Question:1 If  f(x)+2f(1/x)=3x, x!=0  and

S= {x inR: f(x)= f(-x)};then S;

(1) is an empty set

(2) contains exactly one element

(3) contains exactly two element

(2) contains more than two element

Solution:

f(x)+ 2f(1/x)=3x....(1)

Put x=1/x in equation (1)

=> f(1/x)+ 2f(x) = 3/x ...(2)

From equation (1) and (2), we get:

=> f(x) = -x+2/x

Given that f(x)= f(-x)

=> -x+2/x = x-2/x

=> 2x^2 =4

=> x = +- sqrt2

Option (3) is correct.

* Question2:* A value of theta for which (2+3isintheta)/(1-2isintheta) is purely imaginary, is:

(1) pi/3

(2) pi/6

(3) sin^-1 (sqrt3/4)

(4) sin^-1 (1/sqrt3)

Solution:

z= (2+3isintheta)/(1-2isintheta)

Multiple and divide by (1+2isintheta) ;

=> z = (2+3isintheta)/(1-2isinthea) xx (1+2isintheta)/(1+2isintheta)

=>z = ((2+3isintheta)(1+2isintheta))/(1+4sin^2theta)

For pure imaginary Re(z)=0

=> 2- 6sin^2 theta =0

=> sintheta = +- 1/sqrt3

=> theta = sin^-1 (+- 1/sqrt3)

Option (4) is correct.

Question:3 The sum of all real values of x satisfying the equation

(x^2-5x+5)^(x^2+4x-60)=1 is:

(1) 3

(2) -4

(3) 6

(4) 5

Solution: (x^2-5x+5)^(x^2+4x-60)=1. This is true, when:

Case1:x^+4x-60 =0

=> (x+10)(x-6)=0

=> x = -10 0r 6 .....(1)

Case2: x^2-5x+5 =1

=> x^2-5x+4=0 => (x-1)(x-4)=0

=> x = 4 or 1...(2)

Case3: x^2-5x+5 = -1 and x^2+4x-60 is even number

=> x^-5x+6=0 => (x-2)(x-3)=0

=> x= 2 or 3

But for x =3, x^2+4x-60 is not even. Hence x=3 is not an answer.

=> x=2....(3)

Hence sum of all real values of X from the equations (1), (2) and (3). we get

=> -10+6+1+4+2 =3

Option (1) is correct.

Question:4 if A = [(5a, -b),(3,2)] and AadjA =AA^T, then 5a+b is equal to:

(1) -1

(2) 5

(3) 4

(4) 13

Solution:

A = [(5a, -b),(3,2)]

adjA = [(2, b),(-3,5a)] and A^T = [(5a, 3),(-b,2)]

According to question:

AadjA =AA^T

=>[(5a, -b),(3,2)] [(2, b),(-3,5a)] =[(5a, -b),(3,2)] [(5a, 3),(-b,2)]

=> [(10a+3b, 0), (0, 3b+10a)]= [(25a^2+b^2, 15a-2b ), (15a-2b, 13)]

 15a -2b =0

=> 15a = 2b => b = (15a)/2...(1)

and 10a +3b =13

From equation (1), we get:

=> 10a + 3((15a)/2)= 13

=> a = 2/5

value of b is : b = (15a)/2 = 15/ 2 xx 2/5 = 3

Therefore;

5a+b = 5 xx(2/5)+ 3 = 5

Option (2) is correct.

Question:5 The system of linear equations

x+lambda y - z =0

lambda x-y-z=0

x+y-lambdaz=0

has a non-trivial soltion for:

(1) infinitely many values of lambda

(2) exactly one value of lambda

(3) exactly two values of lambda

(4) exactly threevalues of lambda

Solution:

According question :

|(1,lambda, -1),(lambda, -1, -1),(1,1,lambda)|=0

=> (lambda+1)- lambda(-lambda^2+1)-(lambda+1)=0

=>- lambda(-lambda^2+1)=0 

lambda =0 or -lambda^2+1 =0 =>lambda^2= 1 => lambda= +-1

lambda = 0, +-1

Option (4) is correct.

