InterviewSolution
Saved Bookmarks
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.
| 51. |
How ` CO_2 ` makes idies puffy ? |
| Answer» Idlies puffy. The microoganismslike species of Bacillus, Candiada and Saccharomyces areincolved in making idlies. | |
| 52. |
Find k, if one of the lines given by `6x^(2) + kxy + y^(2) = 0` is `2x + y = 0`. |
|
Answer» Let `m_(1)` be the slope of `2x+y=0` `m_(1)=-2` Now, comparing `6x^(2)+kxy+y^(2)=0` we get, `a=6h=k/2,b =1` `thereforem_(1)+m_(2)=(-2h)/(b)=-k` `therefore -2+m_(2)=-kimpliesm_(2)=2-k` Now `m_(1)*m_(2)=a/b` `(-2)(2-k)=6/1` `-4+2k=6` `k=5` |
|
| 53. |
Evaluate : `int_(0)^(pi)(x)/(a^(2)cos^(2)x+b^(*2)sin^(2)x)dx` |
|
Answer» We have, `I =underset(0)overset(pi)int(x)/(a^(2)cos ^(2)x+b^(2)sin^(2)x)dx" "...(1)` `I =underset(0)overset(pi)int(pi-x)/(a^(2)cos^(2)(pi-x)+b^(2)sin^(2)(pi-x))dx` `(because underset(o)overset(a)intf(x)dt=underset(o)overset(a)intf(a-x)dx)` `I=underset(o)overset(pi)int(pi-x)/(a^(2)cos^(2)x+b^(2)sin^(2)x)dx" "...(2)` On adding equations (1) a nd (2), we get `21 =underset(0)overset(pi)int(x+pi-x)/(a^(2)cos^(2)x+b^(2)sin^(2)x)dx` Using the property, `underset(0)overset(pi)intf(x)*dx=underset(0)overset(pi)int[f(x)+(2a-x)]dx` `therefore2I=pi[underset(0)overset(pi)int(1)/(a^(2)cos^(2)x+b^(2)sin^(2)x)dx+underset(0)overset(pi//2)int(1)/(a^(2)cos^(2)(pi-x)+b^(2)sin^(2)(pi-x))dx]` `I=pi underset(0)overset(pi//2)int(dx)/(a^(2)cos^(2)x(1+(b^(2))/(a^(2))tan^(2)x))` `=(pi)/(a^(2))underset(0)overset(pi//2)int(sec^(2)x dx)/(1+(b^(2))/(a^(2))tan^(2)x)` [By putting `u=b/atanx, du =b/asec^(2)x dx` when `x=0, u=0` when `x=pi/2, u=oo` `=(pi)/(a^(2))underset(0)overset(oo)int((a)/(b)du)/(a+u^(2))` `=(pi)/(ab)underset(0)overset(oo)int(du)/(1+u^(2))=(pi)/(ab)[tan^(-1)(u)]_(0)^(oo)+c` `=(pi)/(ab)[tan^(-1),ootan^(-1)0]+c` `=(pi)/(ab)[(pi)/(2)-0]+c` `=(pi^(2))/(2ab)` |
|
| 54. |
Prove that : `intsqrt(a^(2)-x^(2))dx=x/2sqrt(a^(2)-x^(2))+(a^(2))/(2)sin^(-1)((x)/(a))+c` |
|
Answer» Let `I=intsqrt(a^(2)-x^(2))dx` Intergrating by parts `I=sqrt(a^(2)-x^(2))int1dx-int[(d)/(dx)(sqrt(a^(2)-x^(2)))*int1dx]dx` `=xsqrt(a^(2)-x^(2))-int(-2x)/(2sqrt(a^(2)-x^(2)))x*dx` `=xsqrt(a^(2)-x^(2))-int[sqrt(a^(2)-x^(2))-(a^(2))/(sqrt(a^(2)-x^(2)))dx` `=x*sqrt(a^(2)-x^(2))-int[sqrt(a^(2)-x^(2))-(a^(2))/(sqrt(a^(2)-x^(2)))]dx` `=xsqrt(a^(2)-x^(2))-I+a^(2)*sin^(-1)((x)/(a))+c_(1)` `2I=xsqrt(a^(2)-x^(2))+a^(2)*sin^(-1)((x)/(a))+c_(1)` `I=x/2sqrt(a^(2)-x^(2))+(a^(2))/(2)*sin^(-1)((x)/(a))+c_(1)` `therefore intsqrt(a^(2)-x^(2))dx=(x)/(2)sqrt(a^(2)-x^(2))+(a^(2))/(2)sin^(-1)((x)/(a))+c` where, `c=(c_(1))/(2)*` |
|
| 55. |
The probability that a person who undergoes kidney operation will recover is 0.5. Find the probability that of six patients who undergo similar operations. (a) Non will recover. (b) Half of them will recover. |
|
