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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 4002. |
A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails. |
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| 4003. |
Coloured balls are distributed in four boxes as shown in the following table : A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III ? |
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| 4004. |
Evalute the following integrals int (dx)/(sin x + sin 2x ) |
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| 4005. |
The possible percentage error in computing the parallel resistance R of three resistances R_(1), R_(2), R_(3) from the formula (1)/(R) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3)), if R_(1), R_(2), R_(3) are each1.2% in error, is |
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Answer» `1.2%` |
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| 4006. |
Consider the equation sectheta+ cottheta = 31/12 On the basis of above, answer the following Number of values of theta where theta in [0,5pi] is equal to |
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Answer» `4` `implies (1+t^(2))/(1-t^(2))+(1-t^(2))/(2t)=(31)/(12)` `implies6t^(4)+43t^(3)-12t^(2)-19t+6=0` `implies(3t-1)(2t^(3)+15t^(2)+t-6)=0` `implies t=(1)/(3),alpha,beta,gamma` where `alpha in(-8,-7)implies(theta)/(2)in((pi)/(2),(3pi)/(4))` `beta in (-1,(-1)/(2))implies(theta)/(2) in ((3pi)/(4),(7pi)/(8))` `gamma in((1)/(2),1)implies(theta)/(2)in(0,(pi)/(4))` (i) if `0lttheta lt (pi)/(2)` then `t=(1)/(3)` or `t in ((1)/(2),1)` `implies tan theta =(3)/(4) "or if" tan. (theta)/(2)in((1)/(2),1)implies "tan" theta in ((4)/(3),oo)` `implies"Min value of" [tan theta] =0` (ii)No. of values of `theta in ((pi)/(2),(3pi)/(2))`will be 1 |
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| 4007. |
If the altitudes of a Delta are in A.P., then its sdies are in |
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Answer» A.P. |
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| 4008. |
Let A,B and C be three sets of complex numbers as defined below: {:(,A={z:Im(z) ge 1}),(,B={z:abs(z-2-i)=3}),(,C={z:Re(1-i)z)=3sqrt(2)"where" i=sqrt(-1)):} Let z be any point in A cap B cap C. Then, abs(z+1-i)^(2)+abs(z-5-i)^(2) lies between |
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Answer» 25 and 29 |
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| 4009. |
Let A,B and C be three sets of complex numbers as defined below: {:(,A={z:Im(z) ge 1}),(,B={z:abs(z-2-i)=3}),(,C={z:Re(1-i)z)=3sqrt(2)"where" i=sqrt(-1)):} Let z be any point in A cap B cap C " and " omega be any point satisfying abs(omega-2-i) lt 3. Then, abs(z)-abs(omega)+3 lies between |
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Answer» `-6 and 3` |
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| 4010. |
Differentiate the following with respect to x: (e^(x) + log x)/(sin 3x) |
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| 4011. |
A homogeneous polynomial of the second degree in n variables i.e., the expression phi=sum_(i=1)^(n)sum_(j=1)^(n)a_(ij)x_(i)x_(j) where a_(ij)=a_(ji) is called a quadratic form in n variables x_(1),x_(2)….x_(n) if A=[a_(ij)]_(nxn) is a symmetric matrix and x=[{:(x_(1)),(x_(2)),(x_(n)):}] then X^(T)AX=[X_(1)X_(2)X_(3) . . . .X_(n)][{:(a_(11),a_(12)....a_(1n)),(a_(21),a_(22)....a_(2n)),(a_(n1),a_(n2)....a_(n n)):}][{:(x_(1)),(x_(2)),(x_(n)):}] =sum_(i=1)^(n)sum_(j=1)^(n)a_(ij)x_(i)x_(j)=phi Matrix A is called matrix of quadratic form phi. Q. If number of distinct terms in a quadratic form is 10 then number of variables in quadratic form is |
