**SECTION A**

**Question 1: Let L be a line obtained from the intersection of two planes x + 2y + z = 6 and y + 2z = 4. If point P (?, β, γ) is the foot of the perpendicular from (3, 2, 1) on L, then the value of 21 (? + β + γ) equals:**

a. 142

b. 68

c. 136

d. 102

Answer: (d)

Equation of the line is x + 2y + z – 6 = 0 = y + 2z = 4

[x + 2] / 3 = [y – 4] / – 2 = z / 1 = λDr’s of PQ: 3λ – 5, – 2λ – 2, λ – 1

Dr’s of y lines are (3, – 2, 1)

Since PQ is perpendicular to the line.

3 (3λ – 5) – 2 (- 2λ + 2) + 1 (λ – 1) = 0

λ = 10 / 7

P (16 / 7, 8 / 7, 10 / 7)

21 (? + β + γ) = 21 (34 / 7) = 102

**Question 2: The sum of the series ∑ _{n=1}^{∞}∞[n^{2 }+ 6n + 10] / (2n + 1)! is equal to:**

a. (41/ 8) e + (19 / 8) e^{-1} – 10

b. (- 41/ 8) e + (19 / 8) e^{-1} – 10

c. (41/ 8) e – (19 / 8) e^{-1} – 10

d. (41/ 8) e + (19 / 8) e^{-1} + 10

Answer: (c)

∑_{n=1}^{∞}∞[n^{2 }+ 6n + 10] / (2n + 1)!

Put 2n + 1 = r, where r = 3, 5, 7…..

n = [r – 1] / 2

**Question 3: Let f(x) be a differentiable function at x = a with f’ (a) = 2 and f (a) = 4. Then lim _{x→a} [x f (a) – a f (x)] / [x – a] equals:**

a. 2a + 4

b. 2a – 4

c. 4 – 2a

d. a + 4

Answer: (c)

By L-H rule

L = lim_{x→a} [f (a) – a f’ (x)] / [1]

L = 4 – 2a

**Question 4: Let A (1, 4) and B (1, – 5) be two points. Let P be a point on the circle (x – 1) ^{2} + (y – 1)^{2} = 1 such that (PA)^{2} + (PB)^{2} have maximum value, then the points P, A and B lie on:**

a. a parabola

b. a straight line

c. a hyperbola

d. an ellipse

Answer: (b)

PA^{2} = cos^{2} θ + (sin θ – 3)^{2} = 10 – 6 sin θ

PB^{2} = cos^{2} θ + (sin θ + 6)^{2} = 37 + 12 sin θ

PA^{2} + PB^{2}|_{max} = 47 + 6 sin θ|_{max}

θ = π / 2

The points P, A and B lie on the line x = 1.

**Question 5: If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x ^{2 }+ y^{2 }= 1 is a circle of the radius r, then r is equal to :**

a. 14

b. 12

c. 1

d. 13

Answer: (b)

P ≡ (2h – 3), (2k – 2) → on circle

[h – (3 / 2)]^{2}+ (k – 1)

^{2}= 1 / 4

Radius = 1 / 2

**Question 6: Let the slope of the tangent line to a curve at any point P (x, y) be given by [xy ^{2} + y] / x. If the curve intersects the line x + 2y = 4 at x = – 2, then the value of y, for which the point (3, y) lies on the curve, is:**

a. – 18 / 11

b. – 18 / 19

c. – 4 / 3

d. 18 / 35

Answer: (b)

dy / dx = [xy^{2} + y] / x

^{2}= x dx

– d (x / y) = d (x^{2} / 2)

(- x / y) = (x^{2} / 2) + c

The curve intersects the line x + 2y = 4 at x = – 2.

So, – 2 + 2y = 4

y = 3

The curve passes through (-2, 3).

(2 / 3) = 2 + c

c = – 4 / 3

The curve is (- x / y) = (x^{2} / 2) – (4 / 3).

