Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Let A_1,A_2,A_3,….,A_m be the arithmetic means between -2 and 1027 and G_1,G_2,G_3,…., G_n be the gemetric means between 1 and 1024 .The product of gerometric means is 2^(45) and sum of arithmetic means is 1024 xx 171 The n umber of arithmetic means is

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442
342
378
none of these

SOLUTION :`A_(1)+A_(2)+A_(3)+….+A_(m-1)+A_(m)=1025xx771`
`rArrm((-2+1027)/2)=1025xx171`
or m=342
2.

Let A_1,A_2,A_3,….,A_m be the arithmetic means between -2 and 1027 and G_1,G_2,G_3,…., G_n be the gemetric means between 1 and 1024 .The product of gerometric means is 2^(45) and sum of arithmetic means is 1024 xx 171 The value of Sigma_(r=1)^(n) G_ris

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SOLUTION :`G_(1)G_(2)…G_(n)=(SQRT(1xx1024))^(n)=2^(5n)`
GIVEN, `2^(5n)=2^(45)rArrn=9`
Hence, `r=(1024)^(1/(9+1))=2`
`rArrG_(1)=2,r=2`
`rArrG_(1)+G_(2)+..+G_(9)=(2xx(2^(9)-1))/(2-1)=1024-2=1022`
3.

I : If a = I +j - k, b = 2i - 3j + 2k, c = 13 I - 7j + 3k then [a b c] = 0 II : If a,b,c are mutually perpendicular unit vector then [a b c]^(2) = 1

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only I is ture
Only II is ture
both I and II are true
Neither I nor II are true

Answer :C
4.

Evaluate intxsqrt(1 + x - x^(2))dx

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ANSWER :`-(1)/(3)(1+x-x^(2))^(3//2)+((2x-1))/(8)sqrt(1-x-x^(2))+(5)/(16)SIN^(-1)((2x-1))/(sqrt(5))+c`
5.

A box contains 19 screws, 3 of which are defective, Two screws are drawn at random with replacement. Find the probability that neither of the two screws is defective.

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ANSWER :`0.709`
6.

Using elementary transformation find the inverse of the matrix : A=((1, -3,2),(3, 0, 1),(-2, -1, 0))

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SOLUTION :N/A
7.

Number of ways of forming a committee of 6 members out of 5 Indians. 5 Americans and 5 Australians such that there will be atleast one member from each county in the committee is

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3375
4375
3875
4250

Answer :B
8.

bar(a),bar(b) and bar( c ) are three non zero, non planar vectors. bar(p)=bar(a)+bar(b)-2bar( c ),bar(q)=3bar(a)-2bar(b)+bar( c ) and bar( r )=bar(a)-4bar(b)+2bar( c ). The volume of the parallelepiped formed by the vectors bar(a),bar(b) and bar( c ) is V_(1) and the volume of the parallelepiped formed by the vectors bar(p),bar(q) and bar( r ) is V_(2) then V_(2):V_(1) = .............

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`3:1`
`7:1`
`11:1`
`15:1`

ANSWER :D
9.

The mid point of the chord 2x+5y=12 of the ellipse 4x^(2) + 5y^(2) = 20 is

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6,0
1,2
`-3/2,3`
11,2

Answer :B
10.

Evaluate int(x^(2)-1)/(x^(4)+x^(2)+1)dx

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ANSWER :`(1)/(2)LOG|(x^(2)-x+1)/(x^(2)+x+1)|+C`
11.

Evaluate int(x^(2)+1)/(x^(4)-x^(2)+1)dx

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ANSWER :`tan^(-1)((X^(2)-1)/(x))+c`
12.

int (x^(2)+1)/(x^(4)+x^(2)+1)dx=

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Answer :`(1)/(SQRT(3))tan^(-1)((X^(2)-1)/(xsqrt(3)))+c`
13.

Evaluate int(x^(2)-1)/(x^(4)-x^(2)+1)dx

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Answer :`(1)/(2sqrt(2))log|(x^(2)-sqrt(3)x+1)/(x^(2)+sqrt(3)x+1)|+C`
14.

