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.

If A=[{:(1,0,-1),(2,1,3),(0,1, 1):}] then verify that A^(2)+A=(A+I) , where I is 3xx3 unit matrix.

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SOLUTION :We have, `A=[{:(1,0,-1),(2,1,3),(0,1,1):}]`
`THEREFORE A^(2)=A.A`
`=[{:(1,0,-1),(2,1,3),(0,1,1):}][{:(1,0,-1),(2,1,3),(0,1,1):}]=[{:(1,-1,-2),(4,4,4),(2,2,4):}]`
`therefore A^(2)+A=[{:(1,-1,-2),(4,4,4),(2,2,4):}]+[{:(1,0,-1),(2, 1,3),(0,1,1):}]`
`=[{:(2,-1,-3),(6,5,7) ,(2,3,5):}]`
Now, `A+I=[{:(1,0,-1),(2,1,3),(0,1,1):}]+[{:(1,0,0),(0,1,0),(0,0,1):}]=[{:(2,0,-1),(2,2,3),(0,1,3):}]`
and `A(A+I)=[{:(1,0,-1),(2,1,3),(0,1,1):}][{:(2,0,-1),( 2,2,3),(0,1,2):}]=[{:(2,-1,-3),(6,5,7),(2,3,5):}]`
THUS, we see that `A^(2)+A=A(A+I)`
2.

If (ax-1)/((1-x+x^(2))(2x+x))=(x)/(1-x+x^(2))-(1)/(2+x) then a=

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3
-3
2
-2

Answer :A
3.

If a and b are real numbers between 0 and 1 such that z_(1) = a + i and z_(2) = 1 + bi and z_(3) = 0 form an equilateral triangle then find a and b .

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ANSWER :A::B::C
4.

Evaluate (iii) int_(0)^(1)(1-x)/(1+x)dx

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ANSWER :`2 LOG 2 - 1`
5.

Find adjoint of each of the matrices in Exercises 1 and 2 [{:(1,2),(3,4):}]

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ANSWER :`[{:(4,-2),(-3,1):}]`
6.

a,b,c, are positive real numbers forming a. G.P. If ax^(2) + 2bx+ c = 0 and dx^(2) + 2ex + f = 0 have a common root, then prove that d//a, e//b, f//c are in A.P.

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7.

Evaluate the following integrals int(dx)/((x+2)sqrt(x+1))

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ANSWER :`2TAN^(-1)(SQRT(x+1))+C`
8.

If f(x)=x(e|x|+|x|-2)/(|x|+|x|)(where |x| denotes the greaster integer less than or equal to x), then-

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`UNDERSET (xrarr0+)"lim"f(x)=-1`
`underset (xrarr0-)"lim"f(x)=0`
`underset (xrarr0+)"lim"f(x)` does not exist.
underser (xrarr0)"lim"f(x) does not exist.

ANSWER :A,B,D
9.

A : (a, b) = theta rArr (5a, -3b) = pi - theta R :m gt 0, n lt 0, (ma, nb) = pi - (a, b)

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Both A and R are TRUE and R is the correct explanation of A.
Both A and R are ture but R is not the correct explanation of A.
A is ture, but R is false
A is false, but R is true

ANSWER :A
10.

Find the area of the region bounded by the line 6x+5y=30,x-axis and the lines x=-1 and x=3.

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ANSWER :`(96)/(5)`
11.

If alpha = lim_(x to 0) (x*2^(x)-x)/(1 - cos x) and beta= lim_(x to 0) (x*2^(x)- x)/(sqrt(1 + x^(2))-sqrt(1-x^(2))) then

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`ALPHA = 5 BETA`
`alpha = 2beta`
`beta = 2ALPHA^(2)`
`beta = (1)/(6)alpha`

ANSWER :B
12.

obtain the equation of hyperbola in each of the following cases : foci at (+-4,0) and vertices at (+-2,0).

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SOLUTION :Here c=4 ,a=2 `THEREFORE` b^2=c^2-a^2=12` `therefore` EQN of the HYPERBOLA is `x^2/a^2+y^2/b^2=1` or `x^2/4-y^2/12=1`
13.

Evaluate the following lim_(xto2)(x^3-8)/(x^5-32)

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SOLUTION :`lim_(xto2)(x^3-8)/(x^5-32)=lim_(xto2)(x^3-2^3)/(x^5-2^5)`
`lim_(xto2)((x^3-2^3)/(x-2))/((x^5-2^5)/(x-2))=(lim_(xto2)(x^3-2^3)/(x-2))/(lim_(xto2)(x^5-2^5)/(x-2))`
`=(3xx2^2)/(5xx2^4)=3/20 [thereforelim_(xtoa)(x^n-a^n)/(x-a)=NA^(n-1)`
14.

