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5151.

Find the 10^("th")" term of "(3-4x)^(-2//3).

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ANSWER :`(2.5.8……26 )/(ROOT(3)(9.9!) )((4X)/(9))^9`
5152.

Examine whether the function f given by f(x)=x^(3)is continuous at x =0

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Solution :The function is defined at the GIVEN point x =0 and its VALUE is 0.
CLEARLY, ` underset (xto0) lim f(x) = underset (x to 0) x^(3)= 0^(3) =0`
` underset( xto 0) f(x) =0 = f (0) `
Hence, f is CONTINOUS at x = 0
5153.

Examine the consistency of the system of equations 3x-y -2z =2 2y - z =-1 3x- 5y =3

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ANSWER :INCONSISTENT
5154.

Thevalueofxin( 0, pi/2) satisfyingthe equation sinx cosx= 1/4is

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`PI/6`
`pi/3`
`pi/8`
`pi/(12)`

ANSWER :D
5155.

Prove that the points (9, 7), (11, 3) lie on a circle with centre at origin. Find the equation of the circle.

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SOLUTION :Let A and B be two points lie on the circle whose CENTRE at (0, 0).
`THEREFORE abs(bar(AC)) = sqrt(9^2+7^2) = sqrt(81+49) = sqrt(130)`
`abs(bar(BC)) = sqrt(11^2+3^2) = sqrt(121+9)=sqrt(130)`
`therefore` The points A and B lie on the circle whose centre at (0, 0) and radius is `sqrt(130)`
Eqn. of the circle is `x^2+y^2 = 130`.
5156.

The volume of the solids obtained by revolving the area of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 about major and minor axes in the ratio (a gt b).

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`B^2a^2`
`b:a`
`a:b`
`a^2:b^2`.

ANSWER :B
5157.

Show that f(x) = log_a x "on" (0,oo ),(a gt 0 "and" a != 1) functions are injective.

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Solution :`f(X)=log_a x[1,OO]`
` "for" ALPHA,beta in [1, oo]`
`"LET" f(alpha)=f(beta) `
`implieslog_a alpha = log_alpha beta IMPLIES alpha = beta `
`:." F is one -to one".`
5158.

If the solution of (dy)/(dx) = (x-y)/(x+y) is ax^(2) + bxy + cy^(2) = k, k gt 0 then the ascending order of a,b,c is

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B,C,a
a,b,c
c,a,b
a,c,b

Answer :D
5159.

Which of these unclear reactionsis possible ?

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`._(11)^(23)Na+._(1)^(1)H rarr._(10)^(20)NE + ._(2)^(4)He`
`._(5)^(10)B+._(2)^(4)He rarr._(7)^(13)N+._(1)^(1)H`
`._(5)^(10)B+._(0)^(1)nrarr._(5)^(11)B+beta^(-)+bar(v)`
`._(7)^(14)N+._(1)^(1)H rarr ._(6)^(12)C+beta^(-)+bar(v)`

Solution :Charge CONSERVED is violated in [B, C, D] , nucleon conservationis violated in (D), only (A) is balanced
5160.

underset(e)int_(0) ^(r//2) sqrt((1-sin2x)/(1+sin2x))dx=

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1
2log2
`2logsqrt(2)`
2

Answer :D
5161.

Show that the three distinct points (a^2,a) (b^2,b) and (c^2,c) can never be collinear.

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Solution :Area of triangle with VERTICES `(a^2,a) (b^2,b)` and `(c^2,c)` is
`1/2 {a^2(b-c) + b^2(c-a) + c^2(a-b)}`
= (a-b) (b-c) (a-c)
which is NEVER equal to ZERO except when a = b = c, hence the points are not collinear.
5162.

Is the function f defined by f(x)={(x","ifxle1),(5"," ifx>1):} continuous at x=1?

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Solution :`F(X)={(x,ifxle1),(5, ifx>1):}}` For `xle1,f(x)=x,` which is a polynomial function and therefore f(x) is CONTINUOUS. In PARTICULAR f(x) is continuous at x=0.
5163.

If x_(1) epsilonN and x_(1) satisfies the equation. If x^(2)+ax+b+1=0, where a,b!=-1 are integers has a root in natural numbers then prove that a^(2)+b^(2) is a composite.

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Answer :Hence `a^(2)+B^(2)` is COMPOSITE number.
5164.

