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.
| 951. |
Evaluate the following determinants. [[-18,17,19],[3,0,0],[-14,5,2]] |
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Answer» Solution :`[[-18,17,19],[3,0,0],[-14,5,2]]` `-3[[17,19],[5,2]]` (Expanding along 2ND ROW) =-3(34-95) (-3)(-61)=183 |
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| 952. |
Solve the following differential equations. x(x-1)(dy)/(dx)-(x-2)y=x^(3)(2x-1) |
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Answer» `y((X-1))/(x^(2)) = x^(2) - x + C` |
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| 953. |
Let h(x) =f(x)-a(f(x))^(3) for every real number x h(x) increase as f(x) increses for all real values of x if |
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Answer» `a in (0,3)` or `h(X)=f(x)-2af(x)f(x)+3a(f(x))^(2)f(x)` `=f(x)[3a(f(x))^(2)-2af(x)+1]` Now h(x) increase if f(x) increase and `3a(f(x))^(2)-2af(x)+1gt0for all x in R` or `3agt0 and 4A^(2)-12ale0` or `AGT0 and a in [0,3]` or `a in [0,3]` |
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| 954. |
Find intsqrt(3-2x-x^(2))dx |
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| 955. |
If alpha(theta) epsilon R & beta(theta),theta epsilon R-{2n pi-(pi)/2, n epsilin I} are functions satistying (1+x)sin^(2)theta-(1+x^(2))sintheta +(x-x^(2))=0 then which of the following is/are correct? |
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Answer» `lim_(theta to 0^(+)){(alpha(theta))^(1/(SINTHETA))+(beta(theta))^(1/(sintheta))}=1/(e^(2))` `x^(2)-x(sintheta+(1-sintheta)/(1+sintheta))+sintheta((1-sintheta)/(1+sintheta))=0` `impliesx=sintheta , (1-sintheta)/(1+sintheta)` `impliesalpha(theta)=sintheta, beta(theta)=(1-sintheta)/(1+sintheta)` Hence the RESULTS follows. |
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| 956. |
Assertion (A): The number of ways in which 5 boys and 5 girls can sit in a row so that all the girls sit together is 86400. Reason (R) : The number of ways in which m (first type of different) things and n (second type of different) things can be arranged in a row so that all the second type of things come together is n!""^((n+1))P_(m) The correct answer is |
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Answer» Both A and R are true and R is the correct explanation of A |
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| 957. |
Evaluate the following : [[-1,3,2],[1,3,2],[1,-3,-1]] |
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Answer» SOLUTION :`[[-1,3,2],[1,3,2],[1,-3,-1]]` `[[-2,0,0],[1,3,2],[1,-3,-1]] (R_1~~R_1-R_2)` =`-2[[3,2],[-3,-1]]=-2(-3+6)=-6` |
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| 958. |
The term independent of x in the expansion of (1 + x)^n (1 + 1/x)^n is |
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Answer» `C_0^2 + C_1^2 + C_2^2 + ……….+C_n^2` |
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| 959. |
Gross calorific value and physiological value of carbohydrate and liqid is :- |
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Answer» 4.1 KCAL, 4.0 Kcal and 9.45 Kcal, 9.0 Kcal |
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| 960. |
Solve graphically y gt 5 |
Answer» SOLUTION :
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| 961. |
IF f(x) = (cosx)(cos2x)… (cosnx), then f(x)+sum_(t prop1)^(1)(rtanrx)f(x) is equal to |
| Answer» Answer :B | |
| 962. |
Find the angle between the lines, whose direction cosines are given by the equation 3l+m+5n=0 and 6mn-2nl+5lm=0. |
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| 963. |
Let f(x)=(3)/(x^(4)+3x^(2)+9) and g(x) =(x^(2))/(x^(4)+3x^(2)+9) |
