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**SECTION A**

**Question 1: The coefficients a,b and c of the quadratic equation, ax ^{2} + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :**

a. 1/54

b. 1/72

c. 1/36

d. 5/216

Answer: (d)

ax^{2} + bx + c = 0

a,b,c ∈ {1,2,3,4,5,6}

n(s) = 6 × 6 × 6 = 216

D=0 ⇒ b^{2} = 4ac

ac = b^{2}/4, If b = 2, ac = 1 ⇒ a = 1,c = 1

If b = 4, ac = 4 ⇒ a = 1, c = 4

a = 4, c = 1

a = 2, c = 2

If b = 6, ac = 9 ⇒ a = 3, c = 3

Therefore, probability = 5/216

**Question 2: Let α be the angle between the lines whose direction cosines satisfy the equations l +m – n = 0 and l ^{2} + m^{2} – n^{2} = 0. Then the value of sin^{4} α + cos^{4} α is :**

a. 3/4

b. 1/2

c. 5/8

d. 3/8

Answer: (c)

Given that l+m=n ….(1)

l^{2} + m^{2} – n^{2} = 0 ….(2)

Squaring equation (1)

l^{2} + m^{2} + 2lm = n^{2} ….(3)

From equations (2) and (3)

lm=0 ⇒ l = 0 or m = 0

Case (1) : l = 0

⇒ m=n

⇒ l^{2} + m^{2} + n^{2} = 0

⇒ m=n = ±1/(√2 )

∴(l,m,n) = (0, 1/√2, 1/√2) or (0, -1/√2, -1/√2)

Case (2): m = 0

⇒ l = n

⇒ l^{2} + m^{2 }+ n^{2} = 0

**Question 3: The value of the integral **

**(where c is a constant of integration)**

a. 1/18[9 – 2 sin^{6} θ – 3 sin^{4} θ – 6 sin^{2} θ ]^{(3/2)} + c

b. 1/18[11 – 18 sin^{2} θ + 9 sin^{4} θ – 2 sin^{6} θ ]^{(3/2)} + c

c. 1/18[11 – 18 cos^{2} θ + 9 cos^{4} θ – 2 cos^{6} θ ]^{(3/2)} + c

d. 1/18[9 – 2 cos^{6} θ – 3 cos^{4} θ – 6 cos^{2} θ ]^{(3/2)} + c

Answer: (c)

Using trig identities, sin2A = 2sinAcosA and 1-cos2A = 2sin^{2}A

**Question 4: A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform speed. At that point, the angle of depression of the boat with the man’s eye is 30° (Ignore man’s height). After sailing for 20 seconds towards the base of the tower (which is at the level of water), the boat has reached point B, where the angle of depression is 45°. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :**

a. 10(√3-1)

b. 10√3

c. 10

d. 10(√3+1)

Answer: (d)

x + y = √3h ….…….(1)

Also,

h/y = tan 45^{o}

h = y …….(2)

put in (1)

x + y = √3y

x = (√3 – 1)y

x/20=v (speed)

Therefore, time taken to reach

Foot from B

= y/V

= x/(√3-1)x . 20

= 10 (√3 + 1)

**Question 5: **

a. xyz = 4

b. xy – z = (x + y)z

c. xy + yz + zx = z

d. xy + z = (x+y)z

Answer: (d)

**Question 6: The equation of the line through the point (0,1,2) and perpendicular to the line **

**\(\begin{array}{l}\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2}\end{array} \)**

**is :**

a.

b.

c.

d.

Answer: (a)

Any point on this line (2λ + 1, 3λ – 1, -2λ + 1)

Direction ratio of given line

Direction ratio of the line to be found

∴

λ = 2 / 17

Direction ratio of line (21, -28, -21) **≡ **(3, -4, -3) ≡ (-3, 4, 3)

**Question 7: The statement A→ (B → A) is equivalent to:**

a. A → (A ᐱ B)

b. A → (A ᐯ B)

c. A → (A → B)

d. A → (A ↔ B)

Answer: (b)

A → (B → A)

⇒ A → (∼ B ᐯ A)

⇒ ∼ A ᐯ (∼ B ᐯ A)

⇒ ∼ B ᐯ (∼ A ᐯ A)

⇒ ∼ B ᐯ t

= t (tautology)

From options:

(b) A → (A ᐯ B)

⇒ ∼ A ᐯ (A ᐯ B)