Solve the following system of equations bymatrix inversion method. {{:(x + 2y + z = 7 ),(" "x + 3z = 11),(" "2 x - 3y = 1 ):}

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Answer :`X = 2, y = 1, z = 3 `
5165.

Out of (4n + 1) tickets numbered m, m+1, m+2, …., m + 4n, five tickets are chosen at random without replacement. Find the probability that these tickets are in A.P.

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ANSWER :`(2n^(2) - n)/(.^(4n + 1)C_(5))`
5166.

Impressed by Mario's answer, Quacky drags him into a chamber for a challenge.Cham- ber has two cells, A and B, with m dogs in one cell and n dogs in another. He also has a magical Dog Killer, which allows him to kill any number of dogs from any one of the cells or kill equal number of dogs from both the cells. For example, he can kill 4 dogs from A or 7 dogs from B or 3 dogs from both A and B. For winning the challenge, Mario must be the one killing the last dog. However, he realises that Quacky also has the same magical Dog Killer, and would take turns with him in order to try and get him killed.Quacky asks Mario to do the first move.How many dogs should Mario kill in his first move if cell A houses 12 dogs and B 15 dogs so that he can survive?

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10
14
16
12

Solution :If you have the first move to MAKE, what will be your stretagy? One thing is for sure that you can not always win. For example in a configuration (1,2) it is impossible for you to win no matter what move you make. In a sense (1,2) is a losing state. Your job is to find out WHETHER given configuration is a winning state or not and if it is a winning state, what will be the sequence of moves that will lead you to victory. There cannot be any losing state which has a difference of 1 since they can be brought to (1,2) which is a losing state and there is no other losing state with one of the 2 numbers i.e 1,2 since they can be brought tolosing state.
Next (3,5) is a losing state and then (4,7) (5,9) (6,11) (8,14) (10,17) (11,19) (12,21) (13,23)(15,26)
From (12,15), the only LOOSING state you can bring to is (4,7) which REQUIRES Mario to kill 16dogs( 8 from each)
5167.

Toremovethe2^( nd )termof the equationx^4 - 8x^3+ x^2-x +3=0,diminishthe rootsby

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8
`-8`
`-2`
2

Answer :D
5168.

Find the number of ways of arranging the letters of the word. INDEPENDENCE

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ANSWER :`(lfloor12)/(lfloor4lfloor3lfloor2)`
5169.

If inverse of A=[{:(1,-1,1),(2,1,-3),(1,1,1):}]is (1)/(10)[{:(4,2,2),(-5,0,alpha),(1,-2,3):}] then alpha =…..

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5
`-5`
2
`-2`

ANSWER :A
5170.

Solve for x: sin^(-1)(1-x)-2sin^(-1)x=pi/2

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ANSWER :X = 0 or `1/2`
5171.

Find the area in the first quadrant between y^(2)=4x, y6(2)=16x and the straight line x=9.

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ANSWER :36 SQUARE UNIT
5172.

Solve (3(x-2))/(5) le (5(2-x))/(3)

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ANSWER :SOLUTION SET is (`OO`, 2]
5173.

Let h(x)=f(x)=f_(x)-g_(x), where f_(x)=sin^(4)pix and g(x)=In x. Let x_(0),x_(1),x_(2) , ....,x_(n+1_ be the roots of f_(x)=g_(x) in increasing order. In the above question, the value of n is

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1
2
3
4

Answer :B
5174.

The solution of the equation dy/dx= 2y tan x = sin xsatisfying y = 0 whenx = pi/3'is

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`y = 2 SIN ^(2) X + COS x - 2`
`y = 2 sin ^(2) x - cos x - 2`
`y = 2 cos^(2) x - sin x + 2`
`y = 2 cos x- sin ^(2) x - 1`

Answer :A
5175.

