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.
| 401. |
x+2y+4z=12 y+2z=-1 3x+2y+4z=4 |
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| 402. |
xa^3+ya^2+za=0 In the equation above,x,y and z are constants. IF the equations has roots -6,0 and 4, which of the following is a factor of xa^3+ya^2+za? |
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Answer» a-2 |
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| 403. |
A letter is known to have come from either 'MAHARASTRA' or 'MADRAS' on the post mark only conseutive letter 'RA' can be read clearly. What is the chance that the letter came from 'MAHARASTRA'. |
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| 404. |
If f(x)=sqrt("cosec"^2x-2sinx cosx - 1/(tan^2x)) x in ((7pi)/4,2pi), then f.((11pi)/6)= |
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Answer» `(1-sqrt3)/2` `=sqrt((cosx-sinx)^2)` f(x)=cos x -sin x `RARR` f.(x)=-sin x - cos x `f.((11pi)/6)=-[sin (11pi)/6 + cos(11pi)/6]` `rArr f.((11pi)/6)=-[(-1)/2+sqrt3/2]=(1-sqrt3)/2` |
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| 405. |
The longest distance of the point (a, 0) from the curve 2x^2+y^2=2x is |
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Answer» `1+a` |
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| 406. |
The set of points where the function f given by f(x) = |2x – 1| sin x is differentiable is |
| Answer» Answer :B | |
| 407. |
There are 5 bags, each containing 5 white balls and 3 black balls. Also, there are 6 bags, each containing 2 white balls and 4 black balls. A white balls is drawn at random. Find the probability that this white ball is from a bag of the first group. |
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Answer» <P> SOLUTION :Let `E_1` = event of selecting a bag from the first group,`E_2` = event of selecting a bag from the second group, and E = event of drawing a WHITE ball. Then, `P(E_1)=5/11and P(E_2)=6/11`. `P(E//E_1)`= probability of GETTING a white ball, given that it is from a bag of the first group `=5/8`. `P(E//E_2)` = probability of getting a white ball, given that it is from a bag of the second group `=2/6=1/3` Probability of getting the ball from a bag of the first group, given that it is white `P(E_1//E)` `=(P(E//E_1).P(E_1))/(P(E//E_1).P(E_1)+P(E//E_2).P(E_2))`[by Bayes's theorem] `=((5/8xx5/11))/((5/8xx5/11)+(1/3xx6/11))=75/123`. |
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| 408. |
If veca, vecb and vecc are non-zero vectors, then vecaxxvecb = vecaxxvecc |
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Answer» `VECB` = `VECC` |
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| 409. |
Find the vector cartesian equation of the plane passing through the points (1, -2, 3) and (-1, 2, -1) and is parallel to the line vecr=(2hati-hatj+hatk)+t(2hati+3hatj+4hatk) |
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| 410. |
IF a,b,care thenumberof4digitednumberthat canbeformedusingthe digits2,4,5,7,8that aredivisibleby3,4,5 respectively thenascendingorderof a,b,c, is |
| Answer» Answer :C | |
| 411. |
If veca and vecb are two nonzero vectors such that |vecaxxvecb|=veca.vecb then find the angle between veca and vecb. |
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| 412. |
Match the statements/expressions given in List I with the values given in List II. |
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Answer» <P>`{:(a,b,c,d),(s,r,Q,p):}` `rArr""(3 sin (3 cos^(-1)x))/(sqrt(1-x^(2)))` `rArr""sqrt(1-x^(2))y'=3 sin (3 cos ^(-1)x)` `rArr""(-x)/(sqrt(1-x^(2)))y'+sqrt(1+x^(2))y''=3COS (3cos ^(-1) x)CDOT(-3)/(sqrt(1-x^(2)))` `rArr""-xy'+(1-x^(2))y''=-9y` `rArr""(1)/(y)[(x^(2)-1)y''+xy']=9` |
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| 413. |
Let triangle ABC be right angled at C.If tanA+tanB2, then- |
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Answer» `/_A = 30^(@)` |
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| 414. |
If int (3 cos x - 2 sin x )/(4 sin x + 5 cos x )dx = Ax + B log | 5cos x + 4 sin x | then (A, B ) = |
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Answer» `((7)/(41), (22)/(41)) ` |
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| 415. |
If S is defined on R by (x,y) inR hArr xy ge 0 . Then S is ........... |
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Answer» an EQUIVALENCE RELATION |
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| 416. |
If for a continuous function f(x), int_(-pi)^(t) (f(x)+x)dx=pi^(2)-t^(2), forall t ge -pi, thenf(-(pi)/(3)) is equal to : |
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Answer» `PI` |
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| 417. |
