InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2951. |
Find Lt_(n rarr oo){(sqrt(n))/(n^(3//2)) + (sqrt(n))/((n+3)^(3//2)) + …+ (sqrt(n))/([n+3(n-1)]^(3//2))} |
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| 2952. |
For what values of k , f(x)={:{((1-coskx)/(xsinx)," if " x ne 0),(1/2," if " x =0):} is continuous at x = 0 ? |
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| 2953. |
Find the all values of a for which every real root f the equation cos 3x = a cos x + (4-2|a|) cos ^(2) x is a root of the equation cos 3x + cos 2 x =2 cos x cos 2 x-1and vice veirs. |
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| 2954. |
Assertion (A) : If x+(x^(2))/(2)+(x^(3))/(3)+………oo=log((7)/(6)) then x=(1)/(7) Reason (R ) : If |x| lt 1, then x+(x^(2))/(2)+(x^(3))/(3)+…..oo=log((1)/(1-x)) |
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Answer» A is TRUE, R is true and R is correct explanation of A |
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| 2955. |
A is a 3xx4 matrix .A matrix B is such that A'B and BA' are defined . Then the order of B is …….. |
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Answer» `3xx4` |
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| 2956. |
A lightshines fromthe top of apole50 ft. high A ball isdropped from thesame height from a point30 ft. away fromthe light . If the shadow of theball movingat therateof 100lambda ft//sec alongthe ground 1//2sec. later [Assume the ballfalls a distance s= 16t ^(2) ft. in 't' sec ] then | lambda| is : |
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| 2957. |
Let veca = hati- hatk, vecb = x hati + hatj + (1-x) hatk and vec c = y hati + x hatj + (1+ x -y) hatk. Then [veca vecb vec c] dependson: |
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Answer» only X |
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| 2958. |
Ifthe pair of lines given by (x^(2) + y^(2)) cos^(2) theta = (x cos theta + y sin theta)^(2) are pendicular to each other , then theta is equal to |
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Answer» 0 |
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| 2959. |
A cylindrical vessel of volume 25(1)/(7) cu metres, open at the top is to be manufactured from a sheet of metal. If r and h are the radius and height of the vessel so that amount of metal I sused in the least possible then rh is equal to |
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| 2960. |
If x is so small such that its square and higher powers may be neglected, then find the value of ((1-2x)^(1//3) + (1+ 5x)^(-3//2))/((9+x)^(1//2)) |
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| 2961. |
Compute the are length of the involute of a circle x=a (cos t + t sin t), y= a(sin t-t cos t) from t=0 to t= 2pi. |
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| 2962. |
For each positive integer n, consider the highest common factor hn of the two numbers n! + 1 and (n + 1)!. For n lt 100, find the largest value of h_n |
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| 2963. |
Two parabolas y^(2) = 4a(x- lambda_(1) ) and x^(2) = 4a(y- lambda_2) always touch each other, lambda_1 and lambda_2 being varaible parameters. Then show that the locus of their point of contact is a hyperbola. |
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| 2964. |
If I=int2/x(x^("lnx"))(nx)^(3)dx=Ax^("lnx"^(2))-Bx^("lnx")+C, then A/B is equal to: |
| Answer» ANSWER :a | |
| 2965. |
Which of the following statements is tautology ? |
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Answer» <P>`(~~Q^^p)^^q`
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| 2966. |
int_(0)^(a) |x-1|dx |
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Answer» Solution :`" Let I "= int_(0)^(4) |X-1|dx` `:. I= int_(0)^(1) |x-1|dx+INT _(1)^(4) |x-1|dx` `=int_(0)^(1) (1-x)dx+ int_(1)^(4) (x-1) dx` `=[x-(x^(2))/(2)]_(0)^(1) +[(x^(2))/(2)-x]_(1)^(4)` `=(1-(1)/(2))-0+((4^(2))/(2)-x)-((1)/(2)-1)` `=(1)/(2)+4+(1)/(2)=5` |
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| 2967. |
Find the area lying above x-axis and included between the circle x^2+y^2=8x and inside of the parabola y^2=4x. |
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| 2968. |
If the tangents drawn from a point P to the ellipse 4x^(2) + 9 y^(2) - 24x + 36 y = 0 are perpendicular, then the locus of P is |
