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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. |
The length of the L.R. of x2 = -4y is _______. (a) 1 (b) 2 (c) 3 (d) 4 |
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Answer» The correct answer is: (d) 4 |
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| 2. |
Distance from the origin to the plane 3x – 6y + 2z + 7 = 0 is ______. (a) 0(b) 1 (c) 2 (d) 3 |
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Answer» The correct answer is : (b) 1 |
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| 3. |
The circle passing through (1, -2) and touching the axis of x at (3, 0) passing through the point _______. (a) (-5, 2) (b) (2, -5) (c) (5, -2) (d) (-2, 5) |
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Answer» The correct answer is : (c) (5, -2) |
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| 4. |
The distance from the origin to the plane \(\vec {r} .(2\vec{i}-\vec{j}+5\vec{k}) = 7\) is _____.(a) 7/√30(b) √30/7(c) 30/7(d) 7/30 |
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Answer» The correct answer is : (a) 7/√30 |
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| 5. |
If the length of the perpendicular from the origin to the plane 2x + 3y + λz = 1, λ > 0 is 1/5, then the value of λ is _______. (a) 2√3 (b) 3√2 (c) 0 (d) 1 |
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Answer» The correct answer is : (a) 2√3 |
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| 6. |
The number given by the Mean value theorem for the function 1/x, x ∈ [1, 9] is ______. (a) 2(b) 2.5(c) 3(d) 3.5 |
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Answer» The correct answer is : (c) 3 |
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| 7. |
The tangent to the curve y2 – xy + 9 = 0 is vertical when ________. (a) y = 0 (b) y = ± √3(c) y = 1/2(d) y = ± √3 |
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Answer» (b) y = ± √3 |
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| 8. |
f is a differentiable function defined on an interval I with positive derivative. Then f is ______. (a) increasing on I (b) decreasing on I (c) strictly increasing on I (d) strictly decreasing on I |
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Answer» (c) strictly increasing on I |
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| 9. |
The volume of a sphere is increasing in volume at the rate of 3π cm3 /sec. The rate of change of its radius when radius 1/2 cm _______. (a) 3 cm/s (b) 2 cm/s (c) 1 cm/s (d) 1/2 cm/s |
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Answer» The correct answer is: (a) 3 cm/s |
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| 10. |
If v (x, y) = log (ex + ey ), then ∂v/∂x + ∂v/∂y is equal to _____. (a) ex + ey (b) 1/(ex + ey)(c) 2 (d) 1 |
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Answer» The correct answer is : (d) 1 |
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| 11. |
If we measure the side of a cube to be 4 cm with an error of 0.1 cm, then the error in our calculation of the volume is _______. (a) 0.4 cu.cm (b) 0.45 cu.cm (c) 2 cu.cm (d) 4.8 cu.cm |
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Answer» (d) 4.8 cu.cm |
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| 12. |
If we measure the side of a cube to be 4 cm with an error of 0.1 cm, then the error in our calculation of the volume is ________. (a) 0.4 cu.cm (b) 0.45 cu.cm (c) 2 cu.cm(d) 4.8 cu.cm |
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Answer» (d) 4.8 cu.cm |
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| 13. |
If u(x, y) = ex^2 + y^2, then ∂u/∂x is equal to _______. (a) ex^2 + y^2(b) 2 x u(c) x2 u (d) y2 u |
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Answer» The correct answer is : (b) 2 x u |