Answer» Since there are six patients, `n=6.` If a patient recovers, the outcome is a success. Thus, `P=p` (success) `=0.5and q=1-p=0.5` Let X: No. of patents, who recovered out of 6. Then `X~B(n=6,p=0.5)` The p.m.f. of X is given as, `P(X=x)=p(x) ""^(6)C_(x)(0*5)^(x)(0*3)^(6-x), x=0, 1,2,3,4,5,6.` (i) Probability (non will reconver) `=P (X=0)` `=""^(6)C_(0)(0*5)^(0)(0.5)^(6)` `=(1)*(1)*(0*5)^(6)` `=0*015625` (ii) Probability (half of them will recover)=P `(X=3)` `=""^(6)C_(3)(0*5)^(3)(0*5)^(3)` `=20xx0*125xx0*125=0*3125` |
|
| 56. |
Evaluate: `int (sinx)/(sqrt(36-cos ^(2)x))dx` |
|
Answer» We have, `int (sinx)/(sqrt(36-cos^(2)x))dx` Let `cos x=t implies-sin x dx =dt` Putting `sin x dx=-dt and cos x=t.` `=-int (dt)/(sqrt(36-t^(2)))` `=-int(dt)/(sqrt((6)^(2)-t^(2)))` `=-sin^(-1)((t)/(6))+c` `=-sin ^(-1)((cos x)/(6))+c` `[becauset=cosx]` |
|
| 57. |
`cos^(-1)(2xsqrt(1-x^(2)))` |
|
Answer» Let `y=cos ^(-1)(2xsqrt(1-x^(2)))` Put ` x=sin theta` `thereforetheta=sin^(-1)x ` `y=cos^(-1)(2sinthetasqrt(1-sin^(2)theta))` `y=cos^(-1)(2sin thetacostheta)` `y=cos^(-1)(sin 2theta)` `y=cos ^(-1)[cos ((pi)/(2)-2theta)]` `y=pi/2-2theta` `y=pi/2-2sin ^(-1)x` `therefore (dy)/(dx)=(-2)/(sqrt(1-x^(2)))` |
|
| 58. |
The probability distribution of a discrate random variable X is : `{:(X=x, 1,2,3,4,5),(P(X=x),k,2k,3k,4k,5k):}` Find `P (X le 4).` |
|
Answer» Here, we must have `k+2k+3k+4k+5k=1` `15k=1` `k=1/15` Now, `P (Xle4)=k+2k+3k+4k` `=1/15+2/15+3/15+4/15=10/15` `therefore P(Xle4)=2/3` |
|
| 59. |
Find the coordinates of the foot of theperpendicular drawn from the point `A(1,2,1)`to the line joining `B(1,4,6)a n dC(5,4,4)dot` |
|
Answer» The equation of a line joining the points P (1,4, 6) and (5,4,4) is `(x-1)/(5-1)=(y-4)/(4-4)=(z-6)/(4-6)` `(x-1)/(4)=(y-4)/(0)=(z-6)/(-2)" "...(1)` Let M be the foot of the perpendicular drawn from the point `(1,2,1).` Coordinates of M are given by `(x-1)/(4)=(y-4)/(0)=(z-6)/(-2)=lamda` `x=4lamda+1, y=0lamda+4, z=-2lamda+6` `therefore M-=(4lamda+1,4,-2lamda+6)" "...(2)` The direction ratios of AM are `4 lamda+1-1,4-2,-2lamda+6-1` i.e., `4lamda,2,-2lamda+5` Direction ratios of given line are `4,0,-2,` since AM is perpendicular to the given line `therefore4(4lamda)+0(2)+(-2)(-2lamda+5)=0` `therefore 16lamda+4lamda-10=0` `therefore lamda=1/2` Putting `lamda=1/2` in equation (2), we get `M-=(3,4,5)` Hence, the co-ordinates of the foot of the perpendicular are `(3,4,5)` |
|
| 60. |
Solve the differential equation `cos (x+y) dy=dx.` Hence find the particular solution for `x=0 and y=0.` |
|
Answer» The given differential equartion is `(dy)/(dx)=(1)/(cos (x+y))" "...(1)` Let `x+y=v` `1+(dy)/(dx)=(dv)/(dx)` `(dy)/(dx)=(dv)/(dx)-1` Substituting `(dy)/(dx)` in equation (1), we get `(dv)/(dx)-1=(1)/(cos v)` `(dv)/(dx)=(1+cos v)/(cos v)` `therefore` Integrating `int(cos v dv)/(1+cos v)dv=intdx+c` `int(1+cos v)/(1+cos v)dv-int(1)/(a+cos v)dv=intdx+c` `int 1dv-int(1)/(2cos^(2)""(v)/(2))dv=x+c` `v-1/2intcos^(2)""(v)/(2)dv=x+c` `v-tan""(v)/(2)=x+c` `(x+y)-tan((x+y)/(2))=x+c" "...(2)` which is the required general solution of the given equation. For `x=0 and y=0,` equation (2), becomes, `(0+0)-tan((0)/(2))=0+c` `impliesc=0` Putting the value of c in equation (2), we get `x+y-tan((x+y)/(2))=x` `impliesy=tan ((x+y)/(2))` |