⇒ (∼ A ᐯ A) ᐯ B

⇒ t ᐯ B

⇒ t

**Question 8: The integer k, for which the inequality x ^{2} – 2(3k – 1)x + 8k^{2} – 7 >0 is valid for every x in R is :**

a. 3

b. 2

c. 4

d. 0

Answer: (a)

D < 0

(2(3k-1))^{2} – 4(8k^{2} – 7) < 0

4(9k^{2} – 6k + 1) – 4(8k^{2} – 7) < 0

k^{2} – 6 k + 8 < 0

(k–4)(k–2)<0

2< k < 4

then k = 3

**Question 9: A tangent is drawn to the parabola y ^{2} = 6x which is perpendicular to the line 2x + y =1. Which of the following points does NOT lie on it?**

a. (0,3)

b. (-6,0)

c. (4,5)

d. (5,4)

Answer: (d)

Equation of tangent : y = mx + 3/(2m)

m_{T} = (1/2) (Because perpendicular to line 2x + y = 1)

Therefore, tangent is: y = x/2 + 3 ⇒ x – 2y + 6 = 0

Substitute given options in the above equation and check.

We can find that (5, 4) does not lie on the line.

Hence option (d) is the answer.

**Question 10: Let f, g: N→N such that f(n + 1)= f(n)+ f(1) for all n ∈ N and g be any arbitrary function. Which of the following statements is NOT true ?**

a. f is one-one

b. If fog is one-one, then g is one-one

c. If g is onto, then fog is one-one

d. If f is onto, then f(n) = n for all n ∈ N

Answer: (c)

f(n + 1) = f(n) +f(1)

⇒ f(n + 1)- f(n)= f(1) → A.P. with common difference = f(1)

General term = T_{n} = f(1) + (n-1)f(1) = nf(1)

⇒f(n) = nf(1)

Clearly f(n) is one-one.

For fog to be one-one, g must be one-one.

For f to be onto, f(n) should take all the values of natural numbers.

As f(x) is increasing, f(1)=1

⇒f(n)=n

If g is many one, then fog is many one. So “if g is onto, then fog is one-one” is incorrect.

**Question 11: Let the lines (2 – i)z = (2 + i)**

**\(\begin{array}{l}\bar{z}\end{array} \)**

**and (2 + i)z + (i – 2)\(\begin{array}{l}\bar{z}\end{array} \) – 4i = 0, (here i\(\begin{array}{l}\bar{z}\end{array} \) + 1 + i = 0 is tangent to this circle C, then its radius is :**

^{2}= –1) be normal to a circle C. If the line iz +

a. 3/√2

b. 3√2

c. 3/(2√2)

d. 1/(2√2)

Answer: (c)

(2–i)z = (2+i)

⇒(2–i)(x+ iy)=(2+i)(x–iy)

⇒2x–ix + 2iy + y=2x+ ix–2iy+y

⇒2ix –4iy=0

L_{1} ∶ x–2y=0

⇒(2+i)z+(i –2)

⇒(2+i) (x + iy)+(i –2)(x–iy)– 4i = 0.

⇒2x +ix + 2iy – y + ix – 2x + y +2iy – 4i =0

⇒2ix + 4iy – 4i =0

L_{2} ∶ x+2y–2 = 0

Solve L_{1} and L_{2}

4y=2 or y=1/2

∴ x = 1

Centre(1, 1/2)

L_{3}∶ iz +

⇒i(x+iy)+x–iy+1+i=0

⇒ix–y+x–iy+1+i=0

⇒(x–y+1)+i(x–y+1)=0

Radius = distance from (1, 1/2) to x – y + 1 = 0

r = (1-1/2+1)/ √2

r = 3/2√2

**Question 12: All possible values of θ ∈ [0, 2π] for which sin2θ + tan 2θ > 0 lie in**

a.

b.

c.

d.