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Answer» `x^(2) +y^(2) - 6x +4y +13= 0` |
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| 2969. |
If 1,alpha,alpha^2,…,alpha^(n-1) are the n^(th) roots of unity show that sum_(r=1)^(n-1)r(alpha_r+alpha_(n-r))=-n. |
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| 2970. |
Show that the following vector are co-planar. hati+2hatj+3hatk, -2hati-4hatj+5hatk, 3hati+6hatj+hatk |
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Answer» SOLUTION :LET `veca = hati+2hatj+3hatk` `vecb = -2hati-4hatj+5hatk` `vecc = 3hati+6hatj+hatk` = 1(-4-30)-2(-2-15)+3(-12+12) = -34+34 = 0 As we know scalar TRIPLE product`(VECAXXVECB).vecc = 0`, it follows that the given vectors `veca.vecb.vecc` are coplanar. |
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| 2972. |
Find the cartesian equation of the plane through the point (2,-1, 1) and perpendicular to the vector 4hati+2hatj-3hatk. |
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| 2973. |
Find the equation of a curve passing through the point (-2, 3), given that the slopw of the tangent to the curve at any point (x, y) is (2x)/y^(2). |
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| 2974. |
A straight line passes through the points (5, 0) and (0, 3). The length of perpendicular from the point (4, 4) on the line is : |
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Answer» `(15)/(SQRT(34))` |
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| 2975. |
If therootsof theequationx^3 +3px^2+ 3qx-8=0areina geometricprogression, then(q^3)/( p^3)= |
| Answer» ANSWER :D | |
| 2976. |
Let y be an implicit fuction of x defined by : x^(2x)-2x^(x) cot y-1=0. Then f'(1) equals : |
| Answer» ANSWER :A | |
| 2977. |
Find the sum .^(n)C_(0) + 2 xx .^(n)C_(1) + xx .^(n)C_(2) + "….." + (n+1) xx .^(n)C_(n). |
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Answer» Solution :Method I : `.^(n)C_(0)+2xx.^(n)C_(1)+3xx.^(n)C_(2)+"...."+(n+1)xx .^(n)C_(n)` `= UNDERSET(r=0)overset(n)sum(r+1).^(n)C_(r)` `=underset(r=0)overset(n)sum[r.^(n)C_(r)+.^(n)C_(r)]` `=n underset(r=0)overset(n)sum.^(n-1)C_(r-1)+underset(r=0)overset(n)sum.^(n)C_(r)` `= n(.^(n-1)C_(0) + .^(n-1)C_(1) + .^(n-1)C_(2)+"..."+.^(n-1)C_(n-1)) + (.^(n)C_(0)+.^(n)C_(1)+.^(n)C_(2)+"....."+.^(n)C_(n))` `= n2^(n-1) + 2^(n)` `= (n+2)2^(n-1)` Method II : We have `(1+x)^(n) = .^(n)C_(0)+.^(n)C_(1)x+.^(n)C_(2)x^(2)+"....."+.^(n)C_(n)x^(n)` `:. x(1+x)^(n) = .^(n)C_(0)x+.^(n)C_(1)x^(2)+.^(n)C_(2)x^(3) + "....." + .^(n)C_(n)x^(n+1)` Differentiating w.r.t. x, we get `n(n1+x)^(n-1)x+(1+x)^(n)=.^(n)C_(0)+2xx.^(n)C_(1)x+3xx.^(n)C_(2)x^(2)+"..."+(n+1)xx.^(n)C_(n)x^(n)` Putting `x = 1`, we get `n2^(n-1)+2^(n)=.^(n)C_(0) +2xx.^(n)C_(1)+3xx.^(n)C_(2)+"....."+(n+1)xx.^(n)C_(n)` |
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| 2978. |
If x,y,z in R satisfies the system of equations x +(y) + (s) =12.7, [x]+{y}+z=4.1 and {x} +y+[z] =2 where {.} and [.] denotes the fractional and integral parts respectively) then match the following |
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| 2979. |
The p^(th), q^(th) and r^(th)terms of an A.P. are a, b, c, respectively. Show that (q-r) a+(r-p) b+(q-p)c = 0 |
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| 2980. |
Write the negation of the following statements There exists x in N , x + 3 =10 |
| Answer» SOLUTION :For all`X in N, x+3 != 10` | |
| 2981. |
Prove that (3 !)/(2(n + 3)) = sum_(r=0)^(n)(-1)^(r ) ((""^(r )C_(r ))/(""^(r + 3)C_(r ))) |
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Answer» Solution :`sum_(R=0)^(n) (-1) (""^(n)C_(r))/(""^(r+3)C_(r))` `= sum_(r=0)^(n) (-1)^(r)(n!.3!)/((n-r)!(r+3))=3!sum_(r=0)^(n) (-1)^(r)(n!)/((n-r)!(r+3)!)` `= (3!)/((n+1)(n+2)(n+3)).sum_(r=0)^(n) ((-1)^(r).(n+3)!)/((n-r)!(r+3)!)` `=(3!)/((n+1)(n+2)(n+3)) =.sum_(r=0)^(n) (-1)^(r).""^(n+3)C_(r+3)` `= (3!(-1)^(3))/((n+1)(n+2)(n+3))sum_(s=0)^(n) (-1)^(s).""^(n+3)C_(3)` `=(-3!)/((n+1)(n+2)(n+3))(sum_(s=0)^(n+3) (-1)^(s).""^(n+3)C_(s))_(""^(n+3)C_(0)+""^(n+3)C_(1) -""^(n+3)C_(2))` `=(-3!)/((n+1)(n+2)(n+3)){0-1+(n+3)-((n+3)(n+2))/(2!)}` `=(-3!)/((n+1)(n+2)(n+3)).((n+2)(2-n-3))/(2) = (3!)/(2(n+3))` |
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| 2982. |
(C_1)/(2) + (C_3)/(4) + …….+(C_15)/(16)= |
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Answer» `(2^15 - 1)/(16)` |
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| 2985. |
