0 and a = 3, b = 4 and c = 37 , then the angle between a and b is

ππ π [EAMCET 2006]
1) 2) 3) π
4)
42 6 3
Ans: 4
[EAMCET 2006]
Sol. a + b = −c Squaring o.b.s

( )a 2 + b 2 + 2 a b cos a, b = c 2
9 + 16 + 24 cos (a, b) = 37
cos(a, b) = 1 ⇒ (a, b) = π

23

6. a.k = a.(2 i + j ) = a ( i + j + 3k) = 1⇒ a

1) i − k 2) 1 (3i + j − 3k) 3) 1 ( i + j + k) 4) 1 (3i − 3 j + k)
3 3 3

Ans: 4

Sol. Let a = a1i + a2 j + a3k

a.i = 1 ⇒ a1 = 1

a.(2i + j) = 1 ⇒ 2a1 + a2 = 1 ⇒ a2 = 1− 2 = −1

2

Jpacademy Scalar (dot) Product of Vectors

a.(i + j + 3k ) = 1 ⇒ a1 + a2 + 3a3 = 1

⇒ 3a 3 = 1 ⇒ a3 = 1
3

∴ a = 1 [3i − 3j + k]

3

7. If the vector a = 2i + 3 j + 6k and b are collinear and b = 21, then b = [EAMCET 2005]

( ) ( ) ( )1) ± 2i + 3 j + 6k 2) ±3 2i + 3 j + 6k 3) i + j + k ( )4) ±21 2i + 3 j + 6k

Ans: 2

Sol. a = t (b)

a = t b ⇒ t = 7 =1
21 3

t =±1 ∴ b = ±3(a )
3

8. If the vectors i + 3 j + 4k, λ i − 4 j + k are orthogonal to each other, then λ = [EAMCET 2004]

1) 5 2) – 5 3) 8 4) – 8
Ans: 3

Sol. a.b = 0 ⇒ λ −12 + 4 = 0 ⇒ λ = 8

9. If a, b, c are three vectors such that a = b + c and the angle between b and c is π : here
2

a = a ,b = b ,c = c [EAMCET 2003]

1) a2 = b2 + c2 2) b2 = c2 + a2 3) c2 = a2 + b2 4) 2a2 − b2 = c2
Ans: 1

Sol. a2 = (b + c)2

a2 = b2 + c2 + 2(b.c)

⇒ a2 = b2 + c2 (∵(b.c) = π / 2)

( ) ( )10. If a.i = a. i + j = a. i + j + k then a = [EAMCET 2002]

1) i 2) j 3) k 4) i + j + k
Ans: 1
Sol. By verification a = i

3

Jpacademy Scalar (dot) Product of Vectors
11. The orthogonal projection of a on b is [EAMCET 2002]

( a.b ) a ( a.b ) b 3) a 4) b
a a
1) a 2 2) 2
b

Ans: 2

( )Sol.a.bb

2
b

12. If θ is an acute angle and the vector (sin θ) i + (cos θ) j is perpendicular to the vector i − 3 j

then θ = [EAMCET 2000]
4) π
1) π 2) π 3) π
65 4 3

Ans: 4

( )Sol. The given vectors are ⊥ er then (sin θi + cos θj). i − 3j = 0

sin θ − 3 cos θ = 0 ⇒ tan θ = 3 [EAMCET 2000]
sin θ − 3 cos θ = 0 ⇒ tan θ = 3 ⇒ θ = π

3
13. If two out of the three vector a + b + c are unit vectors a + b + c = 0 and

( )2 a.b + b.c + a.c + 3 = 0 , then the third vector is of length

1) 3 2) 2 3) 1 4) 0
Ans: 3

2

( )Sol. a + b + c = 0 ⇒ a + b + c = 0
( )∴ a2 + b2 + c2 + 2 a.b + b.c + c.a = 0

1+1+ c2 − 3 = 0 ⇒ c2 = 1
∴ c =1

“““

4

Jpacademy
TRIPLE PRODUCT AND PRODUCT OF FOUR VECTORS

PREVIOUS EAMG CEG TG BG ITG SG G G G

1. The volume of the tetrahedron having the edges i + 2 j − k, i + j + k, i − j + λk as conterminous, is

2/3 cubic units. Then λ [EAMCET 2009]

1) 1 2) 2 3) 3 4) 4

Ans: 1

Sol. V = 1 ⎣⎡a b c ⎤⎦ = 2 cubc units
6 3

⇒λ =1

2. If a = i + j + k , b = i – j + k, c = i + j + k, d = i – j – k, then observe the following lists

[EAMCET 2008]

List – I List – II

i) a.b A) a.d

ii) b.c B) 3

iii) ⎡⎣a b c ⎦⎤ C) b.d

iv) b × c D) 2 j − 2k

E) 2 j + 2k

F) 4

The correct match of List-I to List – II

i ii iii iv i ii iii iv
C AF E
1) C A B F 2) A CF D
4)
3) A C B F

Ans: 2

Sol. a.b = (i + j + k).(i − j + k) = 1−1+1 = 1

b.c = (i − j + k).(i + j − k) = 1−1−1 = −1

11 1

[abc] = 1 −1 1 = 1(1−1) −1(−1−1) +1(1+1) = 0 + 2 + 2 = 4

1 1 −1

ijk

b × c = 1 −1 1 = i (1−1) − j(−1−1) + k (1+1) = 2 j+ 2k

1 1 −1

a.d = (i + j + k).(i − j − k) = 1−1−1 = −1
b.d = (i − j + k).(i − j − k) = 1+1−1 = −1

3. Let a be a unit vector, b = 2i + j – k and c= i + 3k, the maximum value of [a b c ] is

[EAMCET 2008]

1) – 1 2) 10 + 6 3) 10 − 6 4) 59

Ans: 4

1

Jpacademy Triple product and product of four vectors

ijk

Sol. b × c = 2 1 −1 = i (3 − 0) − j(6 +1) + k (0 −1) = 3i − 7 j− k

10 3

[abc] = a.(b × c) = a.(3i − 7 j − k)
= a 3i − 7 j − k cos θ where θ = (a,3i − 7 j − k)

= 9 + 49 +1.cos θ

= 59 cos θ

∴ Maximum value of ⎣⎡abc ⎤⎦ is 59 G G GG G G G G G

4. The volume (in cubic units) of the tetrahedron with edges i + j + k, i − j + k and i + 2 j − k is

1) 4 2) 2/3 3) 1/6 [EAMCET 2007]
Ans: 2 4) 1/3

11 1
Sol. V = 1 1 −1 1 = 2
63
1 2 −1
G GG G GG
5. i − 2 j,3 j + k and λ i − 3 j are coplanar then = [EAMCET 2006]

1) – 1 2) 1/2 3) –3/2 4) 2

Ans: 3

a1 b1 c1 1 −2 0

Sol. a, b, c are coplanar ⇒ a2 b2 c2 = 0 ⇒ 0 3 1 = 0

a3 b3 c3 λ30

1(0 − 3) + 2(0 − λ) + 0(0 − 3λ) = 0

λ = −3 [EAMCET 2006]
2

6. IfGthe vGoluGmeGof tGhe paraGlleloGpipeGd with coterminous edges
4i + 5 j + k, − j + k and 3i + 9 j + pk is 34 cubic units, then p = ……..

