MATHS COMBIND

Jpacademy⇒ ⎛ 2 c − sin 2 c ⎞ = 0 Properties of Triangles
4R ⎜⎝ cos 2 2 ⎠⎟
[EAMCET 2004]
⇒ 4R cos c = 0 ⇒ c = 90° 4) a2 + b2
∴ A + B = 90°
[EAMCET 2004]
16. In a ΔABC,(a − b)2 cos2 C + (a + b)2 sin C =
22 4) I, II, III
[EAMCET 2003]
1) a2 2) c2 3) b2
4) a < c < b
Ans: 2

Sol. a2 + b2 − 2ab ⎛ cos 2 c − sin 2 c ⎞ = c2
⎜⎝ 2 2 ⎟⎠

17. In a ΔABC, the correct formulae among the following are

I) r = 4R sin A sin B sin C
222

II) r1 = (s − a) tan A
2

III) r3 = Δ
s−c

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

Ans: 3

Sol. I is true, II is false, III is true

18. If in a ΔABC , r1 < r2 < r3 then

1) a < b < c 2) a > b > c 3) b < a < c

Ans: 1

Sol. r1 < r2 < r3

⇒Δ<Δ<Δ
S−a S−b S−c

⇒S−a >S−b>S−c

⇒ −a > −b > −c

∴a<b<c

19. If in a ΔABC,if 3a = b + c , then cot B .cot C = [EAMCET 2003]
22 4) 4

1) 1 2) 2 3) 3

Ans: 2

Sol. cot B cot C = S(S − b) × S(S − c)
22 Δ Δ

= S S a = 2S a) = a + b+ c = 2
− b + c− a
2 (S −

20. In a ΔABC, if b = 20, c = 21 and sinA = 3/5, then a = ……… [EAMCET 2003]

1) 12 2) 13 3) 14 4) 15

Ans: 2

Sol. sin A = 3 ⇒ cos A = 4
55

a2 = b2 + c2 − 2bc cos A

a = 13

5

Jpacademy Properties of Triangles

21. If ΔABC is right angled at A, then r2 + r3 = [EAMCET 2002]
4) R
1) r1 − r 2) r1 + r 3) r − r1

Ans: 1

Sol. A = 90° and r2 + r3

= 4R cos2 A = 2R
2

r2 − r = 4R sin2 A = 2R
2

22. In a ΔABC, cos C + cos A + cos B = [EAMCET 2001]
c+a b

1) 1 2) 1 3) 1 4) c + a
ab c b

Ans: 2

Sol. cos C + cos A + cos B =
c+a b

= b cos C + b cos A + c cos B + a cos B

b(c + a)

(c + a) = 1
b(c + a) b

23. In a ΔABC, a2 sin 2C + c2 sin 2A [EAMCET 2001]

1) Δ 2) 2Δ 3) 3Δ 4) 4Δ

Ans: 4

Sol. a2 sin 2C + c2 sin 2A

= 4R2 sin2 A.2sin C cos C + 4R2 sin2 C.2sin A cos A

= 8R2 sin A sin C (sin A cos C + cos A sin C)

= 8R2 sin A sin Bsin C = 4Δ

24. If in a ΔABC, a, b, c are in arithmetic progression, then the (A/2) tan (C/2) = [EAMCET 2000]

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

Ans: 2

Sol. tan A tan C = s Δ a ) . s Δ c)
2 2
(s − (s −

= s − b = 2(s − b)

s 2s

= a + c − b = 2b − b = 1
a + b + c 2b + b 3

25. If a ΔABC, cos A + cos B + cos B + cos C = [EAMCET 2000]
4) 1+ R
1) 1+ r 2) 1− r 3) 1− R
R R r r

Ans: 1

6

Jpacademy Properties of Triangles

Sol. cos A + cos B + cos C = 1+ 4 R sin A sin B sin C
222

4R sin A sin B sin C r
2 2 2
=1+ =1+
RR

26. In a ΔABC, r + r3 + r1 – r2 = [EAMCET 2000]

1) 4R cosA 2) 4R cosB 3) 4R cosc 4) 4R

Ans: 2

Sol. r1 + r3 = 4R cos2 B
2

r2 − r1 = 4R sin 2 B
2

r + r1 + r3 − r2 = 4R ⎛ cos2 B − sin2 B ⎞ = 4R cos B
⎜⎝ 2 2 ⎠⎟

²²²

7

Jpacademy TRANSFORMATIONS

PREVIOUS EAMCET BITS

1. cos x =λ ⇒ tan (x − y) tan y = [EAMCET 2009]

cos (x − 2y)

1) 1+ λ 2) 1− λ 3) λ 4) λ
1− λ 1+ λ 1+ λ 1− λ

Ans: 2

Sol. cos ( x − 2y) = 1 , using component to and dividendo:

cos x λ

tan (x − y) tan y = 1− λ

1+ λ

2. cos A cos 2A cos 4A......cos 2n−1A = [EAMCET 2009]

sin 2n A 2n sin 2n A 2n sin A sin A
1) 2n sin A 2) 3) sin 2n A 4) 2n sin 2n A

sin A

Ans: 1

Sol. 1 [2sin A cos A cos 2A cos 4A....]

2sin A

( )sin 2n A
( )cos ⎣⎡ 2n−1A ⎤⎦ = 2n sin A

3. If α + β+ γ = 2θ, then cos θ + cos (θ − α) + cos (θ − β) + cos (θ − γ ) = [EAMCET 2008]

α βγ αβγ
1) 4sin cos sin 2) 4 cos cos cos

2 22 2 22

3) 4sin α sin β sin γ 4) 4sinαsinβsinγ
2 22

Ans: 2

Sol. cos θ + cos (θ − α) + cos (θ − β) + cos (θ − γ )

= 2 cos ⎛ 2θ − α ⎞ cos ⎛ α ⎞ + 2 cos ⎛ 2θ −β − γ ⎞ cos β − γ
⎝⎜ 2 ⎟⎠ ⎝⎜ 2 ⎠⎟ ⎝⎜ 2 ⎟⎠ 2

= 2 cos ⎛ β + γ ⎞ cos ⎛ α ⎞ + 2 cos α β− γ
⎝⎜ 2 ⎟⎠ ⎝⎜ 2 ⎠⎟ 2 cos

2

= 2 cos α ⎣⎢⎡cos β + γ + cos β − γ ⎤ = 4 cos α cos β cos γ
2 2 2 ⎥⎦ 2 2 2

4. sin A + sin B = 3 (cos B − cos A) ⇒ sin 3A + sin 3B = [EAMCET 2007]

1) 0 2) 2 3) 1 4) – 1
Ans: 1

Sol. Given sin A + sin B = 3 (cos B − cos A)

1

Jpacademy Transformation

⎛ 3 ⎞ cos A + sin A ⎛ 1 ⎞ = cos B ⎛ 3 ⎞ − sin B ⎝⎜⎛ 1 ⎞
⎝⎜⎜ 2 ⎟⎟⎠ ⎜⎝ 2 ⎠⎟ ⎜⎝⎜ 2 ⎠⎟⎟ 2 ⎠⎟

⇒ cos ⎛ A − π ⎞ = cos ⎛ B + π ⎞
⎝⎜ 6 ⎟⎠ ⎜⎝ 6 ⎟⎠

A = −B ⇒ sin 3A + sin 3B = 0

5. A + C = 2B ⇒ cos C − cos A [EAMCET 2005]
sin A − sin C 4) tan B

1) cotB 2) cot2B 3) tan 2B [EAMCET 2005]
4) 3
Ans: 4
[EAMCET 2004]
Sol. A + C = B 4) 2
2
[EAMCET 2003]
cos C − cos A = 2sin A + C sin A − C = tan B 4) 4cosαcosβcosγ
sin A − sin C 22
[EAMCET 2003]
2 cos A + C sin A − C 4) 2 sin 7°
22

6. A + B = C ⇒ cos2 A + cos2 B + cos2 C − 2 cos A cos B cos C =

1) 1 2) 2 3) 0

Ans: 1

Sol. A + B = C

cos2 A + cos2 B + cos2 C = 1+ cos (A + B) cos (A − B) + cos2 C

= 1+ cos C ⎡⎣cos (A − B) + cos (A + B)⎤⎦
= 1+ cos (2 cos A cos B)

cos2 A + cos2 B + cos2 C − 2 cos A cos B cos C = 1

7. In ΔABC , cos ⎛ Β+ 2C + 3A ⎞ + cos ⎛ A− B⎞ =
⎝⎜ 2 ⎠⎟ ⎜⎝ 2 ⎠⎟

1) – 1 2) 0 3) 1

Ans: 2

Sol. cos ⎛ π + C+ 2A ⎞ + cos ⎛ A − B ⎞
⎝⎜ 2 ⎟⎠ ⎜⎝ 2 ⎠⎟

= cos ⎛ 2π −B + A ⎞ + cos ⎛ A − B ⎞ = 0
⎝⎜ 2 ⎟⎠ ⎝⎜ 2 ⎠⎟

8. cos α,sin (β − γ) + cosβ.sin (γ − α) + cos γ.sin (α − β) = ………..

1) 0 2) 1/2 3) 1
Ans: 1

Sol. Σ cos α sin (β − γ)

= Σ cos α (sin β cos γ − cosβsin γ ) = 0

9. sin 47° − sin 25° + sin 61° − sin11° =

1) cos 7° 2) sin 7° 3) 2 cos 7°

Ans: 1

Sol. sin 47° − sin25° + sin 61° − sin11°

2

Jpacademy Transformation
= sin 47° + sin 61° − (sin 25° + sin11°) [EAMCET 2003]
[EAMCET 2002]
= 2sin 54° cos 7° − 2sin18° cos 7°
[EAMCET 2002]
⎛ 5 +1− 5 −1 ⎞ = cos 7° [EAMCET 2001]
= 2 cos 7°⎜⎝⎜ 4 4 ⎠⎟⎟

10. If A + B + C = 270°, then cos2A + cos2B + cos2C + 4sinAsinBsinC = ….

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

Ans: 2

Sol. cos 2A + cos 2B + cos 2C = 1− 4 sin A sin Bsin C

(or) Put A = B = C= 90°

11. cos2 76° + cos2 16° − cos 76° cos16° =

1) 1 2) 0 3) − 1 4) 3
2 4 4

Ans: 4

Sol. cos2 76° +1− sin2 16° − 1 (2 cos 76° cos16°)

2

= 1+ cos (76° +16°) cos (76° −16) − 1 (cos 92° + cos16°)

2

= 1+ 1 cos 92° − 1 cos 92° − 1 = 3
2 2 44

=3
4

∑12. 3 cos2 ⎛ ( 2k − 1) π ⎞ =
k =1 ⎜⎝ 12 ⎟⎠

1) 0 2) 1 3) – 1 4) 3
2 2 2

Ans: 4

Sol. cos2 π + cos2 3π + cos2 5π
12 12 12

1+ cos 30° + 1+ cos 90° + 1+ cos150° = 3
22 22

13. If cos ecθ = p + q , then cot ⎛ π + θ ⎞
p − q ⎝⎜ 4 2 ⎠⎟

1) p 2) q 3) pq 4) pq
q p

Ans: 2
Sol. 1 = p + q ⇒ 1+ sin θ = p

sin θ p − q 1− sin θ q

1− cos ⎛ π + θ ⎞ 2 sin 2 ⎛ π + θ⎞
⎝⎜ 2 ⎟⎠ ⎜⎝ 4 2 ⎟⎠
⇒ = p ⇒ = p

1+ cos ⎛ π + θ ⎞ q 2 cos2 ⎛ π + θ ⎞ q
⎜⎝ 2 ⎠⎟ ⎝⎜ 4 2 ⎟⎠

3

Jpacademy Transformation

∴ cot ⎛ π + θ ⎞ = p
⎜⎝ 4 2 ⎠⎟ p

14. If sin α + sin β = a, cos α + cosβ = b, then sin (α + β) = [EAMCET 2000]

1) ab 2) a + b 3) 2ab 4) 2ab
a2 − b2 a2 + b2

Ans: 4
Sol. sin α + sin β = a;cos α + cosβ = b

2 sin ⎛ α + β ⎞ cos ⎛ α − β ⎞ = a; 2 cos ⎛ α + β ⎞ cos ⎛ α − β ⎞ = b
⎝⎜ 2 ⎠⎟ ⎜⎝ 2 ⎟⎠ ⎝⎜ 2 ⎠⎟ ⎝⎜ 2 ⎟⎠