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Answer» 4 Hence total number of distinct terms `=n+.^(n)C_(2)=10` `impliesn+(n(n-1))/(2)=10impliesn(n+1)=20impliesn=4` |
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| 4012. |
A homogeneous polynomial of the second degree in n variables i.e., the expression phi=sum_(i=1)^(n)sum_(j=1)^(n)a_(ij)x_(i)x_(j) where a_(ij)=a_(ji) is called a quadratic form in n variables x_(1),x_(2)….x_(n) if A=[a_(ij)]_(nxn) is a symmetric matrix and x=[{:(x_(1)),(x_(2)),(x_(n)):}] then X^(T)AX=[X_(1)X_(2)X_(3) . . . . .X_(n)][{:(a_(11),a_(12)....a_(1n)),(a_(21),a_(22)....a_(2n)),(a_(n1),a_(n2)....a_(n n)):}][{:(x_(1)),(x_(2)),(x_(n)):}] =sum_(i=1)^(n)sum_(j=1)^(n)a_(ij)x_(i)x_(j)=phi Matrix A is called matrix of quadratic form phi. Q. The quadratic form of matrix A[{:(0,2,1),(2,3,-5),(1,-5,8):}] is |
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Answer» `3x_(2)^(2)+8x_(3)^(2)+2x_(1)x_(2)+x_(1)x_(3)-5x_(2)x_(3)` `=2x_(1)x_(2)+x_(1)x_(3)+2x_(1)x_(2)+3x_(2)^(2)-5x_(3)x_(2)+x_(3)x_(1)+5x_(3)x_(2)+8x_(3)^(2)` `=3x_(2)^(3)+8x_(3)^(2)+4x_(1)x_(2)+2x_(1)x_(3)-10x_(2)x_(3)` |
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| 4013. |
Integrate the following functions : int(secx.cosecx)/(log(cotx))dx |
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| 4014. |
If the distance between the points (3,a) and (6,1) is 5, find the value of a. |
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Answer» Solution :Distance between the POINTS (3,a) and (6,1) is `SQRT((3-6)^2+(a-1)^2=sqrt9+(a-1)^2` or, `(a-1)^2=16` or, `a-1=ne4` ` a=1`ne4=5` or, -3. |
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| 4016. |
If f(x) = x^(3) + ax^(2) + bx+c = 0 has three distinct integral roots and (x^(2) + 2x+ 2)^(3) + a(x^(2) + 2x + 2)^(2) + b(x^(2) + 2x + 2) +c=0 has no real roots then {:("Column - I" , " Column- II"), ("(a) a=" , " p 0 ") , ("(B) b =" , "(q) 2 "),("(C) C = ", " (r) 3 ") , ("(D) If the roots of f (x) = k are equal then k= " , "(s) -1"):} |
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| 4017. |
If the tangent at (3,-4) to the circle x^(2)+y^(2)-4x+2y-5=0 cuts the circle x^(2)+y^(2)+16x+2y+10=0 in A and B then the midpoint of AB is |
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Answer» `(-6,-7)` |
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| 4018. |
Write down a unit vector in XY plane, making an angle of 30^(@)with positive direction ofx-axis. |
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| 4019. |
int (1)/(2 sin^(2) x + 3 sin x cos x - 2 cos^(2) x )dx = |
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Answer» `(1)/(5)" log" |(2 tan X- 1)/(2 + tan x)| + C` |
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| 4020. |
bar(x) and bar(y) are unit vectors and (bar(x)""_(,)^(hat)bar(y))=theta. If theta = ………… then bar(x)+bar(y) will becomes unit vector. |
| Answer» Answer :D | |
| 4021. |
Let a = 2i + j + k, b = I + 2j - k and a unit vector c be coplanar. If c is perpendicular to a, then c = |
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Answer» `(1)/(SQRT(2)) (- j + k)` |
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| 4022. |
