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26901.

Integrate the following functions x^3/sqrt(1-x^8)

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SOLUTION :`int x^3/sqrt(1-x^8) DX = int x^3/sqrt(1-(x^4)^2) dx`.
put `t = x^4`. Then `DT = 4x^3 dx`
therefore REQUIRED INTEGRAL
=`int 1/sqrt(1-t^2) 1/4 dt = 1/4 sin^-1 t+c`
=`1/4 sin^-1 x^4 +c`
26902.

If veca,vecb,vecc are unit vectors such that veca+vecb+vecc=vec0 , find the value ofveca*vecb+vecb*vecc+vecc*veca.

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ANSWER :`(-3)/(2)`
26903.

If the magnitude of the vector product of the vector hati+hatj+hatk with aunit vector along the sum of vector 2hati+4hatj-5hatkand lambda hati+2hatj+3hatk is equal to sqrt2, then find the value of 'lambda'

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ANSWER :`lambda=1`
26904.

A and B are two events such that P(A) ne 0. Find P(B|A), if i. A is a subset of B , ii. AnnB = 0

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ANSWER :i. 1, II. 0
26905.

Form the values of the following in terms of a, b, c if alpha, beta are roots of ax^(2)+bx+c=0, c != 0 i) (1)/(alpha)+(1)/(beta) ii) (1)/(alpha^(2))+(1)/(beta^(2)) iii) alpha^(3)+beta^(3) iv) ((alpha)/(beta)-(beta)/(alpha))^(2) v) alpha^(4)beta^(7)+alpha^(7)beta^(4) vi) alpha^(2)+beta^(2) vii) (alpha^(2)+beta^(2))/(alpha^(-2)+beta^(-2)) viii) (alpha)/(beta^(2))+(beta)/(alpha^(2))

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Answer :i) `(-b)/(a)` ii) `(b^(2)-2AC)/(C^(2))` III) `(3abc-b^(3))/(a^(3))` iv) `(b^(2)(b^(2)-4AC))/(a^(2)c^(2))` v) `bc^(4)((3ac-b^(2)))/(a^(7))` VI) `(b^(2)-2ac)/(a^(2))` vii) `(c^(2))/(a^(2))` viii) `(3abc-b^(3))/(ac^(2))`
26906.

If A is 2xx2 matrix such that A^(2) =O then tr (A) is

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1
`-1`
0
none of these

ANSWER :C
26907.

Let f(x)={{:(x^(n)sin(1//x^(2))","xne0),(0","x=0):},(ninI). Then

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`(pi)/(2)`
`(pi)/(2sqrt(2))`
`(pi)/(sqrt(2))`
`sqrt(2)pi`

Solution :For `NGT0, underset(xto0)LIMX^(N)sin(1//x^(2))=0xx(" any value between-1 to 1")=0`
For `nlt0,underset(xto0)limx^(n)sin(1//x^(2))=ooxx(" any value between-1 to 1")=+-oo`
26908.

Evaluate int Sin^(5) x dx

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Answer :`-(sin^(4)xcccosx)/(5)-(4)/(15)sin^(2)xcosx-(8)/(15)cos+c`
26909.

Find the area lying above x-axisand included between the circle x^(2)+ y^(2) = 8x and inside in the parabola y^(2) = 4x.

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ANSWER :`4/3(8+3pi)`
26910.

Which of the following functions is decreasing on (0, (pi)/(2)).

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sin (2x)
TAN X
COS x
cos (3X)

Answer :C
26911.

If the angles of elevation of two towers of heights h_(1) and h_(2) from the mid-point of the line joining their feet be (pi)/(3) and (pi)/(6), respectively, then (h_(1))/(h_(2)) = _________________

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ANSWER :`3.00`
26912.

If the line barr=hati+lambda(2hati-mhatj-3hatk) is parallel to the plane barr.(mhati+3hatj+hatk)=0, then m is equal to

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3
-3
1
-1

Answer :B
26913.