An equation of the ellipse whose length of the major axis is 10 and foci are (pm 2, 0)is |
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Answer» `x^2/25+y^2/21=1` |
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| 2986. |
Evaluation of definite integrals by subsitiution and properties of its : int_(a)^(a+1)|a-x|dx=.........(ainN) |
| Answer» ANSWER :A | |
| 2987. |
int_0^1x(1-x)^100dx |
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Answer» SOLUTION :`I=int_0^1x(1-x)^100dx` =`int_0^1(1-x)x^100dx` =`int_0^1(x^100-x^101)DX` =`[x^101/101-x^102/102]_0^1` =1/101-1/102=1/10302 |
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| 2988. |
If(1+ x+ x ^ 2+ x^ 3 )^5= sum _(k = 0) ^(15)a _k x ^k, thensum _ (k= 0 ) ^(7) a _(2k)isequalto |
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Answer» 128 ` (1 + x + x ^ 2+ x ^ 3 ) ^5 =sum _(k = 0) ^(15 )=a _ k x ^k ` `RARR (1 +x+x ^ 2+x ^3) ^5 =a_ 0+ a _ 1 x ^ 1+ a _2x ^ 2+… +a _ (14)x ^(14 )+a _ (15)x ^(15)`… (1) Substituting`x = 1`,in (1) ` rArr(1 +1+ 1+ 1 ) ^(5)= a_ 0+ a _ 1 + a _ 2+ ... +a _ (14) +a _ (15)` ` rArr4 ^(5)=a _ 0+a _ 1+a _ 2+...+a _(14)+a _ (15) ""... (2) ` Substituting` x =-1`in (1) ` rArr(1 - 1 + (1) ^2+ (1)^3 =a _0- a _1+ a _2-a_3 +... +a _ (14)-a _ (15) ` `0 =a _ 0- a _ 1+a _ 2 +..+ a _ (14 )+a _ (15) `...(3) Adding(2)and(3) `rArr 4 ^ 5= 2 ( a _ 0+a _ 2+a_4+... +a _ (14)) ` `RARRA _ 0+a _ 2 +... +a _ (14 )= ( 4 ^5 )/(2 ) ` `rArra_ 0+a _ 2+... +a _ (14)=2 ^9 ` ` = 512 ` |
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| 2989. |
IF x=1+log_abc,y=1+log_bca,z=1+log_cab, prove that xyz=xy+yz+zx. |
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| 2991. |
If the sixth term in the expansion of [3log_(3sqrt(9^(x-1)+7))+1/(3^(log3(3^(x-1)+1)))]^(7) is 84, ttien sum of the possible values of x is __________. |
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| 2992. |
If A is a square matrix such that A^(2)=I then (A-I)^(3)+(A+I)^(3)-7Ais equal to ……. |
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Answer» A |
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| 2993. |
A straight line 4x +y -1 = 0throughthe point A(2,-7) meets the line BC whose equation is 3x - 4y +1at the point B . Then the equation of the line AC such thatAB = AC , is |
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Answer» 89x - 52y - 162 = 0 |
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| 2994. |
Find the number of ways of arranging the letters of the word SINGING so that they begin and end with I |
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| 2995. |
If [[2x , 5], [4,2]]is singular, then x = |
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Answer» 20 |
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| 2996. |
Discuss the continuity of the functionf (x)= |x| " at "x=0 |
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| 2997. |
The minimum value of the expression sin theta_(1) + sin theta_(2) + sin theta_(3), where theta_(1), theta_(2), theta_(3) are real numbers satisfyingtheta_(1) + theta_(2) + theta_(3) = pi is : |
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Answer» negative |
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| 2998. |
If A = [{:(0,-1,2),(2,-2,0):}], B = [{:(0,1),(1,0),(1,1):}] and M = AB, then thevalue of M^(-1) is |
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Answer» `[{:((2)/(3),(-1)/(3)),((1)/(3),(4)/(5)):}]` |
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| 2999. |
Integrate the functions (e^(x)-e^(-x))/(e^(x)+e^(-x)) |
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| 3000. |
AB and CD are two equal and parallel chords of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2)) =1. Tangents to the ellipse at A and B intersect at P and tangents at C and D at Q. The line PQ |
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Answer» passes through the origin `rArr` Equations of Ab and CD are `(X)/(a) alpha + (y)/(b) beta =1` and `(x)/(a) alpha_(1) +(y)/(b) beta_(1) =1` (Chord of CONTACT) These lines are parallel `rArr (alpha)/(alpha_(1)) = (beta)/(beta_(1)) =k` ALSO `(alpha^(2))/(alpha^(2)) +(beta^(2))/(b^(2)) = (alpha_(1)^(2))/(a^(2)) + (beta_(1)^(2))/(b^(2))` `rArr (alpha)/(alpha_(1)) = (beta)/(beta_(1)) =-1` `rArr PQ` passes through origin and is bisected at the origin. |
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