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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.
| 451. |
Differeniate `cot^(-1)(sqrt(1+x^(2))+x)` w.r.t. x. |
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Answer» Let `y=cot^(-1)(sqrt(1+x^(2))+x)`. Putting `x=cot theta`, we get `y=cot^(-1)("cosec "theta+cot theta)=cot^(-1)((1)/(sin theta)+(cos theta)/(sin theta))` `=cot^(-1)((1+cos theta)/(sin theta))=cot^(-1){(2cos^(2)(theta//2))/(2sin(theta//2)cos(theta//2))}` `=cot^(-1)("cot"(theta)/(2))=(theta)/(2)=(1)/(2)cot^(-1)x.` `therefore(dy)/(dx)=(-1)/(2(1+x^(2)))` Hence, `(d)/(dx){cot^(-1)(sqrt(1+x^(2))+x)}=(-1)/(2(1+x^(2))).` |
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| 452. |
If `y=log_(x^(2)+4)(7x^(2)-5x+1)`, then `(dy)/(dx)` is equal toA. `log_(e)(x^(2)+4).{(14x-5)/(7x^(2)-5x+1)-(2xy)/(x^(2)+4)}`B. `(1)/(log_(e)(x^(2)+4)){(14x-5)/(7x^(2)-5x+1)-(2xy)/(x^(2)+4)}`C. `log_(e)(7x^(2)-5x+1){(2x)/(x^(2)+4)-((14x-5)y)/(7x^(2)-5x+1)}`D. `(1)/(log_(e)(7x^(2)-5x+1)){(2x)/(x^(2)+4)-((14x-5)y)/(7x^(2)-5x+1)}` |
| Answer» Correct Answer - B | |
| 453. |
Find `(dy)/(dx)`if `y=tan^(-1)(4x)/(1+5x^2)+tan^(-1)(2+3x)/(3-2x)` |
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Answer» Correct Answer - `(5)/(1+25 x^(2))` `y=tan^(-1)""(4x)/(1+5x^(2))+tan^(-1)""(2+3x)/(3-2x)` `=tan^(-1)""(5x-x)/(1+5xcdotx)+tan^(-1)""((2)/(3)+x)/(1-(2)/(3)x)` `tan^(-1)5x-tan^(-1)x +tan^(-1)""(2)/(3)+tan^(-1)x` `=tan^(-1)""5x+tan^(-1)""(2)/(3)` `therefore" "(dy)/(dx)=(5)/(1+25x^(2))` |
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| 454. |
If `y=sin^(-1){(5x+12 sqrt(1-x^(2)))/(13)},` find `(dy)/(dx).`A. `(1)/(sqrt(1-x^(2)))`B. `(-1)/(sqrt(1-x^(2)))`C. `(3)/(sqrt(1-x^(2)))`D. `(1)/(sqrt(1+x^(2)))` |
| Answer» Correct Answer - A | |
| 455. |
Find `(dy)/(dx)`, when: `y=(2x+3)^(5)(3x-5)^(7)(5x-1)^(3)` |
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Answer» Correct Answer - `(2x+3)^(5)(3x-5)^(7)(5x-1)^(3).[(10)/((2x+3))+(21)/((3x-5))+(15)/((5x-1))]` `logy=5log(2x+3)+7log(3x-5)+3log(5x-1)` `rArr(1)/(y).(dy)/(dx)={(5)/((2x+3)).2+(7)/((3x-5)).3+(3)/((5x-1)).5}.` |
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| 456. |
if `y=log_(sinx) tanx` then `((dy)/(dx)) _(pi/4)` isA. `(4)/(log 2)`B. `-4 log 2`C. `(-4)/( log 2)`D. none of these |
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Answer» `y=(log tan x)/(log sin x)` `therefore" "(dy)/(dx)=((log sin x)((sec^(2)x)/(tan x))-(log tan x )(cot x))/((log sin x)^(3))` `"or "((dy)/(dx))_(pi//4)=(-4)/(log 2)" (On simplification)"` |
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| 457. |
If `x=e^tsint,y=e^tcost` then `(d^2y)/(dx^2)` at `x=pi` isA. `2e^(pi)`B. `(1)/(2)e^(pi)`C. `(1)/(2e^(pi))`D. `(2)/(e^(pi))` |