Integration using rigonometric identities : int (x sin x)/((x cos x-sin x+5))dx=...+c

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`log|x cosx-sin x+5|`
`-log|x cosx-sin x+5|`
`log|x sinx-sin x+5|`
`-log|x sinx-sin x+5|`

ANSWER :B
26914.

Evaluate the following integrals int (ax^2 + bx + c)dx

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Solution :`int (ax^2 +BX+ c) DX = ax^3/3 +bx^2/2 +CX + c^.` (`c^.` is an arbitrary CONSTANT)
26915.

Fundamental theorem of definite integral : If I_(n)=int_(0)^(pi/4)tan^(n)dx then lim_(ntooo)n(I_(n)+I_(n+2))=.......

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1
`(1)/(2)`
`OO`
0

Answer :A
26916.

If A+ B+C= 180^(@) " then " cos 2A+ cos 2B + cos 2C+ 1=

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`1+ 4 sin A sin B sin C `
`1-2 sin A sin B sin C `
`2+2 sin A sin B sin C `
`4SIN A sin B COS C `

ANSWER :A
26917.

How many factors of 10,000 end with a 5 on the right ?

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SOLUTION :We have `1000 = 2^4xx5^4`
`:." The factors of 10000 ending with 5 are"5,5xx5=25,5xx5xx5xx5=125`
`5xx5xx5xx5=625`
`:. "There are 4 factors ending with 5"`.
26918.

Let 'A' is (4xx4) matrix such that the sum of elements in each row is 1. Find out sum of the all the elements in A^(10).

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ANSWER :4
26919.

For the ellipse given by ((x - 3)^(2))/(25) + ((y - 2)^(2))/(16) = 1, match the equations of the lines given in List I with those on the List Ii. {:("List I","List II"),(i."The equation of the minor axis",p. 3 x = 34),(ii."The equation of a latusrectum",r. x + y = 9),( ,s. x = 6),( ,f. x = 3),( ,u. 3y = 34):}

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<P>I,P,S
Q,U,F
Q,P,F
Q,P,S

Answer :d
26920.

Let A(2 sec theta,3 tan theta) and B (2 sec phi 3 tan phi ) where theta+phi=(pi)/(2), be two points on the hyperbola (x^2)/(4)-(y^2)/(9)=1 If (alpha, beta) is the point of intersection of normals to the hyperbola at A and B, then beta=

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`(-13)/(3)`
`(13)/(3)`
`(3)/(13)`
`(-3)/(13)`

ANSWER :A
26921.

If P is a complex number whose modulus is one, then the equation( (1 + iz)/(1- iz) )^4 = P has

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real and equal ROOTS
real and DISTINCT roots
TWO real and two complex roots
all complex roots

Answer :D
26922.

Find the probability of throwing at most 2 sixes in 6 throws of a single die.

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ANSWER :`(35)/(18)((5)/(6))^(4)`
26923.

IF mx ^2+ 7xy -3y^2+ 4x+ 7y+2is resolvableintotwolinearfactorsthen m=oftwplinearfactors thenthe factorsare

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7
4
2
`-5`

ANSWER :C
26924.

Evaluate sum_(i=0)^(n-1) sum_(j= 1 + i)^(n+1)""^(n)C_(i) ""^(n+1)C_(j) .