∴ tan ⎛ α + β ⎞ = a
⎜⎝ 2 ⎠⎟ b

2 tan ⎛ α + β ⎞
⎜⎝ 2 ⎟⎠
sin (α + β) = = 2ab
⎛ α + β ⎞ a2 + b2
1 + tan 2 ⎝⎜ 2 ⎠⎟

15. If tan θ1 = k cot θ2 , then cos (θ1 − θ2 ) = [EAMCET 2000]
cos (θ1 − θ2 ) 4) k −1

1) 1+ k 2) 1− k 3) k +1 k +1
1− k 1+ k k −1

Ans: 1

Sol. cos (θ1 − θ2 ) = cos θ1 cos θ2 + sin θ1 sin θ2
cos (θ1 + θ2 ) cos θ1 cos θ2 − sin θ1 sin θ2

⇒ 1+ tan θ1 tan θ2 = 1+ k
1− tan θ1 tan θ2 1− k

555

4

Jpacademy TRIGONOMETRIC EQUATIONS

PREVIOUS EAMCET BITS

1. If 3cos x ≠ 2,sin x , then the general solution of sin2 x − cos 2x = 2 − sin 2x is x =

[EAMCET 2009]

1) nπ + (−1)n π , n ∈ Z 2) nπ , n ∈ Z
2
2
4) (2n −1) π, n ∈ Z
3) (4n ±1) π , n ∈ Z

2
Ans: 3

( )Sol. sin2 x − 1− 2sin2 x = 2 − 2sin x cos x

⇒ 3sin2 x + 2sin x cos x − 3 = 0

( )⇒ 3sin2 x + 2sin x cos x − 3 sin2 x + cos2 x = 0

⇒ cos x (2sin x − 3cos x) = 0 [EAMCET 2008]
⇒ cos x = 0(∵ 2sin x ≠ 3cos x)
⇒ x = (4n ±1) π / 2, n ∈ Z

{ }2. x ∈ R : cos 2x + 2 cos2 x − 2 = 0 =

1) ⎧⎨2nπ + π ; n ∈ Z⎫⎬ 2) ⎧⎨nπ ± π ; n ∈ Z⎬⎫ 3) ⎨⎧nπ ± π ; n ∈ Z⎫⎬ 4) ⎨⎧2nπ − π ; n ∈ Z⎬⎫
⎩ 3 ⎭ ⎩ 6 ⎭ ⎩ 3 ⎭ ⎩ 3 ⎭

Ans: 2

Sol. cos 2x + 2 cos2 x − 2 = 0 ⇒ 2 cos2 x −1+ 2 cos2 x − 2 = 0

⇒ 4 cos2 x = 3 ⇒ cos2 x = 3 = cos2 ⎛ π ⎞ ⇒ x = nπ ± π
4 ⎝⎜ 6 ⎟⎠ 6

( )3. +1 ⎛ − 1 ⎞ ≠ 1 ⇒ x∈
cos 2x = 2 ⎝⎜ cos x 2 ⎠⎟ , cos x 2 [EAMCET 2005]

1) ⎨⎧2nπ ± π : n ∈ Z⎫⎬ 2) ⎧⎨2nπ ± π : n ∈ Z⎬⎫
⎩ 3 ⎭ ⎩ 6 ⎭

3) ⎧⎨2nπ ± π : n ∈ Z⎫⎬ 4) ⎨⎧⎩2nπ ± π : n ∈ Z⎬⎫⎭
⎩ 2 ⎭ 4

Ans: 4

( )Sol. 2 2 cos2 x − 2 + 2 cos x +1 = 0

2 2 cos2 x − 2 cos x − 2 cos x +1 = 0

( )(2cos x −1) 2 cos x −1 = 0

cos x ≠ 1 ; cos x = 1
2 2

1

Jpacademy Trigonometric Equations

∴ x ∈ ⎧⎨2nπ ± π / n ∈ Z⎫⎬
⎩ 4 ⎭

4. The solution set of (5 + 4 cos θ)(2 cos θ +1) = 0 in the interval [0, 2π] is [EAMCET 2003]

1) ⎧ π , 2π ⎫ 2) ⎧ 2π , 5π ⎫ 3) ⎧ 2π , 4π ⎫ 4) ⎧ 2π , 5π ⎫
⎨ 3 3 ⎬ ⎨ 3 3 ⎬ ⎨ 3 ⎬ ⎨ 3 3 ⎬
⎩ ⎭ ⎩ ⎭ ⎩ 3 ⎭ ⎩ ⎭

Ans: 3

Sol. (5 + 4 cos θ)(2 cos θ +1) = 0

cos θ = −1 ⇒ θ = 2π , 4π [EAMCET 2001]
2 33
2) two solutions
5. The equation 3 sin x + cos x = 4 has 4) no solutions
1) only one solution
3) infinitely many solutions
Ans: 4

Sol. The max. value of 3 sinx + cosx is ‘2’.

∴ 3 sinx + cosx never equal to ‘4’.

6. If tan θ + sec θ = 3 , then the principal value of θ + π is [EAMCET 2000]
6 4) 3π/4

1) π/4 2) π/3 3) 2π/3

Ans: 2

Sol. tan θ + sec θ = 3

3 cos θ − sin θ = 1

cos ⎛ θ + π ⎞ = 1 ∴ ⎛ θ + π ⎞ = π
⎝⎜ 6 ⎠⎟ 2 ⎜⎝ 6 ⎠⎟ 3

› ››

2

Jpacademy TRIGONOMETRIC FUNCTIONS

PREVIOUS EAMCET BITS

1. If θ lies in the first quadrant and 5 tan θ = 4, then 5sin θ − 3cos θ = [EAMCET 2007]
sin θ + 2 cos θ

1) 5 2) 3 3) 1 4) 0
14 14 14

Ans: 1

Sol. tan θ = 4 ; 5 tan θ − 3 = 5 [EAMCET 2006]
5 tan θ + 2 14

2. sin120° cos150° − cos 240°sin 330° =

1) 1 2) –1 3) 2 ⎛ 3 +1⎞
3 4) − ⎝⎜⎜ 4 ⎟⎟⎠

Ans: 2

Sol. sin120°cos150° − cos 240°sin 330° = 3 ⎛ − 3 ⎞ − ⎛ − 1 ⎞ ⎛ − 1 ⎞
2 ⎝⎜⎜ 2 ⎠⎟⎟ ⎝⎜ 2 ⎟⎠ ⎝⎜ 2 ⎠⎟

= − 3 − 1 = − 4 = −1
44 4

3. If 5cos x +12 cos y = 13 , then the maximum values of 5 sinx + 12 siny is [EAMCET 2006]

1) 12 2) 120 3) 20 4) 13

Ans: 2
Sol. 5cos x +12 cos y = 13

5sin x +12sin y = ksay

squaring and subtracting 169 +120 ⎣⎡cos (x − y)⎦⎤ = 169 − k2

cos (x − y) = −k2

120
−1 ≤ −k2 ≤ 1 ⇒ k2 ≤ 120

120

⇒ k < 120

⇒ 5sin x +12sin y ≤ 120

4. cos θ − 4sin θ = 1 ⇒ sin θ + 4 cos θ = [EAMCET 2005]
4) ±4
1) ±1 2) 0 3) ±2

Ans: 4

Sol. cos θ − 4sin θ = 1

sin θ + 4 cos θ = a2 + b2 − c2

17 −1 = ±4
5. If A, B, C, D are the angles of a cyclic quadrilateral, then cos A + cos B + cos C + cos D =

[EAMCET 2001]
1) 4 2) 1 3) 0 4) –1
Ans: 3

1

Jpacademy Trigonometric Functions

Sol. A + C = 180°; B + D = 180°

cos A + cos (180° − A) + cos B + cos (180° − B) = 0

6. ⎛ 3 + 2 cos A ⎞−3 + ⎛ 1+ 2sin A ⎞−3 = [EAMCET 2000]
⎜⎜⎝ 1− 2sin A ⎟⎟⎠ ⎜⎝ 3 − 2 cos A ⎟⎠

1) 1 2) 3 3) 0 4) – 1

Ans: 3

Sol. Put A = 90°

⎛ 3+0 ⎞−3 + ⎛1+ 2 ⎞−3 = 0
⎜⎝⎜ 1− 2 ⎠⎟⎟ ⎜⎝ 3 ⎠⎟

( )7. If cos A = n and sin A = m , then m2 − n2 sin2 B = [EAMCET 2000]
cos B sin B 4) 1 + n

1) 1 – n2 2) 1 + n2 3) 1 – n

Ans: 1
Sol. cos A = nCosB;sin A = m sin B

cos2 A + sin2 A = n2 cos2 B + m2 sin2 B

( )1 = n2 1− sin2 B + m2 sin2 B

( )⇒ m2 − n2 sin2 B = 1− n2

8. If (sin α + cos ecα)2 + (cos α + sec α)2 = k + tan2 α + cot2 α, then k = [EAMCET 2000]
[EAMCET 2000]
1) 9 2) 7 3) 5 4) 3

Ans: 2

Sol. Put α = 45°

= ⎛ 1+ 2 ⎞2 + ⎛ 1+ 2 ⎞2 = k +1+1
⎜⎝ 2 ⎠⎟ ⎜⎝ 2 ⎟⎠

∴k=7

sin (−660°) tan (1050°)sec (−420°)
9. cos (225°) cos ec(315°) cos (510°) =

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

Ans: 3

− sin (720 − 60) tan (1080 − 30)sec(360 + 60)
Sol. cos (180 + 45) cos ec(360 − 45) cos (360 +150)

Substituting it values and simplifying we get 2 / 3

™™™

2

Jpacademy CHANGE OF AXES

PREVIOUS EAMCET BITS

1. The transformed equation of x2 + y2 = r2 when the axes are rotated through an angle 36° is
[EAMCET 2009]

1) 5X2 − 4XY + Y2 = r2 2) X2 + 2XY − 5Y2 = r2

3) X2 − Y2 = r2 4) X2 + Y2 = r2

Ans: 4

Sol. Equation of circle will not change

2. The transformed equation of 3x2 + 3y2 + 2xy = 2 when the coordinate axes are rotated through an

angle of 45° is [EAMCET 2008]

1) X2 + 2Y2 = 1 2) 2X2 + Y2 = 1 3) X2 + Y2 = 1 4) X2 + 3Y2 = 1

Ans: 2
Sol. Let (X, Y) be the new coordinates of (x, y), when the axes are rotated through an angle 45°. Then

y = Xsin45° + Ycos45° = (X + Y) / 2 and x = X cos 45° − Y sin 45° = (X − Y) / 2

The transformed equation is 3⎝⎜⎛ X −Y ⎞2 + 3⎜⎝⎛ X +Y ⎞2 + 2 ⎛ X −Y ⎞⎛ X +Y ⎞ = 2
2 ⎠⎟ 2 ⎠⎟ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠

( ) ( )⇒ 3 X2 + Y2 + X2 − Y2 = 2 ⇒ 4X2 + 2Y2 = 2 ⇒ 2X2 + Y2 = 1

3. In order to eliminate the first degree terms from the equation [EAMCET 2007]

2x2 + 4xy + 5y2 − 4x − 22y + 7 = 0 , the point to which origin is to be shifted is

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

Ans: 3
Sol. S = 2x2 + 4xy + 5y2 − 4x − 22y + 7 = 0

∂S = 4x + 4y − 4 = 0;
∂x
∂S = 4x +10y − 22 = 0
∂y

(x, y) = (–2, 3)
4. The transformed equation x2+ 6xy + 8y2 = 10 when the axes are rotated through an angle