| Answer» Correct Answer - D | |
| 458. |
Differentiate `cos^(-1)((x-x^(-1))/(x+x^(-1)))` w.r.t. x. |
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Answer» Let `y=cos^(-1)((x-x^(-1))/(x+x^(-1)))=cos^(-1){((x-(1)/(x)))/((x+(1)/(x)))}=cos^(-1)((x^(2)-1)/(x^(2)+1)).` Putting `x=tan theta`, we get `y=cos^(-1)((tan^(2)theta-1)/(tan^(2)theta+1))=cos^(-1)(-cos2 theta)` `=cos^(-1){cos(pi-2 theta)}=pi-2 tan^(-1)x.` `therefore(dy)/(dx)=(d)/(dx)(pi-2 tan^(-1)x)=(d)/(dx)(pi)-2.(d)/(dx)(tan^(-1)x)` `=(0-(2)/(1+x^(2)))=(-2)/((1+x^(2))).` |
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| 459. |
If `y=sin^(-1){(5x+12 sqrt(1-x^(2)))/(13)},` find `(dy)/(dx).` |
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Answer» We have `y=sin^(-1){(5)/(13).x+(12)/(13).sqrt(1-x^(2))}.` Let `(5)/(13)=sin alpha and x = cos theta.` Then, `cos alpha=sqrt(1-(25)/(169))=sqrt((144)/(169))=(12)/(13)` `and sqrt(1-x^(2))=sqrt(1-cos^(2)theta)=sqrt(sin^(2)theta)=sin theta.` `therefore y=sin^(-1){sin alpha cos theta+cos alpha sin theta}` `=sin^(-1){sin(alpha+theta)}` `=alpha+theta = sin^(-1)/(5)/(13)+cos^(-1)x.` `therefore(dy)/(dx)=(d)/(dx){sin^(-1).(5)/(13)+cos^(-1)x}=(d)/(dx){sin^(-1).(5)/(13)}+(d)/(dx)(cos^(-1)x)` `={0-(1)/(sqrt(1-x^(2)))}=(-1)/(sqrt(1-x^(2))).` Hence, `(dy)/(dx)=(-1)/(sqrt(1-x^(2))).` |
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| 460. |
The value of `(dy)/(dx)` at `x=(pi)/(2)`, where y is given by `y=x^(sinx)+sqrt(x)`, isA. `1+(1)/(sqrt(2pi))`B. 1C. `(1)/(2pi))`D. `1-(1)/(sqrt(2pi))` |
| Answer» Correct Answer - A | |
| 461. |
If `x y=c^2`, find `(dy)/(dx)` |
| Answer» Correct Answer - `(-c^(2))/(x^(2))` | |
| 462. |
If `y = tan^(-1) ((sqrta + sqrtx)/(1 - sqrt(ax))) " then " (dy)/(dx) =` ? |
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Answer» Correct Answer - `(1)/(2sqrtx(1+x))` `y=tan^(-1)3x+tan^(-1)2x.` |
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| 463. |
Find `(dy)/(dx)`, when: `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` |
| Answer» Correct Answer - `(-b^(2)x)/(a^(2)y)` | |
| 464. |
Find `(dy)/(dx)`, when: `x^(2//3)+y^(2//3)=a^(2//3)` |
| Answer» Correct Answer - `(-y^(1//3))/(x^(1//3))` | |
| 465. |
If `sqrtx + sqrty = sqrta " then " (dy)/(dx) =` ? |
| Answer» Correct Answer - `-sqrt((y)/(x))` | |
| 466. |
If `y^2=a x^2+b x+c`, then `y^3(d^2y)/(dx^2)`is(a) a constant (b) a function of `x`only(c) a function of `y`only (d) a function of `x`and `y`A. a constantB. a function of x onlyC. a function of y onlyD. a function of x and y |