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Solution :Let ` P = sum_(i=0)^(n-1) sum_(j= 1 + i)^(n+1)""^(n)C_(i) ""^(n+1)C_(j)` .
`sum_(j=1)^(n-1)""^(n)C_(0)""^(n+1)C_(j) +sum_(j= 2)^(n+1)""^(n)C_(i) ""^(n+1)C_(j)+ sum_(j= 3)^(n+1) ""^(n)C_(2)""^(n+1)C_(j) + ...+ sum_(j = n)^(n+1) ""^(n)C_(n-1) ""^(n+1)C_(j)`
`""^(n)C_(0) sum_(j=1)^(n-1)""^(n)C_(j)+""^(n)C_(1) +sum_(j= 2)^(n+1)""^(n+1)C_(j) ""^(n+1)C_(2)+ sum_(j= 3)^(n+1) ""^(n+1)C_(j) + ...+ ""^(n)C_(n-1)sum_(j = n)^(n+1)""^(n+1)C_(j)`
`""^(n)C_(0)(""^(n +1)C_(j) + ""^(n+1)C_(2) + ""^(n+1)C_(3) + ...+ ""^(n+1)C_(n+1))`
` + ""^(n)C_(1) (""^(n+1)C_(2)+""^(n+1)C_(3) + ""^(n+1)C_(4)+ ...+ ""^(n+1)C_(n+1))`
` + ""^(n)C_(2) (""^(n+1)C_(3)+""^(n+1)C_(4) + ""^(n+1)C_(5)+ ...+ ""^(n+1)C_(n+1))`
` + ...+ ""^(n)C_(n-1)(""^(n +1)C_(n) + ""^(n+1)C_(n+1))`
`= ""^(n+1)C_(1)*""^(n)C_(0) + ""^(n+1)C_(2) (""^(n)C_(0) + ""^(n)C_(1)) + ""^(n+1)C_(3) (""^(n)C_(0) + ""^(n)C_(1) + ""^(n)C_(2))`
` + ...+ ""^(n+1)C_(n+1) (""^(n)C_(0) + ""^(n)C_(1) + ""^(n)C_(2)+ ...+ ""^(n)C_(n-1))`
`(""^(n)C_(0) + ""^(n)C_(1)) *""^(n)C_(0)+ (""^(n)C_(1) + ""^(n)C_()) ( ""^(n)C_(2)+ ""^(n)C_(1))+ (""^(n)C_(2) + ""^(n)C_(3)) (""^(n)C_(0) + ""^(n)C_(1) + ""^(n)C_(2))`
`+ ...+ (""^(n)C_(n) + ""^(n)C_(n-1))(""^(n)C_(0) + ""^(n)C_(1)+ ""^(n)C_(2) + ...+ ""^(n)C_(n-1)) + n`
`(""^(n)C_(0))^(2) + (""^(n)C_(1))^(2) + (""^(n)C_(2))^(2) + ...+ (""^(n)C_(n-1))^(2)`
` + {""^(n)C_(0) *""^(n)C_(1) + ""^(n)C_(0) * ""^(n)C_(2) + ""^(n)C_(0) *""^(n)C_(3)`
`+...+""^(n)C_(0) * ""^(n)C_(n-1) + ...+ ""^(n)C_(n-2) + ""^(n)C_(n-1)} + 2^(n) - 1 + n`
` (""^(n)C_(0) + ""^(n)C_(1) + ""^(n)C_(2) + ...+ ""^(n)C_(n-1))^(2) + 2^(n) - 1 + n`
`= (2^(n) -1)^(2) + 2^(n) -1 + n = 2 ^(2N) - 2^(n) + n`
26925.

A hyperbola has one focus at (1, 2) , its corresponding directrix is x + y = 1 and eccentricity is 2. Then

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<P>`{:(P,Q,R,S),(2,1,3,4):}`
`{:(P,Q,R,S),(3,4,1,2):}`
`{:(P,Q,R,S),(1,2,4,3):}`
`{:(P,Q,R,S),(4,3,2,1):}`

SOLUTION :N/A
26926.

Let A=[{:((1)/(6),(-1)/(3),(-1)/(6)),((-1)/(3),(2)/(3),(1)/(3)),((-1)/(6),(1)/(3),(1)/(6))]. If A^(2016l)+A^(2017m)+A^(2018n)=(1)/(alpha)A, for every l,m,n in N, then the value of alpha is

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`(1)/(6)`
`(1)/(3)`
`(1)/(2)`
`(2)/(3)`

ANSWER :B
26927.

int_0^1x(1-x)^ndx

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Solution :`I=int_0^1x(1-X)^ndx`
=`int_0^1(1-x)x^ndx`
=`int(x^n-x^(n+1))DX`
=`[x^(n+1)/(n+1)-x^(n+2)/(n+2)]_0^1`
=`1/(n+1)-1/(n+2)=1/((n+1)(n+2))`
26928.