Question:6 If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is:

(1) 46^(th)

(2) 59^(th)

(3) 52^(nd)

(4) 58^(th)

Solution:

A, L, L , M, S

Probability for A= (4!)/(2!)= 12 words

Probability for L= 4!= 24 words

Probability for M= (4!)/(2!)= 12 words

Probability for S A= (3!)/(2!)= 3 words

Probability for S L= (3!)/(2!)= 3 words

Total 57 words before SMALL => Rank is 58th

Option (1) is correct.

* Question:7* If the number of terms in the exapnsion of (1-2/x+4/x^2)^n,x!=0, is 28, then the sum of the coefficients of all the termsin this expansion, is:

(1) 64

(2) 2187

(3) 243

(4) 729

Solution: (1-2/x+4/x^2)^n= ((x^2-2x+4)/x^2)^n.....(1)

Number of terms = ((n+2)(n+1))/2 = 28

=> (n+2)(n+1) = 56

=> n^2+3n+2=56

=> n^2+3n- 54 =0 

=> (n+9)(n-6)=0 => n = -9, 6

n can't be negative. therefore n=6

Hence the sum of the coefficients is obtained by puttin x=1in equation (1)

=> 3^6 = 729

Option (4) is correct

Question:8 if the 2^(nd),5^(th) and 9^(th)terms of a non-constant A.P. are in G.P, then the common ratio of this G.P is:

(1) 8/5

(2)4/3

(3) 1

(4) 7/4

Solution:

Suppose first term a, and common difference d.

So, T_2 = a+d, T_5 = a+4d and T_9 = a+ 8d

According to question, T_2, T_5, T_9:

(a+4d)^2 = (a+d)(a+8d)

=> a^2+16d^2 +8ad = a^2+9ad+8d^2

=> ad-8d^2 =0

=>d(a-8d)=0

=> d!=0 and a =8d

T_2, T_5, T_9

=> a+d, a+4d, a+8d

=> 8d+d, 8d+4d, 8d+8d

=> 9d, 12d, 16d

Hence the common ratio r= (12d)/(9d)= 4/3

Option (2) is correct.

Question:9 if the sum of the first ten terms of the series

(1 3/5)^2+(2 2/5)^2+(3 1/5)^2+4^2+(4 4/5)^2+.........

is 16/5 m, then m is equal to:

(1) 102

(2) 101

(3) 100

(4) 99

Solution:

(1 3/5)^2+(2 2/5)^2+(3 1/5)^2+4^2+(4 4/5)^2+.........

=> (8/5)^2 +(12/5)^2+ (16/5)^2+ 16^2 + (24/5)^2+...

n^(th) is T_n = 16/25 (n+1)^2

Sum of n terms: S_n = Sum_n (16/25 (n+1)^2)= Sum_n (16/25(n^2+2n+1))

=> S_n = 16/25 [(n(n+1)(2n+1))/6 + n(n+1)+n]

Given that S_10 = 16/5 m

So,  S_10 = 16/25 [385+120]

=> 16/5 m = 16/25 [385+120]

=> m =101

Option (2) is correct.

Question:10 Let p = lim_(x->O^+) (1+tan^2sqrt(x))^(1/(2x)),then logp is equal to:

(1) 2

(2) 1

(3) 1/2

(4) 1/4

Solution:

p = lim_(x->O^+) (1+tan^2sqrt(x))^(1/(2x))

Taking log both the sides,

=> logp = lim_(x->O^+) (1+tan^2sqrt(x))/(2x)=lim_(x->O^+) (sec^2sqrt(x))/(2x) 

=> logp =lim_(x->O^+) (2log secsqrtx)/(2x)

Apply L Hospital rule;

=> log p = lim_(x->O^+) [1/(secsqrtx) xx(secsqrtx tansqrtx)xx1/(2sqrt2)]

=> logp = lim_(x->O^+) [(tansqrtx)/(2sqrtx)]= 1/2 lim_(x->O^+) [(tansqrtx)/(sqrtx)] 

logp = 1/2

Option (3) is correct.