π/4 is [EAMCET 2006]

1) 15x2 −14xy + 3y2 = 20 2)15x2 +14xy − 3y2 = 20

3) 15x2 +14xy + 3y2 = 20 4) 15x2 −14xy − 3y2 = 20

Ans: 3
Sol. θ = π

4
x = X cos θ − Y sin θ = X − Y

2

y = X sin θ + Y cos θ = X + Y
2

transformed equations is

1

Jpacademy Change of Axes

(X − Y)2 (X − Y)(X + Y) (X + Y)2

+ 6 + 8 = 10
222

⇒ 15x2 +14xy + 3y2 = 20

5. The coordinate axes are rotated through an angle 135°. If the coordinates of a point P in the new
system are known to be (4, –3), then the coordinates of P in the original system are
[EAMCET 2003]

1) ⎛ 1, 7⎞ 2) ⎛ 1 , −7 ⎞ 3) ⎛ −1 , −7 ⎞ 4) ⎛ −1 , 7 ⎞
⎜⎝ 2 2 ⎟⎠ ⎜⎝ 2 2 ⎟⎠ ⎝⎜ 2 2 ⎠⎟ ⎜⎝ 2 2 ⎠⎟

Ans: 4

Sol. x = 4 cos135° − (−3)sin135° = − 1

2

y = 4sin135° + (−3) cos135° = 7

2

6. If the axes are rotated through an angle 45° in the positive direction without changing the origin,

( )then the coordinates of the point 2, 4 in the old system are [EAMCET 2002]

( )1) 1− 2 2,1+ 2 2 ( )2) 1+ 2 2,1− 2 2

3) (2 2, 2 ) 4) ( 2, 2 )

Ans: 1

( )Sol. x = X cos θ − y sin θ given (X, Y) = 2, 4

y = X sin θ + Y cos θ and θ = π
4

7. The coordinate axes are rotated about the origin 0 in the counter-clockwise direction through an

angle 60°. If p and q are the intercepts made on the new axes by a straight line whose equation

referred to the original axes is x + y = 1, then 1 +1 = [EAMCET 2000]
p2 q2

1) 1 2) 4 3) 6 4) 8
Ans: 1

Sol. The perpendicular distance from origin to x + y = 1 and x + y = 1 are equal
pq

∴ 1 + 1 =1
p2 q2

›››

2

Jpacademy COORDINATE SYSTEM

PREVIOUS EAMCET BITS

1. If, l, m, n are in arithmetic progression, then the straight line lx + my + n = 0 will pass through

the point [EAMCET 2008]

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

Ans: 2

Sol. l, m, n are in A.P ⇒ m – l = n – m ⇒ l – 2m + n = 0 ⇒ (1, –2) lies on lx + my + n = 0

2. In the triangle with vertices at A (6,3), B(−6,3) and C (−6, −3) , the median through A meets BC

at P, the line AC meets the x-axis at Q, while R and S respectively denote the orthocentre and

centroid of the triangle. Then the correct matching of the coordinates of points in List – I to

List – II is [EAMCET 2007]

List – I List – II

i) P A) (0, 0)

ii) Q B) (6, 0)

iii) R C) (–2, 1)

iv) S D) (–6, 0)

E) (–6, –3)

F) (–6, 3)

i ii iii iv i ii iii iv

1) D A E C 2) D B EC

3) D A F C 4) B A F C

Ans: 3

Sol. i) P is midpoint of BC = (–6, 0) = D

ii) Midpoint of AC is (0, 0) ⇒ AC meets x-axis at Q(0, 0) = A

iii) ΔABC is right angled at B. Orthocentre = R = (–6, 3) = F

iv) Centroid = S = (–2, 1) = C

3. The area (in square units) of the triangle formed by the lines x = 0, y = 0 and 3x + 4y = 12 is

[EAMCET 2005]

1) 3 2) 4 3) 6 4) 12

Ans: 3 B(0,3)
Sol. Area of Δle OAB = 1 base × height 3 x + y =1

2 43
Area = 1 × 4× 3 = 6
O 4 A (4,0)
2 (0,0)

1

Jpacademy Coordinate system

4. If PM is the perpendicular from P(2, 3) onto the line x + y = 3, then the coordinates of M are

[EAMCET 2005]

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

Ans: 3

Sol. P(2, 3), A = x + y = 3 , slope = - 1 by verification product of slopes = –1

from (3) option slope PM = 3 − 2 = 1
2 −1

1(–1) = – 1
5. The point P is equidistant from A(1, 3), B(–3, 5) and C(5, –1). Then PA - . [EAMCET 2003]

1) 5 2) 5 5 3) 25 4) 5 10

Ans: 4
Sol. PA2 = PB2 = PC2

( x −1)2 + ( y − 3)2 = (x + 3)2 + ( y − 5)2

= ( x − 5)2 + ( y +1)2
⇒ P (x, y) = (−8, −10)

∴ PA = 5 10

6. If (–2, 6) is the image of the point (4, 2) with respect to the line L = 0, then L = [EAMCET 2002]

1) 6x − 4y − 7 2) 2x + 3y − 5 3) 3x − 2y + 5 4) 3x − 2y +10

Ans: 3

Sol. L = 0 is perpendicular bisector of line segment joining the roots (–2, 6) (4, 2) L = 3x –2y+5

7. If the altitude of a triangle are in arithmetic progression, then the sides of the triangle are in ..

progression [EAMCET 2002]

1) arithmetic 2) harmonic 3) geometric 4) arithmetico-geometric

Ans: 2

Sol. Δ = 1 P1a ⇒ P1 = 2Δ
2 a

P2 = 2Δ P3 = 2Δ P1, P2 , P3 are in A.P.
b c

⇒ a, b, c are in H.P

8. The lines 2x + 3y = 6, 2x + 3y = 8 cut the x-axis at A, B respectively. A line l drawn through the

point (2, 2) meets the x-axis at C. In such a way that abscissae of A, B and C are in arithmetic

progression. Then the equation of the line l is [EAMCET 2001]

1) 2x + 3y = 20 2) 3x + 2y = 10 3) 2x − 3y = 10 4) 3x − 2y = 10

Ans: 1
Sol. The lines 2x + 3y = 6 and 2x + 3y = 8 cuts x-axis at A and B

∴ Α(3, 0), Β(4, 0)
The point ‘C’ lies on x-axis and the abscissae of the points A, B, C are in A.P.

2

Jpacademy Coordinate system

∴C(5, 0)

∴ The equation of the line passing through (2, 2) and (5, 0) is 2x + 3y = 10

9. The incentre of the triangle formed by the lines x + y = 1 , x = 1, y = 1 is [EAMCET 2001]

1) ⎛⎜⎝1− 1 ,1− 1 ⎞ 2) ⎝⎜⎛1 − 1, 1⎞ 3) ⎛ 1, 1⎞ 4) ⎛ 1 ,1− 1⎞
2 2 ⎠⎟ 2 2 ⎟⎠ ⎝⎜ 2 2 ⎠⎟ ⎜⎝ 2 2 ⎠⎟

Ans: 3

Sol. The vertices of the triangle are (1, 0) (0, 1), (1, 1) and lengths of the sides are 1, 1, 2

∴ Incentre ⎛ 1, 1⎞
⎜⎝ 2 2 ⎟⎠

10. The vertices of a triangle are (6, 6), (0, 6) and (6, 0). The distance between the circumentre and

centroid is [EAMCET 2000]

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

Ans: 3

Sol. Circumcentre = S(3, 3)
JJJG

Centroid = G(4,4 ) ∴ SG = 2

”””

3

Jpacademy
DIRECTION RATIOS AND DIRECTION COSINES

PREVIOUS EAMCET BITS

1. The angle between the lines whose direction cosines satisfy the equation l + m + n = 0,

l2 + m2 – n2 = 0 is [EAMCET 2009]

ππ π 4) π
1) 2) 3) 2

64 3

Ans: 3

Sol. n = − (1+ m)

∴ 2 + m2 − n2 = 0 ⇒ = 0(or) m = 0

= 0 ⇒ n = −m ⇒ = m = n
0 −1 1

m = 0 ⇒ n = −1⇒ = m = n
−1 0 1

∴cos θ = 0 + 0 +1 = 1 ⇒ θ = 60°
22 2

60. If a line in the space makes angles α, β and γ with the coordinates axes, then

cos 2α + cos2 β + cos 2γ + sin2 α + sin2 β + sin2 γ = [EAMCET 2009]

1) –1 2) 0 3) 1 4) 2

Ans: 3

Sol. sin2 α + sin2 β + sin2 γ = 2

∴1− 2 sin2 α +1− 2sin2 β +1− 2 sin2 γ + sin2 α + sin2 β + sin2 γ

= 3−2 =1

52. The angle between the lines whose direction consines are ⎛ 3,1, 3⎞ and ⎛ 3 , 1 , − 3 ⎞ is
⎝⎜⎜ 44 2 ⎠⎟⎟ ⎜⎝⎜ 4 4 ⎟⎠⎟
2

[EAMCET 2008]

1) π 2) π 3) π 4) π
234

Ans: 3

Sol. ⎛ 3 ⎞⎛ 3 ⎞ + ⎛ 1 ⎞ ⎛ 1 ⎞ + ⎛ 3 ⎞ ⎛ − 3 ⎞ = 3 + 1 − 3 = − 1 ⇒ θ = π or 2π
cos θ = ⎝⎜⎜ 4 ⎟⎠⎟⎜⎝⎜ 4 ⎠⎟⎟ ⎝⎜ 4 ⎠⎟ ⎝⎜ 4 ⎟⎠ ⎜⎝⎜ 2 ⎟⎠⎟ ⎜⎝⎜ 2 ⎟⎟⎠ 16 16 4 2 3 3

52. The cosine of the angle A of the triangle with vertices A(1, –1, 2), B(6, 11, 2), C(1, 2, 6) is

[EAMCET 2007]

1) 63 2) 36 3) 16 4) 13
65 65 65 64

Ans: 2

Sol. AB = 5i +12J; AC = 3J + 4k

1

Jpacademy Direction Ratios and Direction Cosines

cos A = AB.AC = 36
AB AC 65

51. If the direction cosines of two lines are such that + m + n = 0, 2 + m2 − n2 = 0 , then the angle

between them is π [EAMCET 2006]
ππ 3) π
4)
1) 2) 4 6
23

Ans: 2

Sol. + m + n = 0 ⇒ = − (m + n)

Substituting

m = 0 or m = – n

1 : m1 : n1 = −1; 0 :1 and 2 : m2 : n2 = 0 : −1:1
cos θ = 1 ⇒ θ = π

23

51. The direction cosines of the line passing through P(2, 3, –1) and the origin are [EAMCET 2005]

1) 2 , 3 , 1 2) 2 , −3 , 1 3) −2 , −3 , 1 4) 2 , −3 , −1
14 14 14 14 14 14 14 14 14 14 14 14

Ans: 3

Sol. Direction cosines are ± x1 ,
x12 + y12 + z12

± y1 , ± z1
x12 + y12 + z22 x12 + y12 + z22

52. If the direction ratio of two lines are given by + m + n = 0 , mn − 2 n + m =0, then the angle

between the lines is [EAMCET 2004]

ππ π
1) 2) 3) 4) 0
43 2

Ans: 3

Sol. f + g + h = 0 ⇒ θ = 90°
abc

22. If the direction ratios of two lines are given by 3 m − 4 n + mn = 0 and + 2m + 3n = 0 , then the

angle between the lines is [EAMCET 2003]

ππ π π
1) 2) 3) 4)

23 4 6

Ans: 1

Sol. 1 = –(2m + 3n)

−3(2m + 3n) m + 4(2m + 3n) n + mn = 0

⇒m=± 2
n

1 = −2 2 − 3 ⇒ 1 = m1 = n1
n1 −3 − 2 2 2 1

2 = m2 = n2
−3 + 2 2 − 2 1

2

Jpacademy Direction Ratios and Direction Cosines

1 2 + m1m2 + n1n2 = 0 ⇒ θ = 90°
21. The acute angle between the two lines whose direction ratios are given by + m − n = 0 and