If a diameter of a hyperbola meets the hyperbola in real points then

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`3x^(2)+4y^(2)-6x-8y+4=0`
`3x^(2)+4y^(2)-2x-8y+4=0`
`4X^(2)+3Y^(2)-8x-6y+4=0`
`4x^(2)+3y^(2)-8x-2y+4=0`

ANSWER :a
15.

Match the following lists:

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Solution :`atos,bto""p,ctoq,d""tor`
(a) `x^(2)ax+B=0` has ROOT `lapha`. HENCE.
`alpha^(2)+aalpha+ b=0""(1)`
`x^(2)+px+q=0` has root `-alpha`. Hence,
`alpha^(2)-palpha+q=""(2)`
ELIMINATING `alpha` from (1) and (2), we get
`(q-b)^(2)=(aq+bp)(-p-a)`
or `(q-b)^(2)=-(aq+pb)(p+a)`
(b) `x^(2)ax+b=0` has root `alpha`. Hence,
`alpha^(2)+aalpha+b=0 ""(1)`
`x^(2)+px+q=0` has root `1//alpha`. Hence,
`qalpha^(2)+palpha+1=0""(2)`
Eliminating `alpha` from (1) and (2), we get
`(1-bq)^(2)=(a-pb)(p-aq)`
(c) `x^(2)+ax+b=0` has roots, `alpha,BETA`. Hence,
`alpha^(2)+aalpha+b=0""(1)`
`x^(2)+px+q=0` has roots `-2//alpha,gamma`. Hence,
`qalpha^(2)-2palpha+4=0""(2)`
Eliminating `alpha` from (1) and (2), we et
`(4-bq)^(2)=(4a+2pb)(-2p-aq)`
(d) `x^(2)+ax+b=0` has roots `alpha,beta`. Hence,
`alpha^(2)+aalpha+b=0""(1)`
`x^(2)+px+q=0` has roots `-1//2alpha,gamma`. Hence,
`4qalpha^(2)-2palpha+1=0""(2)`
Elimintaing `alpha` from (1) and (2), we get
`(1-4bq)^(2)=(a+2bp)(-2p-4aq)`
16.

For what values of x is the rate of increase of x^(3)-2x^(2)+3x+8 is twice the rate of increase of x..

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`(-(1)/(3), -3)`
`((1)/(3), 3)`
`(-(1)/(3),3)`
`((1)/(3),1)`

ANSWER :A::C
17.

If |z-3+i| = 4 then the locus of z = x +iy is

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`X^(2)+ y^(2) -6 x + 2y - 6 = 0`
`x^(2) + y^(2)- 6 = 0`
`x^(2) + y^(2) -3 x + y - 6 = 0`
`x^(2) + y^(2) = 0`

ANSWER :A
18.

Is the function defined by f(x)= |x|, a continuous function?

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ANSWER :HENCE, F is CONTINUOUS at all POINTS.
19.

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

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Answer :Cosine FUNCTION is CONTINUOUS for all `x in R`; cosecant is continuous EXCEPT for `x= npi, N in Z`; secant is continuous except for `x= (2n+1)(pi)/(2), n in Z` and cotangent function is continuous except for `x= npi, n in Z`.
20.

(1-omega)(1-omega^2)(1-omega^4)(1-omega^5)=9

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SOLUTION :L.H.S.`=(1-omega)(1-omega^2)(1-omega^4)(1-omega^5)`
`=(1-omega)(1-omega^2)(1-omega)(1-omega^2)`
(1-omega)^2(1-omega^2)^2`
`={(1-omega)(1-omega^2)}^2`
`=(1-omega^2-omega+omega^3)^2`
`{3-(omega^2+omega+1)^2`
`(3)^2=9=R.H>S` (PROVED)
21.

Show that (vecaxxvecb)^2 = a^2b^2 - (veca.vecb)^2.

