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. |
IfF(x)=underset0overset(x)inte^(2t)cos5tdt, what is F'(x)? |
| Answer» Solution :F(X)=underset0overset(x)inte^(2T)COS(5t)dtifff(x)=E^(2x)cos(5X).` | |
| 2. |
Let the position vectors of vertices of a DeltaABC be OA=3i+j+2k, OB=i+2j+3k" and "OC=2i+3j+k. If length of altitude of DeltaABC from A is p, then 2p^(2)= ________ |
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| 3. |
Find the remainder when 32^(32^(32)) is divided by 7. |
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| 4. |
Let vec(a),vec(b) and vec( c ) be three vectors such that |vec(a)|=3,|vec(b)|=4,|vec( c )|=5 and each one the them being perpendicular to the sum of the other two, find |vec(a)+vec(b)+vec( c )|. |
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| 5. |
Let p: Kiran passed the examination, q: Kiran is sad The symbolic from of a statement "It is not true that Kiran passed therefore he is said" is |
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Answer» <P>`(~prarr`Q) |
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| 6. |
Values of c of Rolle's theorem for f(x)=sin x-sin 2x on [0,pi] |
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Answer» (i) SINCE , we know thatsinefunctionsare continuous functionshence `f(x)=sinx - sin2x` isa continuous function in `[0,pi]` (ii) `f(x) = cosx -COS2X. 2 = cos x - 2 cos 2X ` , which existin `(0,pi)` So, `f(x)`is differentiable in `(0,pi)`.Conditionsof mean value theoremare satisfied. Hence`3c in (0,pi)` such that,`f'(c) = (f(pi)-f (0))/(pi-0)` `rArr cos - 2 cos 2x = (sin pi - sin 2pi - sin 0 + sin 2.0)/(pi-0)` `rArr 2cos2c -cos c = (0)/( pi)` `rArr 2.(2cos^(2)c-1)-cosc= (0)` `rArr 4 cos^(2)-2-cosc=0` `rArr 4 cos^(2)c-cosc-2 = 0` `rArr cosc = (1+-sqrt(1+32))/(8) = (1+-sqrt(33))/(8)` `:. c= cos^(-1)((1+-sqrt(33))/(8))` Also, `cos^(-1) ((1+-sqrt(33))/(8))in (0,pi)` Hence, meanvalue therem has been verified. |
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| 7. |
Using integration find the area of the triangular region whose sides have the equations Y =2x + 1, y = 3x+1 andx=4. |
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| 8. |
If A is square matrix such that A^(2)=A, then (I+A)^(3)-7A is equal to |
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Answer» A |
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| 9. |
The quardritic equation in x such that the arithmetic meanof its roots is 5 and geometric mean of the roots is 4, is given by |
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Answer» `X^(2) + 20X + 16 = 0` |
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| 10. |
Observe the following list {:("List-I","List-II"),(A. [vec(a)vec(b)vec(c )],1. |vec(a)||vec(b)| cos (vec(a).vec(b))),(B.(vec(c ) xx vec(a)) xx vec(b), 2. (vec(a). vec(c )) vec(b) - (vec(a). vec(b)) vec(c)),(C. vec(a) xx (vec(b) xx vec(c )),3.vec(a).vec(b) xx vec(c )),(D. vec(a). vec(b), 4.|vec(a)| |vec(b)|),(,5.(vec(b) - vec(c )) vec(a) - (vec(a) . vec(b)) vec(c)):} Then the correct match for List-I from list II is |
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Answer» 1,2,3,4 |
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| 11. |
Find a point on the curve y = (x – 2)^(2) at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4). |
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| 12. |
Let A, B, C be three sets. If AinBandBsubC then |
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Answer» `AsubC` |
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| 13. |
Examin the following functions for continuity. f(x)= (1)/(x-5), x ne 5 |
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| 14. |
The term independent of x in (2x^(1//2)-3x^(-1//3))^20 |
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Answer» `""^20C_(7) 2^(8) 3^12` |
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| 15. |
Two trains start from the same place. A person travel in a train whose speed is 50 km/h and it go to south direction. Another person travel in another train whose speed is 60 km/h and it go to west direction. After two hours, find the rate of distance between them. |