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Answer» `(1)/(2sqrt(3))LOG |(x^(2)-sqrt(3)x-3)/(x^(2)+sqrt(3)x-3)|+c` |
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| 964. |
Let alpha" and "beta be the roots of the equation px^(2)+qx+r=0. If p,q, r in A.P. and alpha+beta=4, then alpha beta is equal to |
| Answer» ANSWER :A | |
| 965. |
Eco Wildlife preserve contains 5x zebras and 2x lions, where x is a positive integer. If the lions succeed in killing z of t he zebras, is the new ratio of zebras to lions less than 2 to 1 ? (1) z > x (2) z = 4 |
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| 966. |
lim_(x rarr oo) (sqrt(x^(2) + 1) - root(3)(x^(3) + 1))/(root(4)(x^(4) + 1) - root(5) (x^(4) + 1)) equals : |
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Answer» -1 |
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| 967. |
bar(a)=2bar(i)-3bar(j)+6bar(k) and bar(b)=-2bar(i)+2bar(j)-bar(k) then ("Proj"_(bar(b))bar(a))/("Proj"_(bar(a))bar(b))=....... |
| Answer» Answer :B | |
| 968. |
Let sec ^(2)((pi)/(9))+sec^(2)((2pi)/(9))+sec_(2)((4pi)/(9))=S and Sigma_(k-1)^(89) cos^(6)(k^(o))=(a)/(b) ,where a,b are coprime, then which of the following is/are correct? |
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Answer» Number of positive divisors of a+B + S is 4 `=((cos.(2pi)/(9))^(2)(cos.(4pi)/(9))^(2) +(cos.(4pi)/(9))^(2)(cos.(pi)/(9))^(2)+cos^(2)((pi)/(9))cos^(2)((2pi)/(9)))/(cos.(pi)/(9)cos.(2pi)/(9)cos.(4pi)/(9))^(2)` Using CD formulas `=(1)/(4)([(cos.(2pi)/(3)+cos.(2pi)/(9))^(2)+(cos.(5pi)/(9)+cos.(pi)/(3))^(2)+(cos.(pi)/(3)+cos.(pi)/(9))^(2)])/(((1)/(64)))` `= 16 ((3)/(4)+cos^(2).(pi)/(9)+cos^(2).(2pi)/(9)+cos^(2).(4pi)/(9)+cos.(pi)/(9)-cos^(2).(2pi)/(9)-cos.(4pi)/(9))` `=16((3)/(9)+(3)/(2))` `therefore cos .(pi)/(9)-cos^(2).(2pi)/(9)-cos.(4pi)/(9)=0 =36` `underset(k=1)overset(89)Sigmacos^(6)(k^(@))= underset(k=1)overset(89)Sigmasin^(6)(k^(@))=(1)/(2)underset(k=1)overset(89)sum(SIN^(6)(k^(@))+cos^(6)(k^(@)))` `(1)/(2)(underset(k=1)overset(89)Sigma(1-(3)/(4)sin^(2)(2k^(@))))=(89)/(2)-(3)/(8)underset(k=1)overset(89)Sigma(2k)^(@)` `=(89)/(2)-(3)/(8)xx45` `(a)/(b)=(221)/(8)` `a221,b=8` |
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| 969. |
If a pair of dice is thrown 5 times then find the probability of getting three doublets. |
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Answer» Solution :In a single through of a pair of dice p(a DOUBLET) `=6/36=1/6` p (a non doublet) = 1-p9a doublet)`=1-1/6=5/6` CLEARLY the given experiment is a binomial experiment with n=5 `p=1/6` and `q=5/6` p(3 DOUBLETS in 5 THROW) = `^5C_3p^3q^2=10 cdot(1/6)^3(5/6)^2=250/6^5=125/3888` |
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| 970. |
Compute the integral I_(n) = int_(0)^(a) (a^(2) - x^(2))^(n)dx , where n is a natural number. |
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| 972. |
If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is |
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Answer» `(x^(2)+y^(2))^(2)=4R^(2)x^(2)y^(2)` `therefore` Line AB is perpendicular to line OP, so slope of line `AB = -(h)/(k)`[`therefore` product of slope of two perpendicular line s is (-1)] Now, the equation of line AB is `y-k=-(h)/(k)(x-h)rArrhx+ky=h^(2)+k^(2)` or`(x)/(((h^(2)+k^(2))/(h)))+(y)/(((h^(2)+k^(2))/(k)))=1` So, point `A((h^(2)+k^(2))/(h),0) and B(0,(h^(2)+k^(2))/(h))` `therefore Delta AOB` is a RIGHT angled triangle, so AB is ONE of the diameter of the circle having radius R (given). `rArr AB = 2R` `rArrsqrt(((h^(2)+k^(2))/(k))^(2)+((h^(2)+k^(2))/(k)))=2R` `rArr (h^(2)+k^(2))^(2)((1)/(h^(2))+(1)/(k^(2)))=4R^(2)` `rArr(h^(2)+k^(2)) =4R^(2) h^(2)k^(2)` On replacing h by x and k by y, we get `(x^(2)+y^(2))^(3)=4R^(2)x^(2)y^(2)`, which is teh REQUIRED locus. |