Find the number of ways to arrange 5 boys and 5 girl in a row such that all the boys sit together and all girls sit together |
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| 4023. |
Find the dy/dx when sin x = (2t)/(1+t^2), tan y = (2t)/(1-t^2) |
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Answer» Solution :`SIN x = (2t)/(1+t^2)` and `tan y = (2t)/(1-t^2)` `implies x = sin^(-1) frac (2t)(1+t^2) = 2TAN^(-1)t and y = tan^(-1) frac (2t)(1-t^2) = 2 tan^(-1)t implies y = x dy/dx = 1` |
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| 4024. |
Find the ratio of the areas of two regions of the curve C_(1) -= 4x^(2) + pi^(2)y^(2) = 4pi^(2) divided by the curve C_(2) -= y = -(sgn(x - (pi)/(2)))cos x (where sgn (x) = signum (x)). |
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Answer» Solution :`4x^(2) + pi^(2)y^(2) = 4pi^(2)` `therefore` `C_(1) : (x^(2))/(pi^(2)) + (y^(2))/(4) = 1`, which is an ellipise. Curve `C_(2) : y = -sgn (x - (pi)/(2)) cos x` `=({:(-cos x",", x gt (pi)/(2)),(cos x",", x LT (pi)/(2)),(0",", x = (pi)/(2)):}` Graphs of `C_(1)` and `C_(2)` are as shown in the following figure.  From the symmetry of the curves, the areas of regions `A_(1)` and `A_(2)` are EQUAL. Area of ellipse is `pi(pi)(2) = 2pi^(2)`. Area of ellipse above the x - axis is `pi^(2)`. Now the area bounded by y = cos x and the x - axis `(A_(3))` is 2 SQ. units. So the area OFTHE portion of the ellipse above the curve `C_(2)` is `pi^(2) - 2` (as the areas of regions `A_(1)` and `A_(2)` are same) So the area of the portion of the ellipse below the curve `C_(2)` is `pi^(2) + 2`. Hence the required ratio is `(pi^(2) - 2)/(pi^(2) + 2)` |
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| 4026. |
Find the area of the region in the first quadrant enclosed by the x-axis, the line y=x and the circle x^(2)+ y^(2)=32 |
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| 4028. |
If z= (1)/(2) (sqrt3 -i), thentheleast positive integral value of m such that (z^(101) + i^(10 theta))^106= z^(m) is |
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Answer» 8 |
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| 4029. |
If phi(x) is a differentiable function AAx in Rand a inR^(+) such that phi(0)=phi(2a),phi(a)=phi(3a)and phi(0)nephi(a), then show that there is at least one root of equation phi'(x+a)=phi'(x)"in"(0,2a). |
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| 4030. |
Statement I. The four straight lines given by 6x^2+5xy-6y^2=0 and 6x^2+5xy-6y^2-x+5y-1=0 are the sides of a square . Statement II . The lines represented by general equation of second degree ax^2+2hxy+by^2+2gx+2fy+c=0are perpendicular if a+b=0 . |
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Answer» STATEMENT I is TRUE, Statement II is true , Statement II is a correct explanationfor Statement I |
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| 4031. |
Equation of parabola whose vertex is (-1, 2) and focus is (3, 2) is |
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| 4032. |
Match the statements/expressions given in List ( with the value given in List II. |