It also passes through (3, y).

(- 3 / y) = (9 / 2) – (4 / 3)

y = – 18 / 19

**Question 7: Let A _{1} be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A_{2} be the area of the region bounded by the curves y = sinx, y = cosx, the x-axis and x = π / 2 in the first quadrant. Then,**

a. A_{1} = A_{2} and A_{1} + A_{2} = √2

b. A_{1} : A_{2} = 1 : 2 and A_{1} + A_{2} = 1

c. 2A_{1} = A_{2} and A_{1} + A_{2} = 1 + √2

d. A_{1} : A_{2} = 1 : √2 and A_{1} + A_{2} = 1

Answer: (d)

A_{1} + A_{2} = ∫_{0}^{π/2} cosx dx = sin x|_{0}^{π/2} = 1

A_{1 }= ∫_{0}^{π/4} (cosx – sinx) dx = (sinx + cosx)_{0}^{π/4} = √2 – 1

A_{2} = 1 – (√2 – 1) = 2 – √2

A_{1} / A_{2} = (√2 – 1) / (√2 (√2 – 1)) = 1 / √2

**Question 8: If 0 < a, b < 1, and tan ^{-1} a + tan^{-1} b = π / 4, then the value of (a + b) – [a^{2} + b^{2}] / 2 + [a^{3} + b^{3} ] / 3 – [a^{4} + b^{4}] / 4 + …. is**

a. log_{e}2

b. log_{e} (e / 2)

c. e

d. e^{2} – 1

Answer: (a)

tan^{-1} {[a + b] / [1 – ab]} = (π / 4)

a + b = 1 – ab

(1 + a) (1 + b) = 2

(a + b) – [a^{2} + b^{2}] / 2 + [a^{3} + b^{3}] / 3 +…infinity

= [a – (a^{2} / 2) + (a^{3} / 3) …..] + [b – (b^{2} / 2) + (b^{3} / 3) …..]

= log_{e} (1 + a) + log_{e} (1 + b)

= log_{e} (1 + a) (1 + b) = log_{e} 2

**Question 9: Let F _{1} (A, B, C) = (A ∧ ~B) ∨ [~C ∧ (A ∨ B)] ∨ ~A and F_{2} (A, B) = (A ∨ B) ∨ (B → ~A) be two logical expressions. Then:**

a. F_{1} is not a tautology but F_{2} is a tautology

b. F_{1} is a tautology but F_{2} is not a tautology

c. F_{1} and F_{2} both area tautologies

d. Both F_{1} and F_{2} are not tautologies

Answer: (a)

F_{1 }(A, B, C) = (A ∧ ~B) ∨ [~C ∧ (A ∨ B)] ∨ ~A

Using the set theory

(A ∩ B′) ∪ (C′ ∩ (A ∪ B)) ∪ A′ = (A − A ∩ B) ∪ (S − A) ∪ [(S − C) ∩ (A ∪ B)]

= (S − A ∩ B) ∪ [A ∪ B − C ∩ (A ∪ B)]

= S − A ∩ B ∩ C.

Hence not a tautology.

Now F_{2 }(A, B) = (A ∨ B) ∨ (B → ~A) = (A ∨ B) ∨ (~B ∨ A)

Using set theory (A ∪ B) ∪ (B′ ∪ A) = (A ∪ B) ∪ (S − A ∪ B) = S

Hence it is a tautology.