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| 14. |
If f(x, y) = exy , then ∂2f/∂x∂y is equal to ________. (a) xyexy (b) (1 + xy) exy (c) (1 + y) exy (d) (1 + x) exy |
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Answer» (b) (1 + xy) exy |
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| 15. |
If u (x, y) = x2 + 3xy + y – 2019, then (∂u/∂x)(4, - 5) is equal to ______. (a) -4 (b) -3 (c) -7 (d) 13 |
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Answer» The correct answer is : (c) -7 |
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| 16. |
If f(x, y, z) = xy + yz + zx, then fx – fz is equal to ________. (a) z – x (b) y – z (c) x – z (d) y – x |
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Answer» The correct answer is : (a) z – x |
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| 17. |
If w(x,y, z) = x2 (y – z) + y2 (z – x) + z2 (x – y), then ∂w/x + ∂w/∂y + ∂w/∂z is ________. (a) xy + yz + zx (b) x (y + z) (c) y (z + x)(d) 0 |
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Answer» The correct answer is : (d) 0 |
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| 18. |
A circular template has a radius of 10 cm. The measurement of radius has an approximate error of 0.02 cm. Then the percentage error in the calculating area of this template is _______. (a) 0.2% (b) 0.4% (c) 0.04%(d) 0.08% |
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Answer» The correct answer is : (b) 0.4% |
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| 19. |
If /(x, y, z) = xy + yz + zx, then fx – fz is equal to _______. (a) z – x (b) y – z (c) x – z (d) y – x |
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Answer» The correct answer is : (a) z – x |
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| 20. |
The curve y2 = x2 (1 – x2 ) has ______.(a) an asymptote x = -1 (b) an asymptote x = 1 (c) two asymptotes x = 1 and x = -1 (d) no asymptote |
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Answer» (d) no asymptote |
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| 21. |
The minimum value of the function |3 – x | + 9 is ________. (a) 0 (b) 3 (c) 6 (d) 9 |
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Answer» The correct answer is : (d) 9 |
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| 22. |
If the vectors \(\vec {a}\)= (3\(\hat {i}\) + 2\(\hat {j}\) + 9\(\hat {k}\)) and \(\vec {b}\) = (\(\hat {i}\) + m\(\hat {j}\) + 3\(\hat {k}\)) are parallel then m is ___________.(a) 3/2 (b) 2/3(c) -3/2(d) -2/2 |
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Answer» The correct answer is : (b) 2/3 |
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| 23. |
Let all chords of parabola `y^(2)=x+1` which subtends right angle at `(1,sqrt(2))` passes through `(a,b)` then the value of `a+b^(2)` is |
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Answer» Correct Answer - 4 Shifting origin at `(1,sqrt(2))` `(Y+2sqrt(2))^(2)=X+1+1,y=Y+sqrt(2),x=X+1` `Y^(2)+(2sqrt(2)Y-X)((Y-mX)/C)=0` `1+(2sqrt(2))/C+m/C=0` `C+m+2sqrt(2)=0` `Y=mx-=m-2sqrt(2)` `y-sqrt(2)=m(x-1)-m-2sqrt(2)` it is passes through `(a,b)AAm` `impliesa=2` and `b=-sqrt(2)` |
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| 24. |
In a ` A B C ,A-=(alpha,beta),B-=(1,2),C-=(2,3),`point `A`lies on the line `y=2x+3,`where `alpha,beta`are integers, and the area of the triangle is `S`such that `[S]=2`where `[`.`]`denotes the greatest integer function. Then the possible coordinates of `A`can be`(-7,-11)`(b) `(-6,-9)``(2,7)`(d) `(3,9)`A. `(alpha)/(beta)=3/7`B. `3alpha beta=14`C. `2alpha+3beta=18`D. `alpha+6beta=30` |
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Answer» Correct Answer - A::C::D Clearly `R(alpha,beta)` is centroid of `DeltaABC`s `:.R(alpha,beta)=((3+1+2)/3,(9+2+3)/3)=(2,14/3)` `impliesalpha=2` and `beta=14/3` Hence `2alpha+3beta=18` |
