**1.**The value of (2.

^{1}P

_{0}-3.

^{2}P

_{1}+4.

^{3}P

_{2}-….. up to 51

^{th}term) +(1!-2!+3!-…… up to 51

^{th}term) is equal to:

a) 1-51(51)!

b) 1+(52)!

c) 1

d) 1+ (51)!

**Answer: b**

**2.**Let P be a point on the parabola, y

^{2 }= 12x and N be the foot of the perpendicular drawn from P on the axis of the parabola. A line is now drawn through the mid-point M of PN, parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is 4/3 , then:

a) PN = 4

b) MQ = 1/3

c) PN = 3

d) MQ = 1/4

Q (h, 3t) lie on parabola 9t^{2} = 12h

h = 3t^{2}/4

Q = (3t^{2}/4, 3t)

Equation of NQ =

y = 3t(x-3t^{2})/((3t^{2}/4)-3t^{2})

y = -4(x-3t^{2})/3t

put x = 0

y = 4t

4t = 4/3

⇒ t = 1/3

PN = 6t

= 6×(1/3)

= 2

M = (1/3, 1), Q = (1/12,1)

MQ = (1/3)-(1/12)

MQ = 1/4

**Answer: d**

**3.**If Δ =

^{3}+Bx

^{2}+Cx+D, then B+C is equal to:

a) 1

b) -1

c) -3

d) 9

^{3}+Bx

^{2}+Cx+D

R_{2} ? R_{2} -R_{1}

R_{3} ? R_{3} -R_{2}

⇒

^{3}+Bx

^{2}+Cx+D

⇒

^{3}+Bx

^{2}+Cx+D

⇒ (x-1)(x-2){(x-2)(6- 2)-(2x-3)(6-1)+(3x- 4)(2-1)} = Ax^{3}+Bx^{2}+Cx+D

⇒ (x-1)(x -2){4x-8 -10x+15+3x-4} = Ax^{3}+Bx^{2}+Cx+D

⇒ (x-1)(x-2)(-3x +3) = Ax^{3}+Bx^{2}+Cx+D

-3(x-1)^{2} (x-2) = Ax^{3}+Bx^{2}+Cx+D

-3x^{3} +12x^{2}-15x+6 = Ax^{3}+Bx^{2}+Cx+D

Comparing both side

A = -3, B = 12, C = -15

B+C = 12-15 = -3

**Answer: c**

**4.**The foot of the perpendicular drawn from the point (4,2,3) to the line joining the points (1,2,3) and (1,1,0) lies on the plane:

a) x-y-2z = 1

b) x-2y+z = 1

c) 2x+y-z = 1

d) x+2y-z = 1

(-3, 3?-4, -3?).(0,3, -3) = 0

⇒ 0+9?-12+9? = 0

⇒ ?= 12/18 = 2/3

m = (1,0,1) are on 2x+y-z = 1

**Answer: c**

**5.**If y

^{2}+log

_{e}(cos

^{2}x) = y, x?(-π/2, π/2), then

a) |y’(0)|+|y’’(0)| = 1

b) |y’’(0)| = 0

c) |y’(0)|+|y’’(0)| = 3

d) |y’’(0)| = 2

yy’ + 2 (-tanx) = y’ ….(1)

diff. w.r.t.x

2y’’+2(y’)^{2 }-2 sec^{2} x = y” ….(2)

Put x = 0 in given equation we get y = 0, 1

from (1) x = 0, y = 0

⇒ y’(0) = 0

x = 0, y = 1

⇒ y’(0) = 0

from (2) x = 0, y = 0, y’(0) = 0 ⇒ y”(0) = -2

x = 0, y = 1, y’(0) = 0 ⇒ y”(0) = 2

|y”(0)| = 2

**Answer: d**

**6.**2π-(sin

^{-1}(4/5)+sin

^{-1}(5/13)+sin

^{-1}(16/65)) is equal to:

a) 5π/4

b) 3π/2

c) 7π/4

d) π/2

2π-(sin^{-1}(4/5)+sin^{-1}(5/13)+sin^{-1}(16/65)) = 2π-(tan^{-1}(4/3)+tan^{-1}(5/12)+tan^{-1}(16/63))

= 2π-tan^{-1 }[(4/3)+(5/12)/(1-(4/3)(5/12) ]- tan^{-1}(16/63)

= 2π-tan^{-1}(48+15)/(36-20) – tan^{-1}(16/63)

= 2π-tan^{-1}(63/16) – cot^{-1}(63/16)

= 2π-π/2

= 3π/2

**Answer: b**

**7.**A hyperbola having the transverse axis of length v2 has the same foci as that of the ellipse 3x

^{2}+4y

^{2 }= 12, then this hyperbola does not pass through which of the following points ?

a) (v(3/2), 1/v2)

b) (1, -1/v2)

c) (1/v2, 0)

d) (-v(3/2),1)

(x^{2}/4)+(y^{2}/3) = 1

b_{1}^{2} = a_{1}^{2}(1-e_{1}^{2})

3 = 4(1-e_{1}^{2})

e_{1 }= 1/2

focus = (± a_{1}e_{1}, 0)

= (±1, 0)

Length of the transverse axis 2a_{2} = v2

a_{2 } = 1/v2

a_{2}e_{2} = 1

So e_{2} = v2

b_{2}^{2} = a_{2}^{2}(e_{2}^{2}-1)

b_{2}^{2} = (1/2)(2-1)

b_{2}^{2} = 1/2

Equation of hyperbola is x^{2}-y^{2} = 1/2.