If A and B are coefficients of x^(n)in the expansions of (1+x)^(2n) and (1+x)^(2n-1) respectively, then A/B is equal to |
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Answer» 4 |
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| 418. |
Find the direction cosines of the line whose direction ratios are 12, 4, -8. |
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| 419. |
Show that if f :R {(7)/(5)} to R - {(3)/(5)} is defined by f (x) = (3x +4)/( 5x -7) and g :R - {(3/(5)} to R - {(7)/(5)} is defined by g (x) = (7x +4)/(5x -3), then fog = I _(A) andgof =I _(B), where, A = R - {(3)/(5)}, B+R -{(7)/(5)},I_(A)(x) =x, AA x in A, I_(B)|(x) =x , AA x in B are called identity functions on sets A and B, respectively. |
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| 420. |
Given that x is a real number satisfying(5x^(2)-26x+5)/(3x^(2)-10x+3) lt 0, then |
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Answer» `x LT (1)/(5)` |
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| 421. |
Find the eccentricity of the ellipse (a) whose latus rectum is equal to |
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| 422. |
int_(0)^(1) (x^(5)dx)/(1+x^(4))= |
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Answer» `(2-pi)/(4)` |
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| 423. |
Let A and B be independent events with P(A) = 0 . 3 and P(B) = 0 . 4, . FindP(Acap B) |
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| 424. |
Decide whether each statement Must Be True, Could Be True, or Will Never Be True. pqgt t |
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| 425. |
{:("Column A","Kate ate 1/3 of a cake, Fritz ate 1/5 of remaining cake, and what was left was eaten by Emily","Column B"),("Fraction of the cake eaten by Emily",,5//7):} |
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Answer» If COLUMN A is larger |
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| 426. |
|((a+b)^(2),ca,bc),(ca,(b+c)^(2),ab),(bc,ab,(c+a)^(2))|= |
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Answer» `(ab+bc+ca)^(3)` |
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| 427. |
The x-coordinate of the incentre of the triangle where the midpoints of the sides are (0, 1) (1, 1) and (1, 0) is |
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Answer» `2+sqrt2` |
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| 428. |
Ify = A (x) e^(-intpdx) isa solution of (dy)/(dx) +P(x)y = Q(x) ,then A'(x) = |
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Answer» `E^(intpdx)` |
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| 429. |
If the lines (x-1)/-3 = (y-2)/(2k) = (z-3)/2 and (x-1)/(3k) = (y-1)/1 = (z-6)/-5 are perpendicular, then find the value of k. |
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Answer» SINCE the lines are perpendicular, `-3xx3k+2kxx1+2xx-5=0` `rArr` -9k+2k-10=0 `rArr` 7k=-10 `rArr = (-10)/7` |
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| 430. |
Statement 1 : (A hArr ~B)) a tautology to A hArr B Statement 2: A vv (~A ^^ ~B)) a tautology |
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Answer» Statement 1 and statement 2 are both false |
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| 431. |
If I _(1) = int _(0) ^(1) 2 ^(x ^(2)) dx, I _(2) = int _(0) ^(1) 2 ^(x ^(3)) dx, I _(3) = int _(1) ^(2) 2 ^(x ^(3)) dx and I _(4 ) = int _(1) ^(2) 2 ^(x ^(3)) dx then- |
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Answer» `I _(2) gt I _(1)` |
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| 432. |
LeT p(X,Y) be the midpoint of the line joining (1,0) to a point on the curve y^(2)=|(x+1,x+2),(x+3,x+5)|. The locus of P is symmetrical about |
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Answer» X-axis |
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| 433. |
Find number of three digit numbers equal to the sum of the factorials of their digits. |
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| 434. |
When |x|lt (1)/(2) , the coefficient of x^(4)in the expansion of(3x^(2) - 5x + 3)/((x - 1)(2x +1) (x +3))is |
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Answer» `(722)/(27)` |
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| 435. |
Differentiate x^(sin x), x gt 0 w.r.t x |
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| 436. |
Find the number of 4- letter words that can be formed using the letters of the word. MIRACLE. How many of them begin with an vowel |
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| 437. |
Let a=2^(log_35 , b=3^(log_5 2, c=4 the identify the correct option :- |
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| 438. |