2 + m2 − n2 = 0 is [EAMCET 2002]

1) 2) 3) 4)

Ans: 4

Sol. n = l + m substituting in

2 + m2 = n2 ⇒ m = 0

If = 0 ⇒ m = n ⇒ = m = n
011

m=0⇒ =n⇒ = m = n
101

cos θ = a1a2 + b1b2 + c1c2 = 0+0+1 = 1

a12 + b12 + c12 a 2 + b 2 + c22 22 2
2 2

θ= π
3

22. The direction ratios of a normal to the plane passing through (0, 0, 1) (0, 1, 2) and (1, 2, 3) are

[EAMCET 2002]

1) [0,1, −1] 2) [1, 0, −1] 3) [0, 0, −1] 4) [1, 0, 0]

Ans: 1

Sol. A (0, 0,1); B(0,1, 2);C(1, 2,3)

AB = (0,1,1)

AC = (1, 2, 2)

i jk

AB× AC = 0 1 1 = (0,1, −1)

122

21. The direction ratios of two lines are given by a + b + c = 0. 2ab + 2ac – bc = 0. Then the angle

between the lines is [EAMCET 2001]

1) π 2) 2π 3) π 4) π
3 2 3

Ans: 2

Sol. Given a + b + c = 0 and 2ab + 2ac – bc = 0

2a(b + c) = bc ⇒ −2(b + c)2 = bc (∵a = − (b + c))

⇒ 2b2 + 5bc + 2c2 = 0

⇒ (2b + c)(b + 2c) = 0

∴ C = - 2b (or) – b/2
If C = 2b ⇒ a = b
⇒ a : b : c = 1: 1: –2
If C = - b/2 ⇒ a = - b /2
⇒ a : b: c = 1 : –2 : 1

3

Jpacademy Direction Ratios and Direction Cosines

If θ is the angle between the lines then cos θ = (1)(1) + (1)(−2) + (−2)(1) = − 1
1+1+ 4 1+ 4+1 2

∴ θ = 2π
3

22. If a line makes angles π and π with the x–axis and y –axis respectively, then the angle made by
34

the line with z-axis is

[EAMCET 2001]

ππ π 4) 5π
1) 2) 3)
2 3 4 12

Ans: 2

Sol. cos2 α + cos2 β + cos2 γ = 1

⇒ 1 + 1 + cos2 γ = 1 ⇒ cos2 γ = 1 ∴ γ = π
42 43

24. If P = (0, 1, 2). Q = (4, –2, 1), O = (0, 0, 0), then POQ = [EAMCET 2001]

1) π 2) π 3) π 4) π
64 3 2

Ans: 4 [EAMCET 2000]
4) 2
Sol. x1x2 + y1y2 + z1z2 = 0 − 2 + 2 = 0

∴ θ = 90°

24. If 2 + m2 = 1, then the maximum value of + m is

1) 1 2) 2 3) 1
2

Ans: 2
Sol. Let = cos θ and m = sin θ

2 + m2 =1

The max.value of cos θ + sin θ is 2

DDD

4

Jpacademy LOCUS

PREVIOUS EAMET BITS

1. If the sum of the distance of a point P from two perpendicular lines in a planes is 1, then the locus

of P is a [EAMCET 2008]

1) rhombus 2) circle 3) straight line 4) pair of straight lines

Ans: 1

Sol. Let P(x1, y1) be a point such that the sum of the distances of P from two perpendicular lines

x + y = 0, x − y = 0 is 1. Then x1 + y1 + x1 − y1 = 1
22

⇒ ± ( x1 + y1 ) ± ( x1 − y1 ) = 2 ⇒ ( x1 + y1 )2 + ( x1 − y1 )2 ± 2( x1 + y1 )( x1 − y1 ) = 2

( ) ( ) ( )⇒ 2 x12 + y12 ± 2 x12 + y12 ± x12 − y12 = 1 ⇒ 2x12 = 1 or 2y12 = 1

∴ The locus of P is (2x2 – 1) (2y2 – 1) = 0 which represents a rhombus.

2. If a point P moves such that its distances from the point A(1, 1) and the line x + y + 2 = 0 are

equal then the locus of P is [EAMCET 2005]

1) a straight line 2) a pair of straight lines 3) a parabola 4) an ellipse
Ans: 3
Sol. PA2 = PM2

( x −1)2 + ( y −1)2 = (x + y + 2)2

2

x2 + y2 − 8x − 8y − xy = 0

3. If a point (x, y) = (tanθ + sinθ, tanθ - sinθ), then the locus of (x, y) is [EAMCET 2002]

( ) ( )1) x2y 2/3 + xy2 2/3 = 1 2) x2 − y2 = 4xy

3) x2 − y2 = 12xy ( )4) x2 − y2 2 = 16xy

Ans: 4 Eliminating ‘θ’
Sol. x = tan θ + sin θ
( )x2 − y2 2 = 16xy
y = tan θ − sin θ

4. A straight rod of length 9 units slides with its ends A, B always on the x and y axes respectively.

Then the locus of the centroid of ΔOAB is [EAMCET 2000]
4) x2 + y2 = 81
1) x2 + y2 = 3 2) x2 + y2 = 9 3) x2 + y2 = 1

Ans: 2

Sol. Let A (a, 0) B(0, b) and G (x1, y1 )

⎛ a , b ⎞ = ( x1y1 ) ⇒ a = 3x1; b = 3y1
⎜⎝ 3 3 ⎠⎟

a2 + b2 = 81 ⇒ x2 + y2 = 9

›››

1

Jpacademy PAIR OF STRAIGHT LINES

PREVIOUS EAMCET BITS

1. The value of λ with |λ| < 16 such that 2x2 – 10xy + 12y2 + 5x + λy – 3 = 0 represents a pair of

straight lines, is [EAMCET 2009]

1) – 10 2) – 9 3) 10 4)9

Ans: 2

Sol. Δ = 0 ⇒ λ = −9

2. The area (in square units) of the triangle formed by x + y + 1 = 0 and the pair of straight lines
x2 –3xy + 2y2 = 0 is
[EAMCET 2009]

1) 7/12 2) 5/12 3) 1/12 4) 1/6

Ans: 3

Sol. Area = n2 h2 − ab = 1
am2 − 2hAm + bA2 12

3. The pairs of straight lines x2 – 3xy + 2y2 = 0 and x2 – 3xy + 2y2 + x –2=0 form a

[EAMCET 2009]

1) square but not rhombus 2) rhombus

3) parallelogram 4) rectangle but not a square

Ans: 3

Sol. In given two equations homogenious parts are same, hence is a parallelogram.

4. The value of λ such that λx2 −10xy +12y2 + 5x −16y − 3 = 0 represents a pair of straight lines, is

1) 1 2) – 1 3) 2 [EAMCET 2008]
4) – 2

Ans: 3

Sol. λx2 −10xy +12y2 + 5x −16y − 3 = 0 ≡ ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

⇒ a = λ, b = 12, c = −3, f = −8, g = 5 / 2, h = −5

abc + 2fgh − af 2 − bg2 − ch2 = 0

⇒ λ (12)(−3) + 2(−8)(5 / 2)(−5) − λ (−8)2 −12(5 / 2)2 − (−3)(−5)2 = 0

⇒ −36λ + 200 − 64λ − 75 + 75 = 0 ⇒ 100λ = 200 ⇒ λ = 2
5. A pair of perpendicular straight lines passes through the origin and also through the point of

intersection of the curve x2 + y2 = 4 with x + y = a. The set containing the value of ‘a’ is

1) {−2, 2} 2) {−3,3} 3) {−4, 4} [EAMCET 2008]

4) {−5,5}

Ans: 1
Sol. Clearly the point of intersection of x2 + y2 = 4, x + y = a is the origin

1

Jpacademy Pair of straight lines

x2 + y2 = 4, x + y = a ⇒ x2 + (a − x)2 = 4,( x = 0) ⇒ a2 = 4 ⇒ a = ±2

6. The angle between the pair of straight lines formed by joining the points of intersection of

x2 + y2 = 4 and y = 3x + c to the origin is a right angle. Then c2 = [EAMCET 2007]

1) 20 2) 13 3) 1/5 4) 5

Ans: 1

( )Sol. 2c2 = a2 1+ m2

⇒ 2c2 = 40 ⇒ c2 = 20

7. If the lines x2 + 2xy − 35y2 − 4x + 44y – 12 = 0 and 5x + λy − 8 = 0 are cocurrent, then the value

of λ is [EAMCET 2007]

1) 0 2) 1 3) – 1 4) 2

Ans: 4

Sol. Point of intersection of given pair of lines is ⎛ 4 , 2 ⎞
⎝⎜ 3 3 ⎠⎟

It lies on 5x + λy − 8 = 0 ⇒ λ = 2

8. The lines represented by the equation x2 − y2 − x + 3y − 2 = 0 are [EAMCET 2006]

1) x + y −1 = 0, x − y + 2 = 0 2) x − y − 2 = 0, x + y +1 = 0

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

Ans: 4

Sol. By verification

(x − y +1)(x + y − 2) = 0

⇒ x2 + y2 − x + 3y − 2 = 0

9. The centroid of the triangle formed by the pair of straight lines 12x2 – 20xy + 7y2 = 0 and the line

2x – 3y + 4 = 0 is [EAMCET 2006]

1) ⎛ − 7 , 7 ⎞ 2) ⎛ − 8 , 8 ⎞ 3) ⎛ 8 , 8 ⎞ 4) ⎛ 4 , 4⎞
⎝⎜ 3 3 ⎟⎠ ⎜⎝ 3 3 ⎠⎟ ⎜⎝ 3 3 ⎟⎠ ⎜⎝ 3 3 ⎟⎠

Ans: 3 O(0,0)
Sol. Centroid divides median in the ratio of 2 : 1

∴ of ( x1, y1 ) is centroid then ⎛ 3x1 , 3y1 ⎞ must lie on 2 •
⎜⎝ 2 2 ⎠⎟ 1

2x – 3y + 4 = 0 by verification ⎛ 8 , 8 ⎞ is centroid
⎜⎝ 3 3 ⎟⎠

10. The area of the triangle formed by the pair of straight lines (ax + by)2 – 3(bx –ay)2 = 0 and

ax + by + c = 10 is [EAMCET 2005]

2

Jpacademy Pair of straight lines

1) c2 ( )2) c2 ( )3) c2 ( )4) c2
a2 + b2 2 a2 + b2 2 a2 + b2 3 a2 + b2

Ans: 4

Sol. Y

⊥lar distance from 0(0, 0) to ax + by + c = 0 is O
(0,0)
h= c X
a2 + b2

( )Area h2 =
3
c
3 a2 + b2

11. The product of the perpendicular distances from the origin on the pair of straight lines

12x2 + 25xy +12y2 +10x +11y + 2 = 0 is [EAMCET 2005]

1) 1 2) 2 3) 3 4) 4
25 25 25 25

Ans: 2

Sol. Product of perpendiculars = c2

(a − b)2 + (2h)2 = 25

( )12. The angle between the lines represented by y2 sin2 θ − xy sin2 θ + x2 cos2 θ −1 = 0 is

1) π 2) π 3) π 4) π [EAMCET 2004]
3 4 6 2 [EAMCET 2004]

Ans: 4

Sol. a + b = 0 ⇒ θ = 90°

13. Area of the triangle formed by the lines 3x2 − 4xy + y2 = 0, 2x − y = 6 is

1) 16 2) 25 3) 36 4) 49

Ans: 3

Sol. Area of the triangle = n2 h2 − ab = 36 sq.units
am2 − 2hAm + bA2

( )14. If the pair of straight lines given by Ax2 + 2Hxy + By2 = O H2 > AB forms an equilateral

triangle with line ax + by + c = 0, then (A + 3B) (3A + B) = ….. [EAMCET 2003]