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SOLUTION :`vecaxxvecb = ab sinthetahatn`
Where a = `|veca|, B = |vecb|` = `THETA` is the angle between two vectors `veca` and `vecb` and `hatn` is the unit vector perpendicular to `veca.vecb`.
Now `(vecaxxvecb)^2 = (vecaxxvecb).(vecaxxvecb)
= `a^2b^2 sin^2 theta (hatn.hatn)`
= `a^2 b^2 sin^2 theta = a^2b^2- a^2b^2 cos^2 theta`.
= `a^2b^2-(abcostheta)^2 = a^2b^2-(veca.vecb)^2`. (PROVED)
22.

(1)/(2.3)+(1)/(4.5)+(1)/(6.7)+……oo=

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`1+log_(E )2`
`1-log_(e )2`
`(1)/(2)-log_(e )2`
`log_(e )2`

Answer :B
23.

The sum to n terms of the series 1+ 5 ((4n +1)/(4n - 3)) + 9 ((4n +1) /( 4n -3)) ^(2) + 13 ((4n +1)/( 4n -3)) ^(3) + ((4n + 1 )/( 4n -3)) ^(2)+ ……. is .

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ANSWER :`(4N - 3)`
24.

For any value oftheta if the straight lines x sin theta + (1-cos theta ) y = a sin theta and xsin theta -(1+ cos theta)y + a sin theta = 0 intersect at P(theta) thenthe locus of P(theta)is a

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STRAIGHT line
circle
PARABOLA
HYPERBOLA

ANSWER :B
25.

If the length of three sides of a trapezium other than base are equal to 10 cm. Than find the area of trapezium when it is maximum.

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ANSWER :`75sqrt(3)(CM)^(2)`
26.

State whether the following statements are true or false . Justify . If ** is commutative binary operation on N, then a"*" (b"*" c)=(c"*"b)"*"a.

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SOLUTION :N/A
27.

Evaluate int (x cos^(-1)x)/(sqrt(1-x^(2))dx

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ANSWER :`-SQRT(1-x^(2))COS^(-1)x+x+C`
28.

Consider the lines represented by equation (x^(2) + xy -x) xx (x-y) =0 forming a triangle. Then match the following lists:

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Solution :
The GIVEN lines are x (x+y-1) (x-y)=0. So, lines x =0, x+y-1=0, and x-y=0 form triangle OAB as shown in the diagram.
The triangle is right-angled at pont B. Hence, the orthocenter is (1/2, 1/2). Also, the CIRCUMCENTER is the midpoint of OA which is (0, 1/2). The centroid is
`((0+(1//2) +0)/(3), (0+(1//2)+1)/(3)) " or "((1)/(6), (1)/(2))`
`"Also, " OA = 1, OB = OC= 1//2sqrt(2).` Hence, the incenter is
`((0(1//sqrt(2))(1//2)1+0(1//sqrt(2)))/((1//sqrt(2))+1+(1//sqrt(2))),(0(1//sqrt(2))+(1//2)(1)+1(1//sqrt(2)))/((1//sqrt(2))+1+(1//sqrt(2)))) -= ((1)/(2+2sqrt(2)), (1)/(2))`
29.

Let p, q in R. "If " 3 - sqrt(3) is a root of thequadratic equation x^(2) + px + q = 0 , then

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`p^(2) - 4Q + 12 = 0`
`Q^(2) - 4q - 16 = 0`
`q^(2) + 4p + 14 = 0`
`p^(2) - 4q - 12 = 0`

Answer :D
30.

Iff(x)={:{((x -1)/(2x^(2) - 7 x +5) ," for "x ne 1) , (-1/3 , " for "x = 1):} thenf (1)is equal to

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`-1/9`
`-2/9`
`-1/3`
`1/3`.

ANSWER :B
31.

The probability of three mutually exclusive events A, B and C are given by 2/3, 1/4, 1/6 respectively Is this statement a)True? b)False? c)Cannot be said? d)Data not sufficient?

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TRUE?
False?
Cannot be SAID?
DATA not sufficient?

Answer :B
32.