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Answer» Given `int_(0)^(1)f(x)dx=1` `=2a+3b=6` `=(a,b)=(0,2)` and `(3,0)` b. `f(x)=sin(x^(2))+cos(x^(2))` `=sqrt(2)cos(x^(2)-(PI)/4)` For maximum value `x^(2)-(pi)/4=2n pi, n epsilonZ` `impliesx^(2)=2n pi+(pi)/4,n epsilon Z` `IMPLIES x=+-sqrt(x/4)+-sqrt((9pi)/4)` as `x epsilon[-sqrt(13),sqrt(13)]` c. `I=int_(-2)^(2)(3X^(2))/((1+e^(x)))dx`...............1 `=int_(-2)^(2)(3(-x)^(2))/(1+e^(-x))dx` `:. I=int_(-2)^(2)(e^(x)(3x)^(2))/(e^(x)+1)dx`............2 Addding1 and 2 `impliesI+I=int_(-2)^(2)(3x^(2))/((1+e^(x)))dx+int_(-2)^(2)(e^(x)(3x^(2)))/(e^(x)+1)dx` `=int_(-2)^(2)3x^(2)dx=2int_(0)^(2)3x^(2)dx=16` `implies I=8` d. We have `I=(int_(1//2)^(1//2)cos2x.log((1+x)/(1-x))dx)/(int_(0)^(1//2)cos2x.log((1+x)/(1-x))dx)` Let `f(x)=cos2xI((1+x)/(1-x))` `:.f(-x)=cos(-2x)In((1-x)/(1+x))` `=-cos(2x)In((1+x)/(1-x))=-f(x)` Thus, `f(x)` is an odd function. `impliesI=0` |
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| 4033. |
If vec(a)=hat(i)+hat(j)+hat(k),vec(a).vec(b)=1andvec(a)xxvec(b)=hat(j)-hat(k), then vec(b) is |
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Answer» `HAT(i)-hat(J)+hat(k)` |
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| 4034. |
Find the coordinates of the foot of the perpendicular and the length of the perpendicular drawn from the point P(5,4,2) to the line vec(r) = -hat(i) + 3hat(j) + hat(k) + lambda (2hat(i) + 3hat(j) - hat(k)). |
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| 4035. |
int(sin^(3//2)x+cos^(3//2)x)/(sqrt(sin^(3)x.cos^(3)x.sin(x+theta)))dx |
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| 4037. |
the volume generated when the region bounded by y=x, y=1, x=0, is rotated about y-axis is ….. |
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Answer» `(pi)/(4)` |
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| 4038. |
Let f(n) = 20n -n^(2) (n= 1,2,3…), then |
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Answer» `f(n) rarr OO "as " n rarr oo` |
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| 4039. |
Find values of k if area of triangle is 4sq. units and vertices are (-2,0)(0,.4),(0,k) |
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| 4040. |
Write the following functions in the simplest form : tan^(-1)((3a^2x - x^3)/(a^3- 3ax^2)), a gt 0, -a/sqrt3 le x le a/sqrt3 |
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Answer» SOLUTION :`tan^(-1) ((3a^2 X - x^3)/(a^3- 3ax^2)) = tan^(-1)((3a^2 tan THETA - a^3 tan^3 theta)/(a^3 - 3a^3 tan^2 theta)) 9x = a tan theta)` `= tan^(-1)((3 tan theta - tan^3 theta)/(1- 3 tan^2 theta)) = tan^(-1)tan3 theta = 3 theta = 3TAN^(-1)(x/a)` |
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| 4041. |
Obtain mathematical expectation and variance of the discrete random variable X which denotes the minimum of two numbers that appear when a pair of fair dice is tossed once. |
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| 4042. |
Two boxes containing 20 balls each, and each balls is either black or white. The total number of black balls is different from the total numbers of white balls. One ball is drawn at random from each box. The probability that both are white is 0.21, then the probability that both the balls drawn are black is |
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Answer» 0.23 |
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| 4043. |
If the curve x^(2)=-4(y-a) does not intersect the curvey=[x^(2)-x+1] (where [,] denotes the greatest integer function) in [0,(1+sqrt(5))/(2)], then |
| Answer» Answer :C | |
| 4044. |
("^(m)C_(0)+^(m)C_(1)-^(m)C_(2)-^(m)C_(3))+('^(m)C_(4)+^(m)C_(5)-^(m)C_(6)-^(m)C_(7))+..=0 if and only if for some positive integer k, m= |