**Question 10: Consider the following system of equations:**

**x + 2y – 3z = a**

**2x + 6y – 11 z = b**

**x – 2y + 7z = c,**

**Where a, b and c are real constants. Then the system of equations:**

a. has a unique solution when 5a = 2b + c

b. has an infinite number of solutions when 5a = 2b + c

c. has no solution for all a, b and c

d. has a unique solution for all a, b and c

Answer: (b)

= 20a – 2(7b + 11c) – 3(- 2b – 6c)

= 20a – 14b – 22c + 6b + 18c

= 20a – 8b – 4c

= 4(5a – 2b – c)

= 7b+ 11c – a(25) – 3(2c – b)

= 7b+ 11c – 25a – 6c + 3b

= – 25a + 10b+ 5c

= – 5(5a – 2b- c)

= 6c + 2b- 2(2c – b) – 10a

= – 10a + 4b+ 2c

= – 2(5a – 2b- c)

For infinite solution,

Δ = Δ_{1} = Δ_{2} = Δ_{3} = 0

⇒ 5a = 2b + c

**Question 11: A seven-digit number is formed using the digit 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is:**

a. 6 / 7

b. 4 / 7

c. 3 / 7

d. 1 / 7

Answer: (3)

n (S) = 7! / [2! 3! 2!]

n (E) = 6! / [2! 2! 2!]

P (E) = n (E) / n (S) = [6! / 7!] * [[2! 3! 2!] / [2! 2! 2!]] = 3 / 7

**Question 12: **If vectors a_{1} = xi – j + k and a_{2} = i + yj + zk are collinear, then a possible unit vector parallel to the vector xi + yj + zk is :

a. (1 / √2) (- j + k)

b. (1 / √2) (i – j)

c. (1 / √3) (i – j + k)

d. (1 / √3) (i + j – k)

Answer: (3)

(x / 1) = – 1 / y = 1 / z = λ (let)

Unit vector parallel to xi + yj + zk = ± [λi – (1 / λ) j + (1 / λ) k] / √λ^{2} + (2 / λ^{2})

For λ = 1, it is ± [i – j + k] / √3

**Question 13: For x > 0, if f (x) = ∫ _{1}^{x} log_{e} t / [1 + t] dt, then f (e) + f (1 / e) is equal to:**

a. 1 / 2

b. -1

c. 1

d. 0

Answer: (a)

**Question 14: Let f : R → R be defined as**

**If f (x) is continuous on R, then a + b equals:**

a. 3

b. -1

c. -3

d. 1

Answer: (b)

If f is continuous at x = – 1, then f (- 1^{–}) = f (- 1)

2 = |a – 1 + b|

|a + b – 1| = 2 — (i)

Similarly f (1^{–}) = f (1)

|a + b + 1| = 0

a + b = – 1

**Question 15: Let A = {1, 2, 3 ……. , 10} and f : A → A be defined as **

**\(\begin{array}{l}f(k) = \begin{Bmatrix} k+1 & if\: k\: is\: odd\\ k & if\: k\: is\: even \end{Bmatrix}\end{array} \)**

**. Then the number of possible functions g : A → A such that gof = f is:**

a. 10^{5}

b. ^{10}C_{5}

c. 5^{5}

d. 5!

Answer: (a)

g (f (x)) = f (x)

g (x) = x, when x is even.

5 elements in A can be mapped to any 10

So, 10^{5} × 1 = 10^{5}

**Question 16: A natural number has prime factorization given by n = 2 ^{x}3^{y}5^{z}, where y and z are such that y + z = 5 and y^{-1} + z^{-1} = 5 / 6, y > z. Then the number of odd divisors of n, including 1, is: **

a. 11

b. 6x

c. 12

d. 6

Answer: (c)

y + z = 5 . . . (1)

(1 / y) + (1 / z) = 5 / 6

yz = 6

Also, (y – z)^{2} = (y + z)^{2} – 4yz

y – z = ± 1 — (2)

From (1) and (2), y = 3 or 2 and z = 2 or 3

For calculating the odd divisor of p = 2^{x}3^{y}5^{z}, x must be 0.

P = 2^{0}3^{3}5^{2}

Total odd divisors must be (3 + 1) (2 + 1) = 12.