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Answer» `y^(2)=ax^(2)+bx+c` `"or "2y(dy)/(dx)=2ax+b` `"or "2((dy)/(dx))^(2)+2y(d^(2)y)/(dx^(2))=2a` `"or "y(d^(2)y)/dx^(2)=a-((dy)/(dx))^(2)` `=a-((2ax+b)/2y)^(2)` `=(4ay^(2)-(2ax+b)^(2))/(4y^(2))` `"or "4y^(3)(d^(2)y)/(dx^(2))=4a(ax^(2)+bx+c)-(4a^(2)x^(2)+4bax+b^(2))` `"or "4y^(3)(d^(2)y)/(dx^(2))=4ac-b^(2)= constant` |
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| 467. |
`"If "y= sin x +e^(x)," then "(d^(2)x)/(dy^(2))=`A. `(-sin x +e^(x))^(-1)`B. `(sin x - e^(x))/((cos x +e^(x))^(2))`C. `(sin x - e^(x))/((cos x +e^(x))^(3))`D. `(sin x + e^(x))/((cos x +e^(x))^(3))` |
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Answer» `y=sin x +e^(x)or (dy)/(dx)=cos x +e^(x)` `"or "(dx)/(dy)=(cos x +e^(x))^(-1)" (1)"` `therefore" "(d^(2)x)/(dy^(2))=-(cos x +e^(x))^(-2)(-sin x +e^(x))(dx)/(dy)` Substituting the value of `(dy)/(dx)` from (1), `(d^(2)x)/(dy^(2))=((sin x -e^(x)))/((cos x +e^(x))^(2))(cos x +e^(x))^(-1)=(sin x-e^(x))/((cos x +e^(x))^(3))` |
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| 468. |
let `y=t^(10)+1,` and `x=t^8+1,` then `(d^2y)/(dx^2)` isA. `(5)/(2)t`B. `20t^(8)`C. `(5)/(16t^(6))`D. none of these |
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Answer» `"Here, "y=t^(10)+1 and x=t^(8) +1` `therefore" "t^(8)=x-1 or t^(2)=(x-1)^(1//4)` `"So, "y=(x-1)^(5//4)+1` Differentiating both sides w.r.t. x, we get `(dy)/(dx)=(5)/(4)(x-1)^(1//4)` Again, differentiating both sides w.r.t. x, we get `(d^(2)y)/(dx^(2))=(5)/(16)(x-1)^(-3//4)=(5)/(16(x-1)^(3//4))=(5)/(16(t^(2))^(3))=(5)/(16t^(6))` |
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| 469. |
Let `y=in (1+ cos x)^(2).` The the value of `(d^(2)y)/(dx^(2))+(2)/(e^(y//2))` equals |
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Answer» `y=2" In "(1+ cos x)` `(dy)/(dx)=(-2 sin x)/(1+cos x)` `(d^(2)y)/(dx^(2))=-2[((1+cos x)cos x- sin x(-sin x))/((1+cos x)^(2))]` `=-2[(cos x +1)/((1+cos x)^(2))]=(-2)/((1+ cos x))` `"Now, "2e^(-y//2)=2cdote^(-("In "(1+cos x)^(2))/(2))=(2)/((1+cos x))` `therefore" "(d^(2)y)/(dx^(2))+(2)/(e^(y//2))=0` |
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| 470. |
`x=tcost ,y=t+sintdot`Then `(d^(2x))/(dy^2)a tt=pi/2`is`(pi+4)/2`(b) `-(pi+4)/2``-2`(d) none of theseA. `(pi+4)/(2)`B. `-(pi+4)/(2)`C. `-2`D. none of these |
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Answer» `(dx)/(dy)=((dx)/(dt))/((dy)/(dt))=(cos t - t sin t)/(1+ cos t)` `therefore" "(d^(2)x)/(dy^(2))=((d)/(dt)((dx)/(dy)))/((dy)/(dt))` `=(((-2 sin t- t cos t)(1+ cos t )-(cos t - t sin t)(- sin t))/((1+cos t)^(2)))/(1+cos t)` |
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| 471. |
If `y=(1+x)(1+x^2)(1+x^4)...(1+x^(2^n))` then `(dy)/(dx)` at `x=0` isA. 1B. -1C. 0D. none of these |