Question:11 For  x in R, f(x)= |log2-sinx| and g(x)= f(f(x)), then:

(1) g is not differentiable at x=0

(2) g'(0)= cos(log2)

(3) g'(0)= -cos(log2)

(4) g is differentiable at x=0 and g'(0)= -sin(log2)

Solution:

f(x) = log2 -sinx, in the neighbourhood of zero.

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

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

=>g'(0)= f'(f(0)) f'(0)=f'(log2)xx f'(0)

=> g'(0) = -cos(log2) xx(-cos0)= cos(log2)

Option (2) is correct.

Question:12 f(x)= tan^-1 (sqrt((1+sinx)/(1-sinx))), x in (0, pi/2)

A normal to y=f(x) at x= pi/6 also passes through the point:

(1) (0,0)

(2) (0,(2pi)/3)

(3) (pi/6 , 0)

(4) (pi/4, 0)

Solution:

f(x)= tan^-1 (sqrt((1+sinx)/(1-sinx)))

1+sinx = sin^2 (x/2)+ cos^2 (x/2)+ 2sin(x/2)cos(x/2) = (sin(x/2)+cos(x/2))^2

=> f(x)= tan^-1 |(sin(x/2)+cos(x/2))/(sin(x/2)-cos(x/2))|

In the neighbourhood of x = pi/2

=> y = tan^-1((sin(x/2)+cos(x/2))/(sin(x/2)-cos(x/2)))

=> y= tan^-1 ((1+tan(x/2))/(1-tan(x/2)))= tan^-1 (tan(pi/4+x/2))

=> y = x/2+ pi/4 for x in (0, pi/2)

Slope of the line is  m = 1/2 Therefore slope of normal, m_1 = -1

At x = pi/6

=> y = pi/12 + pi/4 = pi/3

Equation of the normal is:

(y - pi/3) = -2 (x - pi/6)

Option (3) satisfy the normal equation.

Question:13 A wire of length 2 units is cut into two parts which are bent respectively to form a square of side =x units and a circle of radius =r units. If the sum of areas of the square and the circle so formed is minimum, then:

(1)  2x = (pi+4)r

(2) (4-pi)x = pir

(3) x = 2r

(4) 2x =r 

Solution: Length of wire is 2 units.

Perimeter of circle+ perimeter of square = 2

=> 2pir+4x =2

Differentiate with respect to x

=> 2pi (dr)/(dx) +4 =0=> (dr)/(dx) = -4/ (2pi).....(1) 

Area of the circle = pir^2 and area of square =x^2

A= x^2 +pir^2

For minimum sum of area (dA)/(dx)=0

(dA)/(dx)= 2x + 2 pir (dr)/(dx)

From equation (1), we get:

(dA)/(dx)= 2x + 2 pi r (-4/(2pi))=0

=> 2x -4r =0

=> x =2r

Option (2) is correct.

Question:14 The integral int (2x^12+5x^9)/(x^5+x^3+1)^3 dx is equal to:

(1) (-x^5)/(x^5+x^3+1)^2+C

(2) x^10/(2(x^5+x^3+1)^2)+C

(3) x^5/(2(x^5+x^3+1)^2)+C

(4) (-x^10)/(2(x^5+x^3+1)^2)+C

Solution:

int (2x^12+5x^9)/(x^5+x^3+1)^3 dx

Take common x^15 from top and x^5 from bottom;

=>int (2/x^3+5/x^6)/(1+1/x^2+1/x^5)^3 dx

Let t = (1+1/x^2+1/x^5)

dt = -2/x^3-5/x^6 dx

=> int (-dt)/t^3 = 1/(2t^2)

Put the value of t, we get

I = x^10/(2(x^5+x^3+1))+c

Option (2) is correct.