1) 4 2) –13 3) 13 4) 6
Ans: 1 or 3

451

Sol. Volume = |[a b c ]|= 0 −1 1 = 34

39p

⇒ 4p +18 = 34 ⇒ p = −13 or 4

7. Observe the following lists [EAMCET 2005]

LAi)st⎡⎣–aG IG cG⎤⎦ G ( )ListG–GII GG
b
)×b 1) a b cos ab
( cG G GG G GG G
B) ×a ( ) ( )2) a.b b− ab c

( )G G G GG G
3) a.b × c
C) a × b × b

2

Jpacademy Triple product and product of four vectors

GG GG
D) a.b 4) a b

GG G GG G
b.c a a.b c
( ) ( )5) −

A BC D A B CD
2) 3 5 21
1) 1 2 3 4 4) 2 3 41

3) 3 2 5 1 [EAMCET 2004]
Ans: 2 GG G
4) a.c × b
Sol. ⎣⎡a b c⎤⎦ = a.( b × c )
( c × a )× b = (b.c ) a − (a.b) c

a ×(b × c ) = (a.c ) b − (a.b) c

a.b = a b cos( a.b)
G G G G G G
( ) ( )8.c.b + c × a + b + c =
GG G G G
1) c.b × a 2) 0 3) GG × b
c.a

Ans: 1

Sol. ( c.b + c.c )× (a + b + c ) = c.( b × a ) = ⎣⎡c b a ⎤⎦

G G GG G G G G [EAMCET 2004]
9. If 3i + 3 j + 3k, i + k, 3i + 3 j + λk are coplanar, then λ = 4) 4

1) 1 2) 2 3) 3

Ans: 1

33 3
Sol. 1 0 1 = 0 ⇒ λ = 1

3 3λ
G G G GG G G G GG G G G
( )10. If a = i + j + k, b = i + j, c = i and a × b × c = λa + μb , then λ + μ.... [EAMCET 2003]

1) 0 2) 1 3) 2 4) 3

AGns:G1 G ( aG.cG ) G GG G G G
a×b c b b.c a = b − a
( ) ( )Sol. × = −

11. IλfG=⎣⎡−aGG1bG;μcG⎤⎦==13⇒, thλe+nμth=e0volume (in cubic units) of the parallelopiped with GGGG and
2a + b, 2b + c

2c + a as coterminous edges is [EAMCET 2002]

1) 15 2) 22 3) 25 4) 27

Ans: 4

Sol. = ⎣⎡2aG + G G + G G + aG ⎦⎤ = 2 1 0
b 2b c 2c 0 2 1 ⎣⎡abc ⎦⎤ = 9.3 = 27

102 [EAMCET 2002]
GG GG GGG
G G
( ) ( ) ( )12. a + b . b + c × a + b + c = b b
⎣⎡aG G G 2 ⎡⎣aG G ⎡⎣aG cG ⎦⎤
1) 0 2) − b c ⎦⎤ 3) c ⎤⎦ 4)

Ans:

3

Jpacademy 1 Triple product and product of four vectors
GG GG GGG 1 1 0 [EAMCET 2001]
a+b . b+c × a+b+c = 0 1 ⎡⎣a b c ⎤⎦
( ) ( ) ( )Sol.

111

= ⎣⎡0 −1(−1) + 0⎤⎦ ⎣⎡a b c ⎦⎤ = ⎡⎣a b c ⎤⎦

13. ⎡⎣ i − j j − k k − i ⎦⎤ =

1) 0 2) 1 3) 3 4) 2
Ans: 1

1 −1 0

Sol. 0 1 −1 = 0

−1 0 1
GGGG G G
If a, b, c, d are coplanar vectors then G b G d
( ) ( )14. a × × c × = [EAMCET 2001]
GG G [EAMCET 2000]
1) 1 2) a 3) b 4) O

AG nGs:G4 G

( ) ( )Sol. ∴aaG,×baGbG,×c,abGdnd×arcGecG×c×dGodGpalra=enOapGrarallel
G G G GG G G G G G G G G GG
( )15. If a = 2i + 3 j − 4k, b = i + j + k and c = 4i + 2 j + 3k and a × b × c =

1) 10 2) 1 3) 2 4) 5

Ans: G4 G aG.cG ) G GG G
G b× c b a.b c
a( ) ( )Sol.× = ( −

= −2i − k = 4 +1 = 5
G
( )16.b×G × ( G × G ) = [EAMCET 2000]
c c a

1) ⎣⎡aG G cG ⎤⎦ G 2) ⎡⎣aG G cG ⎤⎦ G 3) ⎣⎡aG G cG ⎤⎦ G ( )4)G×G × G
b c b b b a a b c

AGns:G1 G G G G G G G G G G
b×c c a b c a c b c c a
( ) ( )( ) ( ) ( )Sol.×(× ) = × − ×

⇒ ⎡⎣aG G cG ⎦⎤ G − 0 = ⎣⎡aG G cG ⎤⎦ G
b c b c

””””

4

Jpacademy VECTOR PRODUCT OF TWO VECTORS

PREVIOUS EAMCET BITS

1. Let a1i + a2 j + a3k [EAMCET 2007]

Assertion (A) : The identify a × i 2 + a × j 2 + a × k 2 = 2 a 2 holds for a

Reason (R) : a × i = a3 j − a2k, a × j = a1k − a3 i , a × k = a2 i − a1 j

Which of the following is correct

1) Both A and R are true and R is the correct reason for A

2) Both A and R are true but R is not the correct reason for A

3) A is true, R is false

4) A is false, R is true

Ans: 1

Sol. axi = a3J − a2k; axJ = a1k − a3i; axk = a2i − a1J

∴ R is true

a = a12 + a 2 + a 2
2 3

( )axi 2 + axJ 2 +
axk 2 = 2 a12 + a 2 + a 2 =2a2
2 3

( ) ( )2. If a and b are unit vectors, then the vector a + b × a × b is parallel to the vector

1) a − b 2) a + b 3) 2a − b [EAMCET 2005]
4) 2a + b

Ans: 2

Sol. By verification

( ) ( ) ( )a + b × a × b . a + b = ⎡⎣a + b a × b a + b⎦⎤ = 0

(∵ [a b a] = 0)

3. Let a, b, c be the position vectors of the vertices, A, B, C respectively of ΔABC. The vector area

of ΔABC is [EAMCET 2003]

{ }( ) ( ) { }1) 1 a × b× c + b×(c× a) + c× a × b
2) 1 a × b + b × c + c × a
22

1

Jpacademy Vector product of two vectors

{ }3) 1 a + b + c { }4) 1 (b.c)a + (c.a) b + (a.b) c
2 2
Ans: 2

( )Sol. Δ = 1 a × b + b× c + c × a
2

4. If θ is the angle between a and b and a × b = a.b , then θ = [EAMCET 2001]

1) 0 2) π 3) π 4) π
Ans: 4 22

Sol. Given a × b = a.b

⇒ a b sin θ = a b cos θ

tan θ = 1 ⇒ θ = π
4

5. If θ is the angle between the vectors 2i − 2 j + 4k and 3 j + j + 2k, then sin θ = [EAMCET 2000]

1) 2 2) 2 3) 2 4) 2
7 7 7 7

Ans: 2

Sol. sin θ = a × b = 2
ab 7

””””

2

Jpacademy COMPLEX NUMBERS

PREVIOUS EAMCET BITS

1. The locus of z satisfying the inequality z + 2i < 1 , where z = x + iy, is [EAMCET 2009]
2z + i

1) x2 + y2 < 1 2) x2 − y2 < 1 3) x2 + y2 > 1 4) 2x2 + 3y2 < 1

Ans: 3

Sol. x + i ( y + 2) 2 < 2x + i (2y +1) 2

⇒ x2 + y2 >1

2. The points in the set ⎨⎧z ∈ C : Arg ⎛ z−2 ⎞ = π⎫ lie on the curve which is a (where C denotes the
⎩ ⎝⎜ z − 6i ⎟⎠ ⎬
2 ⎭

set of all complex numbers) [EAMCET 2008]
4) hyperbola
1) circle 2) pair of lines 3) parabola

Ans: 1

Sol. z−2 = (x − 2) + iy = ⎣⎡( x − 2) + iy⎦⎤ ⎣⎡x −i(y − 6)⎤⎦
z − 6i x + i(y − 6)2
6) x2 −(y−

x (x − 2) + y ( y − 6) xy − (x − 2)( y − 6)
= x2 + (y − 6)2 + x2 + (y − 6)2 i

Arg ⎛ z−2 ⎞ = π ⇒ tan −1 xy − (x − 2)( y − 6) = π
⎝⎜ z − 6i ⎟⎠ 2 ⎡⎣x (x − 2) + y ( y − 6)⎦⎤ 2

⇒ xy − ( x − 2)(y −6) = 1
x(x − 2)+ y(y −6) 0

⇒ x ( x − 2) + y ( y − 6) = 0 ⇒ x2 + y2 − 2x − 6y = 0 ⇒ ( x, y) lies on a circle.