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Answer» Correct Answer - A We have, `y=(1+x)(1+x^(2))(1+x^(4))….(1+x^(2^(n)))` `implies" "y=((1-x)(1+x)(1+x^(2))(1+x^(4))….(1+x^(2^(n))))/((1-x))` `implies" "y=(1-x^(2^(n+1)))/(1-x)` `implies" "(dy)/(dx)=(-2^(n+1)x^(2^(n+1)-1)"("1-x")"+"("1-x^(2^(n+1))")")/((1-x)^(2))` `implies" "((dy)/(dx))_(x=0)=1` |
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| 472. |
If `y=sin^(-1)(sin x),` then `dy/dx` at `x =pi/2` isA. 1B. -1C. non-exisentD. none of these |
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Answer» Correct Answer - C We have, `y=sin^(-1)(sinx)={{:(x" ,",-(pi)/(2)lexle(pi)/(2)),(pi-x",",(pi)/(2)lexle(3pi)/(2)),(x-2pi",",(3pi)/(2)lexle2pi" and so on "):}` We observe that `("LHD at "x=pi//2)=1" and "("RHD at "x=pi//2)=-1` `Hence, `((dy)/(dx))_(x=pi//2)"does not exist"`. |
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| 473. |
If `y=|x-x^(2)|`, then `(dy)/(dx)" at "x=1`.A. -1B. 1C. does not existD. none of these |
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Answer» Correct Answer - C We have, `y=|x-x^(2)|-{{:(x-x^(2)","," if "-1lexle1),(x^(2)-x","," if "|x|le1):}` Dlearly, y is continuous at x=1 but it is not differentiable at x=1, because `("LHD at "x-1)=((d)/(dx)(x-x^(2)))_("at "x=1)=1-2=-1` `("RHD at "x=1)=((d)/(dx)(x^(2)-x))_("at "x=1)=2-1=1`. Hence, `(dy)/(dx)" at "x=1` does not exist. |
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| 474. |
If `y=cos^(-1)(cosx),t h e n(dy)/(dx)`is equal to`x/y`(b) `y/(x^2)``(x^2-y^2)/(x^2+y^2)`(d) `y/x`A. 1B. -1C. `(1)/(sqrt(2))`D. none of these |
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Answer» Correct Answer - B We have, `y=cos^(-1)(cosx)={{:(x" ,"," if "0lexlepi),(2pi-x","," if "pilexle2pi):}` `:." "((dy)/(dx))_(x=5pi//4)=((d)/(dx)(2pi-x))_(x=5pi//4)=-1` |
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| 475. |
If `y=|cosx|+|sinx|`, then `(dy)/(dx)" at "x=(2pi)/(3)` isA. `(1-sqrt(3))/(2)`B. 0C. `(sqrt(3)-1)/(2)`D. none of these |
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Answer» Correct Answer - C In the neighbourhood of `x=(2pi)/(3)`, we have `cosxlt0" and "sinxgt0` `implies" "cosx|=-cosx" and "|sinx|=sinx` `:." "y=|cosx|+|sinx|`. `implies" "y=-cosx+sinx` `implies" "(dy)/(dx)=sinx+cosx` `implies" "((dy)/(dx))_(x=(2pi)/(3))="sin"(2pi)/(3)+"cos"(2pi)/(3)=(sqrt(3)-1)/(2)` |
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| 476. |
`"If "xy+y^(2)=tan x + y," then find "(dy)/(dx).` |
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Answer» The given relation is `xy+y^(2)=tan x + y.` Differentiating both sides with respect to x, we get `(d)/(d)(xy)+(d)/(dx)(y^(2))=(d)/(dx(tan x) +(dy)/(dx)` `"or "[y.1+x.(dy)/(dx)]+2y(dy)/(dx)=sec^(2) x+(dy)/(dx)` |