1) H2 2) – H2 3) 2H2 4) 4H2

Ans: 4

3

Jpacademy
Pair of straight lines

Sol. Equation of one side ax + by + c =0 and other two sides passing through origin then the equation

of other two sides is (ax + by)2 − 3(bx − ay)2 = 0

A = a2 – 3b2; B = b2– 3a2 ; H = 8ab

∴ (A + 3B)(3A + B) = 4H2

15. The area (in square units) of the quadrilateral formed by the two pairs of lines

A2x2 − m2y2 − n (Ax + my) = 0 and A2x2 − m2y2 + n (Azx − my) = 0 [EAMCET 2003]

1) n2 2) n2 3) n2 4) n2
2 A.m A.m A.m 4 A.m

Ans: 1

Sol. A2 x2 − m2 y2 − n (Ax + my) = 0

⇒ (Ax + my)(Ax − my − n) = 0
A2 x2 − m2 y2 + n (Ax − my) = 0
⇒ (Ax − my)(Ax + my + n) = 0

The given lines from a rhombus

∴ Area of rhombus = c2 = n2
2 ab 2 Am

16. If the coordinate axes are the bisectors of the angle between the pair of lines ax2+2hxy+b2 = 0,
where h2 > ab and a ≠ b , then
[EAMCET 2002]

1) a + b = 0 2) h = 0 3) h ≠ 0, a + b = 0 4) a + b ≠ 0

Ans: 2

( )Sol. Equation of pair of angular bisectors is h x2 − y2 − (a − b) xy = 0

Equation of coordinate axes is xy = 0

∴h=0

17. If the angle 2θ is acute then the acute angle between the pair of straight lines x2(cosθ - sinθ) +
2xy cosθ + y2 (cosθ + sinθ) = 0
[EAMCET 2002]

1) 2θ 2) θ/2 3) θ/3 4) θ

Ans: 4

Sol. cos α = a+b

(a − b)2 + 4h2

(α is angle between the pair of lines and a = cosθ + sinθ, b = cosθ- sinθ 2h =2 cosθ)
cosα = cosθ
∴Angle ‘θ’

4

Jpacademy Pair of straight lines

18. If the pair of straight lines xy – x – y + 1 = 0 and the line ax + 2y – 3 = 0 are concurrent, then a =

[EAMCET 2002]

1) – 2 2) 3 3) 1 4) 0

Ans: 3

Sol. xy – x – y + 1 = 0

⇒ (x −1)( y −1) = 0 ∴ (x, y) = (1,1)

ax + 2y – 3 = 0 line passes through (1, 1)

∴a=1
19. The orthocentre of the triangle formed by the lines x + 3y = 10 and 6x2 + xy – y2 = 0 is

1) (1, 3) 2) (3, 1) 3) (–1, 3) [EAMCET 2001]
4) (1, –3)

Ans: 1
Sol. The given lines are x + 3y = 10 , 6x2 + xy – y2 = 0

A = 1, m = 3, n = 10, a = 6, h = 1 ; b = −1
2

∴ Orthocentre = ( (kA; km) where k = am2 n(a + b)

− 2hAm + bA2

20. If one of the lines of the pair of straight lines ax2 + 2hxy + by2 = 0 bisects the angle between the

co-ordinate axes then : [EAMCET 2001]

1) a2 + b2 = h2 2) (a + b)2 = h2 3) a2 + b2 = 4h2 4) (a + b)2 = 4h2

Ans: 4

Sol. The angle bisectors of the axes are x ± y = 0

ax2 ± 2hx2 + bx2 = 0

⇒ (a + b)2 = 4h2

21. If the slope of one line is twice the slope of the other in the pair of straight lines ax2 + 2hxy + by2
= 0, then 8h2 =
[EAMCET 2001]

1) – 9ab 2) 9ab 3) 7ab 4) –7ab

Ans: 2

Sol. If slope of one line is ‘k’ times to other then

4kh2 = (k +1)2 ab ⇒ 8h2 = 9ab

22. The equation of the pair of lines through the point (a, b) parallel to the coordinate axes is

[EAMCET 2000]

1) (x − b)( y − a ) = 0 2) (x − b)( y + b) = 0

3) (x − a)( y − b) = 0 4) (x + a)( y − b) = 0

Ans: 3
Sol. The equation of the coordinate axes is xy = 0

5

Jpacademy
Pair of straight lines
∴ The equation of the pair of lines passing through (a, b) and parallel to xy = 0 is (x – a)
(y – b) = 0

œœœ

6

Jpacademy PLANES

PREVIOUS EAMCET BITS

1. The image of the point (3, 2, 1) in the plane 2x – y + 3z = 7 is [EAMCET 2009]

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

Ans: 3

Sol. Since point lies on the plane, image is itself.

2. If (2, –1, 3) is the foot of the perpendicular drawn from the origin to the plane, then the equation

of the plane is [EAMCET 2004]

1) 2x + y − 3z + 6 = 0 2) 2x − y + 3z −14 = 0

3) 2x − y + 3z −13 = 0 4) 2x + y + 3z −10 = 0

Ans: 2

Sol. O (0, 0, 0), A (2, −1,3), P = (x, y, z)

OA.PA = 0 ⇒ 2x − y + 3z −14 = 0

3. If the plane 3x – 2y – z – 18 = 0 meets the coordinates axes in A, B, C then the centroid of ΔABC

is [EAMCET 2004]

1) (2,3, −6) 2) (2, −3, 6) 3) (−2, −3, 6) 4) (2, −3, −6)

Ans: 4

Sol. x + y + z = 1
6 −9 −18

∴Centroid = (2, −3, −6)

4. A plane π makes intercepts 3 and 4 respectively on Z-axis and X-axis. If π is parallel to

Y-axis, then its equation is [EAMCET 2003]

1) 3x + 4z = 12 2) 3z + 4x = 12 3) 3y + 4z = 12 4) 3z + 4y = 12

Ans: 1

Sol. The equation of the required plane is x + z = 1 ⇒ 3x + 4z = 12
43

5. The equation of the plane passing through (1, 1, 1) and (1, –1, –1) ; and perpendicular to

2x – y + z + 5 = 0 is [EAMCET 2003]

1) 2x + 5y + z − 8 = 0 2) x + y − z −1 = 0

3) 2x + 5y + z + 4 = 0 4) x − y + z −1 = 0

Ans: 2
Sol. The equation of the plane passing through (1, 1, 1) and (1, –1, –1) is x + y – z = 1 and it is ⊥ar to

2x – y + z + 5 = 0
6. If P=(0, 1, 0), Q = (0, 0, 1), then the projection of PQ on the plane x + y + z = 3 is

[EAMCET 2002]

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

Ans: 3

Sol. Perpendicular distances from P, Q to the plane are equal and the point P, Q lies on same side of

the plane. HJJG

∴ The distance between P and Q is the projection of PQ on the given plane = 2

7 In the space the equation by by + cz + d = 0 represents a plane perpendicular to the….. plane
[EAMCET 2002]

1

Jpacademy Planes

1) YOZ 2) ZOX 3) XOY 4) z = k
Ans: 1
Sol. by + cz + d = 0

Plane is parallel to x-axis

∴ It is perpendicular to y o z plane.

8. A plane π passes through the point (1, 1, 1). If b, c , a are the direction ratios of a normal to the

plane, where a, b, c (a < b < c) are the prime factors of 2001, then the equation of the plane π is

[EAMCET 2002]

1) 29x + 31y + 3z = 63 2) 23x + 29y − 29z = 23

3) 23x + 29y + 3z = 55 4) 31x + 37y + 3z = 71

Ans: 3
Sol. By verification 2001 = 23× 29× 3

∴23x + 29y + 3z = 55

9. If the foot of the perpendicular from (0, 0, 0) to a plane is (1, 2, 2) then the equation of the plane

is [EAMCET 2001]

1) – x + 2y + 8z – 9 = 0 2) x + 2y + 2z – 9 = 0

3) x + y + z – 5 = 0 4) x + 2y – 3z + 1= 0

Ans: 2

Sol. The plane passing through (1, 2, 2) with normal D.r.s 1, 2, 2

∴ 1(x – 1) + 2(y – 2) + 2(z – 2) = 0
⇒ x + 2y + 2z = 9

10. A variable plane is at a constant distance k from the origin and meets the co-ordinate axes in A,

B, C. Then the locus of the centroid of ΔABC is [EAMCET 2001]

1) x−2 + y−2 + z−2 = k−2 2) x−2 + y−2 + z−2 = 4k−2

3) x−2 + y−2 + z−2 = 16k−2 4) x−2 + y−2 + z−2 = 9k−2

Ans: 4

Sol. Let A (h, 0,0) B(0, k, 0), C(0, 0, P)

∴ centroid = ⎛ h , k , p ⎞ = G ( x1, y1, z1 )
⎝⎜ 3 3 3 ⎠⎟

The perpendicular distance from origin to x + y + z = 1 is 1 =k
hkp h2 + k2 + p2

⇒ x−2 + y−2 + z−2 = 9k−2

[[[[

2

Jpacademy STRAIGHT LINES

PREVIOUS EAMCET BITS

1. The point on the line 3x + 4y = 5 which is equidistant from (1, 2) and (3, 4) is [EAMCET 2009]

1) (7, −4) 2) (15, −10) 3) (1/ 7,8 / 7) 4) (0,5 / 4)

Ans: 2

Sol. Verification

2. The equation of the straight line perpendicular to the straight line 3x + 2y = 0 and passing
through the point of intersection of the lines x + 3y – 1= 0 and x – 2y + 4=0 is [EAMCET 2009]

1) 2x –3y + 1 =0 2) 2x – 3y + 3 = 0 3) 2x – 3y + 5 = 0 4) 2x – 3y + 7 = 0

Ans: 4

Sol. L1 + λL2 = 0 is perpendicular to 3x + 2y – 0 then find λ

3. The value of k such that the lines 2x – 3y + k = 0, 3x – 4y – 13 = 0 and 8x – 11y – 33 = 0 are

concurrent, is [EAMCET 2008]

1) 20 2) – 7 3) 7 4) – 20

Ans: 2

2 −3 k
Sol. Given lines are concurrent ⇒ 3 −4 −13 = 0

8 −11 −33

⇒ 2(132 −143) + 3(−99 +104) + k (−33 + 32) = 0 ⇒ −22 +15 − k = 0 ⇒ k = −7

4. The angle between the line joining the points ( 1, –2), (3, 2) and the line x + 2y – 7 = 0 is

[EAMCET 2007]

1) π 2) π 3) π 4) π
2 3 6

Ans: 2

Sol. Slope of line joining (1, –2), (3, 2) is = m1= 2

Slope of the line x + 2y − 7 =0 is − 1 = m2
2

m1m2 = −1 ∴θ = 90°

5. If A(2, –1) and B(6, 5) are two points the ratio in which the foot of the perpendicular from

(4, 1) to AB divides it is [EAMCET 2007]

1) 8 : 15 2) 5 : 8 3) –5 : 8 4) –8 : 5

Ans: 2

Sol. Equation of the line perpendicular to AB and passing through (4, 1) is

2x + 3y – 11 = 0 ……..(1)

(1) dividing AB in the ratio

1

Jpacademy Straight lines

− (ax1 + by1 + c);(ax2 + by2 + c)

− (4 − 3 −11) : (12 +15 −11) = 5 : 8

6. The lines x − y − 2 = 0, x + y − 4 = 0 and x + 3y = 6 meet in the common point [EAMCET 2006]

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

Ans: 3

Sol. Point of concurrence is the point of intersection of any two lines

Solving x – y = 2 and x + y = 4

x = 3 , y = 1, (3, 1)

7. The equation of the straight line perpendicular to 5x – 2y = 7 and passing through the point of

intersection of the lines 2x + 3y = 1 and 3x + 4y = 6 is [EAMCET 2005]

1) 2x + 5y +17 = 0 2) 2x + 5y −17 = 0 3) 2x − 5y +17 = 0 4) 2x − 5y = 17

Ans: 1
Sol. Equation of line passing through 2x + 3y = 1, and 3x + 4y = 6 is

2x + 3y −1+ k (3x + 4y − 6) = 0

⊥lar to 5x – 2y = 7, k = – 4/7
∴ 2x + 5y + 1 = 0
8. If the lines 4x + 3y – 1 = 0, x – y +5 = 0 and kx + 5y – 3 = 0 are concurrent, then k =..