( )3. ⎧ π⎫
If ω is a complex cube root of unity, then sin ⎨ ω10 + ω23 π− ⎬ = [EAMCET 2008]
⎩ 4 ⎭

1) 1 2) 1 3) 1 4) 3
22 2

Ans: 1

( ) ( )Sol.⎡ π⎤ ⎡ π ⎤ ⎛ π ⎞ π 1
sin ⎢⎣ ω10 + ω23 π− 4 ⎦⎥ = sin ⎣⎢ ω + ω2 π − 4 ⎥⎦ = sin ⎝⎜ −π − 4 ⎠⎟ = sin 4 = 2

4. If m1, m2, m3 and m4 respectively denote the moduli of the complex numbers 1 + 4i, 3+i, 1-i and

2-3i, then the correct one, among the following is [EAMCET 2008]

1) m1 < m2 < m3 < m4 2) m4 < m3 < m2 < m1

3) m3 < m2 < m4 < m1 4) m3 < m1 < m2 < m4

Ans: 3

Sol. m1 = 1+ 4i = * +16 = 17 , m2 = 3 + i = 9 +1 = 10

m3 = 1− i = 1+1 = 2, m4 = 2 − 3i = 4 + 9 = 13

1

Jpacademy Complex Numbers
∴ m3 < m2 < m4 < m1 [EAMCET 2007]

5. If a = 1− i 3 then the correct matching of List – I from List – II is
2

List – I List – II

i) aa A) 2π
3

ii) arg ⎛ 1⎞ B) −i 3
⎝⎜ a ⎟⎠

iii) a − a C) 2i / 3

iv) Im ⎛ 4 ⎞ D) 1
⎜⎝ 3a ⎠⎟

E) π/3
F) 2

3

i ii iii iv i ii iii iv
2) D A BF
1) D E C B 4) D A BC

3) F E B C

Ans: 2

Sol. i) a.a = ⎛ 1− i 3 ⎞⎛1+i 3 ⎞ = 1 = D
⎜⎝⎜ 2 ⎟⎟⎠⎜⎝⎜ 2 ⎠⎟⎟

ii) Arg ⎛ 1 ⎞ = Arg ⎛ 1−i 3 ⎞ = 2π = A
⎝⎜ a ⎠⎟ ⎜⎝⎜ 2 ⎟⎟⎠ 3

iii) a − a = −i 3 = B

⎛ 4 ⎞ ⎛ 8 ⎞ 2 =F
⎝⎜ 3a ⎟⎠ ⎜ 1− ⎟= 3
( )iv)Im = Im ⎜ ⎟
⎝ ⎠
3 3i

6. The locus of the point z = x + iy satisfying z − 2i = 1 is [EAMCET 2007]
z + 2i 4) x = 2

1) x-axis 2) y-axis 3) y = 2

Ans: 1

Sol. Z − 2i = Z + 2i

x2 + (y − 2)2 = x2 + (y + 2)2 ⇒ y = 0

∴ Locus is x-axis

7. The locus of the point z = x + iy satisfying the equation z −1 = 1 is given by [EAMCET 2006]
z +1

1) x = 0 2) y = 0 3) x = y 4) x + y = 0

Ans: 1

Sol. z −1 2 = z +1 2

( x −1)2 + y2 = ( x +1)2 + y2

2

Jpacademy Complex Numbers

⇒ 4x = 0 ⇒ x = 0 [EAMCET 2006]
8. The product of the distinct (2n)th roots of 1+ i 3 is equal to

1) 0 2) −1− i 3 3) 1+ i 3 4) −1+ i 3

Ans: 2

Sol. by substitution method put n = 1

( )Then 1 ⎛ ⎛ 1 3 ⎞ ⎞1/ 2 = 1 ⎛ π 1
1+ i 3 2 = ⎜⎜⎝ 2 ⎜⎝⎜ 2 + i 2 ⎠⎟⎟ ⎟⎟⎠ ⎝⎜ cis 3
22 ⎞2
⎠⎟

1 1

= 22 cis ⎛ 2kπ + π ⎞ 2
⎜⎝ 3 ⎟⎠

1 cis π

If k = 0, α1 = 22 6

k = 0, α2 = 21/ 2 cis ⎛ π + π ⎞ = 21/2 cis 7π
⎜⎝ 6 ⎟⎠ 6

Product of roots α1α2 = 21/ 221/ 2 cis π .cis ⎛ 7π ⎞
6 ⎝⎜ 6 ⎠⎟

= 2cis ⎛ π + 7π ⎞
⎜⎝ 6 6 ⎠⎟

= 2cis 8π = 2cis ⎛ 4π ⎞
6 ⎝⎜ 3 ⎠⎟

= −1− i 3

9. If α1, α2 , α3 respectively denote the moduli of the complex number – i, 1 (1+ i) and –1 + i, then

3

their increasing order is [EAMCET 2005]

1) α1, α2 , α3 2) α3, α2 , α1 3) α2 , α1, α3 4) α3, α1, α2

Ans: 3

Sol. α1 = −i = 1, 1 1 + i = 2 = α2 , −1 + i = 2 = α3
3 3

α2 , α1, α3

10. If z1, z2 are two complex numbers satisfying z1 − 3z2 = 1, z1 ≠ 3, then |z2| = [EAMCET 2004]
3 − z1z2

1) 1 2) 2 3) 3 4) 4

Ans: 1

Sol. z1 − 3z2 = 3 − z1z2 ⇒ (z1 − 3z2 )

( z1 − 3z2 ) = (3 − z1z2 ) (3 − z1z2 )

⇒ z1z1 + 9z2 z2 = 9 + z1 2 z2 2

⇒ z1 2 + 9 z2 2 − 9 − z1 2 z2 2 = 0

( )( )⇒ 9 − z1 2 1− z2 2 = 0 ⇒ z2 = 1

3

Jpacademy Complex Numbers

∑11. n ⎛ 2i ⎞n [EAMCET 2004]
n=0 ⎝⎜ 3 ⎟⎠ 4) 9 − 6i

1) 9 + 6i 2) 9 − 6i 3) 9 + 6i [EAMCET 2003]
13 13 4) x – y + 1 =0

Ans: 1

∑Sol.n ⎛ 2i ⎞n =1 = 3 = 9 + 6i
n=0 ⎜⎝ 3 ⎠⎟ 1− 2i 3 − 2i 13

3

12. If the amplitude of z − 2 − 3i is π , then the locus of Z = x + iy is
4

1) x + y – 1 = 0 2) x – y – 1 = 0 3) x + y + 1= 0

Ans: 4

Sol. z − 2 − 3i = (x − 2) + ( y − 3)i

tan −1 ⎛ y − 3 ⎞ = π ⇒ x − y +1 = 0
⎝⎜ x − 2 ⎠⎟ 4

( ) ( )13. If ω is a complex cube root of unity, then 225 + 3ω + 8ω2 2 + 3ω2 + 8ω 2 = ... [EAMCET 2003]

1) 72 2) 192 3) 200 4) 248

Ans: 4

( ) ( )Sol. 225 + 3ω + 8ω2 2 + 3ω2 + 8ω 2

= 225 + 73ω4 + 96ω3 + 73ω2 = 248

14. If z = x + iy is a complex number satisfying z+ i 2 z− i 2

= , then the locus of z is
22

[EAMCET 2002]

1) x-axis 2) y-axis 3) y = x 4) 2y = x

Ans: 1

Sol. Z = x + iy;

x + iy + i 2 = x + iy − i 2
22

x2 + ⎛ y + 1 ⎞2 = x2 + ⎛ y − 1 ⎞2
⎝⎜ 2 ⎠⎟ ⎝⎜ 2 ⎠⎟

⇒ y = 0 ∴ x-axis

15. If z = 3 + 5i, then z3 + z +198 = [EAMCET 2002]
4) 3 + 5i
1) –3-5i 2) – 3 + 5i 3) 3 –5i
Ans: 4

Sol. (3 + 5i)3 + (3 − 5i) +198 = 3 + 5i

16. If 3 + 2i sin θ is a real number and 0 < θ < 2π, then θ = [EAMCET 2002]
1− 2i sin θ

1) π 2) π/2 3) π/3 4) π/6

Ans: 1

Sol. 3 + 2i sin θ × 1+ 2i sin θ
1− 2i sin θ 1+ 2i sin θ

4

Jpacademy Complex Numbers

Purely real ⇒ Imag. Part = 0
ima . part = 8i sin θ = 0

1+ 4sin2 θ
sin θ = 0

∴θ = π

17. If α is a complex number and b is real number then the equation : az + az + b = 0 represents a

[EAMCET 2001]

1) Straight line 2) Parabola 3) Circle 4) Hyperbola

Ans: 1

Sol. Let a = p + iq and z = x + iy

az + az + b = 0

⇒ (p − iq)(x + iy) + (p + iq)(x − iy) + b = 0

equating real parts (or) imaginary parts on both sides then the locus of ‘z’ is straight line.