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| 477. |
Find `(dy)/(dx)`, when: `logsqrt(x^(2)+y^(2))=tan^(-1).(y)/(x)` |
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Answer» Correct Answer - `(x+y)/(x-y)` `(1)/(2)log(x^(2)+y^(2))=tan^(-1)((y)/(x))` `rArrlog(x^(2)+y^(2))=2tan^(-1).(y)/(x)rArr(x^(2)+y^(2))=e^(2tan^(-1)y//x)." ...(i)"` On differentiating (i) w.r.t. x, we get `2x+2y.(dy)/(dx)=e^(2tan^(-1)y//x).(2)/((1+(y^(2))/(x^(2)))).((x(dy)/(dx)-y))/(x^(2))` `rArrx+y(dy)/(dx)=(x^(2)+y^(2)).(x^(2))/((x^(2)+y^(2))).((x(dy)/(dx)-y))/((x^(2)))." [using (i)]"` `rArr x+y(dy)/(dx)=x(dy)/(dx)-y` `rArr(x-y)(dy)/(dx)=(x+y)rArr(dy)/(dx)=((x+y))/((x-y)).` |
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| 478. |
Find `(dy)/(dx)`, when: `xylog(x+y)=1` |
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Answer» Correct Answer - `(-(x+y+x^(2)y))/(x^(2){y+(x+y)log(x+y)})` `ylog(x+y)=(1)/(x)` `rArry.(1)/((x+y)).(1+(dy)/(dx))+log(x+y).(dy)/(dx)=(-1)/(x^(2))` `rArr {(y)/((x+y))+log(x+y)}.(dy)/(dx)=-((1)/(x^(2))+(y)/(x+y)).` |
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| 479. |
`" If "y=cot^(-1)[(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))](0ltxltpi//2)," then "(dy)/(dx)=`A. `(1)/(2)`B. `(2)/(3)`C. 3D. 1 |
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Answer» `y=cot^(-1)[(sqrt(1+ sin x)+sqrt(1-sin x))/(sqrt(1+sin x)-sqrt(1-sin x))]` `=cot^(-1)[(2+2cos x)/(2x)]=cot^(-1)[(1+ cos x)/(sin x)]` `=cot^(-1)[cos""(x)/(2)]=(x)/(2)` `therefore" "(dy)/(dx)=(1)/(2)` |
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| 480. |
If`x=3sint-sin3t ,y=3cos t-cos3t ,"f i n d"(dy)/(dx)"a t"t=pi/3dot` |
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Answer» `x = 3sint-sin3t` `:. dx/dt = 3cost-3cos3t = 3(cost-cos3t)` `y= 3cost - cos3t` `:. dy/dt = -3sint +3sin3t = 3(sin3t-sint)` Now, `dy/dx = (dy/dt)/(dx/dt) = (3(sin3t - sint))/(3(cost-cos3t))` At `t = pi/3`, `dy/dx = (3(0-sqrt3/2))/(3(1/2+1)) = (-sqrt3/2)/(3/2) = -1/sqrt3` `:. dy/dx|_(t = pi/3) = -1/sqrt3.` |
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| 481. |
Differentiate: ex cos x |
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Answer» By Product Rule: (uv)′ = u′v + uv′ Here u = ex and v = cos x (ex cosx )’ = ex (cos x) + ex (-sin x) = excos x – exsin x = ex (cos x – sin x) |
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| 482. |
Differentiate: x2 sin x |
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Answer» By Product Rule: (uv)′ = u′v + uv′ Here u = x2 and v = sin x (x2 sin x)’ = 2x (sin x) + x2 (cos x) = 2x sin x + x2cos x |
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