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. |
Everybody in a room shakes hand with everybody else. The total number of handshakes is 66. The total number of persons in the room is : |
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Answer» 9 |
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| 2. |
Examine the consistency of the system of linear equtions in 1 to 6 5x-y+4z=5 2x+3y+5z=2 5x-2y+6z=-1 |
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| 3. |
Evalute the following integrals int e^(x) ((cos x - sin x)/(1 - cos 2 x ))dx |
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| 4. |
If N denotes the set of all positive integers and if f : N to N is defined byf(n) = the sum of positive divisors of n then, f(2^(k)3), where k is a positive integer, is |
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Answer» `2^(k+1)-1` |
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| 5. |
Let X be a random variable such that P(X=-2)=P(X-1)=P(X=2)=P(X=1)=(1)/(6) and P(X=0)=(1)/(3). Find the mean and variance of X. |
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| 6. |
The measure of dispersion which is used to find more consistent data is |
| Answer» Answer :A | |
| 7. |
13 persons sit a round a table. Find the odds in favour of two specified persons sitting together. |
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| 8. |
Match column I & II. {:(,"Column-I",,"Column-II"),((a),"Monophyodont teeth",(1),(2)/(2)","(1)/(1)","(0)/(0)","(2)/(2)),((b),"Adolescence teeth",(2),(2)/(2)","(1)/(1)","(2)/(2)","(2)/(2)),((c),"Adult teeth",(3),(2)/(2)","(1)/(1)","(2)/(2)","(3)/(3)),((d),"Child teeth",(4),(0)/(0)","(0)/(0)","(2)/(2)","(1)/(1)):} |
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Answer» a-1, b-3, c-2, d-4 |
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| 9. |
Find the numerically greatest term in the expansion of (2+5x)^(21) when x =2/5. |
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| 10. |
Calculate the area of the triangle ABC (by vector method) where A(1,1,2), B(2,2,3), C(3,-1,-1) |
Answer» Solution : =`-hati+5hatj-4hatk` Area of `TRIANGLE ABC` =`1/2 sqrt(1+25+16) = sqrt(42)/2` SQ. UNITS. |
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| 11. |
L (1, 3) and L^(1) = (1, -1) are the ends of latus rectum of a parabola, then area of quadrilateral formed by tangents and normals at L and L^(1) ( in Square Units ) is |
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Answer» 2 |
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| 12. |
Evaluate lim_(n rarr oo)(2^(k) + 4^(k) + 6^(k)+…+(2n)^(k))/(n^(k+1)) by using the method of finding definite integral as the limit of a sum. |
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| 13. |
Match the following lists: |
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Answer» `UNDERSET(xtooo)lim("("sqrt("("x^(2)-x+1")")-ax-b")""("sqrt(x^(2)-x+1")")+ax+b")")/("("sqrt("("x^(2)-x+1")")+ax+b")")` `=0` or `underset(xtooo)lim((x^(2)-x+1)-(ax+b)^(2))/(sqrt((x^(2)-x+1))+ax+b)=0` or `underset(xtooo)lim((1-a^(2))x^(2)-(1+2ab)x+(1-b^(2)))/(sqrt((x^(2)-x+1))+ax+b)=0` or `underset(xtooo)lim((1-a^(2))x-(1+2ab)+((1-b^(2)))/(x))/(sqrt(1-(1)/(x)+(1)/(x^(2))+a+(b)/(x)))=0` This is POSSIBLE only when `1-a^(2)=0" and "1+2ab=0` `:.""a=+-1` or `a=1""(becauseagt0)""(1)` `:.""b=-1//2` `:.""(a,2b)-=(1,-1)` b. Divide numerator and denominator by `e^(1//x)`. Then, `underset(xtooo)lim((1+a^(3))e^(-(1)/(x))+8)/(e^(-(1)/(x))+(1-b^(3)))=2` or `(0+8)/(0+1-b^(3))=2` or `""1-b^(3)=4` `:.""b^(3)=-3" or "b=-3^(1//3)` Then, `ainR.` Therefore, `(a,b^(3))-=(a,-3)` c. `underset(xtooo)lim"("sqrt("("x^(4)-x^(2)+1)")"-ax^(2)-b")"=0` Put `s=(1)/(t)`.Then `underset(t TO0)lim(sqrt(((1)/(t^(4))-(1)/(t^(2))+1))-(a)/(t^(2))-b)=0` or `underset(t to0)lim(sqrt((1-t^(2)+t^(4)))-a-bt^(2))/(t^(2))=0""(1)` Since R.H.S. is finite, numerator must be equal to 0 at `t to0.` Therefore,`1-a=0" or "a=1.` From equation (1), `underset(t to0)lim(sqrt((1-t^(2)+t^(4)))-1-bt^(2))/(t^(2))=0` `underset(t to0)lim(-1+t^(2))(((1-t^(2)+t^(4))^(1//2)-(1)^(1//2))/((1-t^(2)+t^(4))-1))=b` or `(-1)((1)/(2))=b" or "a=1,b=-(1)/(2)" or "(a,-4b)-=(1,2)` d. `underset(xto-a)lim(x^(7)-(-a)^(7))/(x-(-a))=7" or "7a^(6)=7" or "a^(6)=1" or "a=-1` |