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| 16. |
If the two circles (x-1)^(2)+(y-3)^(2)=r^(2) and x^(2)+y^(2)-8x+2y+8=0 intersect in two distinct points, then |
| Answer» ANSWER :B | |
| 18. |
The point of contact of the circle x^(2)+y^(2)+2x+2y+1=0 and x^(2)+y^(2)-2x+2y+1=0 |
| Answer» ANSWER :B | |
| 19. |
If sqrt(1-x^(2)) + sqrt(1 -y^(2))= a(x-y), then prove that (dy)/(dx)= sqrt((1-y^(2))/(1-x^(2))). (Where |x| le 1, |y| le 1) |
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| 20. |
The value of a for which the points (9, 5), (1, 2), (a, 8) are collinear is equal to |
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Answer» 17 |
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| 21. |
int((x^(2)-1)dx)/((x^(4)+3x^(2)+1)Tan^(-1)((x^(2)+1)/(x)))= |
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| 22. |
Which of the following undergoes partial hydrolysis? |
| Answer» Solution :Due to Back bonding. | |
| 23. |
Internal angle bisecotors of DeltaABC meets its circum circle at D, E and F where symbols have usual meaning. Q. The ratio of area of triangle ABC and triangle DEF is : |
| Answer» ANSWER :B | |
| 24. |
Match the following lists: |
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Answer» `alphax^(3)+betax` is an odd function `I=0+2int_(0)^(2)gamma dx=2xx2gamma=4gamma` b. `I=1/2 int_(0)^(1)2sin alpha x sin beta x dx` `=1/2int_(0)^(1)(cos(alpha-beta)x-cos(alpha+beta)x)dx` `=1/2[(sin(alpha-beta)x)/(alpha-beta)-(sin (alpha+beta)x)/(alpha+beta)]_(0)^(1)` `=1/2[(sin (alpha-beta))/(alpha-beta)-(sin (alpha+beta))/(alpha+beta)]`...........1 Also `2 alpha=tan alpha` and `2beta=tan beta` `:. 2 (alpha-beta)=tan alpha-tan beta ` and `2(alpha+beta)=tan alpha+tan beta` `2(alpha-beta)=(sin(alpha-beta))/(cos alpha cos beta)` and `2(alpha+beta)=(sin (alpha +beta))/(cos alpha cos beta)` substituting these values, we get C.`f(x+alpha)+f(x)=0` or `(x+2alpha)+f(x)+alpha)=0` or `f(x+2alpha)=f(x)` Thus, `f(x)` is periodic with periodic `2 alpha`. Hence `int_(beta)^(beta+2gamma alpha) f(x)dx=gamma int_(0)^(2alpha) f(x)dx` d. Let `I=int_(0)^(alpha)[SINX]dx, alpha epsilon[(2beta+1)pi,(2beta+2)pi],beta epsilon N`, ( wher [.] denotes the greatest INTEGER function) `=int_(0)^(2betapi) [sinx]dx+int_(2betapi)^((2beta+1)pi) [sinx]dx+int_((2beta+1)pi)^(alpha) [sinx] dx` `=betaint_(0)^(2PI) [sinx]dx+0+int_((2beta+1)pi)^(alpha) (-1)dx` `=-betapi+(2beta+1)pi-alpha` `=(beta+1)pi-alpha` Thus, `gamma int_(0)^(alpha) [sinx]dx` depends on `alpha beta` and `gamma`. |
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| 25. |
Match the following lists: |
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Answer» `=lim_(nto oo) (1+2/(n+1))^(n+1)-1` `=E^(2)-1` `f'(x)=f(x)impliesf(x)=Ce^(x)` and since `f(0)=1, 1=f(0)=C` Therefore `f(x)=e^(x)` and HENCE `g(x)=x^(2)-e^(x)`. THUS, `int_(0)^(1)f(x)g(x)DX=int_(0)^(1)(x^(2)e^(x)-e^(2x))dx` `=x^(2)e^(x)|_(0)^(1)-2int_(0)^(1)xe^(x) dx=int_(0)^(1)(x^(2)e^(x)-e^(2x))dx` `=x^(2)e^(x)|._(0)^(1)-2int_(0)^(1)xe^(x)dx-(e^(2x))/2|_(0)^(1)` `=(e-0)-2xe^(x)|._(0)^(1)+2e^(x)|_(0)^(1)+2e^(x)|_(0)^(1)-1/2(e^(2)-1)` `=(e-0)-2e+2e-2-1/2(e^(2)-1)` `=e-1/2e^(2)-3/2` `I=int_(0)^(1)e^(e^(x))(1+xe^(x))dx` Let `e^(x)=t` `:. int_(1)^(e)(1+tlogt)(dt)/t int_(1)^(e)e^(t)(1/t +logt)dt` `=[e^(t)logt]_(1)^(e)=e^(e)` d. `L=lim_(kto0)(int_(0)^(k)(1+sin2x)^(1/x)dx)/k` (form `0/0`) `=lim_(kto0) (1+sin2k)^(1/k)` `=e^(lim_(kto0)1/k(sin2k))=e^(2)` |
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| 26. |
Find the projection of the vector hati+3hatj+hatk on thevector 7hati-hatj+8hatk |
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| 27. |
Match the following lists: |