**Question 17: Let f (x) = sin ^{-1} x and g (x) = [x^{2} – x – 2] / [2x^{2} – x – 6]. If g (2) = lim_{x→2 }g (x), then then the domain of the function fog is:**

a. (- ∞, – 2] ? [- 4 / 3, ∞]

b. (- ∞, – 1] ? [2, ∞)

c. (- ∞, – 2] ? [- 1, ∞]

d. (- ∞, 2] ? [- 3 / 2, ∞)

Answer: (a)

g (2) = lim_{x→2} [(x – 2) (x + 1)] / [(2x + 3) (x – 2)] = 3 / 7

For domain of f (g (x))

|(x^{2} – x – 2) / (2x^{2} – x – 6)| ≤ 1 [because of f (x) is [- 1, 1])

(x + 1)^{2} ≤ (2x +3)^{2}

3x^{2} + 10x + 8 ≥ 0

(3x + 4) (x + 2) ≥ 0

x belongs to (- ∞, – 2] ? [- 4 / 3, ∞]

**Question 18: If the mirror image of the point (1, 3, 5) with respect to the plane 4x – 5y + 2z = 8 is (?, β, γ), then 5 (? + β + γ) equals:**

a. 47

b. 39

c. 43

d. 41

Answer: (a)

Image of (1, 3, 5) in the plane 4x – 5y + 2z = 8 is (?, β, γ).

[? – 1] / 4 = [β – 3] / – 5 = [γ – 5] / 2 = – 2 [4 (1) – 5 (3) + 2 (5) – 8] / [4^{2}+ 5

^{2}+ 2

^{2}] = 2 / 5

? = 1 + 4 (2 / 5) = 13 / 5

β = 3 – 5 (2 / 5) = 1 = 5 / 5

γ = 5 + 2 (2 / 5) = 29 / 5

5 (? + β + γ) = 5 [(13 / 5) + (5 / 5) + (29 / 5)] = 47

**Question 19: Let f (x) = ∫ _{0}^{x} e^{t} f (t) dt + e^{x} be a differentiable function for all x ∈ R. Then f (x) equals**

a. 2e^{ex-1} – 1

b. e^{(e^x-1)}

c. 2e^{e^x} – 1

d. e^{e^x} – 1

Answer: (a)

Given, f (x) = ∫_{0}^{x} e^{t} f (t) dt + e^{x} —- (1)

f’ (x) = e^{x} . f (x) + e^{x} [using Newton Leibnitz Theorem]
[f’ (x)] / [f (x) + 1] = e^{x}

ln (f (x) + 1) = e^{x} + c

Put x = 0, ln 2 = 1 + c

ln [f (x) + 1] = e^{x} + ln 2 – 1

f (x) + 1 = 2 e^{ex-1}

f (x) = 2 e^{ex-1} – 1

**Question 20: The triangle of the maximum area that can be inscribed in a given circle of radius ‘r’ is:**

a. A right-angle triangle having two of its sides of length 2r and r.

b. An equilateral triangle of height 2r / 3.

c. isosceles triangle with base equal to 2r.

d. An equilateral triangle having each of its side of length √3r.

Answer: (d)

The triangle of the maximum area that can be inscribed in a circle is an equilateral triangle.

Let ?ABC be inscribed in the circle,

Now, in ?OBD

OD = r cos 60^{o} = r / 2

Height = AD = 3r / 2

Again in ?ABD

Now sin 60^{o} = (3r / 2) / AB

AB = √3r

**Section – B**

**Question 1: The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is _______**

Answer: 1000

Since the required number has G.C.D with 18 as 3. It must be an odd multiple of ‘3’ but not a multiple of ‘9’.

(i) Now, a 4-digit number which is an odd multiple of ‘3’ are,

1005, 1011, 1017, ? , 9999 → 1499

(ii) The 4-digit number which is an odd multiple of 9 are,

1017, 1035, ? , 9999 → 499

Required numbers = 1499 − 499 = 1000

**Question 2: Let ? and β be two real numbers such that ? + β = 1 and ? β = – 1. Let P _{n} = (?)^{n} + (β)^{n}, P_{n-1} = 11 and P_{n+1} = 29 for some integer n= 1. Then, the value of P_{n}^{2} is______________.**

Answer: 324

Given, ? + β = 1 and ? β = – 1

A quadratic equation with roots ?, β is x^{2} – x – 1 = 0.