Question:15 lim_(n->oo) (((n+1)(n+2)......3n)/n^(2n))^(1/n) is equal to:

(1) 18/e^4

(2) 27/e^2

(3) 9 /e^2

(4) 3log3-2

Solution:

y = lim_(n->oo) (((n+1)(n+2)......3n)/n^(2n))^(1/n)

Take a ln both the side;

=> ln y = lim_(n->oo) [1/n(ln(1+1/n)+ln(1+2/n)+.....+ln(1+2n/n))]

=> ln y = lim_(n->oo) 1/n sum_i ln(1+i/n)

=> ln y = int_0^2 ln(1+x) dx

=> ln y = xln(1+x)|_0^2- [x+ ln(1+x)]|_0^2

=> ln y = 2ln3-2+ln3

=> ln y = 3ln3 -2

=> lny = ln27 - lne^2

=> ln y = ln (27/e^2)

=> y = 27/e^2

Option (2) is correct.

Question:16 The area (in sq. units) of the region {(x,y):y^2 >= 2x and x^2+y^2 <= 4x, x>=0 , y>=0}is:

(1) pi - 4/3

(2) pi - 8/3

(3) pi- (4sqrt2)/3

(4) pi/2 - (2sqrt2)/3

Solution:

Area = Area of quarter circle - Area under parabola

A = pi - int_0^2 sqrt(2x) dx

=> A = pi - (2sqrt2)/3 x^(1/2)|_1^2

=> A = pi - 8/3

Option (2) is correct.

Question:17 If a curve y= f(x) passes through the point (1,-1) and satisfies the differential equation,y(1+xy)dx =x dy, then f(-1/2) is equal to:

(1) -2/5

(2) -4/5

(3) 2/5

(4) 4/5

Solution:

y(1+xy)dx = x dy

=> y dx + xy^2 dx = x dy

=> x dy - y dx = xy^2 dx 

=> (-xdy+ydx)/y^2 = -x dx 

=> (x/y) = -x^2 /2 +c 

It passes through the point (1,-1), therefore:

=> (1/(-1)) =-1/2+c

=> c = -1/2

x/y = - x^2/ 2 -1/2

y = (2x)/(-(x^2+1))

y = f(-1/2 )=(2 xx (-1/2))/(- (1/4+1)) = 4/5

Option (4) ic correct.

Question:18 Two sides of a rhombus are along the lines, x-y+1=0 and 7x-y-5=0. If its diagonals intersect at (-1,-2), then which one of the following is a vertex of this rhombus?

(1) (-3,-9)

(2) (-3,-8)

(3) (1/3, -8/3)

(4) (-10/3, - 7/3 )

Solution:

Equation of angle bisector of lines x-y+1=0 and 7x=y-5=0 is given by:

(x-y+1)/sqrt2 = +- (7x-y-5)/(5sqrt2)

=> 5(x-y+1)= 7x -y-5

and

5(x-y+1)= -7x +y+5

2x+4y-10 =0 =>x+2y-5=0

and

12x-6y=0 => 2x-y=0

Now equation of diagonals are :

(x+1)+2(y+2)=0=> x+2y+5 =0....(1)

and

2(x+1)-(y+2)=0 => 2x-y =0

Clearly, (1/3, -8/3 ) lies on (1)

Question:19 The centres of those circles which touch the circle, x^2+y^2-8x-8y-4=0, externally and also touch the x-axis, lie on:

(1) a circle

(2) an ellipse

(3) a hyperbola

(4) a parabola

Solution:

x^2+y^2-8x-8y-4=0

=> (x-4)^2 + (y-4)^2 = 36.....(1)

Equation of circle which touches X-axis and center (h,k) is:

(x-h)^2+(y-k)^2 = k^2....(2) 

As circle (2) touches circle (1) extermelly

=> C_1C_2 = r_1 +r_2

=> (h-4)^2 +(k-4)^2 = (6+k)^2

=> (h-4)^2 = 20k+16 

Hence locus is part of parabola.

Option (4) is correct.

Question:20 If one of the diameters of the circle, given by the equation, x^2+y^2-4x+6y-12=0 is a chord of circle S, whose center is at (-3,2), then the radius of S is:

(1) 5sqrt2

(2) 5sqrt3

(3) 5

(4) 10

Solution:

Option (2) is correct.