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Answer» Applying property `int_(a)^(b)f(a+b-x)dx=int_(a)^(b)f(x)dx` `I_(1)=int_(pi//6)^(pi//3)sec^(2)((pi)/2-theta)f(2sin2((pi)/2-theta))d theta` `=int_(pi//6)^(pi//3)cosec^(3) theta f(2sin 2 theta) d theta=I_(2)` b. `f(x+1)=f(x+3)` or `f(x)=f(x+2)` Thus, `f(x)` is periodic with period 2. Then `int_(a)^(a+b)f(x)dx` is independent of `a` for which `b` is multiple of 2. Thus, `b=2,4,6`........... c. Let `I=int_(1)^(4)(tan^(-1)[x^(2)])/(tan^(-1)[x^(2)]+tan^(-1)[25+x^(2)-10x])`.............1 Appling `int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`, we get `I=int_(1)^(4)(tan[(5-x)^(2)])/(tan^(-1)[(5-x)^(2)]+tan^(-1)[x^(2)])dx`.............2 Adding equatiions 1 and 2 we get `2I=int_(1)^(4)dx` or `2I=3` or `I=3//2` d.Let` y=SQRT(x+sqrt(x+sqrt(x+)))..............=sqrt(x+y)` or `y^(2)-y-x=0` or `y=(1+-sqrt(1+4x))/(2.1)=(1+sqrt(1+4x))/2( :' YGT1)` `:.I=int_(0)^(2)(1+sqrt(1+4x))/2 dx` `=[x/2+((1+4x)^(3//2))/(3/2xx2xx4)]_(0)^(2)` `=[(1+27/12)=(0+1/12)]` `=1+26/12=19/6` `:. [I]=3` `[I]=3` |
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| 28. |
The value of the expression (1+cot7^(@))(1+cot9^(@))(1+cot11^(@))......(1+cot45^(@))(1-cot52^(@))(1-cot54^(@))......(1-cot88^(@)) is equal to |
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Answer» `2^(18)` ` (1+cot45^(@))(2^(19))=(2^(19))=20^(20)` |
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| 30. |
The locus of the centre of a circle which touches externally the circle x^(2)+y^(2)-6x-6y+14=0 and also touches the y-axis is given by the equation. |
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Answer» `y^(2)-6x-10y-14=0` |
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| 31. |
A channel 27m wide falls at a right angle into another channel 64m wide. The greatest length of the log that can be floated along this system of channels is |
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Answer» 120 |
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| 33. |
Thenormalat thepoint(2,-2)on thecurve3x^2-y^2 =8 is _____ |
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Answer» `x+y =0` |
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| 34. |
If (barb xx barc) xx (barb xx bara) = 3barc, then find [barb xx barc barc xx bara bara xx barb]. |
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| 35. |
Which of the following is/are periodic? |
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Answer» f(x) = x - [x], where [x] DENOTES the greatest integral function |
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| 38. |
The solution of tany(dy)/(dx)=sin(x+y)+sin(x-y) is |
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Answer» `secy=2cosx+c` |
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| 40. |
In triangleABC,a(cosB+sinBcot((A)/(2)))= |
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Answer» `a+b+c` |
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| 41. |
Find the numerically greatest terms in the expansion of(3y + 7x)^10 whenx = 1/3, y = 1/2 |
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| 42. |
Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean, variance and standard deviation of the number of kings. |
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| 43. |
Find the approximate value of f(2.01), where f (x) = 4x^(2) + 5x + 2. |
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| 46. |
Solve the following linear programming problem graphically: Maximise Z=4x+y ………….1 subject to the constraints: x+yle50 ……………2 3x+yle90…………..3 xge0, yge0…………..4 |
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| 47. |
Find the equation of plane passing through the point (-1,1,2) and parallel to the plane x + 2y + 3z - 5 = 0. |
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| 48. |
A : If x=sin(alpha - beta ) sin (gamma - delta), y=sin(beta - gamma ) sin (alpha - delta) , z=sin(gamma - alpha ) sin ( beta - delta) " then " x+y+z=0 R :2 sin A sin B = cos(A-B)+cos(A+B) |
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Answer» A is TRUE , R is true and R is CORRECT explanation of A |
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| 50. |
Prove that : Find the numerically greatest term in the binomial expansion of (1-5x)^(12) when x=(2)/(3). |
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