Write relations in tabular form and determine their type for R={(x,y):2x-y=0} on A={1,2,3,…,13}

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SOLUTION :R`={(X,y):2x-y =0}"on" A`
`={(x,y): y=2x} "on" A`
`={(1,2),(2,4),(3,6),(4,8),(5,10),(6,12)}`
` R is neither REFLEXIVE nor SYMMETRIC nor transitive.
5176.

lim_(xrarroo) root3(x)(root3((x+1)^(2))-root3((x-1)^(2)))=

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`(1)/(3)`
`(2)/(3)`
1
`(4)/(3)`

Solution :`UNDERSET(xrarroo)(lim)root3x(ROOT3((x+1)^(2))-root3((x-1)^(2)))`
`=underset(xrarroo)(lim)x^(1//3){(x+1)^(1//3)+(x-1)^(1//3)}{(x+1)^(1//3)-(x-1)^(1//3)}`
`""underset(xrarroo)(lim)(x^(1//3){(x+1)^(1//3)+(x-1)^(1//2)}2)/({(x+1)^(2//3)+(x^(2)-1)^(1//3)+(x-1)^(2//3)})`
`""=underset(xrarroo)(lim)(2.x^(2//3){(1+(1)/(x))^(1//3)+(1-(1)/(x))^(1//3)})/(x^(2//3){(1+(1)/(x))^(2//3)+(1-(1)/(x^(2)))^(1//3)+(1-(1)/(x))^(2//3)})=(4)/(3)`
5177.

If p,q,r are three positive real numbers then the value of (p + q) (q + r) (r + p) is

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`GT` 8 pqr
`lt` 8 pqr
8pqr
8( p + q + R )

ANSWER :A
5178.

If the normal subtends a right angle at the focus of the parabola y^(2)=8x then its length is

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`2sqrt(5)`
`10sqrt(5)`
`5sqrt(5)`
`4sqrt(5)`

ANSWER :B
5179.

Thesolutionset of(x-1)(X-3) (x-5) (x-7) =9is

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`{-4,4,4 +- SQRT(10)}`
`{4,4,4 +- sqrt(10)}`
` {-4,-4, 4 +- sqrt(10)}`
none

Answer :B
5180.

7 boys and 3 girls sit in a row at random, Match the following

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a,c,d,B
d,c,b,a
b,c,d,a
c,b,a,d

Answer :B
5181.

If f(x) = (1)/( 2^n)when (1)/( 2^(n+1) ) lt x le (1)/( 2^n), n= 0,1,2… then "lt"_(n to oo)int_(t//2^n) ^(1) f(x) dx equals

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`1/3`
`2/3`
`0`
`1/2`

ANSWER :B
5182.

If 2sinh^(-1) (a/(sqrt(1-a^(2))))= log ((1+x)/(1-x))then x =

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a
`1/a`
`SQRT(1-a^(2))`
`1/(sqrt(1-a^(2)))`

ANSWER :a
5183.

Find the number of values of x satisfyingint_(0)^(x)t^(2)sin (x-t)dt=x^(2) in [0, 100].

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ANSWER :16
5184.

Let f(x)=sinx+cos(sqrt(4-a^(2)))x. Then, the integral values of 'a' for which f(x) is a periodic function, are given by

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{2,-2}
(-2,2]
[-2,2]
NONE of these

Answer :D
5185.

The remainder obtained when 5^(124) is divide by 124 is

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5
0
2
1

Answer :A
5186.

The condition that the circle x^(2) +y^(2) + 2gx+ 2fy + c = 0 to bisect the circumference of the circle x^(2) + y^(2) + 2g^(1)x + 2f^(1)y + c^(1) = 0 is

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16
24
`-42`
-62

Answer :D
5187.

The sum of the fourth powers of the roots of the equation x^(5) + px^(3) + qx^(2) + s = 0is

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`2P^(2)`
`3P^(2)`
2p
3p

Answer :1
5188.

{:("Quantity A","Quantity B"),("The tenths digits of the product","The tenths digit of the product"),("of two even integers","of an even and an odd integer"),("divided by 4","divided by 4"):}

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ANSWER :The RELATIONSHIP cannot be DETERMINED from the INFORMATION GIVEN.
5189.

If Q(x) = (1)/(2) a_(0) + a_(1) cos x + b_(1) sin x+ a_(2) cos 2x +…+ a_(n) cos nx + b_(n) sin nx, then the value of int_(0)^(2pi) Q(x) sin kx dx (k=1,2,…n)

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`pia_(K)`
`pi a_(0)`
`pib_(N)`
`pib_(k)`

Answer :D
5190.

Evaluate the integrals by using substitution int_(0)^(1)sin^(-1)((2x)/(1+x^(2)))dx

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ANSWER :`(PI)/(2) - LOG 2`
5191.

lim_(n to oo)(int_(1//(n+1))^(1//n)tan^(-1)(nx)dt)/(int_(1//(n+1))^(1//n)sin^(-1)(nx)dx) is equal to

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SOLUTION :N//A
5192.