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| 14. |
Match the following lists: |
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Answer» `underset(xto-1)lim(root(3)((7-x))-2)/((x+1))=underset(hto0)lim((8-h)^(1//3)-2)/(h)` `=underset(hto0)lim(2(1-(h)/(8))^(1//3)-2)/(h)` `=2underset(hto0)lim((1-(1)/(3)(h)/(8))-1)/(h)` `=-(1)/(12)` b. We have `underset(xto pi//4)lim(tan^(3)x-TANX)/(cos(x+pi//4))` `=underset(xto pi//4)lim(tanx(tanx-1)(tanx+1))/(cos(x+pi//4))` `=underset(xto pi//4)lim(tanx(sinx-cosx)(tanx+1))/(cosxcos(x+pi//4))` `=-underset(xto pi//4)lim(tanx(cosx-sinx)(tanx+1))/(cosxcos(x+pi//4))` `=-sqrt(2)underset(xto pi//4)lim(tanx((1)/(sqrt(2))cosx-(1)/(sqrt(2))sinx)(tanx+1))/(cosxcos(x+pi//4))` `=-sqrt(2)underset(xto pi//4)lim(tanx(tanx+1))/(cosx)` `=-sqrt(2)xx2xxsqrt(2)=-4` `underset(xto1)lim((2x-3)(sqrt(x-1)))/(2x^(2)+x-3)=underset(xto1)lim((2x-3)(sqrt(x)-1))/((2x+3)(x-1))` `=underset(xto1)lim((2x-3)(sqrt(x)-1))/((2x+3)(sqrt(x)-1)(sqrt(x)+1))` `=underset(xto1)lim((2x-3))/((2x+3)(sqrt(x)+1))` `=(2-3)/((2+3)(sqrt(1)+1))=-1//10` d. `underset(xtooo)lim(LOGX^(n)-[x])/([x])=underset(xtooo)lim(logx^(n))/([x])-underset(xtooo)lim([x])/([x])` `=0-1=-1` |
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| 15. |
Match the following lists: |
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Answer» or `[underset(xto0)lim100(sinx)/(x)]=99` and `[underset(xto0)lim100(x)/(sinx)]=100` ALSO, `underset(xto0)lim(sin^(-1)x)/(x)=1""`(but a value is more than 1) or `[underset(xto0)lim100(sin^(-1)x)/(x)]=100` and `[underset(xto0)lim100(x)/(sin^(-1)x)]=99` `underset(xto0)lim(tanx)/(x)=1""`(but a value is bigger than 1) or `[underset(xto0)lim100(tanx)/(x)]=100` and `[underset(xto0)lim100(tan^(-1)x)/(x)]=99` Hence, a. `underset(xto0)lim([100(sinx)/(x)]+[100(tanx)/(x)])=199` b. `underset(xto0)lim([100(x)/(sinx)]+[100(tanx)/(x)])=200` c. `underset(xto0)lim([100(sin^(-1)x)/(x)]+[100(tan^(-1)x)/(x)])=199` d. `underset(xto0)lim([100(x)/(sin^(-1)x)]+[100(tan^(-1)x)/(x)])=198` |
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| 16. |
Three lines px + qy+r=0, qx + ry+ p = 0 and rx + py + q = 0 are concurrent of |
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Answer» p+q+r=0 |
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| 17. |
iff(x)= (x-1)/(x+1) thef(2x )is |
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Answer» `(F(X) +1)/(f(x) +3)` |
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| 18. |
Using the identity sin^(4) x=3/8-1/2 cos 2x+1/8 cos 4x or otherwise, if the value of sin^(4)(pi/7)+sin^(4)((3pi)/(7))+sin^(4)((5pi)/(7))=a/b, where a and b are coprime, find the value of (a-b). |
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| 19. |
Find the probability of drawing 2 red balls in succession from a bag containing 4 red balls and 5 black balls when the ball that is drawn first is (i) not replaced (ii) replaced. |
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| 20. |
int(e^(exlog_(e)x)+(log_(e)x)(e^(exlog_(e)x)))dx=...+c |
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Answer» `(x^(-EX))/(e)` |
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| 21. |