?^{2} = ? + 1

Multiplying both sides by ?^{n-1}

?^{n+1} = ?^{n} + ?^{n-1} — (1)

Similarly,

β^{n+1} = β^{n} + β^{n-1} — (2)

Adding (1) and (2)

?^{n+1} + β^{n+1} = (?)^{n} + (β)^{n} + [?^{n-1} + β^{n-1}]

P_{n+1} = P_{n} + P_{n-1}

29 = P_{n} + 11

P_{n} = 18

P_{n}^{2} = 18^{2} = 324

**Question 3: Let X _{1}, X_{2}, …. X_{18} be eighteen observation such that ∑_{i=1}^{18} (X_{i} – ?) = 36 and ∑_{i=1}^{18} (X_{i} – β)^{2} = 90, where ? and β are distinct real numbers. If the standard deviation of these observations is 1, then the value of |? – β| is ___________.**

Answer: 4

Given, ∑_{i=1}^{18} (X_{i} – ?) = 36

∑x_{i} – 18? = 36

∑x_{i} = 18 (? + 2) —- (1)

∑_{i=1}^{18} (X_{i} – β)^{2} = 90

∑x_{i}^{2} + 18β^{2} – 2β ∑x_{i} = 90

∑x_{i}^{2} + 18β^{2} – 2β * 18 (? + 2) = 90 (using equation (1))

∑x_{i}^{2} = 90 – 18β^{2} + 36β (? + 2)

Now, ??^{2} = 1 ⇒ (1 / 18) ∑x_{i}^{2} – (∑x_{i} / 18)^{2} = 1

(1 / 18) [90 – 18β^{2} + 36?β + 72β] – [18 (? + 2) / 18]^{2} = 1

5 − β^{2} + 2?β + 4β − (? + 2)^{2} = 1

⇒ 5 − β^{2} + 2?β + 4β − ?^{2} − 4 − 4? = 1

⇒ ?^{2} − β^{2} + 2?β + 4β − 4? = 0

⇒ (? − β)(? − β + 4) = 0

⇒ ? − β = −4

Hence |? − β| = 4 (? ≠ β).

**Question 4: In I _{m,n} = ∫_{0}^{1} x^{m-1} (1 – x)^{n-1} dx, for m, n ≥ 1 and ∫_{0}^{1} [x^{m-1} + x^{n-1}] / [(1 + x)^{m+n}] dx = ?I_{m,n}, ? belongs to R, then ? =**

Answer: 1

**Question 5: Let L be a common tangent line to the curves 4x ^{2} + 9y^{2} = 36 and (2x)^{2} + (2y)^{2} = 31. Then the square of the slope of the line L is ______________.**

Answer: 3

E : (x^{2} / 9) + (y^{2} / 4) = 1

C : x^{2} + y^{2} = 31 / 4

The equation of the tangent to the ellipse is y = mx ± √(9m^{2} + 4) — (1)

The equation of the tangent to the circle is y = mx ± √[(31 / 4) m^{2} + (31 / 4)] — (2)

Comparing equation (1) & (2), 9m^{2} + 4 = (31 / 4) m^{2}+ (31 / 4)

36m^{2} + 16 = 31m^{2} + 31

5m^{2} = 15

m^{2} = 3

**Question 6: If the matrix **

**\(\begin{array}{l}A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 3 & 0 & -1 \end{bmatrix}\end{array} \)**

**satisfies the equation A\(\begin{array}{l}\begin{bmatrix} 1 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 1 \end{bmatrix}\end{array} \) for some real numbers ? and β, then β -? is equal to_________.**

