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. |
W(x,y,z) = xy + yz + zx, x = u - v, y = uv, z = u + v, u, v inR. Find (del w)/(del u) , (del w)/(del v)and evaluate then at ((1)/(2), 1). |
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| 2. |
The position vectors of the points A,B,C are given to be 2overset^^i+overset^^j-overset^^k,overset^^3i-overset^^2j+overset^^k"and" 1overset^^i+4overset^^j-3overset^^k respectively.Prove that A,B,C are collinear |
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Answer» SOLUTION :From `vec(AB)=-vec(AC)` `:. A,B,C` are collinear |
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| 3. |
Find the equation of the circum circle of the triangle formed by the straight lines given in each of the following: x-y-2=0, 2x - 3y + 4 = 0, 3x - y +6 = 0 |
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| 4. |
Find the equation of the circum circle of the triangle formed by the straight lines given in each of the following: 5x-3y + 4 = 0, 2x+ 3y - 5 = 0, x + y = 0 |
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| 5. |
If A and B are two events of a sample space then P(A)-P(B)= |
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Answer» <P>`P(ANNB')+P(A'nnB)` |
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| 6. |
Find the equation of the circum circle of the triangle formed by the straight lines given in each of the following: 2x+y = 4, x+y=6, x+2y = 5 |
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| 7. |
Differentiate the functions (x+ 3)^(2).(x+ 4)^(3).(x+ 5)^(4) |
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Answer» `(1)/(y) sum_(i=1)^(3) (i-1)/( (X+1- i) )` |
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| 8. |
The sides BC, CA, AB of a triangle ABC have 3, 4 and 5 interior points respectively on them. The number of triangles that can be constructed using these points as vertices is |
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Answer» 220 |
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| 9. |
{:("List- I" , " List - II") , ( "(P) At any point on the curve " y=a^(x) " then lengthof subnormal " , "(1) Varies as cube of the ordinate of the point "),("(Q) At any point on the curve " xy = a^(2) " then length of subnormal" , "(2) Varies as square of the ordinate of the point"),("(R) At any point on the curve " x^(2)y^(3) = a^(5) " the length of subtangent" , "(3) Constant"),("(S) At any point on the curve " Y =be^(x/a) " the length of subtangent " , "(4) Varies as abscissa of the point "):} |
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Answer» <P>(I) (II) ( R) |
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| 10. |
Choose the correct answer int dx/sqrt(9x-4x^2)= |
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Answer» `1/9 sin^-1((9x-8)/8)+C` |
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| 11. |
cot theta - tan theta - 2 tan 2theta- 4 tan 4theta= |
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Answer» `4cot8theta-tan6theta` |
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| 12. |
A letter is known to have come from either 'MAHARASTRA or MADRAS on the post mark only consecutive letters 'RA' can be read clearly. What is the chance that the letter came from 'MAHARASTRA'. |
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| 13. |
IF2x^2 -10 xy+2 lamda y^2 +5x-16y-3=0 representsa pairof straightlines, thenpointof intersectionof those linesis |
| Answer» Answer :C | |
| 14. |
Assertion(A ):theequationwhoseroosare multipled ofby 2 ofthoseofx^5 - 2x^4 +3x^3 -2x^2 +4x +3=0isx^5 - 4x^4 +12 x^3 -16 x^2 +64 x+96=0 Reason (R ) :theequationwhoserootsare mulipliedby k of thoseoff(x ) =0isf((x )/(k ))=0 |
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Answer» bothA and RaretrueR isthe correctexplanationof A |
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| 15. |
If cosalpha + cosbeta + cosgamma =0= sinalpha + sinbeta + singamma, then |