Find the number of 4- letter words that can be formed using the letters of the word. MIRACLE. How many of them begin and end with vowels |
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| 439. |
Find the number of 4-letter words that can be formed using the letters of the word MIRACLE. How many of them end with a consonant ? |
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| 440. |
The xy-plane above shows the two points of intersection of thegraphs of a linear function and a quadratic function. The leftmost point of intersection has coordinates (a, b) and the rightmost point of intersection has coordinates (c, d) . If the vertex of the graph of the quadraticfunction is at (2, -27) , what is the value of b-d ? |
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| 441. |
Let f(x) be a non-positive continuous function and F(x)=int_(0)^(x)f(t)dt AA x ge0 and f(x) ge cF(x) where c lt 0 and let g:[0, infty) to R be a function such that (dg(x))/(dx) lt g(x) AA x gt 0 and g(0)=0 The total number of root(s) of the equation f(x)=g(x) is/ are |
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Answer» `infty` Now, `f(x) ge cF(x)` or `e^(-cx)F^(')(x) -ce^(-cx)F(x) ge0` THUS, `e^(-cx)F(x)` is an increasing function `therefore e^(-cx) F(x) ge e^(-c(0))F(0)` or `e^(-cx)F(x) ge` or `F(x) ge0` [as `f(x) ge cF(x)` and c is POSITIVE. Also, `(dg(x))/(dx) lt g (x) AA x gt 0` or `(d/(dx)) e^(-x)g(x) lt 0` Thus, `e^(-x)g(x)` has ONE solution, `x=0` |
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| 442. |
Let f(x) be a non-positive continuous function and F(x)=int_(0)^(x)f(t)dt AA x ge0 and f(x) ge cF(x) where c lt 0 and let g:[0, infty) to R be a function such that (dg(x))/(dx) lt g(x) AA x gt 0 and g(0)=0 The number of solution(s) of the equation |x^(2)+x-6|=f(x)+g(x) is/are |
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Answer» 2 Thus, no solution exists. |
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| 443. |
Let f(x) be a non-positive continuous function and F(x)=int_(0)^(x)f(t)dt AA x ge0 and f(x) ge cF(x) where c lt 0 and let g:[0, infty) to R be a function such that (dg(x))/(dx) lt g(x) AA x gt 0 and g(0)=0 The solution set of inequation g(x)(cos^(-1)x-sin^(-1)) le0 |
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Answer» `[-1,1/sqrt(2)]` or `(cos^(-1)x-sin^(-1)x) ge0` or `x in [-1, 1/sqrt(2)]` |
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| 444. |
There are 50 stations on a railway line. The number of different kinds of single 2nd class tickets must be printed so as to enable a passenger to go from one station to another station is |
| Answer» ANSWER :C | |
| 445. |
If the vertices A, B, C of a DeltaABC have position vectors (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively, then angleABC angleABC is the angle between the vectors BA and BC), is equal to |
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Answer» `(pi)/(2)` `=( HATI + 2 hatj+3 hatk )- (-hati+ 0 hatj + 0 hatk )` `=[ hati-(-hati )+(2hatj -0 )+ (3 hatk -0 )] = 2 hati+ 2 hatj+ 3hatk` ` IMPLIES| BA|= sqrt((2) ^(2)+(2)^(2)+(3)^(2))=sqrt(4+4+9)=sqrt(17)` `implies BC=PVofC-PV of B ` `=(0 hati +1 hatj + 2 hatk)-(-hati +0hatj+0hatk)=hati +hatj+2hatk` `implies |BC|=sqrt((1)^(2) +(1)^(2)+(2)^(2))=sqrt(1+1+4)= sqrt(6)` Now , `BA.BC =(2 hati + 2 hatj + 3 hatk ).(hati + hatj + 2 hatk )` `=2xx1+2xx1+3xx2=10` ` thereforecos theta= (BA.BC)/(|BA||BC|)implies cos (angle ABC)=(10)/(sqrt(17)sqrt(6))` `impliesangleABC= cos ^(-1) ((10)/(sqrt(102)))` |
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| 446. |
In the figure, the area of the quadrilateral ABCD is 75. What is the area of parallelogram EFGH? |
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Answer» 96 |
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| 447. |
Integrate the following functions : int(dx)/(x(x^(5)+1)) |
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| 448. |
Vertices of a parallelogram are (0, 0), ((1)/(m-n), (m)/(m-n)), ((-1)/(m-n), (m)/(m-n)) and (0, 1) are m,n are the roots of the equation 441x^(2)+42x-8=0. If area of the parallelogram is A then A is equal to |
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| 449. |
Four rotten oranges are accidentally mixed with 16 good ones. Three oranges are drawn at random from the mixed lot. Let X be the number of rotten oranges drawn. Find the mean and variance of X. |
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| 450. |
Equation of circle which touch X-axies at (3,0) and making Y-intercept of length 8 units is |
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Answer» `X^(2)+y^(2)-6x+-10y+9=0` |
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