Let x in R and let [{:(1,,1,,1),(0,,2,,2),(0,,0,,3):}],Q=[{:(2,,x,,x),(0,,4,,0),(x,,x,,6):}]and R=PQR^(-1) which of the following options is/are correct?

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There exists a real, number x such that `PQ= QP`
For `x=0,if R[{:(1),(a),(b):}]=6[{:(1),(a),(b):}],then""a+b =5`
For `x=1`, there exists a unit vector `alphahati+betahatj+gamma hatk` for which
`R[{:(alpha),(BETA ),(gamma):}]=[{:(0),(0),(0):}]`
det `R=Ddet[{:(2,,x,,x),(0,,4,,0),(x,,x,,5):}]+8, "forall " x in R `

SOLUTION :It is given, then matrices
`P=[(1,1,1),(0,2,2),(0,0,3)],Q=[(2,x,x),(0,4,0),(x,x,6)]`
`therefore""p^(-1)=(adj(P))/(|P|)`
`as |P|=6 and " adj P"=[(6,0,0),(-3,3,0),(0,-2,2)]`
`RARR""p^(-1)=(1)/(6)[(6,-3,0),(0,3,-2),(0,0,2)]`
`therefore""|R|=|PQP^(-1)|""[because R=PQP^(-1)"(given)"]`
`rArr""|R|=|P||Q||P^(-1)|=|Q|""[because|P||P^(-1)|=|I|=1]`
`=|(2,x,x),(0,4,0),(x,x,6)|=|(2,x,x),(0,4,0),(x,x,5)|+|(2,x,0),(0,4,0),(x,x,1)|`
`=|(2,x,x),(0,4,0),(x,x,5)|+2(4-0)-x(0-0)+0(0-4x)`
`=|(2,x,x),(0,4,0),(x,x,5)|+8 " for all " x inR`
`because""PQ=[(1,1,1),(0,2,2),(0,0,3)][(2,x,x),(0,4,0),(x,x,6)]`
`=[(2+x,4+2x,x+6),(2x,2x+8,12),(3x,3x,18)]`
`"and"QP=[(2,x,x),(0,4,0),(x,x,6)][(1,1,1),(0,2,2),(0,0,3)]`
`=[(2,2+2x,2+5x),(0,8,8),(x,3,3x+18)]`
There is no common value of 'x' for which each corresponding element of matrices PQ and QP is equal.
For `x=0,Q=[(2,0,0),(0,4,0),(0,0,6)]`
then, if `R[(1),(a),(b)]=6[(1),(a),(b)]`
`rArrPQP^(-1)[(1),(a),(b)]=6[(1),(a),(b)]""[because R=PQP^(-1)]`
`rArr(1)/(6)[(1,1,1),(0,2,2),(0,0,3)][(2,0,0),(0,4,0),(0,0,6)][(6,-3,0),(0,3,-2),(0,0,2)][(1),(a),(b)]=6[(1),(a),(b)]`
`rArr""(1)/(6)[(2,4,6),(0,8,12),(0,0,18)][(6,-3,0),(0,3,-2),(0,0,2)][(1),(a),(b)]=6[(1),(a),(b)]`
`rArr""[(12,6,4),(0,24,8),(0,0,36)][(1),(a),(b)]=36[(1),(a),(b)]`
`rArr""[(12+6a+4b),(0+24a+8b),(0+0+36b)]=[(36),(36a),(36b)]`
`rArr""6a+4b=24 and 12a=8b`
`rArr""3a+2b=12 and 3a=2b`
`rArr a=2 and b=3`
So `a+b=5`.
Now, `R[(alpha),(beta),(gamma)]=[(0),(0),(0)] and alphahati+betahatj+gammahatk" is a unit vector, so det "(R)=0`
`rArrdet(Q)=0""[becauseR=PQP^(-1)" So, "|R|=|Q|]`
`rArr""|(2,x,x),(0,4,0),(x,x,6)|=0`
`rArr""2(24-0)-x(0-0)+x(0-4x)=`
`rArr""48-4x^(2)=0`
`rArr""x^(2)=12 rArr x= pm2sqrt3`
So, for `x=1`, there does not exist a unit vector `alpha hat (i) +betahat(j) +gammahat(k)`, for which `R [{:(alpha,),(beta,),(gamma,):}]=[{:(0,),(0,),(0,):}]`
Hence, options (b) and (d) are correct.
15.