[EAMCET 2003]
1) 4 2) 5 3) 6 4) 7
Ans: 3

4 3 −1
Sol. 1 −1 5 = 0 ⇒ k = 6

k 5 −3

9. If a straight line perpendicular to 2x – 3y + 7 = 0 forms a triangle with the coordinate axes whose

area is 3 sq. units, then the equation of the straight line(s) is [EAMCET 2002]

1) 3x + 2y = ±7 2) 3x + 2y = ±6 3) 3x + 2y = ±8 4) 3x + 2y = ±4

Ans: 2

Sol. Perpendicular line form is 3x + 2y + K = 0

Area of Δle = c2 = K2 = 3 ⇒ K = ±6
2 ab 12

10. For all values of a and b the line (a + 2b) x + (a − b) y + (a + 5b) = 0 passes through the point :

1) (−1, 2) 2) (2, −1) 3) (−2,1) [EAMCET 2001]

Ans: 3 4) (1, −2)

2

Jpacademy Straight lines

Sol. (a + 2b) x + (a − b) y + (a + 5b) = 0

a (x + y +1) + b (2x − y + 5) = 0

It always passes through, the point of intersection of the lines

x + y +1 = 0 and 2x − y + 5 = 0 i.e,(−2,1)

11. The number of circles that touch all the straight lines x + y – 4 =0 , x – y + 2 =0 and y = 2 is

[EAMCET 2001]
1) 1 2) 2 3) 3 4) 4
Ans: 4
Sol. The given lines form a triangle
∴ the No, circles are 4, i.e. incircle and three ex-circles of the triangle.
12. If the point (1, 2) and (3, 4) were to be on the same side of the line 3x – 5y + a = 0 , then

1) 7< a < 11 2) a = 7 3) a = 11 [EAMCET 2000]
Ans: 4 4) a < 7 or a > 11

Sol. 3(1) − 5(2) + a = a − 7

9 − 20 + a = a −11
∴ a − 7, a −11 having same sign (a -7) ( a- 11) > 0

∴ a < 7 (or) a > 11
13. The coordinates of the image of the origin O with respect to the straight line x + y + 1 = 0 are

[EAMCET 2000]

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

Ans: 4

Sol. x = y = −2 (1) ⇒ (x, y) = (−1, −1)
1 1
2

14. The area of the triangle formed by the axes and the line (coshα - sinhα)x + (coshα + sinhα) y =

2, in square units is [EAMCET 2000]

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

Ans: 3

Sol. Area = c2
2 ab

( )4 =2

2 cosh2 α − sinh2 α

”””

3

Jpacademy THREE DIMENSIONAL GEOMETRY

PREVIOUS EAMCET BITS

1. The perimeter of the triangle with vertices at (1, 0, 0), (0, 1, 0) and (0, 0, 1) is [EAMCET 2009]

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

Ans: 4

Sol. AB = BC = CA = 2

∴Perimeter = 3 2

2. In ΔABC the midpoints of the sides AB, BC and CA are respectively (l, 0, 0), (0, m, 0) and (0, 0,

AB2 + BC2 + CA2 [EAMCET 2008]
n). Then A2 + m2 + n2

1) 2 2) 4 3) 8 4) 16

Ans: 3

Sol. (A, 0, 0)(0, m, 0)(0, 0, n) are the midpoint of AB, BC, CA

⇒ A = (A, −m, n), B = (A, m, −n), C = (−A, m, n)

AB2 + BC2 + CA2 = (A − A)2 + (−m − m)2 + (n + n)2 + (A + A)2 + (m − m)2 + (−n − n)2

+ (−A − A)2 + (m + m)2 + (n − n)2 = 0 + 4m2 + 4n2 + 0 + 4n2 + 4A2 + 4m2 + 0

( )= 8A2 + 8m2 + 8n2 = 8 A2 + m2 + n2

⇒ AB2 + BC2 + CA2 =8
A2 + m2 + n2

3. The ratio in which yz-plane divides the line segment joining (–3, 4, –2), (2, 1, 3) is

[EAMCET 2007]

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

Ans: 2

Sol. −x1 : x2 = 3 : 2

4. If OA is equally inclined to OX, OY and OZ and If A is 3 units from the origin, then A is

[EAMCET 2006]

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

Ans: 4

Sol. α = β = γ

cos α = cosβ = cos γ = ± 1
3

OA = 3

A = ( OA cos α, OA cosβ, OA cos γ)

1

Jpacademy Three Dimensional Geometry

= ( ±1, ±1, ±1)
= (1,1,1) or (−1, −1, −1)

5. The point collinear with (1, –2, –3) and (2, 0, 0) among the following is [EAMCET 2005]

1) (0, 4, 6) 2) (0, –4, –5) 3) (0, –4, –6) 4) (0, –4, 6)

Ans: 3

Sol. Three points are collinear

if x1 − x2 = y1 − y2 = z1 − z2 verification
x2 − x3 y2 − y3 z2 − z3

6. XOZ plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio : [EAMCET 2003]

1) 3 : 7 2) 2 : 7 3) –3 : 7 4) –2 : 7

Ans: 3

Sol. −y1 : y2 = −3 : 7
7. If the plane 7x +11y +13z = 3003 meets the coordinate axes in A, B, C then the centroid of the

ΔABC is [EAMCET 2002]

1) (143,94, 77) 2) (143, 77,91) 3) (91,143, 77) 4) (143, 66,91)

Ans: 1

Sol. A (429, 0, 0)

B(0, 282,0)
C(0, 0, 231)

are vertices of Δle ABC

then centroid is ⎛ ∑ x1 , ∑ y1 , ∑ z1 ⎞ = (143,94, 77)
⎜⎝ 3 3 3 ⎟⎠

8. If the extremities of a diagonal of a square are (1, –2, 3), (2, –3, 5) then the length of its side is

[EAMCET 2001]

1) 6 2) 3 3) 5 4) 7
Ans: 2

Sol. Length of the diagonal = 6
∴ Length of the side = d = 3
2

”””

2

Jpacademy APPROXIMATIONS AND SMALL ERRORS

PREVIOUS EAMCET BITS

1. There is an error of ± 0.04 cm in the measurement of the diameter of a sphere. When the radius is

10 cm, the percentage error in the volume of the sphere is [EAMCET 2009]

1) ± 1.2 2) ± 1.0 3) ± 0.8 4) ± 0.6

Ans: 4
Sol. r = 10cm;δr = 0.02

∴ δr ×100 = ±0.2
r

∴ δV ×100 = 3× (±0.2) = ±0.6

V

2. The circumference of a circle is measured as 56 cm with an error 0.02 cm. The percentage error

in its area is [EAMCET 2007]

1) 1 2) 1 3) 1 4) 1
7 28 14 56

Ans: 3

Sol. radius = r, circumference = x; Area = A

∴ x = 2πr ⇒ r = x ;δx = 0.02
2π

A = πr2 = x2
4π

δA = x .δx
2π

Percentage error in A = δA ×100
A

= x .δx 1
2π ×100 =
⎛ x2 ⎞ 14
⎜ ⎟
⎝ 4π ⎠

3. The radius of a circular plate is increasing at the rate of 0.01 cm/sec when the radius is 12 cm.

Then the rate at which the area increases is [EAMCET 2005]

1) 0.24 π sq.cm /sec 2) 60 π sq.cm /sec 3) 24 π sq.cm /sec 4) 1.2 π sq.cm /sec

Ans: 1

Sol. r = 12, dr = 0.01/ sec
dt

A = πr2

dA = 2πr dr = 24π× 0.01
dr dt
= 0.24π sq.cm / sec

4. The approximate value of (1.0002)3000 is [EAMCET 2002]

1) 1.2 2) 1.4 3) 1.6 4) 1.8

1

Jpacademy Errors and Approximation

Ans: 3

Sol. Let y = f ( x ) = x3000

here x = 1, δ = 0.0002

δy = f ′( x ) δx = 3000x2999δx
= (3000)(0.0002)

= 0.6

∴ f (x + δx) = y + δy = 1+ 0.6 = 1.6

””””

2

Jpacademy CONTINUITY

PREVIOUS EAMCET BITS

⎧ 2sin x − sin 2x if (x ≠ 0)
⎪
1. If f = IR → IR is defined by f ( x ) = ⎨ 2x cos x

⎪⎩ a if (x = 0)

Then the value of a so that f is continuous at 0 is [EAMCET 2009]

1) 2 2) 1 3) – 1 4) 0

Ans: 4

⎪⎧ cos 3x − cos x for x≠0
⎨ x2 and if f is continuous at x = 0,
2. If f :R →R is defined by f ( x ) =

⎪⎩ λ for x = 0

then λ = [EAMCET 2008]

1) – 2 2)– 4 3) – 6 4) – 8

Ans: 2

Sol. f is continuous at

x = 0 ⇒ Lt f (x) = f (0)
x→0

⇒ λ = Lt cos 3x − cos x = Lt −3sin 3x + sin x
xx →0 2 x→0 2x

= Lt −9 cos 3x + cos x = −9 +1 = −4
x→0 2 2

⎧ x − 5 for x ≤ 1 [EAMCET 2007]

( ) ( )3. If f (x) = ⎪⎨4x2 − 9 for 1 < x < 2 then f 2+ − f 2− =
⎩⎪ 3x + 4 for x ≥ 2

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

Ans: 3

Sol. f (2+ ) − f (2− )

( )= (3(2) + 4) − 4(2)2 − 9 = 3

⎧1− 2 sin x if x ≠ π
⎪⎪ 4
4. If f (x) = ⎨ π − 4x is continuous at, π then a = [EAMCET 2006]

⎪ a if x=π 4
⎩⎪
4

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

Ans: 4

Sol. f(x) is continuous at x = π
4

lim f ( x ) = f ⎛ π ⎞
⎜⎝ 4 ⎠⎟
x→π
4

1

Jpacademy Continuity

a = lim 1− 2 sin x = lim f′(x)
x→π π − 4x g′ ( x )
x→π
44

= lim − 2 cos x = 2 . 1 = 1
x→π −4 4 24
4

⎧ x−2 if x ∈ R − (1, 2)
⎪
⎪ x 2 − 3x + 2 f (x)−f (2)
2
5. If f :\ → \ is defined by f (x) = ⎨ if x = 1 then lim =
if x = 2 x→2 x − 2
⎪ 1
⎪
⎩

[EAMCET 2005]

1) 0 2) –1 3)1 4) – 1/2

Ans: 2

f (x)−f (2) = x 2 x −2 2 −1
− 3x +
Sol. Lt Lt
x→2 x − 2 x→2 x − 2

= Lt 1 −1 = Lt −(x − 2) = −1
x −1

x→2 x − 2 x→2 x − 2

⎧ x+2 if x ∈ \{−1, −2}
⎪
⎪ x 2 + 3x + 2 if x = −2 then f is continuous on the set
⎨ −1
6. If f :\ → \ is defined by f ( x ) =

⎪ 0 if x = −1
⎪
⎩

[EAMCET 2005]

1) \ 2) \ −{−2} 3) \ −{−1} 4) \ −{−1, −2}

Ans: 3

Sol. ( )Lt f x = Lt x + 2 =
x → −1− x→−1− x2 + 3x + 2

Lt 1 = −∞, Lt f (x) = ∞
x→−1− x + 1 x → −1+

∴ Lt ≠ Lt
x→−1− x→−1+

∴ f(x) is not continuous at x = – 1

⎧ 1+ kx − 1− kx for −1 ≤ x < 0 is continuous at x = 0, then k = … [EAMCET 2003]
⎪ x
7. If f(x) =- ⎨