18. If 1−i i = x + iy , then x = [EAMCET 2001]

1+ 2i −1

1) 1 2) – 1 3) 2 4) –2
Ans: 1

1−i i = x + iy
Sol .
1+ 2i −1

−1(1− i) − i (1+ 2i) = x + iy

−1+ i − i + 2 = x + iy

∴x =1 [EAMCET 2000]
19. The locus of the point Z in the Argand plane for which z +1 2 + z −1 2 = 4 is a

1) Straight line 2) Pair of straight line 3) Circle 4) Parabola
Ans: 3
Sol. z +1 2 + z −1 2 = 4

( x +1)2 + y2 + ( x −1)2 + y2 = 4
∴ x2 + y2 = 1(circle)

20. If θ is real, then the modulus of 1 is [EAMCET 2000]
4) sec −θ
(1+ cos θ) + i sin θ
2
1) 1 sec θ 2) 1 cos θ 3) sec θ
22 22 2

Ans: 1

Sol. 1

(1+ cos θ) + i sin θ

= 1 =1

(1+ cos θ)2 + sin2 θ 2 + 2 cos θ

= 2 = 1 sec θ
2 cos θ / 2 2 2

5

Jpacademy Complex Numbers

3 aω2 + bω 3 =
( ) ( )21.
If 1, ω, ω2 are the cube roots of unity, then (a + b)2 + aω + bω2 +

1) a3 + b3 ( )2) 3 a3 + b3 3) a3 − b3 [EAMCET 2000]
4) a3 + b3 + 3ab

Ans: 2

( ) ( )Sol. (a + b)3 + aω + bω2 3 + aω2 + bω 3

= 3(a + b)(aω + bω2 )(aω2 + bω)

( )= 3 a3 + b3

22. In the Argand plane the area in square units of the triangle formed by the points 1+ i,1− i, 2i is

[EAMCET 2000]

1) 1/2 2) 1 3) 2 4) 2

Ans: 2

Sol. A(1,1) B(1, –1) C(0, 2)

Area ΔABC = 1 1−1 1− 0 = 1 (2) = 1sq.unit

2 1+1 1−2 2

23. If 3 + i is a root of x2 + ax + b = 0, then a = [EAMCET 2000]

1) 3 2) – 3 3) 6 4) –6

Ans: 4

Sol. One root is 3 + i then other roots is 3 – i sum of roots = 6 = – a

⇒ a = −6

””””

6

Jpacademy COMPOUND ANGLES

PREVIOUS EAMCET BITS

1. 2 cos ec20°sec 20° = [EAMCET 2008]
1) 2 2) 2sin20° cosec40° 3) 4 4) 4sin45° cosec40°
Ans: 4

Sol. 2 cos ec20°sec 20° = 2 = 22 =22
sin 20° cos 20° 2sin 20°cos 20° sin 40°

= 2 2 cos ec40° = 4sin 45° cos ec40°

2. If cos (A − B) = 3 and tanA tanB=2, then which one of the following is true? [EAMCET 2007]

5

1) sin (A + B) = 1 2) sin (A + B) = − 1 3) cos (A − B) = 1 4) cos (A + B) = − 1

5 55 5

Ans: 4

Sol. tan A tan B = 2 ⇒ sin A sin B = 2
cos A cos B 1

By using compounds and dividends cos (A − B) = −3 ⇒ cos (A + B) = −1
cos (A + B) 5

3. tan 80° − tan10° = [EAMCET 2007]
tan 70°

1) 0 2) 1 3) 2 4) 3

Ans: 3

Sol. tan 70° = tan (80° −10°)

tan 70° = tan 80° − tan10° [EAMCET 2006]
1+ tan 80° tan10° 4) 6 + 2

⇒ tan 80° − tan10° = 2
tan 70°

4. cos ec15° + sec15° =

1) 2 2 2) 6 3) 2 6

Ans: 3

Sol. sin15 = 3 −1 ; cos15 = 3 +1
22 22

cos ec15° + sec15° = 2 2 + 2 2
3 −1 3 +1

( )2 2 3 +1+ 3 −1 = 2 2 × 2 3 = 2 6
( )( )3 +1 3 −1 2

5. cos12° + cos84° + cos132° + cos156° = 3) – 1 [EAMCET 2004]
1) 1 2) 1 4 4) − 1
24
Ans: 4 2

Jpacademy Compounds Angles

Sol. cos132° + cos12° + cos156° + cos84° = 2 cos 72° cos 60° + 2 cos120°cos36° = − 1
2

6. If cos (α + β) = 4 sin (α − β) = 5 and α, β lie between 0 and π , then tan2α= [EAMCET 2002]
5 13 4

1) 56 2) 33 3) 16 4) 60
33 56 65 61

Ans: 1

Sol. 2α = (α + β) + (α − β)

tan 2α = tan (α + β) + tan (α − β)
1− tan (α + β) tan (α − β)

= 3+5 = 56
42 33
1− 3. 5
4 12

7. cos2 ⎛ π + θ ⎞ − sin 2 ⎛ π − θ ⎞ = [EAMCET 2001]
⎝⎜ 6 ⎟⎠ ⎜⎝ 6 ⎟⎠ 4) 1

1) 1 cos 2θ 2) 0 3) −1 cos2 θ 2
2 2

Ans: 1

Sol. cos2 ⎛ π + θ ⎞ − sin 2 ⎛ π − θ ⎞
⎜⎝ 6 ⎟⎠ ⎜⎝ 6 ⎟⎠

= cos ⎛ π + θ + π − θ ⎠⎟⎞.cos ⎛ π + θ − π + θ ⎞
⎜⎝ 6 6 ⎜⎝ 6 6 ⎟⎠

= 1 cos 2θ
2

DDDD

Jpacademy DEMOVIRE’S THEOREM AND

TRIGONOMETRIC EXPANSION

PREVIOUS EAMCET BITS

1. A value of a such that ⎛ 3 + i ⎞n =1 is [EAMCET 2007]
⎝⎜⎜ 2 2 ⎟⎟⎠ 4) 1

1) 12 2) 3 3) 2 [EAMCET 2005]
4) α
Ans: 1
[EAMCET 2004]
Sol. ⎛ 3 + i ⎞n = 1 ⇒ cis nπ = 1 4) i
⎝⎜⎜ 2 2 ⎟⎠⎟ 6
2
∴ n = 12
[EAMCET 2003]
2. If α is a non-real root of x6 = 1, then α5 + α3 + α +1 = 4) sin 2 θ
α2 +1

1) α2 2) 0 3) – α2

Ans: 3

Sol. x6 = 1 ⇒ let α = ω

α5 + α3 + α +1 = ω5 + ω3 + ω +1
α2 +1 ω2 +1

= −ω2 = −α2

∏3. If = π + π ∞ =
xn cos 2n i sin 2n , then
xn

n =1

1) – 1 2) 1 3) 1
2

Ans: 1

Sol. xn = cis π ⇒ x1x 2 x 3 .....
2n

= cis (π/ 2) = cisπ = −1

1− 1

2

4. If sin 6θ = 32 cos2 θsin θ − 32 cos3 θsin θ + 3x, then x = .........

1) cos θ 2) cos 2θ 3) sin θ

Ans: 4
Sol. sin 6θ = 2sin 3θ cos 3θ

( )( )= 2 3sin θ − 4sin3 θ 4 cos3 θ − 3cos θ

= 32 cos5 θsin θ − 32 cos3 θsin θ + 3sin 2θ
∴ x = sin2θ

5. If xn = cos ⎛ π ⎞ + i sin ⎛ π ⎞ , x1x2x3…….. ∞ = [EAMCET 2002]
⎝⎜ 4n ⎠⎟ ⎝⎜ 4n ⎠⎟

1

Jpacademy Demovire’s Theorem and Trigonometric Expansions

1) 1+ i 3 2) −1+ i 3 3) 1− i 3 4) −1− i 3
2 2 2 2

Ans: 1

Sol. x1x2.............∞
= cis π .cis π .cis π ......∞
4 42 43

= cis ⎛ π + π + ......∞ ⎞ = cis ⎛ 1 π/4 ⎞
⎜⎝ 4 42 ⎟⎠ ⎜⎝ −1/ 4 ⎠⎟

= cis π = 1+ i 3
32

6 If θ = π/6, then the tenth term of 1+ (cos θ + i sin θ) + (cos θ + i sin θ)2 + ... is [EAMCET 2001]

1) i 2) – 1 3) 1 4) –1

Ans: 4

1+ c i s θ + (ci s θ)2 + ......,10th term = cis9θ

Sol.
= cos 9θ + i sin 9θ

at θ = π ⇒ 10th term = −i
6

7 sin 5θ = [EAMCET 2001]
sin θ

1) 16 cos4 θ −12 cos2 θ +1 2) 16 cos4 θ +12 cos2 θ +1

3) 16 cos4 θ −12 cos2 θ −1 4) 16 cos4 θ +12 cos2 θ −1

Ans: 1

Sol. sin 5θ = 16 cos4 θ.sin θ −12 cos2 θ.sin θ + sin θ

∴ sin 5θ = 16 cos4 θ −12 cos2 θ +1
sin θ

UUUU

2

Jpacademy HEIGHTS AND DISTANCES

PREVIOUS EAMCET BITS

1. P is a point on the segment joining the feet of two vertical poles of heights a and b. The angles of

elevation of the tops of the poles from P are 45° each. Then the square of the distance between

the tops of the poles is [EAMCET 2009]