Answer: (b)

tan 2θ(1+cos 2θ) >0 (since 1+cos 2θ>0)

2θ ∈ (0, π/2) ⋃(π, 3π/2) ⋃(2π, 5π/2)⋃(3π, 7π/2)

=> (0, π/4) ⋃(π/2, 3π/4) ⋃(π, 5π/4)⋃(3π/2, 7π/4)

**Question 13: The image of the point (3,5) in the line x – y + 1 = 0, lies on :**

a. (x – 2)^{2} + (y – 4)^{ 2} =4

b. (x – 4)^{ 2} + (y + 2)^{2} =16

c. (x – 4)^{2} + (y – 4)^{2} = 8

d. (x – 2)^{2} + (y – 2)^{2} =12

Answer: (a)

Image of P(3, 5) on the line x – y + 1 = 0 is

x = 4, y = 4

Image is (4, 4)

Which lies on (x – 2)^{2} + (y – 4)^{2} = 4

**Question 14: If Rolle’s theorem holds for the function f(x) = x ^{3} – ax^{2} + bx – 4, x ∈ [1, 2] with f'(4/3) = 0, then ordered pair (a, b) is equal to :**

a. (–5, 8)

b. (5, 8)

c. (5, –8)

d. (–5, –8)

Answer: (b)

f(1) = f(2)

⇒ 1 – a + b –4 = 8 – 4a + 2b –4

3a – b = 7 …(1)

f’(x) = 3x^{2} – 2ax + b

⇒f’(4/3) = 0 ⇒ 3 x 16/9 – 8a/3 + b =0

⇒–8a + 3b = –16 …(2)

a = 5, b = 8

**Question 15: If the curves, **

**\(\begin{array}{l}\frac{x^2}{a}+\frac{y^2}{b}\end{array} \)**

**and\(\begin{array}{l}\frac{x^2}{c}+\frac{y^2}{d}\end{array} \) intersect each other at an angle of 90°, then which of the following relations is true ?**

a. a + b = c + d

b. a- b = c – d

c. ab = (c+d)/(a+b)

d. a-c = b+d

Answer: (b)

diff :

Diff :

m_{1}m_{2} = –1 ⇒

⇒ bdx^{2} = – acy^{2}…….(5)

(1)–(3) ⇒ (1/a – 1/c)x^{2} + (1/b – 1/d) y^{2} = 0

Solve above equation using 5

⇒ (c – a) – (d – b) = 0

⇒ c – a = d – b

⇒ c – d = a – b

**Question 16: **

a. 1/2

b. 1/e

c. 1

d. 0

Answer: (c)

**Question 17: The total number of positive integral solutions (x, y, z) such that xyz = 24 is**

a. 36

b. 45

c. 24

d. 30

Answer: (d)

x.y.z = 24

x.y.z=2^{3}.3^{1}

x=2^{(a_1 )}⋅3^{(b_1 )}

y = 2^{(a_2 )}⋅3^{(b_2 )}

z = 2^{(a_3 )}⋅3^{(b_3 )}

a_{1}, a_{2}, a_{3} ∈ {0,1,2,3}

b_{1}, b_{2}, b_{3} ∈ {0,1}

Case 1: a_{1} + a_{2} + a_{3} =3

Non negative solution = ^{(3+3-1) }C_{(3-1)} = ^{5}C_{2 }= 10

Case 2: b_{1} + b_{2} + b_{3} = 1

Non negative solution = ^{(1+3-1) }C_{(3-1)} = ^{3}C_{2} = 3

∴ Total solutions =10×3=30

**Question 18: If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is **

**\(\begin{array}{l}\frac{x^2-4x+y+8}{x-2}\end{array} \)**

**, then this curve also passes through the point :**

a. (4, 5)

b. (5, 4)

c. (4, 4)

d. (5, 5)

Answer: (d)

Let x – 2 = t ⇒ dx = dt

and y + 4 = u ⇒dy = du

dy/dx = du/dt

du/dt = t + u/t ⇒ du/dt – u/t = t

Here, IF = 1/t

u. (1/t) = ∫ t.(1\t) dt

⇒ u/t = t + c

⇒ (y+4)/(y-2) = (x – 2) + c

Passing through (0, 0)

c = 0

⇒ (y + 4) = (x – 2)^{2}

**Question 19: The value of **

**\(\begin{array}{l}\int_{-1}^1 x^2 e^{x^3}dx\end{array} \)**

**where [t] denotes the greatest integer ≤ t,is :**

a. (e+1)/3

b. (e-1)/3e

c. (e+1)/3e

d. 1/3e

Answer: (c)

**Question 20: When a missile is fired from a ship, the probability that it is intercepted is 1/3 and the probability that the missile hits the target, given that it is not intercepted, is 3/4. If three missiles are fired independently from the ship, then the probability that all three hit the target, is: **

a. 1/8

b. 1/27

c. 3/4

d. 3/8

Answer:(a)

Probability of not getting intercepted = 2/3

Probability of missile hitting target = 3/4

Probability that all 3 hit the target = (2/3 x3/4)^{3} = 1/8

**Section B**

**Question 1: Let A _{1}, A_{2}, A_{3},…. be squares such that for each n ≥ 1, the length of the side of A_{n} equals the length of diagonal of A_{n+1}. If the length of A_{1} is 12 cm, then the smallest value of n for which area of A_{n} is less than one is ____________.**

Answer: (9)

Side lengths are in GP.