Question:21 Let P be the point on the parabola, y^2=8x which is at a minimum distance from the center C of the circle, x^2+(y+6)^2 =1. Then the equation of the circle, passing through C and having its centre at P is:

(1) x^2 +y^2-4x+8y+12=0

(2) x^2 +y^2-x+4y-12=0

(3) x^2 +y^2-x/4 +2y-24=0

(4) x^2 +y^2-4x+9y+18=0

Solution:

x^2+(y+6)^2 =1 center of the circle C(0, -6)

y = mx-4m-2m^3

Point C satisfy:

-6 = -4m-2m^3

=> 2m^3+4m-6 =0

=> 2m^3-2m+6m-6=0

=> 2m(m^2-1)+6 (m-1)=0

=> (m-1)(2m^2+2m+6)=0

m=1

we will get P= (2,-4)

Equation of circle:

(x-2)^2+(y+4)^2 =(0-2)^2+(-6+4)^2

=> x^2+y^2-4x+8y+12=0

Option (1) is correct.

Question:22 The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length od its conjugate axis is equal to half of the distance between its foci, is:

(1) 4/3

(2) 4/sqrt3

(3) 2/sqrt3

(4) sqrt3

Solution:

Given that  (2b^2)/a = 8  and  2b = ae

=> 4b^2 = a^2e^2

We know that b^2 = a^2 (e^2- 1)

=> 4 a^2(e^2 -1) = a^2e^2

=> 4e^2- e^2 = 4

=> e = 2/sqrt3

Option (3) is correct.

Question:23 The distance of the point (1,-5,9) from the plane x-y+z=5 measured along the line  x=y=z is:

(1) 3sqrt10

(2) 10sqrt3

(3) 10/sqrt3

(4) 20/3

Solution:

A(1,-5,9)

Equation of line parallel to x=y=z through (1,-5,9) is

(x-1)/1 = (y+5)/1 = (z-9)/1 = lambda

=> P(lambda+1, lambda-5, lambda+9)

If P be poin tof intersection of line and plane.

x-y+z=5

=> lambda+1 -lambda+5 +lambda +9 =5

=> lambda = -10

Point is B(-9, -15, -1)

Hence the distance required is :

AB=sqrt((1+9)^2 + (-5+15)^2 +(9+1)^2)= 10 sqrt3

Option (2) is correct.

Question:24 If the line, (x-3)/2 = (y+2)/-1 = (z+4)/3 lies in the plane, lx+my-z=9, then l^2+m^2 is equal to:

(1) 26

(2) 18

(3) 5

(4) 2

Solution:

Given line:

(x-3)/2 = (y+2)/-1 = (z+4)/3

and given plane is lx+my-z=9,

Now, it is given that line lies on plane

2l-m-3 =0.....(1) 

Also, (3,-2,-4) lies on plane

3l-2m+4 =9.....(2)

From equation (1) and (2), we get

l=1, m=-1

=> l^2+m^2=2

Option (4) is correct

Question:25 Let veca,vecb and vecbbe three unit vectors such that veca xx(vecbxxvecc)= sqrt3/2 (vecb+vecc). If vecb is not parallel to vecc, then the angle between veca and vecb is:

(1) (3pi)/4

(2) pi/2

(3) (2pi)/3

(4) (5pi)/6

Solution:

veca xx(vecbxxvecc)= sqrt3/2 (vecb+vecc)

=> (veca.vecc)vecb- (veca-vecb)vecc = sqrt3/2 vecb + sqrt3/2 vecc

=>(veca.vecc- sqrt3/2)vecb- (veca.vecb- sqrt3/2)vecc =0

=> veca. vec c = sqrt3/2

and  veca. vec b = -sqrt3/2

Angle between vec a and vec b is theta

veca. vec b = -sqrt3/2

=> |veca||vecb| costheta =-sqrt3/2

=> costheta =-sqrt3/2

theta = (5pi)/6

Option (4) is correct.

Question:26 If the standard deviation of the numbers 2,3,a and 11 is 3.5, then which of the following is true?