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Answer» `sin^(2)alpha + sin^(2)BETA + sin^(2)GAMMA = 3/2` |
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| 16. |
Match the following. I. int (sin x)/(1+cos 2x)dx= "" a) tan x-(1)/(2) x+c II int (1)/(1-cos2x)dx= "" b)- cot x-(1)/(2)x+c III. int (1+cos^(2)x)/(1-cos2x)dx= "" c) (1)/(2)secx+c IV. int (1+sin^(2)x)/(1+cos2x)dx= "" -d) (1)/(2)cotx+c |
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Answer» a,b,C,d |
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| 17. |
Evaluate the following:lim_(xtoinfty)(x^4-5x+2)/(x^3-3x+1) |
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Answer» SOLUTION :`lim_(xtoinfty)(x^4-5x+2)/(x^3-3x+1)` `lim_(xtoinfty)(x-5/x^2+2/x^3)/(1-3/x^2+1/x^3)=INFTY` |
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| 18. |
Let n be positive integer such that, (1+x+x^(2))^(n)=a_(0)+a_(1)x+a_(2)x^(2)+….+a_(2n)x^(2n), then (n-r)a_(r)+(2n-r+1)a_(r-1),0ltrlt2n is : |
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Answer» `(r+1)a_(r+1)` |
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| 20. |
int e^(x)(1+sin x)/(1+cos x) dx is equal to |
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Answer» `E^(x) sin x + C` |
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| 21. |
Write the simplest form of tan^(-1)((cosx-sinx)/(cosx+sinx)), 0 lt x lt (pi)/(2) |
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| 22. |
A point on the hypotenuse of a right - angled triangle is at distance a and b from sides of the triangle . Showthat the minimum length of the hypotenuse is(a^((2)/(3))+b^((2)/(3)))^((3)/(2)) |
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| 23. |
If k and l respectively are the order and degree of the differentialequation whose general solution represents the family of circles of constant radius, then k^(2)+ l^(2)= |
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Answer» 2 |
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| 24. |
int(sinx)/(sqrt(9-sin^(2)x))dx= |
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Answer» `LOG(sinx+SQRT(9-sin^(2)x))+C` |
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| 25. |
If y=3 cos(log x)+4sin(logx) show that x^(2)y_(2)+xy_(1)+y=0 |
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| 26. |
The numberof trianglewhosevertices are atthe verticesof anoctagonbutnoneofwhosesideshappento comefrom the sidesof theoctagonis |
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Answer» 24 |
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| 28. |
State which of the following statements are true (T) or false(F) The line (x+3)/(-1)=(y-2)/3=(z-1)/4 is perpendicular to the plane 3x-3y+3z-1=0 |
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| 29. |
The number of circles that touch all the 3 lines 2x+y=3, 4x-y=3, x+y=2 is |
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Answer» 1 |
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| 31. |
Solution set of the inequality 2x+y gt 5 is ……… |
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Answer» The half PLANE containing origin |
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| 32. |
Prove that the sequence with the general term x_(n)=(1)/(3+1) +(1)/(3^(2)+2)+...+(1)/(3^(n)+n) has a finite limit. |
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| 33. |
The random variable X has probability distribution P(X) of the following form. P(X)= {(k, if X=0),(2k,if X=1),(3k, if X=2),(0, "otherwise"):} Find P(X lt 2) |
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| 34. |
The random variable X has probability distribution P(X) of the following form. P(X)= {(k, if X=0),(2k,if X=1),(3k, if X=2),(0, "otherwise"):} Determine value of K |
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| 35. |
If Delta_(k)=|(2(3^(k-1)),3(4^(k-1)),4(5^(k)-1)),(alpha, beta, gamma),(3^(n)-1,4^(n)-1,5^(n)-1)| then the value of sum_(k=1)^(n)Delta_(k) depends |
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Answer» only on `ALPHA` and `BETA` not on `GAMMA` |