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| 25. |
∫sin9x x∈[-π/2,π/2] dx=?(a) -1 (b) 0 (c) 1 (d) π/2 |
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Answer» correct option (b) 0 |
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| 26. |
If `[.]` denotes the greatest integer function and `x,yepsilonRr,"n" epsilonN` then which of the following is true?A. `[x+y]ge[x]+[y]`B. `[x+y]le[x]+[y]`C. `[([x])/n]=[x/n]`D. `[x+1/2]=[2x]-[x]` |
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Answer» Correct Answer - A::C::D (A) & (B) `x+y=[x]+[y]+alpha+beta` where `alpha={x},beta={y}` `:.0ltalpha+betalt2implies[alpha+beta]=0` or `1` `:.[x+y]ge[x]+[y]` (C) By division a lgorithm, `[x]=qn+r(0lerlen-1)` Let `x=[x]+alpha,0ltalphalt1` Then `x=1n+ralpha` `:.[x/n]=q ( :.0ltr+alphaltn)` (D) let `x=n+alpha "n" epsilonN` Consider Case-1 `0lealphale1/2` Case II `1/2lealphale1` |
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| 27. |
vector (a . b)=?(a) - vector(b.a) (b) 1 (c) 1 (d) vector(b . a) |
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Answer» (d) vector(b . a) |
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| 28. |
If 2[(3,4),(5,x)]+[(1,y),(0,1)]=[(7,10),(10,5)], then(a) (x = -2, y = 8) (b) (x = 2,y = -8) (c) (x=3,y=-6) (d) (x=-3,y=6) |
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Answer» (b) (x = 2,y = -8) |
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| 29. |
If A and B can do a piece of work in 10 days and 15 days respectively, then the part of work they can do together in 3 days is?1. 1/22. 13. 1/64. 1/3 |
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Answer» Correct Answer - Option 1 : 1/2 Given: A and B can do a piece of work in 10 days and 15 days respectively Concept Used: One day work = 1/number of days taken by the worker to complete the whole work Calculation: One day work of A = 1/10 One day work of B = 1/15 One day work of A + B = 1/10 + 1/15 ⇒ (3 + 2)/30 ⇒ 1/6 Now, 3 days work done by (A + B) is 3 × 1/6 ⇒ 1/2 Hence, the part of work they can do together in 3 days is 1/2 of the whole work |
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| 30. |
Evaluate [(1,-1)(y,x)](a) x+y (b) x-y (c) -y-x (d) 1-x |
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Answer» option: (a) x+y |
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| 31. |
From the given options complete the series - 10, 100, 20, 400, 30, (____), 40, 1600.1. 12002. 9003. 10004. 800 |
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Answer» Correct Answer - Option 2 : 900 Here, we have to fill the blank of the given series 10, 100, 20, 400, 30, (____), 40, 1600. Calculation: Here, we can see that the terms present at even place is the square of its previous term Like 2nd term 100 which is square of 10 (1st term) Like 4th term 400 which is square of 20 (3rd term) Like 8th term 1600 which is square of 40 (7th term) Similarly, 6th term will be the square of 30 6th term = (30)2 = 900 Hence, the required blank can be filled by 900. |
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| 32. |
∫xndx,(n≠ 0)=?(a) xn-1/n-1+k (b) xn+1/n+1+k (c) xn+1+k (d) xn-1+k |
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Answer» (b) xn+1/n+1+k |
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| 33. |
P(A)+P(A')=?(a) 0 (b) 1 (c) -1 (d) P(E) |
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Answer» Option: (b) 1 |
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| 34. |
∫dx/x x∈[1,2] =?(a) log2/3 (b) log3/2 (c) log1/2 (d) logx/2 |
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Answer» Option (c) log1/2 |
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| 35. |
tan-1x+cot-1x=?(a) 0 (b) 1 (c) π/2 (d) -π/2 |
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Answer» Option:(c) π/2 |
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| 36. |