(1) 3a^2-26a+55 =0

(2) 3a^2-32a+84 =0

(3) 3a^2-34a+91 =0

(4) 3a^2-23a+44 =0

Solution: barx = (2+3+a+11)/4 = (16+a)/4

=> SD = sqrt ((sum x_i^2)/n- (barx)^2

Given that SD = 3.5

=> 3.5 = 7/2 = sqrt [(4+9+a^2+121)/4- ((16+a)/4)^2]

=> 49/4 = (134+a^2)/4- (16+a)^2/16

=> 3a^2 - 32a +84 =0

Option (2) is correct.

Question:27 Let two fair six-faced dice A and B be thrown simultaneously. If E_1 is the event that die A shows up four, E_2 is the event that die Bshows up two and E_3 is the event that the sum of numbers on both dice is odd, then which of the following statemnets is NOT true?

(1) E_1 and E_2 are independent

(2) E_2 and E_3 are independent

(3) E_1 and E_3 are independent

(4) E_1,E_2 and E_3 are independent

Solution:

S = {(1,1),(1,2),...,(1,6),(2,1),(2,2),...,(2,6),....(6,6)}

E_1 = {(4,1),(4,2),.....(4,6)}

E_2 = {(1,2),(2,2),....,(6,2)}

E_3 = {(1,2),(1,4),(1,6),(2,1),(2,3),(2,3).....(6,1),...,(6,5)}

P(E_1)= 6/36 = 1/6; P(E_2)= 6/36 =1/6; P(E_3)= 18/36= 1/2

P(E_1nnE_2nnE_3)= 0/36 =0!=P(E_1)P(E_2)P(E_3)=> E_1, E_2, E_3 are dependent

=>  statement (1) is not true

P(E_1nnE_2) = 1/36 = P(E_1)P(E_2) => E_1 and E_2 are independent =>  statement (2) is true

P(E_2nnE_3) = 1/36 = P(E_2)P(E_3) => E_2 and E_3 are independent =>  statement (3) is true

P(E_1nnE_3) = 1/36 = P(E_1)P(E_3) => E_1 and E_3 are independent =>  statement (4) is true

Option (4) is correct.

Question:28 If 0<= x <2pi, thenthe number of real values of x, which satisfy the equation cosx+cos2x+cos3x+cos4x =0, is:

(1) 3

(2) 5

(3) 7

(4) 9

Solution:

cosx+cos2x+cos3x+cos4x =0

=> 2cos2x.cosx+2cos3x.cosx =0

=> 2cosx[cos2x+cos3x]=0

=> 2cosx[2 cos((5x)/2) cos (x/2)]=0

=> x = pi/2 , (3pi)/2, pi/5, (3pi)/5, pi, (7pi)/5, (9pi)/5

Total 7 values.

Option (3) is correct.

Question:29 A man is walking towards a vertical pillar in a staright path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 3060. After walking for 10 mins from A in the same direction, at a point B, he observes that the angle of elevationof the top of the pillaris 60^0. Then the time taken (in mins) by him, from B to reach the pillar, is:

(1) 6

(2) 10

(3) 20

(4) 5

Solution:

tan30^0 = h/(x+y)= 1/sqrt3....(1)

tan60^0 = h/y = sqrt3...(2)

=> sqrt3 h = x+y

3y = x+y

x = 2y

=> (dx)/(dt)= 2 (dy)/(dt)

Given (dx)/(dt)=10

=> 10 = 2 (dy)/(dt)

=> (dy)/(dt) = 5

Option (4) is correct.

Question:30 The Boolean Experssion (p Lambda ~q)VqV(~p Lambda q) is equivalent to:

(1) ~p Lambda q

(2) p Lambda q

(3) p V q

(4)  p V ~q

Solution:

Given boolean expression is:

(p Lambda ~q)VqV(~p Lambda q)

(p Lambda ~q)Vq = (pVq) Lambda (~qVq)=(pVq) Lambda t = (pVq)

Now,

(pVq)V(~p Lambda q)= pVq!!

Option (3) is correct