I= int "arc" sin x dx.

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Answer :`X "ARC" SIN x+ sqrt( 1-x^(2) ) +C`.
26929.

The value of ""^(2)P_(1) + ^(3)P_(1) + ...... + ^(n)P_(1) equal to

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`(N^(2) - n + 2)/2`
`(n^(2) + n + 2)/2`
`(n^(2) + n - 2)/2`
`(n^(2) - n - 2)/2`

ANSWER :D
26930.

If x sin (a + y) + sin a cos (a + y)= 0, then prove that (dy)/(dx)= (sin^(2) (a + y))/(sin a)

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ANSWER :`(SIN^(2) (a + y))/(sin a)`
26931.

Let f(x)=min{x,x^2}, for every x in R. Then

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f (x) is CONTINUOUS for all x
f (x) is diffentiable for all x
`f'(x) =1` for `x GT 1`
f (x) is not DIFFERENTIABLE at TWO VALUES of x

Answer :B
26932.

int_(0)^([x])(2^(x))/(2^[x])dx=

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`- ([2X])/(LN 2)`
`([2x])/(ln 2)`
`- ([X])/(ln 2)`
`([x])/(ln 2)`

ANSWER :D
26933.

Match the statement in Column I with those in Column II. [Note : Here z takes values in the complex plane and Im z and Re z denote, respectively, the imagi- nary part and the real part of z]. {:(,Column -I, Column -II),((A),"The set of points z satisfying"|z-iz||=|z+i|z||,(P)" an ellipse with eccentricity"4/5),((B),"The set of points z satisfying"|z+4|+|z-4|=10" is contained in or equal to",(Q)"the set of points z satisfying "Im z = 0 ),((C),"If " |w|=2",then the set of points "z=w-1/w" is contained in or equal to",(R)"the set of points z satisfying "|Im z |ge1),(,,(S)"the set of points z satisfying "|Re z |le2),((D),"If" |w|=1",then the set of points"z=w+1/w" is contained in or equal to ",(T) " the set of points z satisfying"|z|le3):}

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ANSWER :A::B::C::D
26934.

From the set of all families having three children, a family is picked at random If one child of the family is a son. find the probablilty that he has two sisters.

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Solution :The ONE child of the FAMILY is a son. We have to find the probability that he has two sisters. We have the following mutually exclusive EVENTS : BGG,GBG,GGB.
`THEREFORE` The required probability `=P(B)xxP(G)xxP(G)+P(G)xxP(B)+P(G)+P(G)xxP(G)xxP(B)`
`=1/2xx1/2xx1/2+1/2xx1/2xx1/2+1/2xx1/2xx1/2=3/8`
26935.

A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

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ANSWER :`(2)/(5)`
26936.

The value of n for which 704 + (1)/(2) (704) + (1)/(4)(704) + ….. Upto n terms = 1984 - (1)/(2) ( 1984) + (1)/(4) (1984) …. Upto n terms is

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5
4
4
10

Answer :A
26937.

Integrate the following function : intsqrt((1-x)/(1+x))dx

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ANSWER :`SIN^(-1)x+sqrt(1-x^(2))+C`
26938.

Iff(x) = x+|x|+ cos([ pi ^(2) ]x) and g(X)=sin x,where [.]denotesthegreatestintegerfunction, then

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f(X)+ G(x)iscontinuouseverywhere , then
f(x) + g(x) is differentiableeverywhere
`f(x) xxg(x) ` is DIFFERENTIABLE EVERYWHERE
`f(x) xx g(x) ` is continuousbut not differentiableat x=0

ANSWER :A::C
26939.

Let A = R - {3} and B =R -{1}. Consider the function f : A to B defined by f (x) = ((x -2)/(x -3)). Is f one-one and onto ? Justify your answer.

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ANSWER :YES
26940.