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| 973. |
If 1,alpha,alpha^2,…...,alpha^(n-1) are the n^(th) roots of unity, then the value of (3-alpha)(3-alpha^2)…...(3-alpha^(n-1)) is |
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Answer» n |
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| 974. |
Using problems are of the Inequality Type. Examples of this type are as follows: |
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Answer» Solution :Let `P(n): tan n alpha gt n tan alpha` Step I For `n=2, tan 2 alpha gt 2 tan alpha ` `rArr (2 tan alpha)/(1-tan^2alpha)-2tan alpha gt 0` `rArr 2 tan alpha ((1-(1-tan^2alpha)/(1-tan^2alpha))gt0` `rArr tan^2alpha.tan 2alpha gt 0 [because 0 lt alpha lt (pi)/(4) "for"n=2]` `rArr tan 2 alpha gt 0 [because 0 lt 2 alpha lt (pi)/(2)]` Which is true (`because` in first quadrant , `tan2alpha ` is POSITIVE) Therefore , P(2)is true. Step II ASSUME that P(k) is true , then `P(k): tan k alpha gt tan alpha` Step III For `n=k+1`, we shall prove that `tan(k+1)alpha gt (k+1)tan alpha` `becausetan (k+1)alpha=(tan k alpha +tan alpha)/(1-tank alpha tan alpha )` .......(i) When `0 lt alpha lt (pi)/(4k)or 0 lt kalpha lt (pi)/(4)` i.e., `0 lt tan k alpha lt1`, also `0 lt tan alpha lt 1` `therefore tan k alpha tan alpha lt1` `1-tan k alpha tan alpha gt 0 and 1-tan k alpha tan alpha lt 1`......(ii) From Eqs. (i) and (ii) , we GET `tan(k+1alpha gt (tan kalpha +tanalpha)/(1) gt tan kalpha +tanalpha gt k tan alpha +tan alpha ` [ by assumption step ] `therefore tan (k+1)alpha gt (k+1)tan alpha` Therefore , `P(k+1)` is true , Hence by the PRINCIPLE of mathematical induction P(n) is true for all `n in N`. |
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| 975. |
By using the properties of definite integrals evaluate the integrals in exercise. overset(5)underset(-5)int|x+2|dx |
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| 976. |
The sumof the numbers formedfromthe digits 2,3,4,5 is |
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Answer» 1,03,124 |
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| 977. |
The plane x-2y + 3z= 2 makes an angle ... With Y-axis. |
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Answer» `cos^(-1)""(2)/(sqrt(14))` |
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| 979. |
If A =[(1,2,1),(0,1,-1),(3,-1,1)] and AA' =I , then x+y is equal to |
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Answer» O |
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| 980. |
If P, Q, R lie on xy=c^(2), then the orthocentre of DeltaPQR lies on |
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Answer» `x+y=0` |
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| 981. |
Choose the correct answer from the bracket. The antiderivative of (sqrtx + 1/sqrtx) equals |
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Answer» `1/3 X^(1/3) +2X^(1/2) + C` |
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| 982. |
Two tangentsare drawfroma pointP, 40 units fromthe centreof the circleand inclinedto eachother at60^(@). Whatis the area , to the nearet integer, of theregionAXBP ? |
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| 983. |
One plane is parallel to the vectors hat i + hat j + hat k and 2 hat i Other plane is parallel to the vectroshat i + hat j andhat i - hat k . The angle between the line of intersection of both the planes and the vector 2 hat i - hat j is ................ |
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Answer» `COS^(-1)"" (3/(sqrt(50)))` |