`cot^(-1)(-1/2)+cot^(-1)(-1/3)` is equal toA. `(3pi)/4`B. `(5pi)/4`C. `(pi)/4`D. `(-3pi)/4` |
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Answer» Correct Answer - B `pi-cot^(-1)(1/2)+pi-cot^(-1)(1/3)` `2pi-[cot^(-1)(1/2)+cot^(-1)(1/3)]` `=pi+tan^(-1)(1/2)+tan^(-1)(1/3)` `=pi+"tan"^(-1) [(1//2+1//3)/(1-1//6)]=pi+(pi)/4=(5pi)/4` |
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| 37. |
If the points `(2-x, 2, 2), (2, 2-y, 2) and (2, 2, 2-z)` are coplanar then prove `2/x+2/y+2/z=1`A. `1/x+1/y+1/z=1`B. `x+y+z=1`C. `1/(1-x)+1/(1-y)+1/(1-z)=1`D. `1/(1-x)+1/(1-y)+1/(1-z)=2` |
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Answer» Correct Answer - A `vec(AB)=vec(OB)-vec(OA)={2hati+(2-y)hatj+2hatk}-` `{(2-x)hati+2hatj+2hatk}=xhati-yhatj` `vec(AC)=vec(OC)-vec(OA)` `={(2hati+2hatj+(2-z)hatk}-{(2-x)hati+2hatj+2hatk}` `=xhati-zhatk` `vec(AD)=vec(OD)-vec(OA)` `=(hati+hatj+hatk)-{(2-x)hati+2hatj+2hatk}=(x-1)hati-hatj-hatk` As these vectors are coplanar, `|(x, -y, 0),(x, 0, -z),(x-1, -1, -1)|=0` On simplication we get `1/x+1/y+1/z=1` |
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| 38. |
∫dx/1+x2=?(a) tanx+c (b) tan2x+c (c) cotx+c (d) -cot-1x+c |
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Answer» Option: (a) tanx+c |
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| 39. |
The coplanar points `A,B,C,D` are `(2-x,2,2),(2,2-y,2),(2,2,2-z)` and `(1,1,1)` respectively thenA. `1/x+1/y+1/z=1`B. `x+y+z=1`C. `1/(1-x)+1/(1-y)+1/(1-z)=1`D. `1/(1-x)+1/(1-y)+1/(1-z)=2` |
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Answer» Correct Answer - A `vec(AB)=vec(OB)-vec(OA)={2hati+(2-y)hatj+2hatk}-` `{(2-x)hati+2hatj+2hatk}=xhati-yhatj` `vec(AC)=vec(OC)-vec(OA)` `={(2hati+2hatj+(2-z)hatk}-{(2-x)hati+2hatj+2hatk}` `=xhati-zhatk` `vec(AD)=vec(OD)-vec(OA)` `=(hati+hatj+hatk)-{(2-x)hati+2hatj+2hatk}=(x-1)hati-hatj-hatk` As these vectors are coplanar, `|(x, -y, 0),(x, 0, -z),(x-1, -1, -1)|=0` On simplication we get `1/x+1/y+1/z=1` |
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| 40. |
If P(A)=3/8, P(B)=1/2, P(A∩B)= 1/4, then P(A∪B)= ?(a) 2/3 (b) 1/3 (c) 1/2 (d) 5/8 |
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Answer» Option:(b) 1/3 |
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| 41. |
vector(k x k)=?(a) 1 (b) -1 (c) k2 (d) 0 |
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Answer» option: (d) 0 |
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| 42. |
In a `Delta ABC, A -= (alpha, beta), B -= (1, 2), C -= (2,3)` and point A lies on the line y = 2 x + 3 where `alpha, beta in l`. If the area of `Delta ABC` be such that `[Delta]=2`, where [.] denotes the greatest integer function, find all possible coordinates of A.A. `(alpha)/(beta)=3/7`B. `3alpha beta=14`C. `2alpha+3beta=18`D. `alpha+6beta=30` |
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Answer» Correct Answer - A::C::D Clearly `R(alpha,beta)` is centroid of `DeltaABC`s `:.R(alpha,beta)=((3+1+2)/3,(9+2+3)/3)=(2,14/3)` `impliesalpha=2` and `beta=14/3` Hence `2alpha+3beta=18` |
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| 43. |
Find the locus of the point of intersection of the perpendiculartangents of the curve `y^2+4y-6x-2=0`.A. `2x-1=0`B. `2x+3=0`C. `2y+3=0`D. `2x+5=0` |
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Answer» Correct Answer - D Given equation reduces to `Y^(2)=6x` where `x+1=X` and `y+2=Y` `implies` Locus is directrix |
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| 44. |
Which of the following is True?A. if `f(x)` is continuous at `x=c` and `g(x)` is discontinuous at `x=c` then `(f.g)(x)` must be discontinuousB. If `f(x)` is continuous at `x=c` and `g(x)` is discontinuous at `x=c` then `(f.g)(x)` may be continuous.C. If `f(x)` and `g(x)` are discontinuous at `x=c`, then the product function must be discontinuous.D. If `f(x)` and `g(x)` are discontinuous at `x=c`, then the product function may be continuous. |