(v(3/2), 1/v2) does not lie on it.

**Answer: a**

**8.** For the frequency distribution

Variate (X): x_{1} x_{2} x_{3}…x_{15}

Frequency (f): f_{1 }f_{2} f_{3}…f_{15}

where 0<x_{1}<x_{2}< x_{3}<… x_{15} = 10 and

a) 1

b) 4

c) 6

d) 2

s^{2 }= (1/4)(M-m)^{2}

M = upper bound of value of any random variable

m = Lower bound of value of any random variable

s^{2 }= (1/4)(10-0)^{2}

s^{2 }= 25

-5< s < 5

s ≠ 6

**Answer: c**

**9.**A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is:

a) 1/3

b) 1/4

c) 1/8

d) 1/9

Total Possibilities = (1, 3), (3, 1), (2, 2), (2, 6), (6, 2) (4, 4), (3, 5), (5, 3) (6, 6)

(A n B) = (4,4)

n(A n B) = 1

Required probability = P(B/A)

= P(A n B)/P(A)

= 1/9

**Answer: d**

**10.**If the number of integral terms in the expansion of (3

^{1/2 }+5

^{1/8})

^{n}is exactly 33, then the least value of n is:

a) 128

b) 248

c) 256

d) 264

T_{r+1} = ^{n}C_{r}(3^{1/2})^{n-r}(5^{1/8})^{r}

(n-r)/2 -> n-r = 0, 2, 4, 6, 8,…

r/8 -> r = 0, 8, 16, 24….

common ratio = 0, 8, 16, 24…….

no. of integral term = 33.

L = 0+(33-1)×8

L = 32×8

= 256

**Answer: c**

**11.**

a) π

^{2}

b) π

^{2}/2

c) v2π

^{2}

d) 2π

^{2}

= 2[πx-x^{2}/2]_{0}^{π}

= 2[π^{2}/2]

= π^{2}

**Answer: a**

**12.**Consider the two sets:

A = {m ∈ R : both the roots of x^{2}-(m+1)x+m+4 = 0 are real} and B = [-3,5).

Which of the following is not true ?

a) A-B = (-∞,-3) ? (5,∞)

b) A n B = {-3}

c) B-A = (-3,5)

d) A U B = R

D = 0

(m+1)^{2} -4(m+4) = 0

m^{2}-2m-15 = 0

(m-5)(m+3) = 0

m ∈(-∞, -3] U [5, ∞)

A = (-∞, -3] U [5, ∞)

B = [-3, 5]

A-B = (-∞, -3] U [5, ∞)

A U B = R

Only first option is not true.

**Answer: a**

**13.**The proposition p -> ∼(p^q) is equivalent to:

a) (∼p) ?(∼ q)

b) (∼ p) ^q

c) q

d) (∼ p) ?q

∼(p^ ∼q) ? ∼ p?q

p ? (p~?q)

⇒ ~ p?(~p?q)

⇒ ~ p?q

**Answer: d**

**14.**The function, f(x) = (3x-7)x

^{2/3}, x ∈ R is increasing for all x lying in:

a) (-∞, -14/15) U (0,∞)

b) (-∞, 14/15)

c) (-∞,0)U(14/15, ∞)

d) (-∞,0)U(3/7, ∞)

f’(x) = (3x-7)(2/3x^{1/3})+(x^{2/3}×3)

= (6x-14+9x)/3x^{1/3}

= (15x-14)/3x^{1/3}

f(x) >0 ⇒ x ∈ (-∞,0)U(14/15, ∞)

**Answer: c**

**15.**If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of

**its next 15 terms, then the common difference of this A.P. is:**

a) 1/6

b) 1/5

c) 1/4

d) 1/7

a = 3

(25/2)(2a+24d) = (15/2)(2(a+25d)+14d)

50a+600d = 15[2a+50d+14d]

20a+600d = 960d

60 = 360d

d = 1/6

**Answer: a**

**16.**The solution curve of the differential equation, (1+e

^{-x})(1+y

^{2})dy/dx = y

^{2}, which passes

**through the point (0,1), is:**

a) y

^{2}= 1+y log

_{e}(1+e

^{-x})/2

b) y

^{2}+1 = y ((log

_{e}(1+e

^{-x})/2 )+2)

c) y

^{2}+1 = y ((log

_{e}(1+e

^{x})/2 )+2)

d) y

^{2}= 1+y log

_{e}(1+e

^{x})/2

⇒ (-1/y)+y = lne^{x}+1 + C

x = 0, y = 1

⇒ -1+1 = ln 2+C

⇒ C = -ln 2

⇒ (-1/y)+y = lne^{x}+1 – ln 2

⇒ y^{2} = 1+y[ln(e^{x}+1)/2]