T_{n} = 12/(√2)^{n-1}

Area = 144/2^{n-1} <1

=> 2^{n-1} > 144

Smallest n = 9

**Question 2: The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then A ^{4} is equal to ___________**

Answer: (64)

**Question 3: The locus of the point of intersection of the lines (√3)kx + ky – 4√3 = 0 and √3x – y – 4√3 k = 0 is a conic, whose eccentricity is _________.**

Answer: (2)

(√3)kx + ky – 4√3 = 0 …(1)

√3kx -ky= 4√3 k^{2} …(2)

Adding equation (1) & (2)

2√3 kx = 4√3(k^2 + 1)

x = 2 (k + 1/k) ….(3)

Subtracting equation (1) & (2)

y = 2√3(1/k – k) ………(4)

x^{2}/4 – y^{2}/12 = 4

x^{2}/16 – y^{2}/48 = 1 – Hyperbola

e^{2} = 1 + 48/16

or e = 2

**Question 4:**

Answer: 13

|I_{2} + A|=|I_{2} – A|

From equation (1)

∴a^{2} + b^{2} = 1

⇒13(a^{2} + b^{2}) = 13

**Question 5: Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x ^{6} is unity and it has extrema at x = –1 and x = 1. If **

**\(\begin{array}{l}\lim_{x \to 0}\frac{f(x)}{x^3}=1\end{array} \)**

**, then 5.f(2) is equal to _________.**

Answer: (144)

f(x) = x^{6} + ax^{5} + bx^{4} + x^{3}

f’(x) = 6x^{5} + 5ax^{4} + 4bx^{3} + 3x^{2}

Roots 1 & – 1

Therefore, 6 + 5a + 4b + 3 = 0 & – 6 + 5a – 4b + 3 = 0 solving

a = -3/5 and b = -3/2

f(x) = x^{6} – (3/5)x^{5} – (3/2)x^{4} + x^{3}

5.f(2) = 5 [64 – 96/5 – 24 + 8] = 144

**Question 6: The number of points, at which the function f(x) = |2x + 1| – 3|x+2|+|x ^{2} + x–2|, x ∈ R is not differentiable, is ________.**

Answer: 2

Check at 1, –2 and -1/2

Non-Differentiable at x = 1 and -1/2

**Question 7: If the system of equations**

**kx + y + 2z = 1**

**3x – y – 2z = 2**

**–2x – 2y – 4z = 3**

**has infinitely many solutions, then k is equal to __________.**

Answer: 21

D = D_{1} = D_{2} = D_{3} = 0

Choose D_{2} = 0

k(–8+6)–1(–12–4)+2(9+4)=0

–2k + 16 + 26 = 0

2k = 42

k = 21

**Question 8: Let **

**\(\begin{array}{l}\vec{a} = \hat{i} + 2 \hat{j} – \hat{k}, \vec{b} = \hat{i} – \hat{j}\end{array} \)**

**and\(\begin{array}{l}\vec{c} = \hat{i} – \hat{j} – \hat{k}\end{array} \) be three given vectors. If \(\begin{array}{l}\vec{r}\end{array} \) is a vector such that \(\begin{array}{l}\vec{r} \times \vec{a} = \vec{c} \times \vec{a}\end{array} \) and \(\begin{array}{l}\vec{r} . \vec{b} = 0\end{array} \), then \(\begin{array}{l}\vec{r} . \vec{a} = 0\end{array} \) is equal to _______**

Answer: 12

**Question 9: Let A = **

**\(\begin{array}{l}\begin{bmatrix} x &y &z \\ y &z &x \\ z & x &y \end{bmatrix}\end{array} \)**

**, where x, y and z are real numbers such that x + y + z > 0 and xyz = 2. If A**

^{2}= l_{3}, then the value of x^{3}+ y^{3}+ z^{3}is ______.

Answer: 7

**Question 10: The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4,5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5 is ________.**

Answer: 32

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