^{20}+ ?A^{19}+ βA =

Answer: 4

**Question 7: If the arithmetic mean and the geometric mean of the p ^{th} and q^{th} terms of the sequence – 16, 8, – 4, 2, … satisfy the equation 4x^{2} – 9x + 5 = 0, then p + q is equal to ______________.**

Answer: 10

Given, 4x^{2} – 9x + 5 = 0

(x – 1) (4x – 5) = 0

AM = 5 / 4, GM = 1 (AM ≥ GM)

Again for the series, – 16, 8, – 4, 2, …

p^{th }term t_{p} = – 16 (- 1 / 2)^{p-1}

q^{th }term t_{q} = – 16 (- 1 / 2)^{q-1}

AM = [t_{p} + t_{q}] / 2 = 5 / 4

GM = √(t_{p}t_{q}) = 1

16^{2} (- 1 / 2)^{p+q-2} = 1

(- 2)^{8} = (- 2)^{p+q-2}

p + q = 10

**Question 8: Let the normals at all the points on a given curve pass through a fixed point (a, b). If the curve passes through (3, – 3) and (4, – 2 √2), and given that a – 2√2b = 3, then (a ^{2} + b^{2} + ab) is equal to______________.**

Answer: 9

Let the equation of normal is Y – y = (- 1 / m) (X – x), where m = dy / dx

As it passes through (a, b)

b – y = (- 1 / m) (a – x) = – dx / dy [a – x]

(b – y) dy = (x – a) dx

by – (y^{2} / 2) = (x^{2} / 2) – ax + c — (i)

It passes through (3, – 3) & (4, – 2 √2)

-3b – (9 / 2) = (9 / 2) – 3a + c

3a – 3b – c = 9 — (ii)

Also – 2√2b – 4 = 8 – 4a + c

4a – 2 √2b – c = 12 — (iii)

Also a – 2 √2b = 3 — (iv) (given)

From (ii) – (iii), – a + (2√2 – 3) b = – 3 —- (v)

From (iv) + (v), b = 0, a = 3

a^{2} + b^{2} + ab = 9

**Question 9: Let z be those complex number which satisfies |z + 5| ≤ 4 and z (1 + i) + **

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

**(1 – i) ≥ – 10, i = √-1. If the maximum value of |z + 1|**

^{2}is ? + β√2, then the value of ? + β is ________.

Answer: 48

Given, |z + 5| ≤ 4

(x + 5)^{2} + y^{2} ≤ 16

Also, z (1 + i) +

x – y ≥ – 5 —- (2)

From (1) and (2), the locus of z is the shaded region in the diagram.

|z + 1| represents the distance of ‘z’ from Q (- 1, 0).

Clearly, p is the required position of z when |z + 1| is maximum.

P ≡ (- 5 – 2 √2, – 2 √2)

PQ^{2} = 32 + 16 √2

? = 32

β = 16

? + β = 48

**Question 10: Let a be an integer such that all the real roots of the polynomial 2x ^{5} + 5x^{4} + 10x^{3} + 10x^{2} + 10x + 10 lie in the interval (a, a + 1). Then, |a| is equal to _______.**

Answer: 2

Let f (x) = 2x^{5} + 5x^{4} + 10x^{3} + 10x^{2} + 10x + 10

f’ (x) = 10 (x^{4} + 2x^{3} + 3x^{2} + 2x + 1)

= 10x^{2} [x^{2} + (1 / x^{2}) + 2 [x + (1 / x)] + 3]

= 10x^{2} [(x + (1/x))^{2} + 2 (x + (1 / x)) + 1]

= 10x^{2} [[x + (1 / x)] + 1]^{2} > 0, for all x belongs to R

f (x) is strictly increasing function. Since it is an odd degree polynomial it will have exactly one real root.

By observation, f (- 1) = 3 > 0

f (- 2) = – 64 + 80 – 80 + 40 – 20 + 10

= – 34 < 0

f (x) has at least one root in (- 2, – 1) = (a, a + 1)

|a| = 2

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