If barb and barc are two non-collinear vectors, then number of solution (x,y) in x,y in [0,10], satisfying the equation bara.[barb+barc] = 5 and bara xx (barb xx barc) = (x^2 -2x+7) barb+(siny)barc is :

Answer»

ONE
two
zero
infinite

Solution :`bara XX (BARB xx barc) = (bara. ""barc)barb - (bara-barb).barc`
`impliesbara-barc =x^2 - 2x + 7, bara .""barb=-siny`
`bara.(barb + barc) = 5`
`impliesx^2 - 2x + 7 -siny=5`
`implies (x-1)^2 + 1 = sin y`
`x=1, y=(4n+1) (pi)/(2) = (pi)/(2),(5pi)/(2)`
33.

Match the relation for derivatives given in List II with the relation given in List I and then choose the correct code.

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<P>`{:(a,b,c,d),(q,p,s,r):}`
`{:(a,b,c,d),(s,p,q,r):}`
`{:(a,b,c,d),(r,q,s,p):}`
`{:(a,b,c,d),(q,p,r,s):}`

Solution :a. `"If"xy-log y=1" then "xy_(1)+y=y_(1)//y`
`"so "y^(2)+(xy-1)y_(1)=0`
b. `ysqrt(1-x^(2))=sin^(-1)x`
`RARR" "-(yx)/(sqrt(1-x^(2)))+y_(1)sqrt(1-x^(2))=(1)/(sqrt(1-x^(2)))`
`"so "-xy+y_(1)(1-x^(2))=1`
c. `y^(2)=2x-x^(2)rArr yy_(1)=1-x`
`rArr" "y_(1)^(2)+yy_(2)=-1`
`"so "(1-x)^(2)/(y^(2))+yy_(2)=-1`
`rArr""y^(3)y_(2)=-[1+x^(2)-2x+y^(2)]=-1`
d. `y=e^(sqrt(x))+e^(-sqrt(x))`
`rArr" "y_(1)=(e^(sqrt(x))-e^(-sqrt(x)))/(2sqrt(x))`
`rArr" "y_(1)^(2)x=(e^(sqrt(x))-e^(-sqrt(x)))^(2)=y^(2)-4`
`rArr""8y_(1)y_(2)x+4y_(1)^(2)=2yy_(1)`
`rArr" "4y_(2)x+2y_(1)=y`
`rArr" "xy_(2)-(1)/(2)y_(1)-(1)/(4)y=0`
34.

If 3hat(i)+hat(j)-2hat(k)andhat(i)-3hat(j)+4hat(k) are the diagonals of a parallelogram, then the area of the parallelogram is

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`10sqrt(3)sq.units`
`5sqrt(3)sq.units`
`5sqrt(2)sq.units`
`10sqrt(2)sq.units`

ANSWER :B
35.

Phenol and benzoic acid can not be separated by

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`NaHCO_(3)`
`NAOH`
NA
`NaNH_(2)`

ANSWER :B::C::D
36.

n^(th) term of log_(e )(6//5) is

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`((-1)^(n-1))/(5^(n))`
`((-1)^(n-1))/(n*5^(n))`
`((-1)^(n-1))/(6^(n))`
`((-1)^(n-1))/(n*6^(n))`

ANSWER :B
37.

If the relation R:A rarr B where A={1,2,3,4} and B={1,3,5}is defined by R={(x,y),xlty,x in A,y in B} then ROR^(-1) is

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`{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)}`
`{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)}`
`{(3,3),(3,5),(5,3),(5,5)}`
`{(3,3),(3,4),(4,5)}`

ANSWER :C
38.

Evaluate the determinants below in examples number 1 and 2 |{:(2,4),(-5,-1):}|

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ANSWER :18
39.

A monument ABCD stands at A on a level ground. At a point P on the ground the portions AB, AC, AD subtend alpha, beta, gamma respectivelyl. If AB = a, AC = b, AD = c, AP = x and alpha + beta + gamma = 180^(@) "then x"^(2) is equal to

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`(a)/(a+B+C)`
`(b)/(a+b+c)`
`(c)/(a+b+c)`
`(ABC)/(a+b+c)`

ANSWER :D
40.