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Answer» `4k` `(costheta-isintheta)^(m)` `=.^(m)C_(0)cos^(m)theta-^(m)C_(1)cos^(m-1)thetaisintheta+...+^(m)C_(m)(-isintheta)^(m)`.......`(i)` `(costheta+isintheta)^(m)` `=^(m)C_(0)cos^(m)theta+^(m)C_(1)cos^(m-1)thetaisintheta+...+^(m)C_(m)(isintheta)^(m)`.......`(II)` Adding `(1)` and `(2)` , we GET `2cosmtheta=2['^(m)C_(0)cos^(m)theta-^(m)C_(2)cos^(m-2)thetasin^(2)theta....]`.......`(iii)` Subtracting `(1)` from `(2)`, we get `2isinmtheta=2i['^(m)C_(1)cos^(m-1)thetasintheta-^(m)C_(3)cos^(m-3)thetasin^(3)theta.....]`.....`(iv)` Adding `(3)` and `(4)`, we get `cosmtheta+sinmtheta` `=['^(m)C_0)cos^(m)theta+^(m)C_(1)cos^(m-1)thetasintheta-^(m)C_(2)cos^(m-2)thetasin^(2)theta-^(m)C_(3)cos^(m-3)thetasin^(3)theta....]` `impliessqrt(2)sin(mtheta+(pi)/(4))` `=['^(m)C_(0)cos^(m)theta+^(m)C_(1)cos^(m-1)thetasintheta-^(m)C_(2)cos^(m-2)thetasin^(2)theta-^(m)C_(3)cos^(m-3)thetasin^(3)theta...]` Putting `theta=(pi)/(4)`, we get `sqrt(2)sin"(((m+1)pi)/(4))` `=(1)/(2^(m//2))` `[('^(m)C_(0)+^(m)C_(1)-^(m)C_(2)-^(m)C_(3))+('^(m)C_(4)+^(m)C_(5)-^(m)C_(6)-^(m)C_(7))+...+('^(m)C_(m-3)+^(m)C_(m-2)-^(m)C_(m-1)-^(m)C_(m))]` Hence , `m+1=4k`, for given quantity to be `0`. `implies m=4k-1`, where `K in N` |
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| 4045. |
int (1)/(x^(4)+1)dx=A tan^(-1)((x^(2)-1)/(sqrt(2)x))-B log|(x^(2)-sqrt(2)x+1)/(x^(2)+sqrt(2)x+1)|+Cthen |
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Answer» A) `A=2B` |
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| 4046. |
Let fbe a continuous function satisfying int_(- pi)^(t^2) (f (x) + x^(2) ) dx= pi + (4)/(3) t^(3) for allt, then f(pi^(2) //4) is equal to |
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Answer» `PI - (pi^4)/( 8)` |
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| 4047. |
Let vec(a)={:[(1),(0),(-3)]:},vec(b)={:[(2),(1),(0)]:},vec(c)={:[(1),(-1),(1)]. If the numbers alpha,betaandgamma are such that alphavec(a)+betavec(b)+gammavec(c)={:[(-2),(-5),(6)], then |
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Answer» `alpha=-1` |
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| 4049. |
Draw the graph of y=sin^(-1)|sin x|and y=(sin^(-1)|sinx|)^(2),0lexle2pi |
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Answer» Solution :We have `y=f(X)=SIN^(-1)|sinx|,0lexle2pi` `{{:(sin^(-1)(sin x),0lexltpi),(-sin^(-1)(sin x),pilexle2pi):}` `{{:(x, 0lexltpi/2),(pi-x,pi/2lexltpi):}` `{{:(x, 0lexltpi/2),(pi-x,pi/2lexltp),(-(pi-x),pilexlt(3pi)/(2)),(-(x-2pi),(2pi)/2(3pi)/(2)lexlt2pi):}` `{{:(x, 0lexltpi/2),(pi-x,pi/2lexltp),(-(pi-x),pilexlt(3pi)/(2)),(x-2pi,(3pi)/(2)lexlt2pi):}` The graph of `y=f(x)` is as SHOWN in the following figure.  `y=g(x)=(sin^(-1)|sin x|)^(2), 0 lexle2pi` `{{:(x^(2),0lexltpi/2),((pi-x)^(2),pi/2lexltpi),((x-pi)^(2),pilexlt(3pi)/2):}={{:(x^(2),0lexltpi/2),((pi-x)^(2),pi/2lexltpi),((2pi-x)^(2),(2pi)/2lexlt2pi):}` `y=x^(2)` is parabola having VERTEX at (0,0) `y=(x-pi)^(2)` is a parabola having vertex at `(pi,0)` `y=(x-2pi)^(2)`is a parabola having vertex at `(2pi,0)` The graph of `y=g(x)` is as shown in the followin figure. 
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| 4050. |
A plane prod passes through the point (1, 1, 1). If b, c, a are the direction ratios of a normal to the plane, where a, b, c (a |
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Answer» 29x + 31Y + 3z = 63 |
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