⎪⎩ 2x2 + 3x − 2 for 0 ≤ x ≤ 1

1) –4 2) –3 3) – 2 4) –1
Ans: 3

1+ kx − 1− kx
Sol. lim

x→0 x

= (lim 2x2 + 3x − 2)
x→0

2

Jpacademy Continuity

= lim 2k = −2
x→0 1+ kx + 1− kx

k=–2

8. If f : R → R defined by f (x) = ⎪⎧a2 cos2 x + b2 sin 2 x (x ≤ 0)
⎨ (x > 0) is a continuous function, then
⎪⎩ eax + b

[EAMCET 2002]

1) b = 2 log a 2) 2b = log a 3) b = log 2a 4) b2 = log a

Ans: 1

Sol. Lt f (x) = f (0)
x→0
= Lt eax+b = a2 (1) + b2 (0)
x →0+

eb = a2 ⇒ b = 2 log a

9. If f (x) = ⎧⎪2x + b (x < α) is such that lim f (x) = A, then A = [EAMCET 2001]
⎨ (x ≥ α) 4) b – 2d
⎩⎪ x + d x→α

1) 2d – b 2) 2b – d 3) 2d + b

Ans: 1

Sol. lim f (x) = lim f (x)
x→α− x→α+

lim (2x + b) = lim (x + d)
x→α− x→α+

2α + b = α + d ⇒ α = d − b

∴ lim (2x + b) = A ⇒ 2α + b = A
x→α

⇒ 2(d −b)+ b = A

∴ A = 2d − b

⎧sin 3x (x ≠ 0)⎪⎪⎫
⎪⎪
10. If the function f (x) = ⎨ x ⎬ is continuous at x = 0 then k = [EAMCET 2000]
k
⎪ (x = 0)⎪⎪⎭
⎩⎪ 2

1) 3 2) 6 3) 9 4) 12

Ans: 2

Sol. lim ⎛ sin 3x ⎞ = k ⇒ 3 = k
⎜⎝ x ⎠⎟ 2 2
x→0

⇒∴ k =6

]]]]

3

Jpacademy DIFFERENTIATION

PREVIOUS EAMCET BITS

1. x = 1− y ⇒ dy = [EAMCET 2009]
1+ y dx

4 4(x −1) 3) x −1 4) 4
2) (1+ x)3
1) ( x +1)2 (1+ x)3 (1+ x)3

Ans: 2

Sol. Using Componendo and dividendo, then find y and dy
dx

2. x = cos−1 ⎛ 1 ⎞ , y = sin −1 ⎛ t ⎞ ⇒ dy = [EAMCET 2009]
⎜ ⎟ ⎜ ⎟ 4) sint cost
⎝ 1+ t2 ⎠ ⎝ 1 + t2 ⎠ dx

1) 0 2) tan t 3) 1

Ans: 3
Sol. x = tan−1 t, y = tan−1 t

⇒ y = x ⇒ dy = 1
dx

3. d ⎡ tan −1 + b log ⎛ x −1 ⎞⎤ = x 1 ⇒ a − 2b = [EAMCET 2009]
dx ⎣⎢a ⎝⎜ x +1 ⎠⎟⎦⎥ 4 −1

1) 1 2) –1 3) 0 4) 2

Ans: 2

Sol. a + b − b = 1
1+ x2 x −1 x +1 x4 −1

⇒ a + 2b = 1
x2 +1 x2 −1 x4 −1

Put x = 0; a – 2b = 1

4. If x = ⎨⎧cos θ + log tan ⎛ θ ⎠⎟⎞⎬⎭⎫ and y = a sin θ then dy = [EAMCET 2008]
⎩ ⎜⎝ 2 dx

1) cotθ 2) tanθ 3) sinθ 4) cosθ

Ans: 2

dy

Sol. dy = dθ = a cos θ

dx dx a ⎡ sin θ + 1 × sec2 ⎛ θ ⎞ . 1 ⎤
dθ ⎢− ⎝⎜ 2 ⎟⎠ 2 ⎥
⎣ tan (θ / 2) ⎦

= cos θ 1 = cos θ = cos θsin θ = cos θ.sin θ = tan θ
− sin θ + 1 1− sin2 θ 1− sin2 θ
− sin θ +
⎛ θ ⎞ ⎛ θ ⎞ sin θ
2 sin ⎝⎜ 2 ⎟⎠ cos ⎝⎜ 2 ⎠⎟

1

Jpacademy x ) then x2 d2y + x dy Derivatives
5. If y = sin (loge dx 2 dx = [EAMCET 2008]

1) sin (loge x) 2) cos (loge x) 3) y2 4) – y

Ans: 4

Sol. y = sin (log x) ⇒ dy = cos (log x) 1 ⇒ x dy = cos (log x)

dx x dx

⇒ x d2y + dy = − sin (log x) 1 ⇒ x2 d2y + x dy = −y
dx 2 dx x dx 2 dx

6. If 2x2 − 3xy + y2 + x + 2y − 8 = 0 then dy [EAMCET 2007]
dx 4) 3y − 4x +1

1) 3y − 4x −1 2) 3y + 4x +1 3) 3y − 4x +1 2y + 3x + 2
2y − 3x + 2 2y + 3x + 2 2y − 3x − 2

Ans: 1

Sol. dy = − ∂f = 3y − 4x −1
∂x

dx ⎛ ∂f ⎞ 2y − 3x + 2
⎜ ⎟
⎝ ∂y ⎠

⎧⎪⎛ 1 + x ⎞1/ 4 ⎫⎪ 1 tan−1 ( x ), then dy
⎨⎪⎩⎝⎜ 1 − x ⎠⎟ ⎬ dx
7. If y = log ⎭⎪ − 2 = [EAMCET 2007]

1) x 2) x2 3) x 4) x
1− x2 1− x4 1+ x4 1− x4

Ans: 2

Sol. y = log ⎛1+ x ⎞1/ 4 − 1 tan −1 ( x)
⎜⎝ 1− x ⎠⎟ 2

= 1 (log (1+ x) − log (1− x)) − 1 tan−1 x

42

1⎛ 1 −1 1 = x2
4 ⎝⎜ 1+ x 1− x 1+ 1− x4
( )dy = ⎞ − 2 x2
⎠⎟
dx

( )8. d2y dy
x = cos θ, y = sin 5θ ⇒ 1− x2 dx 2 − x dx = [EAMCET 2007]
4) – 25 y
1) – 5 y 2) 5 y 3) 25 y
[EAMCET 2006]
Ans: 4
4) x ( y − x log y)
Sol. dy = −5cos 5θ = −5 1− sin2 5θ
dx sin θ 1− cos2 θ

y1 = −5 1− y2
1− x2

( ) ( ) ( )1− x2 y12 = 25 1− y2 ⇒ 1− x2 y2 − xy1 = −25y

9. xy = yx ⇒ x ( x − y log x) dy = 3) x (x + y log x)

dx

1) y ( y − x log y) 2) y ( y + l og y)

2

Jpacademy Derivatives

Ans: 1
Sol. xy = yx

⇒ y log x = x log y

⇒ x (x − y log) dy = y ( y − x log y)

dx

10. If f : \ → \ is an even function which is twice differentiable on \ and f ′′ (π) = 1, then f ′′(−π)

= [EAMCET 2005]

1) – 1 2) 0 3) 1 4) 2

Ans: 3

Sol. Consider f (x) = x2

2

f ′(x) = 2x = x,f′(x) =1

2

f ′′(π) = 1 = f ′′(−π)

11. Observe the following statements : [EAMCET 2005]

I : f (x) = ax 41 + bx−40 ⇒ f ′′(x) = 1640x−2
f (x)

II : d tan −1 ⎛ 2x ⎞ = 1
dx ⎝⎜ 1− x2 ⎟⎠ 1+ x2

Which of the following is correct ?

1) I is true, but II is false 2) Both I and II are true
4) I is false, but II is true
3) Neither I nor II is true

Ans: 1

Sol. I) f ′′( x ) = 1640ax39 +1640x−42.b

( )= 1640 1640 ( )
x2 x2
ax41 + bx−40 = f x

f ′′(x) = 1640x−2 True
f (x)

II) ∴ tan −1 ⎛ 2x ⎞ = 2 tan−1 x
⎝⎜ −x ⎠⎟
1 2

( )d 2 tan−1 x = 2 false

dx 1+ x2

12. f (x) = 10cos x + (13 + 2x)sin x ⇒ f ′′(x) + f (x) = [EAMCET 2005]
4) 4 sinx
1) cosx 2) 4cosx 3) sinx

Ans: 2

Sol. f ′( x) = −10sin x + (13 + 2x) cos x + 2snix

f ′′(x) = −10 cos x − (13 + 2x)sin x + 2 cos x + 2 cos x

= −f (x) + 4 cos x

f ′′(x) + f (x) = 4cos x

3

Jpacademy Derivatives

13. x 1+ y + y 1+ x = 0 ⇒ dy = [EAMCET 2005]
dx 1
4) 1+ x2
1 −1 1
3) 1+ x2
1) (1+ x)2 2) (1+ x)2

Ans: 2
Sol. x 1+ y = −y 1+ x

x2 + x2y = y2 + y2x

x2 − y2 = −xy ( x − y)

x + y = −xy

y = −x ⇒ y1 = −1
1+ x
(1+ x)2

14. If f : \ → \ is an even function having derivatives of all orders, then an odd function among the

following is 2) f ′′′ 3) f ′ + f ′′ [EAMCET 2004]
1) f ′′ 4) f ′′ + f ′′′

Ans: 2
Sol. f ′′′ is odd, since ‘f’ is even

15. x > 0, xy = ex−y ⇒ dy = [EAMCET 2004]
dx

1) 1 2) log x 3) ⎛ log x ⎞2 (log x)2
⎜ ⎟
(1+ log x)2 (1+ log x)2 ⎝ 1 + log x ⎠ 4)
1+ log x

Ans: 2

Sol. x − y = y log x ⇒ y = x
1+ log x

⇒ dy = log x
dx
(1+ log x)2

16. If f (x) = ⎧⎪ 3x 2 x −1 + 5 for x ≠ 0 then f 1 (1) = [EAMCET 2003]
⎨ − 7x , 4) 1

⎪⎩ 1/ 3 for x = 1 3

1) − 1 2) − 2 3) − 1 [EAMCET 2003]
9 9 3 4)

Ans: 2

Sol. f ′(1) = lim f (x) − f (1) = −2 / 9

x→1 x −1

17. If f ( x) = x for x ∈ then f 1 (0) = ....

1+ x

1) 0 2) 1 3) 2

Ans: 2

Sol. f ′(0) = lim f (x) − f (0) = 1

x→0 x

4

Jpacademy h (x) = f (g (x)), then h′(x) Derivatives
18. Let f (x ) = ex , g (x ) = sin−1 x and h(x) = [EAMCET 2002]
4) esin−1 x
1) sin−1 x 2) 1 3) 1
1− x2 1− x2

Ans: 2

( )Sol. h ( x ) = f ⎡⎣g ( x )⎤⎦ = f sin−1 x = esin−1 x

( ) ( )h x = esin−1 x ⇒ h1 x = esin−1 x

1− x2

h1 (x) 1
∴ h(x) = 1− x2

19. If h(x) = eex then h′(x) = [EAMCET 2001]
h(x) 4) –log h(x)

1) h(x) 2) 1 3) log h(x)

h(x)

Ans: 3

Sol. Given h ( x) = eex ⇒ log (h (x )) = ex

⇒ h′(x) = ex = log h (x)
h(x)

20. If f (x) = x2 then f ′′(a ) = [EAMCET 2001]

x+a

1) 4a 2) 1 3) 1 4) 8a
8a 4a

Ans: 3

Sol. f (x ) = x2 = x − a + a2

x+a x+a

f ′ ( x ) = 1 − ( a2 )2
+a
x

f ′′ ( x ) = ( 2a 2 )3
x+a

∴ f ′′(a ) = 2a 2 = 1
4a
(a + a)3

21. If y = 22x , then dy = [EAMCET 2000]
dx 4) y loge 2

1) y (log10 2)2 2) y (loge 2)2 ( )3) y2x loge2 2

Ans: 3

Sol. y = 22x

⇒ log y = 2x log 2
e

5

Jpacademy Derivatives
1 dy = 2x. 2
y dx [EAMCET 2000]
( )⇒ 2 4) −1
log e
9− x2
( )∴dy 2
dx = y.2x. log 2
e