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Answer» Correct Answer - B::D (A) & (B) Let `f(x)=x,xepsilonR` & `g(x)={("sin"(pi)/x,x!=0),(0,x=0):}` `f(x)` is continuous at `x=0` but `g(x)` is not. Now, `(f.g)(x)={(x"sin"(pi)/x,x!=0),(0, "at" x=0):}` is continuous at `x=0` (C) & (D) Let `f(x)={(0, xepsilonQ),(1,x!inQ):} , g(x) {(1,xepsilonQ),(0,x!inQ):}` `f(x)` & `g(x)` both are discontinuous everywhere. but `f(x).g(x)=0,xepsilonR` |
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| 45. |
A curve is represented paramtrically by the equations `x=e^(t)cost` and `y=e^(t)sint` where `t` is a parameter. Then The value of `(d^(2)y)/(dx^(2))` at the point where `t=0` isA. `1`B. `-2`C. `2`D. `3` |
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Answer» Correct Answer - C `(d^(2)y)/(dx^(2))=(sec^(3)((pi)/4+t))/(e^(1)(cost-sint))` `(d^(2)y)/(dx^(2)):|_(t=0)=2` |
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| 46. |
If both `Lim_(xrarrc^(-))f(x)` and `Lim_(xrarrc^(+))f(x)` exist finitely and are equal, then the function `f` is said to have removable discontinuity at `x=c`. If both the limits i.e. `Lim_(xrarrc^(-))f(x)` and `Lim_(xrarrc^(+))f(x)` exist finitely and are not equal, then the function `f` is said to have non-removable discontinuity at `x=c`. Which of the following function not defined at `x=0` has removable discontinuity at the origin?A. `f(x)=1/(1+2^(1/x))`B. `f(x)="tan"^(-1) 1/x`C. `f(x)=(e^(1/x)-1)/(e^(1/x)+1)`D. `f(x)=(|sinx|)/(|x|)` |
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Answer» Correct Answer - D (A) `lim_(xrarr0^(-))f(x)=1 Lim_(xrarr0^(+))f(x)=0` (B) `Lim_(xrarr0^(+))f(x)=-(pi)/2 Lim_(xrarr0^(+))f(x)=(pi)/2` (C) `Lim_(xrarr0^(-))f(x)=-1 Lim_(xrarr0^(+))f(x)=1` `Lim_(xrarr0^(-))f(x)=Lim_(xrarr0^(+))f(x)=1` |
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| 47. |
`lim_(xrarrc)f(x)` does not exist for wher `[.]` represent greatest integer function `{.}` represent fractional part functionA. `f(x)=[x]-x,c=0`B. `f(x)=[|x|]-[2x-1],c=3`C. `f(x)={x}^(2)-{-x}^(2),c=0`D. `f(x)=(tan(sgn x))/((sgn x)),c=0` |
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Answer» Correct Answer - A::C::D (A) `lim_(xrarr3)[|x|]-[2x-1]` R.H.L `x=3-impliesf(3^(+))=3-5=-2` L.H.L `x=3^(-)impliesf^(3^(-))=2-4=-2` (B) `f(0^(+))=0-0=0,f(0^(-))=-1` (C) `f(x)=(x-[x])^(2)-(-x-[-x])^(2)` `f(0^(+))=0-(+1)^(2)=-1` `f(0^(-))=(0-1)-(0-0)^(2)=1` (D) `f(0^(+))=(tan1)/1=tan1` `f(0^(-))=(tan(-1))/(-1)=tan1` |
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| 48. |
The average of four consecutive even numbers is 27. Find the largest of these numbers. |
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Answer» Let the four consecutive even numbers be x, x + 2, x + 4 and x + 6. Then, sum of these numbers = (27 x 4) = 108. So, x + (x + 2) + (x + 4) + (x + 6) = 108 or 4x = 96 or x = 24. ∴ Largest number = (x + 6) = 30. |
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| 49. |
If the sum of two numbers is 42 and their product is 437, then find the absolute difference between the numbers. |
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Answer» Let the numbers be x and y. Then, x + y = 42 and xy = 437 x - y = sqrt[(x + y)2 - 4xy] = sqrt[(42)2 - 4 x 437 ] = sqrt[1764 – 1748] = sqrt[16] = 4. Required difference = 4. |
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| 50. |
The sum of two numbers is 16 and the sum of their squares is 113. Find the numbers. |
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Answer» Let the numbers be x and (15 - x). Then, x2 + (15 - x)2 = 113 => x2 + 225 + X2 - 30x = 113 => 2x2 - 30x + 112 = 0 => x2 - 15x + 56 = 0 => (x - 7) (x - 8) = 0 => x = 7 or x = 8. So, the numbers are 7 and 8. |
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