Compute the magnitude of the following vectors :veca=hati+hatj+k,vecb=2hati-7hatj-3hatk,vecc=(1)/(sqrt(3))hati+(1)/(sqrt(3))hatj-(1)/(sqrt(3))hatk |
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| 23. |
If ""^(12)P_r=1320 find r |
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| 24. |
If vecu=veca-vecb,vecv=veca+vecb and |veca|=|vecb|=2, then |vecuxx vecc| is : |
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Answer» `2sqrt(16 - (veca.vecb)^(2))` |
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| 25. |
sin alpha=sinbeta, cos alpha=cos beta then |
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Answer» `SIN((alpha+beta)/2=0` |
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| 26. |
Let x > - 1, then statement P(n) : (1 + x)^(n) > 1 + nx is true for |
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Answer» all `N in N` |
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| 27. |
Observe the following statements Assertion (A) : X is binomial variate with parameters 2n+1 and p = 1/2 then P(x = odd values) = 1/2 Reason ( R ) : If ""^(n)C_(r)=C_(r) then C_(1)+C_(3)+C_(5)+……=2^(n-1). Then |
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Answer» A is FALSE, R is TRUE |
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| 28. |
if 2^((log_3 5)(log_x 81)-6(25)^(log_x^2)+8=0, then :- |
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Answer» NUMBER of solutions is 3 |
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| 29. |
If I_(a) = int_(0)^(pi//4) tan^(n) theta d theta for n=1,2,3,…, then I_(n-1) + I_(n+1) is equal to |
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Answer» 0 |
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| 30. |
The two curves x^(3)-3xy^(2)+2=0 and 3x^(2)y-y^(3)-2=0: |
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Answer» 1)Touch each other |
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| 31. |
The slope of tangent to the curve x=t^(2)+3t-8, y=2t^(2)-2t-5 at the point (2,-1) is ………….. |
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Answer» `(22)/(7)` |
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| 32. |
If f:R to R is a differentible function such that f(x+y) = f (x). F(y) for all x,y in R and if f'(4) =24 and f'(0) = 3 then f(4) = |
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Answer» 72 |
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| 33. |
Total number of 6 digited numbers in which all the odd digits and only odd digits appear is |
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Answer» `(5)/(2)ANGLE6` |
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| 34. |
Let f:[0,1] to Rbe an injective continuous function that satisfies the condition -1 lt f(0) lt f(1) lt 1 Then the number of functions g:[-1,1] to [0,1]such that (gof)x =xfor all x in [0,1] is |
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Answer» 0 |
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| 35. |
Statement-I: The area bounded by x=2 cos theta,y=3sin theta is 36pi sq. units. Statement-II : The area bounded by x=2cos theta, y=2 sin theta is 4pi sq. units. Which of the above statement is correct. |
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Answer» only I |
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| 36. |
If the median and the range of four numbers{x,y, 2x+ y,x -y} , "where"0 lt y lt x lt 2yare 10 and 28 respectively, then the mean of numbers is |
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Answer» 18 |
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| 37. |
Let veca=hati+hatj+hatk,vecb=2hati+3hatk,vecc=hati-hatj and vecd=6hati+2hatj+3hatk. If vecd=x(vecbxxvecc)+y(veccxxveca)+z(vecaxxvecb) where x, y and z are scalars then the value of x+y+z is k. the number of prime factors of 5k is. |