1) a2 + b2 2) a2 + b2 ( )3) 2 a2 + b2 ( )4) 4 a2 + b2
2

Ans: 3 A

2 2 a 2a
a
2a + 2b
( ) ( )Sol. AC2 = B C

( )2 a2 + b2 b
2b
bD

2. From the top of the hill h meters high the angles of depressions of the top and the bottom of a

pillar are α and β respectively. The height (in metres) of the pillar is [EAMCET 2008]

h (tan β − tan α) h (tan α − tan β) h (tan β + tan α) h (tan β + tan α)

1) 2) 3) 4)
tan β tan α tan β tan α

Ans: 1 βα

Sol. x = h cot β, x = (h − d) cot α

⇒ h cot β = (h − d) cot α

⇒ h tan α = (h − d) tan β h α
⇒ d tan β = h (tan β − tan α) h
⇒ d = h (tan β − tan α)
d
tan β
β

x

3. The angle of elevation of an object from a point P on the level ground is α. Moving d metres on

the ground towards the object, the angle of elevation is found to be β. Then the height (in metres)

of the object is [EAMCET 2007]

1) d tan α 2) d tan β 3) d 4) d
cot α + cot β cot α − cot β

Ans: 4

Sol. h cot α = d + x h
h cot β = x αβ

d = h (cot α + cot β)

∴ h= d

cot α − cot β dx

4. The locus of the point z = x + iy satisfying the equation z −1 = 1 is given by [EAMCET 2006]
z +1

1

Jpacademy Heights and Distances

1) x = 0 2) y = 0 3) x = y 4) x + y = 0
Ans: 1
Sol. z −1 2 = z +1 2

( x −1)2 + y2 = ( x +1)2 + y2

⇒ 4x = 0 ⇒ x = 0 [EAMCET 2006]
5. The product of the distinct (2n)th roots of 1+ i 3 is equal to

1) 0 2) −1− i 3 3) 1+ i 3 4) −1+ i 3

Ans: 2

Sol. by substitution method put n = 1

( )Then 1 = ⎛ ⎛ 1 3 ⎞ ⎞1/ 2 = 1 ⎛ π 1
1+ i 3 2 ⎜⎜⎝ 2 ⎜⎝⎜ 2 + i 2 ⎟⎟⎠ ⎟⎠⎟ ⎜⎝ cis 3
22 ⎞2
⎟⎠

1 1

= 22 cis ⎛ 2kπ + π ⎞ 2
⎜⎝ 3 ⎠⎟

1 cis π

If k = 0, α1 = 22 6

k = 0, α2 = 21/ 2 cis ⎛ π + π ⎞ = 21/ 2 cis 7π
⎜⎝ 6 ⎟⎠ 6

Product of roots α1α2 = 21/ 221/ 2 cis π .cis ⎛ 7π ⎞
6 ⎜⎝ 6 ⎟⎠

= 2cis ⎛ π + 7π ⎞
⎜⎝ 6 6 ⎟⎠

= 2cis 8π = 2cis ⎛ 4π ⎞
6 ⎝⎜ 3 ⎟⎠

= −1− i 3

6. A tower, of x meters high, has a Flagstaff at its top. The tower and the Flagstaff subtend equal

angle at a point distant y metres from the foot of the tower. Then the length of the Flagstaff in

metres is ( )x y2 + x2 ( )x x2 + y2 [EAMCET 2005]
( )2) ( )3)
( )y x2 − y2 ( )x x2 − y2
( )1) y2 − x2 x2 − y2 ( )4)

x2 + y2 x2 + y2

Ans: 2 D
Sol. tan 2α = x + h , tan α = x h

yy B

Use tan(2α) formula x 2α α
α
( )x y2 + x2
( )Then h = Ay C

y2 − x2

7. An aeroplane flying with uniform speed horizontally one kilometer above the ground is observed

at an elevation of 60°. After 10 seconds if the elevation is observed to be 30°, then the speed of

the plane (in km/hr) is [EAMCET 2004]

1) 240 2) 200 3 3) 240 3 4) 120
3 3

2

Jpacademy Heights and Distances
Ans: 3 BD
Sol. In ΔAPD ⇒ tan 60° = 1 ⇒ AP = 1 E
60°
AP 3 30° 1km
d
⇒ AP + PQ = 3 AP
PQ = 3 − 1 = 2 km Q

33
10sec− 2 km

3
1hr − 2 × 3600 = 240 3km / hr

3 10

8. A tower subtends angles, α, 2α and 3α respectively at points A, B and C, all lying on a

horizontal line through the foot of the tower. Then AB =……. [EAMCET 2003]
BC

1) sin 3α 2) 1+ 2 cos 2α 3) 2 cos 2α 4) sin 2α
sin 2α sin α

Ans: 2 P
Sol. Let height of the tower OP = h

AB = OA – OB = h(cotα - cot2α)

BC = OB – OC = h(cot2α - cot3α) h

AB = cot α − cot 2α α 2α 3α
BC cot 2α − cot 3α A BC O

= cos α sin 2α − cos 2α sin α × sin 2α.sin 3α

sin α sin 2α cos 2α sin 3α − cos 3α sin 2α

= sin 3α = 1+ 2 cos 2α
sin α

9. From a point on the level ground, the angle of elevation of the top of a pole is 30°. On moving 20

mts nearer, the angle of elevation is 45°. Then the height of the pole in mts is

[EAMCET 2002]

( )1) 10 3 −1 ( )2) 10 3 +1 3) 15 4) 20

Ans: 2

Sol. tan 30° = h
h + 20

( )⇒ h = 10 3 +1

10. The shadow of the two standing on a level ground is found to be 60 metres longer when the sun’s

altitude is 30° then when it is 45°. The height of the tower is [EAMCET 2001]

1) 60 m 2) 30 m 3) 60 3 m ( )4) 30 3 +1 m

Ans: 4

Sol. h = 60

cot 30° − cot 45° 45° 30°
h
( )h = 60 = 30 3 +1
3 −1

30°3 45°

60mt

Jpacademy Heights and Distances

11. If two towers of height h1 and h2 subtend angles 60° and 30° respectively at the midpoint of the

line joining their feet, then h1: h2 = [EAMCET 2000]

1) 1 : 2 2) 1 : 3 3) 2 : 1 4) 3 : 1

Ans: 4

Sol. h1 : h2 = tan α : tan β
= tan 60° : tan 30°

= 3:1

DDD

4

Jpacademy HYPERBOLIC FUNCTIONS

PREVIOUS EAMCET BITS

1. sinh−1 2 + sinh−1 3 = x ⇒ cosh x = [EAMCET 2009]

3) 1 12 + 2 50 ( )4) 1 12 − 2 50
( ) ( ) ( )1) 1 3 5 + 2 10 2) 1 3 5 − 2 10 2
22 2

Ans: 3

( )Sol. cosh x = cosh sinh−1 2 + sinh−1 3

( )= 1 12 + 2 50 [EAMCET 2008]
2

1+ tanh (x / 2)
2. 1− tanh (x / 2)

1) e−x 2) ex 3) 2ex / 2 4) 2e−x / 2

Ans: 2

1+ tanh ⎛ x ⎞ cosh ⎛ x ⎞ + sinh ⎛ x ⎞ ⎣⎢⎡cosh ⎛ x ⎞ + sinh ⎛ x ⎞⎤2
⎜⎝ 2 ⎠⎟ ⎜⎝ 2 ⎟⎠ ⎝⎜ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠⎦⎥
Sol. = =
⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞
1 − tanh ⎝⎜ 2 ⎠⎟ cosh ⎝⎜ 2 ⎠⎟ − sinh ⎜⎝ 2 ⎟⎠ cosh 2 ⎜⎝ 2 ⎠⎟ − sinh 2 ⎝⎜ 2 ⎟⎠