A coin is biased so that the probability of falling head when tossed is (1)/(4) . If the coin is tossed 5 times , the probability of obtaining 2 heads and 3 tails is

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`(135)/(512)`
`(75)/(512)`
`(1)/(512)`
`(1)/(256)`

ANSWER :A
16.

Find the derivative of tan^(-1) [(sqrt(1+x^(2))-1)/(x)] with respect to tan^(-1) ((2x)/(1-x^(2)))

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ANSWER :`(1)/(4)`
17.

The set N of positive natural number set is finite set or infinite set ?

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Solution :"The set N of POSITIVE NATURAL NUMBERS" is an infinite set.
18.

Define a binary operation ** on the set {0,1,2,3,4,5} as a**b={{:(a+b","," if " a+blt6),(a+b-6","," If " a + b ge6):}Show that zero is the identity for this operation and each element a ne 0 of the set is invertible with 6 - a being the inverse of a.

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SOLUTION :N/A
19.

Solve the following linear programming problems graphically : Maximize : Z = x + 170y subject to the constraints 3x+2y le 6, x+4y le 16, x, y ge 0.

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ANSWER :The maximum profit of Rs. 48,000 COMPANY has to produce 150 SWEATER.
20.

Of the following four types of savings account plans, which option would yield exponential growth of the money in the account?

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Each successive year, 2% of the initial savings is added to the VALUE of the account.
Each successive year, 1.5% of the initial savings and $100 is added to the value of the account.
Each successive year, 1% of the CURRENT value is added to the value of the account.
Each successive year, $100 is added to the value of the account.

Answer :C
21.

Find the area enclosed between y= sin 2x,y= sqrt3 sin x, x = 0 and x= (pi)/(6)

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Answer :`((7)/(4)- SQRT3)` SQ UNITS
22.

The range of the function f(x) = 3x^2+7x+10 is

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`[10,PROP)`
`[(71)/(12),prop)`
`[1,prop)`
None of these

Answer :B
23.

The number of ways of selecting 6 books from a library which has 8 books each on History ,Civies,Economics and Geography is

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36
84
66
None of these

Answer :B
24.

Find the probability of drawing and ace or a spade from a well suffled pack of 52 cards ?

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ANSWER :`(4)/(13)`
25.

int (sinx)/(3+4 cos^(2)x)dx

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Answer :`I=(-1)/(2sqrt(3))tan^(-1)((2cos X)/(SQRT(3)))+C`
26.

Find the equation of the circle orthogonal to each of the circles x^(2)+y^(2)+2x+4y+1=0,2(x^(2)-y^(2))+6x+8y-3=0, x^(2)+y^(2)-2x+6y-3=0

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ANSWER :`X^(2)+y^(2)-5x-14y-34=0`
27.

int ( sin^(6) x + cos ^(6) x + 3 sin ^(2) x cos ^(2) x ) dx is equal to

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`x +C`
` ( 3)/( 2) SIN 2X +C`
` - ( 3)/( 2) cos 2x + C`
NONE of these

Answer :A
28.

The scatterplot above shows the number of people diagnosed with melanoma, in ten-thousands, from 1940 to 1970. Based on the line of best fit to the data, as shown in the figure, which of the followingvalues is closest to the average yearlyincrease in the number of incidences of melanoma?

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1300
330
0.33
0.13

Answer :A
29.

Evaluate : int_(0)^((pi)/(2)) (x sin x.cosx)/(sin^(4)x+cos^(4)x)dx

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ANSWER :`(PI^(2))/(16)`
30.

if 0 < a < b> 0 nadc > 0thenboththe rootsof theequation2ax ^2+ 3bx + 5c=0

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are REAL andnegative
havenegativerealparts
havepositiverealparts
none

Answer :B
31.

If x^(2)y=2x-y then y^(2)+(y^(4))/(2)+(y^(6))/(3)+…oo=

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`2LOG((1+x)/(1-x))`
`2log((1+x^(2))/(1-x^(2)))`
`LOG((1+x)/(1-x))`
`log((1+x^(2))/(1-x^(2)))`

ANSWER :B
32.

Solve 2(x+(1)/(x))^(2)-7(x+(1)/(x))+5=0," when "x ne 0

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`{2 +-1/2,(1+- isqrt(3))/(2)}`
`{2,-1/2,(1+I SQRT(3))/(2)}`
`{-2,1/2,(1+-isqrt(3))/(2)}`
none

Answer :A
33.

If ""^(56)P_((r+6)):""^(54)P_((r+3))=30800:1, find r.

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ANSWER :r=40
34.