20. d ⎧⎨⎪cos−1 ⎛ 4x3 − x ⎞⎪⎫ =
dx ⎩⎪ ⎜ 27 ⎟⎬
⎝ ⎠⎭⎪

1) 3 2) 1 3) −3
9− x2 9− x2 9− x2

Ans: 3

Sol. y = cos−1 ⎛ 4x3 − x ⎞
⎜ 27 ⎟
⎝ ⎠

= cos−1 ⎡ ⎛ x ⎞3 − 3⎛⎝⎜ x ⎞⎟⎠⎦⎥⎤⎥ = 3 cos−1 ⎛ x ⎞
⎢4 ⎜⎝ 3 ⎟⎠ 3 ⎝⎜ 3 ⎠⎟
⎣⎢

∴ dy = −3
dx 9 − x2

YYZZ

6

Jpacademy LIMITS

PREVIOUS EAMCET BITS

1. lim ⎛ x + 5 ⎞x+3 = [EAMCET 2009]
⎝⎜ x + 2 ⎟⎠
x→∞ [EAMCET 2008]
[EAMCET 2008]
1) e 2) e2 3) e3 4) e5 [EAMCET 2008]

Ans: 3

+3)⎢⎣⎡ x + 5 −1⎦⎥⎤ Lt 3x+9
x + 2
= ex→∞ x+2
( )e lt (x = e3 ⎢⎣⎡∵ ⎣⎡f ⎤⎦ g( x ) = e Lt g( x )⎣⎡f ( x )−1⎤⎦ ⎤
x→∞ x→∞ ⎥⎦
Sol. Lt x

x→∞

2. lim (1− ex )sin x

x→0 x2 + x3

1) – 1 2) 0 3) 1 4) 2

Ans: 1

( )Sol.
Lt 1− ex sin x = Lt 1− ex × sin x = Lt 1− ex = Lt −ex = −1
x2 + x3 x + x2 x x + x2 x→0 1+ 2x
x→0 x→0 x→0

3. If f : R → R is defined by f (x ) = [x − 3] + x − 4 for x ∈ R then lim f (x) =
x →3−

1) – 2 2) – 4 3) – 6 4) – 8

Ans: 3

Sol. Lt f (x) = Lt − {[x − 3] + x − 4} = −1+ 1 = 0
x →3− x→3−

4. If f(2) = 4 and f ′(2) = 1 then lim xf (2) − 2f (x) =

x→2 x − 2

1) – 2 2) 1 3) 2 4) 3

Ans: 3

Sol. Lt xf (2) − 2f (x) = Lt f (2) − 2f ′(2) = 4 − 2 = 2
x→2 x − 2 x→2 1 1

5. lim ex − esin x = [EAMCET 2007]
−
x→0 2(x sin x)

1) − 1 2) 1 3) 1 4) 3
2 2 2

Ans: 2

( )Sol.
esin x ex−sin x −1 =1
lim
( )x→0 2 x − sin x 2

⎧sin (1+ [x]) for[x] ≠ 0
⎪
6. If f (x) = ⎨ [x] where [x] denotes the greatest integer not exceeding x, then

⎪ 0 for[x] = 0
⎩

lim f (x) = [EAMCET 2007]

x →0−

1) –1 2) 0 3) 1 4) 2

Ans: 2

1

Jpacademy Limits

Sol. lim f (x)
x →0−

sin (1+ [x]) 1+ [x]
= lim ×1+[x]
x →0− [x]

= 1× 0 = 0 3) q [EAMCET 2006]
−1 4) 0

( )7. If 0 < p < q, then lim qn + pn 1/n =
n→∞

1) e 2) p
Ans: 3
Sol. 0 < p < q, 0 < p < 1

q

⎛⎜1 p ⎞1/ n
⎝ q ⎟
( )lim 1/ n ⎠

n→∞
qn + pn = lim q + = q x 1 = q

n→∞

8. lim ⎡ x2 + 2x −1 − x ⎤ = [EAMCET 2006]
⎣ ⎦ 4) 1
n→∞

1) ∞ 2) 1/2 3) 4

Ans: 4

Sol. lim 2x −1

x→∞ x2 + 2x −1 + x

⎛ 2 − 1 ⎞ = 2 =1
⎜⎝ x ⎟⎠
lim
x→∞ 2 1 2
x x2
1+ − +1

9. If lim ⎛ cos 4x + a cos 2x + b ⎞ is finite, then the values of a, b, are respectively [EAMCET 2006]
⎝⎜ x4 ⎠⎟
x→0

1) 5, –4 2) –5, –4 3) –4, 3 4) 4, 5

Ans: 3

Sol. lim ⎛ cos 4x + a cos 2x + b ⎞ = lim f ( x ) exists
⎜⎝ x4 ⎠⎟ g ( x )
x→0 x→0

f(0) = 0

1+ a + b = 0 ⇒ a + b = −1

lim f′(x) = lim −4 sin 4x − 2a sin 2x
g′ ( x ) 4x3
x→0 x→0

f ′′(0) = 0

−16 − 4a = 0 ⇒ a = −4

⇒ b = 4 −1 = 3 –4, 3

( ) ( )10. cos x
If I1 = lim [ ]x + x , I2 = lim 2x + [x] and I3 = lim ⎛ x−π ⎞ , then [EAMCET 2006]
⎜⎝ ⎟⎠
x→2+ x→2− x→π 2
2

1) I1 < I2 < I3 2) I2 < I3 < I1 3) I3 < I2 < I1 4) I1 < I3 < I2
Ans: 3

2

Jpacademy Limits
Sol. [ ]1= + =4
lim x x [EAMCET 2005]
[EAMCET 2004]
x →2+ [EAMCET 2003]
[EAMCET 2003]
[ ]2= 2x − x
lim

x→2−

= lim{2(2 − h) −[2 − h]}
h→0

= 4−2 = 2

sin ⎛ π − x ⎞
⎜⎝ 2 ⎞ ⎟⎠
= (−1) lim ⎟⎠ = −1
3 ⎛ π ⎞ ⎛ π
⎜⎝ 2 ⎟⎠ ⎝⎜ 2
−x − x

3< 2< 1 2) 0 3) does not exist 4) ∞

11. lim x2 sin π =
x→0 x
1) 1
Ans: 2

Sol. Lt x2 sin ⎛ π ⎞ = Lt x 2 . ⎝⎜⎛ Lt sin π ⎞ = 0 × (finite value between – 1 to 1) = 0
⎜⎝ x ⎟⎠ x ⎟⎠
x→0 x→0 x→0

∑( )12. 1 n
lim n3 k =1 k2x =

n→∞

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

Ans: 3

Sol. x Lt 1 ⎡ n (n + 1) ( 2n +1) ⎤ = x Lt 2n3 + .... = x × 2 = x
n3 ⎢ 6n3 6 3
n→∞ ⎣ 6 ⎥ n→∞
⎦

13. lim ⎛ 3 sin x− 3 cos x ⎞ =
⎝⎜⎜ 6x π ⎟⎠⎟
x→π −
6

1) 3 2) 1 3) – 1 4) −1
3 3 3

Ans: 2

Sol. lim ⎛ 3sin x − 3 cos x ⎞
⎜⎜⎝ ⎠⎟⎟
x→π 6x − π
6

lim 3co s x + 3 sin x = 1
x→π 6 3
6

14. If a > 0 lim ax − xa = −1, then a = ……..
x→a xx − aa

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

Ans: 2

Sol. lim ax − xa = −1

x→a xx − aa

Apply L-hospital rule

3

Jpacademy Limits

⇒ lim ax log a − axa−1 = −1

x→a xx (1+ log x)

⇒ log a −1 = −1 ⇒ log a = 0 ⇒ a = 1
1+ log a

15. 4x − 9x = [EAMCET 2002]
lim
( )x→0 x 4x + 9x

1) log 2 2) log 3 3) 1 log 2 4) 1 log 3
3 2 23 22

Ans: 1
4x − 9x

Sol. Lt

( )x→0 x 4x + 9x

(4x −1) (9x −1)
−
xx
Lt 4x + 9x

x→0

log 4 − log 9 = 1 log ⎛ 2 ⎞2 = log 2
2 2 ⎜⎝ 3 ⎠⎟ 3

16. The quadratic equation whose roots are l and m where

⎛ 3sin θ − 4 sin2 θ ⎞ ⎛ 2 tan θ ⎞
θ ⎟ = lim ⎜ 1− tan2 θ ⎟
( )l =lim ⎜ and m θ→0 ⎜⎝ θ ⎟⎠ is [EAMCET 2002]
⎝ 4) x2 + 5x − 6 = 0
θ→0 ⎠

1) x2 + 5x + 6 = 0 2) x2 − 5x + 6 = 0 3) x2 − 5x − 6 = 0

Ans: 2

⎛ 3 sin θ −4 sin 2 θ ⎞
⎜ θ ⎟
Sol. = Lt ⎝ ⎠

θ→0

= Lt sin θ (3 − 4sin θ) = 3

θ→0 3θ

( )m = Lt 2 tan θ = 2
θ→0 θ 1− tan2 θ

∴ The quadratic equation required is x2 − ( + m) x + m = 0

x2 −(3+ 2)x + 6 = 0

⇒ x2 − 5x + 6 = 0

17. If f : R → R is defined by f(x) = x – [x], where [x] is the greatest integer not exceeding x, then

the set of discontinuities of f is [EAMCET 2002]

1) the empty set 2) R 3) Z 4) N

Ans: 3

Sol. [x] is discontinuous at inter values of x hence

f(x) = x – [x] is discontinuous on Z

18. lim ⎛ x + a ⎞x+b = [EAMCET 2001]
⎜⎝ x + b ⎠⎟
x→∞

1) 1 2) eb−a 3) ea−b 4) eb

4

Jpacademy Limits

Ans: 3

( b) ⎡ x − a −1⎤⎦⎥ ⎣⎢⎡∵ Lt g(x )⎡⎣f (x)−1⎤⎦
⎣⎢ x + b
x→∞
[ ]e Lt x + x = e ⎦⎤⎥g(x)
x→∞
Sol. = ea−b Lt f

x→∞

19. lim x.10x − x = [EAMCET 2001]
x→α 1− cos x 4) 4 log 10

1) log 10 2) 2log 10 3) 3 log 10

Ans: 2

Sol. lim x (10x −1)

x→0 2sin2 x

2

lim ⎛ 10x − 1 ⎞ . Lt x2 = 2 log10
⎜ x ⎟ 2 sin 2
x→0 ⎝ ⎠ x→0 x

2

20. If f (x) = x2 −10x + 25 x ≠5 and f is continuous at x = 5 then f(5) = [EAMCET 2001]
x2 − 7x +10 for

1) 0 2) 5 3) 10 4) 25

Ans: 1

Sol. lim x2 −10x + 25 = f (5)
x2 − 7x +10
x→5

⇒ f (5) = 0

21. lim 1− sin θ = [EAMCET 2000]
π ⎛ π ⎞ 4) 1
θ→ 2 cos θ ⎜⎝ 2 − θ ⎟⎠
2
1) 1 2) –1 3) −1
2

Ans: 4

Sol. lim 1− sin θ
π ⎛ π ⎞
θ→ 2 cos θ ⎝⎜ 2 − θ ⎟⎠

⎛ π − θ ⎞
⎜ 2 2 ⎟
2 sin 2 ⎜ ⎟
⎜ ⎟
⎝⎠ =1
lim
θ→ π ⎛ π ⎞ ⎛ π ⎞⎛ π ⎞ 2
2 ⎜ 2 − θ ⎟ ⎜ 2 − θ ⎟⎜ 2 − θ ⎟
⎜ 2 ⎟ ⎜ 2 ⎟ .2 ⎜ 2 ⎟
2 sin ⎜ ⎟ cos ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

log (x +1)

22. lim e = [EAMCET 2000]

x→0 3x −1

1) log3e 2) 0 3) 1 4) log3e

Ans: 4

5