= cosh 2 ⎛ x ⎞ + sinh 2 ⎛ x ⎞ + 2 cosh ⎛ x ⎞ sinh ⎛ x ⎞ = cosh x + sinh x = ex
⎝⎜ 2 ⎠⎟ ⎜⎝ 2 ⎠⎟ ⎝⎜ 2 ⎟⎠ ⎜⎝ 2 ⎠⎟

3. sec h−1 (sin θ) = [EAMCET 2007]

1) log tan θ 2) log sin θ 3) log cos θ 4) log cot θ
2 2 2 2

Ans: 4

Sol. sec h −1 ( sin θ ) = log ⎛ 1 + 1− sin 2 θ ⎞
⎜⎜⎝ sin θ ⎠⎟⎟

= log ⎛ 1+ cos θ ⎞ = log ( cot θ / 2)
⎜⎝ sin θ ⎟⎠

4. ( )e =log cosh−1 2 ( )2) log 3 − 2 ( )3) log 2 + 3 [EAMCET 2006]

( )1) log 2 − 3 4) log (2 + 5)

Ans: 3

( )Sol. eloge f (x) = f x

( )e = cosh 2logecosh−1(2) −1

{ }= log 2 + 22 −1

( )= log 2 + 3

1

Jpacademy Hyperbolic Functions
5. 2 tanh−1 1 = [EAMCET 2005]
2 3) log 3 4) log 4

1) 0 2) log 2 [EAMCET 2004]
Ans: 3 4) –cot2θ
Sol. tanh−1 x = 1 log 1+ x
[EAMCET 2003]
2 1− x
( )4) log 8 + 27
1+ 1
∴ 2 tanh−1 x = log 2 = log 3 [EAMCET 2002]
4) i sin(ix)
1− 1
2 [EAMCET 2001]
4) 20
6. x = log ⎡ ⎛ π + θ ⎞⎤ ⇒ sinh x =
⎣⎢cot ⎜⎝ 4 ⎠⎟⎦⎥ [EAMCET 2000]
4) e2
1) tan2θ 2) – tan2θ 3) cot 2θ
Ans: 2 [EAMCET 2000]
4) 5
cot ⎛ π + θ ⎞ − tan ⎛ π + θ ⎟⎞⎠
⎝⎜ 4 ⎟⎠ ⎜⎝ 4
Sol. sinh x = = − tan 2θ
2

( )7. sinh−1 23/2 = ……

( ) ( )1) log 2 + 18 2) log 3 + 8 ( )3) log 3 − 8

Ans: 2

( ) ( )( )Sol. sinh−1 23/2 = log 8 + 1+ 8 = log 3 + 8

8. sin h ix =

1) i sinx 2) sin (ix) 3) – sinx

Ans: 1

Sol. sinh (ix) = eix − e−ix = (cos x + i sin x ) − (cos x − i sin x) = i sin x

22

( ) ( )9. sec2 tan−1 2 + cos ec2 cot−1 3 =

1) 5 2) 10 3) 15
Ans: 3

Sol. Let tan−1 (2) = α and cot−1 (3) = β

tan α = 2; cot β = 3

⇒ sec α = 5;cos ecβ = 10

∴ sec2 α + cos ec2β = 5 +10 = 15

10. cosh 2 + sinh 2 =

1) 1/e 2) e 3) 1/e2
3) 3
Ans: 4

Sol. cosh 2 + sinh 2 = e2 + e−2 + e2 − e−2 = e2
22

( )11. If cosh−1 x = loge 2 + 3 ,then x =

1) 2 2) 1

2

Jpacademy Hyperbolic Functions

Ans: 1

( )Sol. log ⎡⎣x + x2 −1⎦⎤ = log 2 + 3

log ⎡⎣x + x2 −1⎦⎤ = log ⎣⎡⎢2 + ( 2)2 − 1 ⎤
⎦⎥

∴x=2

””””

3

Jpacademy INVERSE TRIGONOMETRIC FUNCTIONS

PREVIOUS EAMCET BITS

1. cos−1 ⎛ −1 ⎞ − 2 sin −1 ⎛ 1 ⎞ + 3 cos−1 ⎛ −1 ⎞ − 4 tan −1 ( −1) = [EAMCET 2009]
⎜⎝ 2 ⎟⎠ ⎝⎜ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ 4) 43π

1) 19π 2) 35π 3) 47π 12
12 12 12 [EAMCET 2008]

Ans: 4 4) 11

Sol. 2π − 2× π + 3× 3π + 4× π = 43π [EAMCET 2007]
3 6 4 4 12 4) 2

2. If sin −1 ⎛ 3 ⎞ + sin −1 ⎛ 4 ⎞ = π then x = 3
⎜⎝ x ⎟⎠ ⎝⎜ x ⎠⎟ 2
[EAMCET 2005]
1) 3 2) 5 3) 7 4) 0

Ans: 2 [EAMCET 2004]

Sol. sin −1 ⎛ 3 ⎞ + sin −1 ⎛ 4 ⎞ = π ⇒ sin −1 3 = cos−1 4
⎝⎜ x ⎟⎠ ⎝⎜ x ⎠⎟ 2 x x

⇒ sin−1 3 = sin−1 1 − 16 ⇒ 9 = 1 − 16 ⇒ x2 = 25
x x2 x2 x2

( )3. ⎛ ⎛ 1 ⎞ ⎞
The value of x where x > 0 and tan ⎝⎜ sec−1 ⎜⎝ x ⎟⎠ ⎟⎠ = sin tan−1 2 is

1) 5 2) 5 3) 1
3

Ans: 2

( )Sol. ⎛ 1 ⎞
tan ⎜⎝ sec−1 x ⎠⎟ = sin tan−1 2

⎛ 1− x2 ⎞ = sin ⎛ sin −1 2⎞
tan ⎜⎜⎝ tan−1 x ⎟⎟⎠ ⎜ ⎟
⎝ 5 ⎠

⇒ 1− x2 = 2 ⇒ x = 5
x5 3

4. sin−1 4 + 2 tan−1 1 =
53

1) π 2) π 3) π
34 2

Ans: 3

Sol. sin−1 4 + 2 tan−1 1 = sin−1 4 + tan−1 3
5 354

= sin−1 4 + cos−1 4 = π
5 52

5. sin−1 x + sin−1 (1− x ) = cos−1 x ⇒ x ∈

1

Jpacademy ⎧⎨0, 1 ⎫ Inverse Trigonometric Functions
1) {1, 0} 2) {−1,1} 3) ⎩ 2 ⎬
⎭ 4) {2,0}

Ans: 3 [EAMCET 2003]
4) 4
Sol. sin−1 (1− x ) = π − 2sin−1 ( x)
9
2
[EAMCET 2002]
1− x = 1− 2x2 ⇒ x = 0, 1 4) − 3
2
2
6. cos ⎢⎡⎣cos−1 ⎛ 1 ⎞ + sin −1 ⎛ − 1 ⎞⎤ = …….. [EAMCET 2001]
⎝⎜ 7 ⎟⎠ ⎝⎜ 7 ⎠⎟⎥⎦
4) 20
1) − 1 2) 0 3) 1
3 3 [EAMCET 2000]
4) 14/5
Ans: 2

Sol. cos ⎣⎢⎡cos−1 ⎛ − 1 ⎞ + sin −1 ⎛ − 1 ⎞⎤
⎝⎜ 7 ⎟⎠ ⎝⎜ 7 ⎟⎠⎦⎥

= cos π = 0 ⎣⎡⎢∵ sin −1 x + cos−1 x = π⎤
2 2 ⎦⎥

7. If sin−1 x − cos−1 x = π , then x =
6

1) 1 2) 3 3) −1
2 2 2

Ans: 2 3) 15

Sol. By verification x = 3 / 2 3) 5/14

( ) ( )8. sec2 tan−1 2 + cos ec2 cot−1 3 =

1) 5 2) 10
Ans: 3

Sol. Let tan−1 (2) = α and cot−1 (3) = β

tan α = 2; cot β = 3

⇒ sec α = 5;cos ecβ = 10

∴ sec2 α + cos ec2β = 5 +10 = 15

9. If tan−1 3 + tan−1 x = tan− 8 , then x =

1) 5 2) 1/5

Ans: 2
Sol. tan−1 3 + tan−1 x = tan−1 8

3+x =8⇒ x = 1
1− 3x 5

\\[[

2

Jpacademy MULTIPLE AND SUBMULTIPLE ANGLES

PREVIOUS EAMET BITS

1. If A = 35°, B = 15° and C = 40°, then tanAtanB+tanBtancC + tanC tanA [EAMCET 2008]

1) 0 2) 1 3) 2 4) 3

Ans: 2

Sol. A + B + C = 35° +15° + 40° = 90°

⇒ tan A tan B + tan B tan C + tan C tan A = 1

2. If tan θ + tan ⎛ θ + π ⎞ + tan ⎛ θ + 2π ⎞ = 3, then which of the following is equal to 1? [EAMCET 2008]
⎝⎜ 3 ⎠⎟ ⎝⎜ 3 ⎟⎠