**Answer: d**

**17.**The area (in sq. units) of the region {(x,y):0 =y =x

^{2}+1, 0 =y =x+1, 1/2 =x=2} is

a) 23/16

b) 79/16

c) 23/6

d) 79/24

A=

=

= (1/3)+1-[(1/24)+(1/2)]+[2+2-(3/2)]

= (4/3)-(13/24)+(5/2)

= (19+60)/24

= 79/24

**Answer: d**

**18.**If α and β are the roots of the equation x

^{2}+px+2 = 0 and (1/α) and (1/β) are the roots of the equation 2x

^{2}+2qx+1 = 0, then (α-1/α)(β-1/β)(α+1/α)(β+1/β) is equal to :

a) (9/4)(9+p

^{2})

b) (9/4)(9+q

^{2})

c) (9/4)(9-p

^{2})

d) (9/4)(9-q

^{2})

α+β = -p

αβ = 2

(1/α)+(1/β) = -q

1/αβ = 1/2

(α+1/β)(β+1/α) = αβ+(1/αβ)+2

= 2+(1/2)+2

= 9/2

(α-1/α)(β-1/β) = αβ+(1/αβ)-(α/β)-(β/α)

= 2+(1/2)-(α^{2}+β^{2})/αβ

= (5/2)-[((α+β)^{2}-2αβ)/αβ)]

= (5/2)-[(p^{2}-4)/2]

= (9-p^{2})/2

(α-1/α)(β-1/β)(α+1/α)(β+1/β) = ((9-p^{2})/2)(9/2)

= (9/4)(9-p^{2})

**Answer: c**

**19.**The lines

a) do not intersect for any values of l and m

b) intersect when l = 1 and m = 2

c) intersect when l = 2 and m = 1/2

d) intersect for all values of l and m

Let two lines

and

= 1/v(-1)^{2}+(-1)^{2}+2^{2}

= 1/v6

Since shortest distance between these lines exist

Hence lines do not intersect.

**Answer: a**

**20.**Let [t] denote the greatest integer =t. If for some ? ∈ R -{0, 1},

Then L is equal to:

a) 0

b) 2

c) 1/2

d) 1

RHL:

= 1/(-1)

Since |?| = |?-1|, [-h] = -1

?^{2} = ?^{2}-2?+1

⇒ ?= 1/2

L = 2

**Answer: b**

**21.**If

⇒

⇒ 1/256 = 2^{-k}

⇒ 2^{-8} = 2^{-k}

⇒ k = 8

**Answer: 8**

**22.**The diameter of the circle, whose centre lies on the line x+y = 2 in the first quadrant and which touches both the lines x = 3 and y = 2, is ……

p = r

for y = 2

r = |(2-α-2)/1| = |α|

for x = 3

r = |(α-3)/1| = |α-3|

| α | = |α – 3 |

⇒ α^{2} = α^{2}-6α+9

⇒ α = 3/2

2α = 3 = 2r

**Answer: 3**

**23.**The value of

(1/3)+(1/3^{2})+(1/3^{3})+….∞ = 1/3(1-1/3) = 1/2

log_{2.5}(1/2) ⇒ log_{5/2} (1/2)

0.16 = 16/100 = 4/25 = (2/5)^{2}

⇒

⇒

= 4

**Answer: 4**

**24.**Let A =

^{4}= [a

_{ij}]. If a

_{11}= 109, then a

_{22}is equal to:

A =

A^{2} =

A^{3} =

=

A^{4} =

=

a_{11 }⇒ x^{4}+3x^{2}+1 = 109

x^{4 }+3x^{2}-108 = 0

⇒ (x^{2}+12)(x^{2}-9) = 0

x = ±3

a_{22 }= x^{2}+1 = 10

**Answer: 10**

**25.**If ((1+i)/(1-i))

^{m/2}= ((1+i)/(i-1))

^{n/3}= 1, (m, n∈N) then the greatest common divisor of the least values of m and n is :

[(1+i)(1+i)/(1+i)(1-i)]

^{m/2}= [(1+i)(-1-i)/(-1+i)(-1-i)]

^{n/3}= 1 (i = v-1)

Solving LHS of above equation, we get

(2i/2)^{m/2} = 1

⇒ m = 8

Solving RHS,

[(-1-i-i+1)/(1+1)]^{n/3}= 1

(-2i/2)^{n/3} = 1

(-i)^{n/3} = 1

⇒ n = 12

Greatest common divisor of m and n is 4.

**Answer: 4**

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