Solution of Maths paper (JEE 2016: Mains)

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=1!!in 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 vecb!!be 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