1) tan2θ 2) tan3θ 3) tan2θ 4) tan3θ

Ans: 2

Sol. tan θ + tan ⎛ θ + π ⎞ + tan ⎛ θ + 2π ⎞ = 3
⎝⎜ 3 ⎠⎟ ⎝⎜ 3 ⎟⎠

tan θ + tan ⎛ π ⎞ tan θ + tan ⎛ 2π ⎞
⎜⎝ 3 ⎟⎠ ⎝⎜ 3 ⎟⎠
⇒ tan θ + + =3
⎛ π ⎞ ⎛ 2π ⎞
1− tan θ tan ⎝⎜ 3 ⎟⎠ 1− tan θ tan ⎝⎜ 3 ⎟⎠

⇒ tan θ + tan θ + 3 + tan θ − 3 = 3
1− 3 tan θ 1− 3 tan θ

⇒ tan θ + tan θ + 3 tan2 θ + 3 + 3 tan θ + tan θ − 3 tan2 θ + 3 tan

( )( )1− 3 tan θ 1+ 3 tan θ

⇒ tan θ + 8 tan θ = 3⇒ tan θ − 3 tan3 θ + 8sin θ = 3
− 3 tan2 1− 3 tan2 θ
1 θ

⇒ 3 ⎡ 3 tan θ − tan3 θ ⎤ = 3 ⇒ tan 3θ = 1
⎢ 1− 3 tan2 θ ⎥
⎣ ⎦

3. If x = tan15°, y = cosec 75° and z = 4sin18°, then [EAMCET 2006]
4) x < z < y
1) x. > y > z 2) y < z < x 3) z < x < y
[EAMCET 2005]
Ans: 1 4) a

Sol. x = tan15° = 2 − 3 = 0.3 a −1

y = cos ec75° = 1 = 2 2
sin 75° 3 +1

= 6 − 2 ≅ 0.9

z = 4sin18° = 5 −1 = 1.05

x<y<z

4. tan 3A = α ⇒ sin 3A =
tan A sin A

1) 2a 2) 2a 3) a
a +1 a −1 a +1

1

Jpacademy Multiple and Submultiple Angles

Ans: 2

Sol. a = tan 3A , putA = 45° ⇒ a = −1
tan A

Verification sin 3A = 1 = 2a
sin A a −1

5. tan 9° − tan 27° − tan 63° + tan 81° = [EAMCET 2004]

1) 4 2) 3 3) 2 4) 1

Ans: 1

Sol. tan 9° − tan 27° − cot 27° + cos 9° = ⎛ sin 9° + cos 9° ⎞ − ⎛ sin 27° + cos 27° ⎞
⎜⎝ cos 9° sin 9° ⎠⎟ ⎜⎝ cos 27° sin 27° ⎠⎟

⇒ 2 − 2 = 2×4 − 2×4 = 4
sin18° sin 54° 5 −1 5 +1

6 cos 6°sin 24° cos 72° = [EAMCET 2000]
4) 1/4
1) –1/8 2) –1/4 3) 1/8
777
Ans: 3

Sol. 1 (2 cos 6°sin 24°) cos 72°

2

= 1 (sin 30° + sin18°) cos 72°

2

= 1⎛1 + 5 −1⎞⎛ 5− 1⎞ = 1
2 ⎜⎝⎜ 2 4 ⎠⎟⎟⎜⎝⎜ 4 ⎟⎠⎟ 8

2

Jpacademy PERIODICITY AND EXTREME VALUES
PREVIOUS EAMCET BITS

1. The period of sin4 x + cos4 x is [EAMCET 2009]

π4 π2 3) π 4) π
1) 2) 4 2

2 2

Ans: 4

Sol. f ⎛ π + x ⎞ = cos4 x + sin4 x = f (x)
⎝⎜ 2 ⎠⎟

∴ Period = π/2

2. For all values of θ, the values of 3 − cos θ + cos ⎛ θ + π⎞ lie in the interval [EAMCET 2006]
⎜⎝ 3 ⎟⎠

1) [−2,3] 2) [−2,1] 3) [2, 4] 4) [1,5]

Ans: 3

Sol. cos ⎛ θ + π ⎞ = cos θ cos π − sin θ.sin π
⎜⎝ 3 ⎟⎠ 3 3

= 1 cos θ − 3 sin θ
22

3− cos θ + cos ⎛ θ + p ⎞ = 3− 1 cos θ − 3 sin θ
⎜⎝ 3 ⎟⎠ 2 2

Min. value = C − a2 + b2 = 3 − 1 + 3 = 2
44

Max. value = C + a2 + b2 = 3 +1 = 4

( )3. x2 ⎛ π + x2 ⎞ ⎛ π − x2 ⎞ \ [EAMCET 2005]
The extreme value of 4 cos cos ⎜⎝ 3 ⎠⎟ cos ⎜⎝ 3 ⎟⎠ over are

1) – 1, 1 2) –2, 2 3) –3, 3 4) –4, 4

Ans: 1

Sol. cos A cos (60 − A) cos (60 + A) = 1 cos 3A

4

∴ 4 cos x2 cos ⎛ π + x2 ⎞ cos ⎛ π − x2 ⎞ = cos 3x2 = [−1 1]
⎝⎜ 3 ⎠⎟ ⎝⎜ 3 ⎟⎠

4. If n ∈ N , and the period of cos nx is 4π, then n = [EAMCET 2004]
⎛ x ⎞
sin ⎝⎜ n ⎠⎟

1) 4 2) 3 3) 2 4) 1

Ans: 3

Sol. Period of cos nx = 2π
n

1

Jpacademy ⎛ x ⎞ = 2nπ Periodicity and Extreme values
Period of sin ⎝⎜ n ⎠⎟
[EAMCET 2004]
2nπ = 4π ⇒ n = 2
4) [2,7]
5. For x ∈ \,3cos (4x − 5) + 4 lies in the interval
[EAMCET 2003]
1) [1,7] 2) [4,7] 3) [0,7] 4) 12π

Ans: 1 [EAMCET 2002]
4) 2π
Sol. Maximum and Minimum values of cosθ are 1 an – 1

∴ [−3 + 4,3 + 4] = [1, 7]

6. The period of the function f (θ) = sin θ + cos θ is

32

1) 3π 2) 6π 3) 9π

Ans: 4

Sol. The period of sin ⎛ θ ⎞ is 2π = 6π
⎜⎝ 3 ⎟⎠ 1/ 3

The period of θ 2π = 4π
cos is
2 1/ 2

The period of sin ⎛ θ ⎞ + cos ⎛ θ ⎞ is L.C.M of 6π, 4π = 12π
⎜⎝ 3 ⎠⎟ ⎜⎝ 2 ⎠⎟

7. If f (x) = sin 2 ⎛ π + π ⎞ − sin 2 ⎛ π − π ⎞ , then the period of f is
⎜⎝ 8 2 ⎟⎠ ⎜⎝ 8 2 ⎠⎟

1) π 2) π/2 3) π/3

Ans: 4

Sol. Sin2A − sin2 B = sin (A + B)Sin (A − B)

∴ sin 2 ⎛ π + x ⎞ − sin 2 ⎛ π − x ⎞ = sin π sin x
⎜⎝ 8 2 ⎟⎠ ⎜⎝ 8 2 ⎟⎠ 4

∴ period = 2π

777

2

Jpacademy PROPERTIES OF TRIANGLES

PREVIOUS EAMCET BITS [EAMCET 2009]

1. In any ΔABC, a (b cos C − c cos B) =

1) b2 + c2 2) b2 − c2 3) 1 + 1 4) 1 −1
bc b2 c2

Ans: 2

Sol. 1 (2ab cos C − 2ca cos B)

2

1 ⎣⎡a 2 + b2 − c2 − c2 − a2 + b2 ⎦⎤
2

= b2 − c2

2. In a ΔABC (a + b+ c)(b + c−a)(c+ a − b)(a + b−c) [EAMCET 2009]
4) sin2B
4b2c2

1) cos2A 2) cos2B 3) sin2A

Ans: 3

2s (2s − 2a )(2s − 2b)(2s − 2c)

Sol.
4b2c2

= 4Δ2 = ⎛ 2Δ ⎞2 = sin2 A
b2c2 ⎝⎜ bc ⎟⎠

3. In ΔΑBC if 1 + 1 = 3 then C = [EAMCET 2008]
b+c c+a a+b+c

1) 90° 2) 60° 3) 45° 4) 30°

Ans: 2

Sol. Given 1 + 1 = 3
b+c c+a a+b+c

⇒1+ b +1+ a = 3
a+c a+c

⇒ b(b + c)+ a(a + c) = (a + c)(b+ c)

⇒ b2 + bc + a2 + ac = ab + ac + bc + c2

⇒ a2 + b2 − c2 = ab [EAMCET 2008]
Now, cos C = a2 + b2 − c2 = ab = 1

2ab 2ab 2
⇒ C = 60°

4. Observe the following statements :
I) In ΔABC b cos2 C + c cos2 B = s
22
II) In ΔABC, cot A = b + c ⇒ B = 90°
22

1

Jpacademy Properties of Triangles

Which of the following is correct?