|
| 61. |
Find the general solution for each of the following equation: `sin x + sin 3x + sin 5x = 0` |
|
Answer» We have, `sin x+sin 3x+sinn 5x=0` `therefore (sin x+sin 5x) +sin 3x=0` `therefore 2 sin ((x+5x)/(2))cos ((5x-x)/(2))+sin3x=0` `therefore2 sin 3x cos 2x+sin 3x=0` `therefore sin3x(2 cos2x+1) =0` Either `sin 3x=0 or 2 cos 2x+1=0` i.e., `sin 3x=0 or cos 2x=-1/2` Now, `cos 2x=-cos ""(pi)/(2)` `cos 2x=cos(pi-(pi)/(3))` `cos 2x=cos ""(2pi)/(3)` `therefore sin 3x=0or cos 2x=cos ""(2pi)/(3)` `3pi=npi, n ne Zor 2x=2mpipm(2pi)/(3)` where` m inZ` Hence, `x=(npi)/(3)or x=mpipm (pi)/(3), wheren, m in Z` |
|
| 62. |
The following is the p.d.g. (Probability Density Function) of a continuous random varible X: `f(x)=(x)/(32), 0ltxlt8 =0,` otherwise (a) Find following the expression for c.d.f. (Cumulative Distribution Function) of X. (b) Also find its value at `x=0.5 and 9.` |
|
Answer» (a) c.d.f. of a continous random variable X is given by `f(x)=underset(-oo)overset(x)intf(y)dy` `therefore f(x)=underset(0)overset(x)intf(y)dy=underset(0)overset(x)int(y)/(32)dy` `=[(y^(2))/(64)]_(0)^(x)=(x^(2))/(64)` Thus , `F(x)=(x^(2))/(64),x inR.` (b) Values of `F(x)` at different values of x. At `x=0.5,F(x)=F(0.5)=((0.5)^(2))/(64)=(0.25)/(64)=(1)/(256)` For any value ofx greater than or equal to `8.F(x)=1` `therefore F(9)=1` |
|
| 63. |
If `y=x^(x), "find" (dy)/(dx).` |
|
Answer» We have, `y=x^(2)` `therefore log y=x log x` On differentiating w.r.t. x, we get `1/y (dy)/(dx)=x/x +log x` `(dy)/(dx)=y+ylog x` `(dy)/(dx)=x^(x)(1+logx)` `(becaue y=x^(3))` |
|
| 64. |
If the function `f(x)={{:(k +x, "for " x lt1),(4x+3, "for " x ge1):} ` is continuous at `x=1` then `k=`A. 7B. 8C. 6D. -6 |
|
Answer» f (x) is continuous at `x=1.` `therefore underset(x to 1)lim f(x) underset(x to 1^(+))lim f(x)=f" "...(1) ` Hence `underset(x to 1 ^(-1))lim (k+x)=underset(x to 1^(+))lim(4x+3)` `therefore k+1=4(1)+3` `k=7-1` `k=6` Hence, the correct answer from the given alternative is (c ). |
|
| 65. |
Given that X ~ B( n=10, p) .If E (x) = 8, find the value of p .A. `0.6`B. `0.7`C. `0.8`D. `0.4` |
|
Answer» Given, `X~B(n=10,P).` `E(X)=8` `E(X)=nP` `therefore nP=8` `10P=8` `P=0.8` Hence the correct answer from the given alternative is (c ). |
|
| 66. |
The displacement s of a moving particle at a time t is given by `s=5+20t-2t^(2)`. Find its acceleration when the velocity is zero. |
|