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| 36. |
I :tan 20^@+ tan 40^(@) + sqrt(3) tan 20^(@) tan 40^(@) =1 II : ((1+ tan 21^(@))(1+ tan 24^(@)))/((1+ tan 22^(@) )(1+ tan 23^(@)))=2 |
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Answer» only I is TRUE |
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| 37. |
If (a^(2) , a-2) be a point interior to the region of the parabola y^(2) = 2x bounded by the chord joining the points (2,2) and (8,-4) then find the set of all possible real values of a. |
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| 38. |
If a=hati+hatj+hatk, b=hati-hatj+2hatk and c=xhati+(x-2)hatj-hatk and if the vector c lies in the plane of vectors a and b, then x equals |
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Answer» 0 and `c=xhati+(x-2)hatj-hatk` Since, c lies in the plane of VECTORS a and b, THEREFORE a, b and c are coplanar. `therefore |(1,1,1),(1,-1,2),(x,(x-2),-1)|=0` `implies 1(1-2x+4)-1(-1-2x)+1(x-2+x)=0` `implies 5-2x+1+2x+2x-2=0` `implies x = - 2 ` |
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| 39. |
Evaluate the following definite integrals : int_(0)^(pi/4)sqrt(1+sin2x)dx |
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| 40. |
Let vecam vecvm vecc be the position vectors of the vertices A,B,C respectively of triangleABC. The vector area of triangleABC is |
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Answer» `1/2 {veca XX (vecb xx VECC)+vecb xx (vecc xx veca)+vecc xx (veca xx vecb)}` |
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| 42. |
If alpha is an n^(th) root of unity then prove that 1+2alpha+3alpha^2+4alpha^3+….+nalpha^(n-1)=-(n)/(1-alpha) |
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Answer» `(N)/(1-alpha)` |
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| 43. |
Does the following graph pass the vertical or horizontal line test ? |
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Answer» SOLUTION :CLEARLY, the graph passes the vertical line test as any vertical line, including x = 1 andx = 2, meets the graph only once. HOWEVER, it does not pass the horizontal test as the LINEY = 2 meets the graph at two POINTS(1, 2) and (2, 2).
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| 44. |
Let f(x)=int_(0)^(x)f(t) dt equals |
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Answer» `5int_(X+5)^(5)g(t)dt` On replacingx by (-x) , we GET `f(x)=int_(0)^(-x)g (t)dt` Now , PUT t =- u , so `int(-x)=-int_(0)^(x)g(-u)du=-int_(0)^(x)g(u)=-f(x)` `[:' `g is an even function] `rArrf(-x)=-f(xrArr f` an odd function. Now , it is given that f (x +5) = g (x) `:.f(5-x)=g(-x)=g(x)=f(x+5)` [ `:.` g is an even function] `rArrf(5-x)=f(x+5)`. . .(i) Let `I=int_(0)^(x)f(t)dt` Putt = u +5 `rArrt -5=u rArrdt = du` `:.u=-trArrdu=-dt ` , we get `I=-int_(5)^(5-x)g(-t)dt=int_(5-x)^(5)f (t) dt` `[:'-int_(a)^(b)f(x)dx=int_(b)^(a)f (x)` dx andg is an even function] `I=int_(5-x)^(5)f'(t)dt` [ by Leibnitz RULE f ' (x) = g (x) ] `=f(5)-f(5-x)=f(5)-f(5+x)`[ from Eq . (i)] `=int_(5+x)^(5) f ' (t) dt = int_(5+x)^(5)g (t) dt` |
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| 46. |
Let S(M) denote the sum of the digits of a positive integer M written in base 10. Let N be the smallest positive integer such that S(N) = 2017. Find the value of S(5N + 2017) |
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| 47. |
If f(x) = (1)/(x^(2)) int_(3)^(x)(2t-3f'(t)dt, then f'(3) is equal to |
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Answer» `-(1)/(2)` |
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| 49. |
The matrices P[(u_(1),v_(1),w_(1)),(u_(2),v_(2),w_(2)),(u_(3),v_(3)w_(3))] and Q=(1)/(9)[(2,2,1),(12,-5,m),(-8,1,5)] are such that PQ=l, an identify matrix. Solving the equation [(u_(1),v_(1),w_(1)),(u_(2),v_(2),w_(2)),(u_(3),v_(3),w_(3))][(x),(y),(z)]=[(1),(1),(5)], the value of y comes out to be -3, then the value of m is equal to |
| Answer» ANSWER :D | |