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| 985. |
Digits 1, 2, 3, ….. 9 are written in a random order to form a 9 digit number. If the number happens to be divisible by 4. Then probability that it will be divisible by 36 is given by |
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Answer» `2/9` |
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| 988. |
Find least non negative integer r such that 1936xx8789 -= r"(mod4)" |
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Answer» SOLUTION :`1936xx8789-= R ("MOD "4) ` `1936xx8789 -= 0 "mod" 4 ` `r=0` |
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| 989. |
int(sin^(3)x)/((cos^(4)x+3cos^(2)x+1)tan^(-1)(secx+cosx))dx= |
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Answer» `tan^(-1)(SECX+cosx)+C` `I=int_(sin^(3)x)/((COS^(4)x+3cos^(2)x+1)tan^(-1)(secx+cosx))dx` `impliesI=int((sin^(3)x)/(cos^(2)x))/((cos^(2)x+3+sec^(2)x)tan^(-1)(secx+cosx))dx` `impliesI=int1/(1+(secx+cosx)^(2))XX(sinx(1-cos^(2)x))/(cos^(2)x)xx1/(tan^(-1)(secx+cosx))dx` `impliesI=int1/(tan^(-1)(secx+cosx))xx1/(1+(secx+cosx))^(2)xx(tanx secx -sinx)dx` `impliesI=int1/(tan^(-1)(secx+cosx))d{tan^(-1)(secx+cosx)}` `impliesI=log|tan^(-1)(secx+cosx)|+C` |
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| 992. |
Evalute : int (cos 2 x - cos2 alpha)/(cos x - cos alpha)dx. |
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Answer» `2(sin X + x cos theta)+C` |
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| 993. |
Differntiate the following functions by proper substitution.sin^(-1)((2x)/(1+x^2)) |
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Answer» Solution :`y=sin^(-1)frac(2x)(1+x^2)`[PUT x=`tantheta` `sin^(-1)frac(2 TAN theta)(1+tan^2theta)=sin^(-1)sin2theta` `=2theta=2 tan^(-1)x` `thereforedy/dx=2/(1+x^2)` |
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| 994. |
The random variable X has a probability distribution P(X) of the following form, where k is some number.Determine the value of k P(x)={k, if x=0 ""2k, if x=1 ""3k if x=2 ""0, otherwise} |
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Answer» SOLUTION :`SUMP(X)=1` `RARR k+2k+3k+0=1 rArr k=1/6` |
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| 995. |
Let A=[[2,4],[3,2]] , B=[[1,3],[-2,5]] , C=[[-2,5],[3,4]] Find each of the folowing AB |
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Answer» SOLUTION :`AB=[[2,4],[3,2]],[[1,3],[-2,5]]=[[2xx1+4xx-2, 2xx3+4xx5],[3xx1+2xx-2, 3xx3+2xx5]]` `=[[2-8, 6+20],[3-4, 9+10]]=[[-6,26],[-1,19]]` |
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| 996. |
Solve the following Linear Programming Problems graphically : Minimise Z = 3x + 5y such that x+3y ge 3, x+y ge 2, x, y ge 0. |
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Answer» <P> |
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| 997. |
Find the area of the region bounded by the curves y = x^(2) + 2, y = x, x = 0 and x = 3. |
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| 998. |
Read of the following two statements I: sqrt(3)x-y+4=0 is tangent to the circle x^(2)+y^(2)=4 II: y=(sqrt(m^(2)-1))x+-mr is tangent to the circle x^(2)+y^(2)=r^(2) |
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Answer» I is true, II is true, II is CORRECT explanation of I. |
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| 999. |
For what value of k, the function f(x) ={:{(2x+1", "x gt2),(""k ", " x=2),(3x-1 ", "x lt2):}, is continuous at x=2. |
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| 1000. |
Evaluate the following integrals inte^(x)((x-1))/((x+1)^(3))dx |
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