The logically equivalent statement of p toqis

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<P>`(p ^^ Q) VV ( q to p)`
`(p ^^ q) to (p vv q)`
`( p to q) ^^ (q to p)`
`(p ^^ q) vv (p ^^ q)`

ANSWER :C
41.

Show that the position vector of the point P, which divides the line joining the points A and B having position vectors veca and vecb internally in ratio m:n is (mvecb+nveca)/(m+n)

Answer»
42.

If the mapping f(x) = mx + c, m gt 0maps [-1,1] onto [0,2], thentan ( tan^(-1). 1/7 + cot ^(-1) 8 + cot ^(-1) 18) is equal to

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`F(2/3)`
`f(1/3)`
`f((-1)/3)`
`f((-2)/3)`

ANSWER :D
43.

Solve(dy)/(dx)=(3x+y+4)^(2)

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ANSWER :x+c
44.

Show that the position vector of the point P, which divides the line joining the points A and B having position vector a and b internally in the ratio: m:n is (mvecb + nveca)/(m+n)

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ANSWER :`VECR = (mvecb + nveca)/(m+n)`
45.

A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package on screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacturer a package of screws B. Each machine is available for at the most 4 hours on any day.The manufacturer can sell a package of screws A at a profit of Rs. 7 and screws B at a profit of Rs. 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day order to maximise his profit? Determine the maximum profit.

Answer»

Solution :Let the manufacturer MAKES `x` packets of screw `A` and `y` packets of screw B per DAY. Then

Maximise `Z=7x+10y`………………1
and constraints
`4x+6yle240implies2x+3yle120`…………………2
`6x+3yle240implies2x+yle80`.....................3
and `xge0,yge0`

First draw the graph of the line `2x+3y=120`

Put `(0,0)` in the INEQUATION `2x+3yle120`
`2xx0+3xx0le120`
`implies 0le120` (True)
Therefore, half plane contains the origin,
Now draw the graph of the line `2x+y=80`

Put `(0,0)` in the inequation `2x+yle80`
`2xx0+0le80implies0le80` (True)
Therefore half plane contains the origin.Since, `x,yge0`. So the feasible region will be in first quadrant.
From EQUATIONS `2x+3y=120` and `2x+y=80` the point of intersection is `B(30,20)`.
`:.` Feasible region is OABCE.
The vertices of the feasible region are `O(0,0), A(40,0),B(30,20)` and `C(0,40)` we find the value of `Z` at these points.

Therefore, the maximum value of `Z` is RS. 410 at point `B(30,20)`. Therefore, the maximum profit of Rs. 410 will be obtained when the factory produces 30 packets of screw A and 20 packets of screw B per day.
46.

If the square of the length of the tangents from a point P to the circles x^(2)+y^(2)=a^(2), x^(2)+y^(2)=b^(2), x^(2)+y^(2)=c^(2) are in A.P. then a^(2),b^(2),c^(2) are in

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A.P.
G.P.
H.P.
A.G.P.

Answer :A
47.

Represent graphically : (i) A displacement of 50 km., 30^(@) West of Sout. (ii) A displacement of 70 km., 40^(@) West of North. (iii) A displacement of 50 km., 45^(@) North of East.

Answer»


ANSWER :`(##KPK_AIO_MAT_XII_P2_C10_E01_001_A01##)`
48.

Two ships A and B are sailing straight away from the foot of a tower OP along routes such that ul(|AOB) is always 120^(@). At a certain instance, the angles of depression of the ships A and B from the top P of the towers are 60^(@) and 30^(@) respectively. The distance between the ships when the height of the tower is 15m is

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`5 SQRT(39)` m
`5 sqrt(30)` m
`5 sqrt(21)` m
`5 sqrt(3)` m

ANSWER :A
49.

x+(x^(3))/(3)+(x^(5))/(5)+(x^(7))/(7)+….oo=

Answer»

`1/2log_(E )((1+X)/(1-x))``
`log_(e )(1-x)`
`tan h^(-1)x`
`SIN h^(-1)y`

Answer :A
50.

Solve :cos ^(-1) x + sin ^(-1) "" (x)/( 2) = (pi)/(6)

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ANSWER :1