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Answer» `x=(veca.vecd)/([vecavecbvecc]),y=(vecbvecd)/([vecavecbvecc]),z=(veccvecd)/([vecavecbvecc])` `((veca+vecb+vecc).vecd)/([vecavecbvecc])=(34)/(5)` |
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| 38. |
If f (x) = min [ x ^(2), sin ""(x)/(2), (x -2pi) ^(2)], the area bounded by the curve y=f (x), x-axis, x=0 and x=2pi is given by Note: x_(1) is the point of intersection of the curves x ^(2) and sin ""(x)/(2),x _(2) is the point of intersection of the curves sin ""x/2 and (x-2pi)^(2)) |
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Answer» `int_(0)^(x _(1))(SIN ""(x)/(2)) dx + int _(x_(1))^(pi) x ^(2) dx + int _(pi)^(x _(2)) (x-2pi)^(2) dx + int _(x _(2))^(2pi) (sin ""(x)/(2)) dx` |
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| 39. |
Prove the equality int_(-a)^(a) f(x) dx = int_(0)^(a) [f(x) + f(-x)] dxfor any continuous function f(x) |
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| 40. |
5 balls of different colours are to be placed in 3 boxes of different sizes. Each box can hold all 5 balls. No. of different ways the balls can be placed so that no box remain empty is |
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Answer» 50 |
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| 41. |
Statement 1: F a,b,c are real positive numbers with abc =1 and AA^(T)=1 where A= [{:( a,b,c),( b,c,a),( c,a,b) :}] then a^(3) +b^(3) + c^(3) =4. Statement 2: (1)/(a) +( 1)/(b) +(1)/( C) = 0. |
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Answer» STATEMENT -1 is True ,Statement -2 is True ,Statement -2 is CORECT explanation for Statement -1 |
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| 42. |
If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find (dy)/(dx). x= a(cos t+ log tan (t)//(2))y= a sin t. |
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| 43. |
If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find (dy)/(dx). x= (sin^(3)t)/(sqrt(cos 2t)), y= (cos^(3)t)/(sqrt(cos 2t)). |
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| 44. |
Find k if the following pairs of circles are orthogonal x^(2)+y^(2)-16y+k-0, x^(2)+y^(2)+4x+8=0 |
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| 45. |
Match the following . {:("Circles"," Radical centre"),(I.x^(2) + y^(2) = 1" ," x^(2) + y^(2) - 2x = 1 "," x^(2) + y^(2) - 2y = 1 , "a)(0,0)"),(II. x^(2) + y^(2) - x + 3y - 3 = 0 ", " x^(2) + y^(2) - 2x+ 2y + 2 = 0 ", " x^(2) + y^(2) + 2x + 3y - 9 = 0,"b)(2,3)"),(III.x^(2) + y^(2) - 8x + 40 = 0 "," x^(2) + y^(2) - 5x + 16 = 0"," x^(2) + y^(2) - 8x + 16y + 160 = 0 ,"c)(8,-15/2)"):} |
| Answer» Answer :A::B | |
| 46. |
If theroots ofx^4+5x^3 - 30 x^2- 40x+64=0are inG.Pthen roots ofx^4 - 5 x^3- 30 x^2 + 40 x+ 64 =0 arein |
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Answer» A.P |
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| 47. |
For what x is the 4th term in the expansion of [(5^(1/3))^(-1/2log_(10)(6-sqrt8x))+(5^(log_(10)(x-1))/(25^(log_(10)5)))^(1/6)]^(m) is equal to 84/5, if it is known that 14/9 of binomial coefficient of 3rd term, binomial coefficient of 4th term and binomial coefficient of 5th term in the expansion constitute a G.P. |
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| 48. |
If the straight lines (x-1)/k = (y-2)/2 = (z-3) and (x-2)/3 = (y-3)/k = (z-1)intersect at a point, then the integer k is equal to |
| Answer» ANSWER :A | |
| 49. |
int_(-1)^(1) (x-[x])dx= |
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Answer» 0 |
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