1) Both I and II are true 2) I is true, II is false

3) I is false , II is true 4) Both I and II are false

Ans: 2

Sol. I) b cos2 C + c cos2 B = b s (s − c) + c s (s − b) = s (s − c + s − b) = s

2 2 ab ac a

II) If A = 45°, B = 90°, C = 45°, then a = 2R, b = 2R, c = 2R

( )But b + c = 2R + 2R = 2 +1 R = R cot 22 1 ° ≠ cot A
22 22 2 2

5. In a triangle, if r1 = 2r2 = 3r3 , then a+b+c = [EAMCET 2008]
bca

1) 75 2) 155 3) 176 4) 191
60 60 60 60

Ans: 4

Sol. r1 = 2r2 = 3r3 ⇒ s Δ = 2Δ = 3Δ = Δ (say)
−a s−b s−c k

⇒ s − a = k,s − b = 2k,s − c = 3k ⇒ s − a + s − b + s − c = 6k

⇒ s = 6k ⇒ a = 5k, b = 4k, c = 3k

∴ a + b + c = 5k + 3k = 5 + 4 + 3 = 75 + 80 − 36 = 191
b c a 4k 5k 4 3 5 60 60

(or)

If xr1 = yr2 = zr3 , then
a:b:c=y+z:z+x:x+y

∴a:b:c=5:4:3
6. If two angles of ΔABC are 45° and 60°, then the ratio of the smallest and the greatest sides are

( )1) 3 −1 :1 2) 3 : 2 3) 1: 3 [EAMCET 2007]
4) 3 :1

Ans: 1
Sol. Angles are 45°, 60° and 75°.

The ratio of smallest and greatest sides = sin45°; sin75°= 3 −1:1

7. In ΔABC, (a + b + c) ⎛ tan A + tan B ⎞ = [EAMCET 2007]
⎜⎝ 2 2 ⎟⎠

1) 2c cot C 2) 2a cot A 3) 2b cot B 4) tan C
2 2 2 2

Ans: 1

Sol. ( a + b + c) ⎛ tan A + tan B ⎞ = 2s ⎛ s Δ a ) + s Δ b ) ⎞
⎝⎜ 2 2 ⎠⎟ ⎜⎜⎝ ⎟⎟⎠
(s − (s −

= 2Δc = 2c s(s −a)(s − b)(s −c)
(s −a)(s − b)
(s − a)(s − b)

= 2c cot c
2

2

Jpacademy Properties of Triangles

8. In ΔABC, with usual notation, observe the two statements given below [EAMCET 2007]

I) rr1r2r3 = Δ2

II) r1r2 + r2r3 + r3r1 = s2

Which of the following is correct

1) Both I and II are true 2) I is true, II is false

3) I is false, II is true 4) Both I and II are false

Ans: 1

Sol. i) rr1r2r3 = Δ . s Δ . Δ . Δ = Δ2
s −a s −b s −c

ii) r1r2 + r2 r3 + r3 r1 = s Δ . Δ + s Δ. Δ + s Δ. Δ
−a s −b −b s −c −c s −a

= s(s − c) + s(s − a) + s(s − b) = s2

9. If, in a ΔABC, tan A = 5 and tan C = 2 then a, b, c are such that : [EAMCET 2006]
26 25

1) b2 + ac 2) 2b = a + c 3) 2ac = b(a + c) 4) a + b = c

Ans: 2

Sol. tan A .tan C = 5 . 2 = 1
2 2 65 3

(s − b)(s −c) (s −a)(s − b) = 1
s(s −a) s(s −c) 3

⇒ s − b = 1 ⇒ 3s − 3b = s
s3

2s = 3b
a + b + c = 3b
⇒ a + c = 2b

10. The angles of a triangle are in the ratio 3 : 5 : 10. Then the ratio of the smallest side to the

greatest side is [EAMCET 2006]

1) 1: sin 10° 2) 1 : 2 sin 10° 3) 1 : cos 10° 4) 1 : 2 cos10°

Ans: 4

Sol. Let angles as 3x, 5x, 10x

∴ 18x = 180° ⇒ x = 10°

∴ angle aer 30°, 50°, 100°

a : c = sinA : sinC = sin30 : sin100

1 : sin (90 +10) = 1: 2 cos10°

2

11. In a triangle ABC, s − a = 1 , s − b = 1 , s − c = 1 then b = [EAMCET 2006]
Δ 8 Δ 12 Δ 24

1) 16 2) 20 3) 24 4) 28

Ans: 1

Sol. 1 = s − a = 1 ; 1 = s−b = 1 ; 1 = s−c = 1
r1 Δ 8 r2 Δ 12 r3 Δ 24

1=1+1+1 = 6 =1
r r1 r2 r3 24 4

3

Jpacademy Properties of Triangles
(r2 − r )(r1 + r3 ) = b2
(12 − 4)(24 + 8) = b2 ⇒ 16×16, b = 16 [EAMCET 2005]
4) a + b +c
( )12. In a ΔABC, a cos2 B + cos2 C + cos A (c cos C + b cos B) =

1) a 2) b 3) c

Ans: 1

Sol. = a ⎡⎢⎜⎛ a2 + c2 − b2 ⎞2 + ⎛ a2 + b2 − c2 ⎞2 ⎤
⎣⎢⎝ 2ac ⎟ ⎜ 2ab ⎟ ⎥
⎠ ⎝ ⎠ ⎦⎥

+ ⎛ b2 + c2 − a2 ⎞⎡ ⎛ b2 + a2 − c2 ⎞ + ⎛ a2 + c2 − b2 ⎞⎤
⎜ 2bc ⎟ ⎢c⎜ 2ab ⎟ b⎜ 2ac ⎟⎥
⎝ ⎠⎣ ⎝ ⎠ ⎠⎦
⎝

a2 + c2 − b2 + a2 + b2 − c2 = 2a2 = a
2a 2a 2a

13. In a ΔABC, ∑(b + c) tan A tan ⎛ B − C ⎞ = [EAMCET 2005]
2 ⎝⎜ 2 ⎠⎟

1) a 2) b 3) c 4) 0

Ans: 4

Sol. From Napier’s formula tan ⎛ B − C ⎞ = b − c cot A
⎜⎝ 2 ⎟⎠ b+c 2

∑ (b + c) tan A tan ⎛ B − C ⎞ = ∑ ( b + c ) tan A . b − c cot A
2 ⎜⎝ 2 ⎠⎟ 2 b + c 2

∑(b−c) = 0

14. Two sides of a triangle are given by the roots of the equation x2 – 5x + 6 = 0 and the angle

between the sides is π/3. Then the perimeter of the triangle is [EAMCET 2005]

1) 5 + 2 2) 5 + 3 3) 5 + 5 4) 5 + 7

Ans: 4

Sol. c2 = a2 + b2 − 2ab cos C A

= 9 + 4 −12× 1 = 7 2
2 π/3

=c= 7 C3 B

2S = a + b + c = 5 + 7

15. If , in a ΔABC, r3 = r1 + r2 + r , then A + B = [EAMCET 2004]

1) 120° 2) 100° 3) 90° 4) 80°

Ans: 3

Sol. r3 − r = 4R sin 2 c
2

r1 + r2 = 4R cos2 c
2

r1 + r2 + r − r3 = 0

4