Answer» Given, `s=5+20t-2t^(2)` velocity `(v) =(ds)/(dt)=20-4t` Now, `v=0.` `therefore 20-4t=0` `therefore t=5` and acceleration `(a) =(dv)/(dt)=-4` `((dv)/(dt))_(t=5)=-4` Hence, the acceleration is-4 unit/`sec^(2).` |
|
| 67. |
The equation of tangent to the curve `y=x^(2)+4x+1` at (-1, -2) isA. `2x-y=0`B. `2x+y-5=0`C. `2x-y-1=0`D. `x+y-1=0` |
|
Answer» The given equation of the curve is, `y=x^(2)+4x+1.` On differentiating w.r.t.x, we get `(dy)/(dx)=2x+4` `therefore ((dy)/(dx))_(x=-1)=2(-1)+4=-2+4=2.` The equation of the tangent at `(-1,2)` is, `y-(-2)=2(x-(1))` `y+2=2(x+1)` ` y+2=2x+2` `2x-y=0` Hence, the correct answer from the given alternative is (a). |
|
| 68. |
If `barp=hati-2hatj+hatkand barq =hati+4hatj-2hatk` are position vector (P,V.) of points P and Q find the position internally in the ratio 2:1. |
|
Answer» Let R `(vecr)` is the point which divides th line segment joining the points P and Q internally in the ratio `2:1.` `vecr=(2(hati+4hatj-2hatk)+1(hati-2hatj+hatk))/(2+1)` `vecr=(3hati+6hatj-3hatk)/(3)` `thereforevecr=hati+2hatj-hatk` and the coordinates of the point R are `(1,2-1).` |
|
| 69. |
Find the vector equation of the plane passing thrugh a point having position vector `3hati-3hatj+hatk` and perpendicular to the vector `4hati+3hatj+2hatk.` |
|
Answer» We know that the vector equation of the plane passing through a point `A(veca)` and normal to `vecn` is `vecr*vecn=veca*vecn.` Here, `veca=3hati-3hatj+hatkand vecn=4hati+3hatj+2hatk.` `therefore` The equation of the required plane is `vecr*(4hati+3hatj+2hatk)=(3hati-2hatj+hatk)(4hati+3hatj+2hatk)` `vecr*(4hati+3hatj+2hatk)=12-6+2` `vecr*(4hati+3hatj+2hatk)=8` This is the required vector equation. |
|
| 70. |
Find the vector equation of the plane passing through the point `hati + hatj -2hatk` and `hati + 2hatj + hatk, 2hati - hatj + hatk.` |
|
Answer» `veca, vecb, vecc` are the position vectors i.e., `veca=hati+hatj-2hatk, vecb=hati+2hatj+hatk, vecc=2hati-hatj+hatkand vecb-veca=0hati+hatj+3hatk` `vecc-veca=hati-2hatj+3hatk` Now, `(vecb-veca)xx(vecc-veca)=|{:(hati, hatj, hatk),(0,1,3),(1,-2,3):}|` `=hati(3+6)-hatj(0-3)+hatk(0-1)=9hati+3hatj-hatk` Let `vecn =(vecb-veca)xx(vecc-veca)=9hati+3hatj-hatk` Then, the vector equation of the required plane is `vecr*vecn=veca*vecn` `vecr(9hati+3hatj-hatk)=(hati+hatj-2hatk)(9hati+3hatj-hatk)` `vecr(9hati+3hatj-hatk)=9+3+2` `vecr(9hati+3hatj-hatk)=14` The cartensian equation of the plane is given by `(x hati+yhatj+zhatk)(9hati+3hatj-hatk)=14` `because"["vecr=x hati+yhatj+z hatj"]"` `9x+3y-x=14` |
|
| 71. |
If the lines `(x-1)/(-3)=(y-2)/(2k)=(z-3)/2`and `(x-1)/(3k)=(y-1)/1=(z-6)/(-5)`are perpendicular, find the value of k. |
|
Answer» We have, `(x-1)/(-3)=(y-2)/(2k)=(z-3)/(2)" "...(1)` and ` (x-1)/(3k)=(y-5)/(1)=(x-6)/(-5)" "...(2)` Now, directin ratios of line (1) are -3, 2k, and line (2) are k,1, -5. If lines are perpendicular, than `(-3)(3k)+(2k)(1)+(2)(-5)=0` `-9k+2k-10=0` `-7k=10` `k=-10/7` |
|