If two dice are thrown the probability that atleast one of the dice shows a number greater than or equal to 4 is

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`9//36`
`3//4`
`1//9`
`4//9`

ANSWER :B
26941.

Integrate the functions 1/(1+cotx)

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ANSWER :`x/2-1/2logabs(cosx+sinx)+C`
26942.

f (x) = lim _(x to oo) (x ^(2) + 2 (x+1)^(2n))/((x+1) ^(2n+1) + x^(2) +1),n in N and g (x) =tan ((1)/(2)sin ^(-1)((2f (x))/(1+f ^(2) (x)))), then The number of points where g (x) is non-differentiable AA x in R is:

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1
2
3
4

Answer :D
26943.

If |(z_1 z- z_2)/(z_1 z+z_2)|=k, (z_1 , z_2 ne 0) then

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for `K =1` LOCUS of Z is straight line
for `k cancel(in) {1,0}z` lieson a circle
for `k=0` z represents a point
for `k=1,z` lies on the perpendicular bisector of the segment JOINING `(z_2)/(z_1) and (z_2)/(z_1)`

Answer :A::B::C::D
26944.

Solve the inequation sqrt(x^(2)-3x-10) gt (8-x).

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ANSWER :`((74)/(13), INFTY)`
26945.

A and B are two points on the hyperbola O is the centre. If OA is perpendicular to OB then (1)/(OA)^(2)+(1)/(OB)^(2) is equal to

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`(1)/(a^(2))+(1)/(B^(2))`
`(1)/(a^(2))-(1)/(b^(2))`
`(1)/(b^(2))-(1)/(a^(2))`
`a^(2)+b^(2)`

ANSWER :b
26946.

Evaluate int_(0)^((pi)/(4)) [sin x + [cos x + [tan x + [sec x]]]] dx

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ANSWER :`(PI)/(4)`
26947.

sintheta= ( x + Y)/( 2 sqrt(xy))is possible

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all realx,y
`X , ygt 0`
`x=y ne 0`
no realx,y

Answer :C
26948.

To double the covering range of a TV transmittion tower, its height should be multiplied by :

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`(1)/(sqrt(2))`
2
4
`sqrt(2)`

SOLUTION :NA
26949.

The total number of 5 digit numbers formed using the digits 2, 4, 6, 7, 8 if the digits are no repeated is

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5!
`5^(5)`
6!
`5^(2)`

ANSWER :B
26950.

Analyze the roots of the following equations: (i)2x^(3) - 9x^(2) + 12x - (9//2) = 0 (ii) 2x^(3) - 9x^(2) + 12x - 3 = 0

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Solution :Let `f(x) = 2X^(3) - 9x^(2) + 12x - (9//2)`. Then
`f(x) = 6X^(2) - 18x +12`
`= 6(x^(2)-3x+2) = 6(x- 1)(x-2)`
Now `f'(x) = 0 rArr` x = 1 and x = 2
Also `f(1) = 2 - 9 + 12 - (9//2) gt 0`
and`f(2) = 16 - 36 + 24 - (9//2) lt 0`
HENCE, the graphs of the function `y= f(x)is as shown in the figure.
(##CEN_ALG_C02_SLV_026_S01.png" width="80%">
As shown in the figure, the graph CUTS the x-axis at three distinct points
Hence, equation`f(x) = 0` has three distinct roots.
(ii) For `2x^(3) - 9x^(2) + 12x - 3 = 0,`
`f(x) = 2x ^(3) - 9x + 12x - 3`
`f'(x)= 0`
`rArr6x^(2) - 18x + 12 = 0`
or6 (x - 1) (x - 2) = 0
`rArr`x = 1and x = 2
Also ` f(1)= 2 - 9 + 12 - 3 = 2`
and `f(2) = 16 - 36 + 24 - 3 = 1`
Hence, the graphof y = f(x) is as shown in the figure.

Thus , from the graph, we can see thatf(x) = 0 has only one real root, though y = f(x) has two turning points .