Fundamental theorem of definite integral : int_(0)^(pi)sqrt(1+4sin^(2)""(x)/(2)-4sin""(x)/(2))dx=......

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`4sqrt3-4`
`4sqrt3-4-(PI)/(3)`
`pi-4`
`(2PI)/(3)-4-4sqrt3`

ANSWER :B
35.

Find the co-ordinates of a point on the parabola y=x^(2)+7x+2 which is closed to the straight line y=3x-3.

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ANSWER :`(-2,-8)`
36.

Match the following

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

Answer :D
37.

Five balls are placed in three boxes. Each box can hold all the five balls. In how many different ways can we place the balls in the boxes so that no box remains empty if the balls are identical but the boxes are different

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ANSWER :6
38.

Let f(x)=2tan^(3)x-6tan^(2)x+1+sgn(e^(x)),AA x in [-(pi)/(4),(pi)/(4)], Then the positive difference between the least value and the local maximum value of the function is (where sgn (f(x)) represents the signum function)

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ANSWER :4
39.

Five balls are placed in three boxes. Each box can hold all the five balls. In how many different ways can we place the balls in the boxes so that no box remains empty if the balls and the boxes are different

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ANSWER :150
40.

Differentiate the following functions with respect to x: sin x + sin y= tan (xy)

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ANSWER :`(y SEC^(2)(xy)-COS X)/(cosy-x.sec^(2)(xy))`
41.

The solution of int_(sqrt(2))^(x) (dt)/(tsqrt(t^(2)-1))=pi/12 is

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4
2
6
`SQRT(3)`

ANSWER :B
42.

The solution of (x + y +1) dx + (3x + 4y + 4) dy = 0 is

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`x-2y -2=ce^(x//(2X + 4y + 4))`
`x + 2y - 3 =ce^(x//(2x + 4y + 4))`
`x + 2y + 2 =ce^(x//(2x + 4y + 4))`
`x-2y + 2 =ce^(x//(2x + 4y + 4))`

Answer :C
43.

Match the following

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a,b,c
b, c,a
c, a,b
c, b, a

Answer :C
44.

Match the following

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

Answer :B
45.

If bar(a)=bar(i)-2bar(j)+bar(k), bar(b)=2bar(i)+bar(j)+bar(k), bar(c ) = bar(i) + 2bar(j)-bar(k), then find bar(a) xx (bar(b) xx bar(c )) and |(bar(a)xxbar(b))xxbar(c )|.

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ANSWER :9
46.

In the interval (-3, 3) the function f(x)= x/3 + 3/x, xne0 is

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INCREASING
Decreasing
Neither increasing nor decreasing
PARTLY increasing and partly decreasing

Answer :B
47.

A tangent to the parabola y^(2)+4bx=0 meets the parabola y^(2)=4ax in P and Q . The locus of the middle point of PQ is

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`y^(2)(2A+B)=4A^(2)X`
`y^(2)(2a-b)=4a^(2)x`
`y^(2)(2a +b)=4ax`
`y^(2)(2a-b)=4ax`

ANSWER :A
48.

Number of points from where perpendicular tangents can be drawn to the curve x^(2)/(16)- y^(2)/(25)=1 is

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1
2
0
infinite

Answer :C
49.

If overset(-)a, overset(-)b, overset(-)c are three vectors such that |overset(-)a|=1, |overset(-)b|=2, |overset(-)c|=3, and overset(-)a. overset(-)b. =overset(-)b. overset(-)c=overset(-)c. overset(-)a=0," then "|[overset(-)a overset(-)b overset(-)c]|=

Answer»

0
2
3
6

Answer :D
50.

A man owns a fielfd of area 1000 m^(2). He wants to plant fruit trees in it. He has a sum of Rs. 1400 to purchase young thees. He has the choice of two types of trees. Type A requires 10m^(2) of ground per tree and costs Rs.20 per tree, and type B requires 20m^(2) of ground per tree and costs Rs. 25 per tree. When full grown, a type-A tree produces on average of 20 kg of fruit which can be sold at a profit of Rs.2 per kg and a type-B tree produces an average of 40 kg of fruit which can be sold at a profit of Rs72 per kg and a type-B three produces on average of 40 kg of fruit which can be sold at a profit of Rs 1.50 per kg. How many of each type should be planted to achieve maximum profit when trees are fully frown? What is the maximum profit ?

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Solution :LET x plants of type A and y plants of type B be planted. Then `XGE0,yge0,20x+25yle1400. 10+20yle1000.`
PROFIT FUNCTION is `Z=2xx20x+3/2xx40y,i.e., Z=40x+60y.`