Integrate the following function : int(x-1)/(sqrt((x+1)(x-2)))dx

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ANSWER :`sqrt(x^(2)-x-2)-(1)/(2)LOG|(x-(1)/(2))+sqrt(x^(2)-x-2)+c|`
5193.

Consider an ellipse E:x^(2)/a^(2)+y^(2)/b^(2)=1, centered at point 'O' and having AB and CD as its major and minor axes respectively if S_1be one of the focus of the ellipse, radius of the incircle of ∆OCS_1 be 1 unit and OS_1=6units.Q. The equation of the director circle of (E) is

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`X^(2)+y^(2)=48.5`
`x^(2)+y^(2)=97`
`x^(2)+y^(2)=sqrt48.5`
`x^(2)+y^(2)=sqrt97`

ANSWER :A::B::D
5194.

A purse contains 4 copper and 3 silver coins while another purse contains 6 copper and 2 silver coins. A purse is drawn at random and coin is drawn. Then ....... is the probability that selected coin is of copper.

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`(3)/(7)`
`(4)/(7)`
`(3)/(4)`
`(37)/(56)`

Answer :D
5195.

Parametric form of the equation of the planeis bar r =(2hati + hatk ) + lambda hati + mu (hati + 2hati-3 hatk) lambda and mu are parameters. Find normal to the plane and hence equation of the plane in normal form. Write its Cartesion form.

Answer»

Solution :The EQUATION `barr = BARA + lambdabarb + mubarc` represents a plane passing through a point having position vector a and parallel to vectors `barb` and `BARC.`
Here, `bara = 2hati + hatk, barb = hati, barc = hati + 2hatj - 3hatk`
`therefore` The normal vector `barn` to the plane is give by
`barnxxbarbxxbarc=|{:(hati,,hatj,,hatk),(1,,0,,0),(1,,2,,"-3"):}|`
` = (0-0) hati -(-3-0) hatj + (2-0)hatk`
i.e., `barn = 3hatj + 2hatk`
The equation of the plane is scalar product form is
` barr * (barb xx barc) = bara * (barb xx barc)`
i.e., `barr * barn = bara * barn`
where `bara * barn = (2hati + hatk) * (3hati + 2hatk)`
` = 2(0) + 0(3) + 1(2)`
` = 0+ 0 + 2 = 2`
`therefore` the equation of the plane is scalar product form is
`barr*(3hatj + 2hatk) =2`
i.e., `barr * barn = 2`, where `barn = 3hatj + 2hatk`
`therefore |barn| = sqrt(3^(2) + 2^(2)) = sqrt(13)`
The equation `barr * barn = 2`can be written as
`barr*(barn)/(|barn|)=(2)/(|barn|)`
i.e., `barr*((3hatj+2hatk)/(sqrt(13)))=(2)/(sqrt(13))`
i.e., `barr*((3)/(sqrt(13))hatj+(2)/(sqrt(13))hatk)=(2)/(sqrt(13))`
This is the normal form of the equation of plane.
If `barr = XHATI + yhatj + zhatk`, then equation (1) BECOMES
`(xhati+yhatj+zhatk)*(3hatj+2hatk)=2`
`therefore x(0)+y(3)+z(2)=2`

` therefore 3y + 2z = 2`
This is the cartasian equation of the plane.
5196.

If for x in (0,1/4) , the derivation of tan^(-1)((6xsqrtx)/(1-9x^3)) is sqrt(xg(x) , then find g(x).

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ANSWER :`G(X)=9/(1+9x^3)`
5197.

The truth table of p vv (~q) rArr p is

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ANSWER :C
5198.

int3x^2dx' is equal to :

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x+C
x^2+c
x^3+c
x^4+c

Answer :C
5199.

int_(-1)^(1){((x+2)/(x-2))^(2)+((x-2)/(x+2))^(2)-2}^(1//2)dx=

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`8"LN" (4)/(3)`
`8"ln"(3)/(4)`
`4"ln"(4)/(3)`
0

Answer :1
5200.

If a point P is moving such that the lengths of tangents drawn from P to the circles x^(2) + Y^(2) -4x -6y -12 = 0 and x^(2) +y^(2) + 6x + 18 y + 26 = 0are in the ratio 2 : 3, then find jthe equation of the locus of P.

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Answer :`5X^(2) + 5Y^(2) - 60 X - 126 y - 212 = 0`