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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.
| 351. |
prove that `(d(sec^(-1)x))/(dx) =1/(|x|(sqrt(x^2-1)))` |
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Answer» Let `y=sec^(-1)x`, where `x in R-[-1,1]` and `y in [0,pi]-{(pi)/(2)}`. Then, `x = secy` `rArr (dx)/(dy)= secy tan y gt 0` `rArr(dy)/(dx)=(1)/(secy tany)=(1)/(secy.sqrt(sec^(2)y-1))` `rArr(dy)/(dx)=(1)/(|x|sqrt(x^(2)-1))` Hence, `(d)/(dx)(sec^(-1)x)=(1)/(|x|sqrt(x^(2)-1)).` |
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| 352. |
`y=sqrt(sinsqrt(x))` |
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Answer» Correct Answer - `(cos sqrt(x))/(4sqrt(x sin sqrt(x)))` `(dy)/(dx)=(1)/(2sqrt(sin)sqrt(x))xxcossqrt(x)xx(1)/(2sqrt(x))` |
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| 353. |
Differentiate `x^(sin^(-1)x)` w.r.t. x. |
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Answer» Let `y=x^(sin^(-)x)" …(i)"` Taking logarithm on both sides of (i), we get `log y=(sin^(-1)x)(logx)." …(ii)"` On differentiating both sides of (ii) w.r.t. x, we get `(1)/(y).(dy)/(dx)=(sin^(-1)x).(d)/(dx)(logx)+(logx).(d)/(dx)(sin^(-1)x)` `=(sin^(-1)x)(1)/(x)+(logx).(1)/(sqrt(1-x)^(2))` `rArr" "(dy)/(dx)=y.[(sin^(-1)x)/(x)+(logx)/(sqrt(1-x^(2)))]` `rArr" "(dy)/(dx)=x^(sin^(-1)x).{(sin^(-1)x)/(x)+(logx)/(sqrt(1-x^(2)))}.` |
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| 354. |
The differential cofficient of `sec^(- 1)(1/(2x^2-1)) ` w.r.t `sqrt(1-x^2)` is- |
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Answer» Correct Answer - 4 `"Let "y=sec^(-1)((1)/(2x^(2)-1))and z= sqrt(1-x^(2))` Put `x= cos theta. ` Therefore, `y=sec^(-1) (sec 2theta)=2theta and z=sqrt(1-cos^(2)theta)=sin theta` `therefore" "(dy)/(d""theta)=2, (dz)/(d""theta)=c os theta or (dy)/(dz)=(2)/(cos theta)=(2)/(x)` `"At "x=(1)/(2), (dy)/(dz)=4` |
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| 355. |
Differentiate `tan^(-1)((a cosx-b sinx)/(b cosx + a sinx))`w.r.t. x. |
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Answer» Let `y=tan^(-1)((acosx-bsinx)/(b cosx+asin x)).` Putting `a=r sin theta and b = r cos theta,` we get `y=tan^(-1){(r(sin thetacosx-cos thetasin x))/(r(cos thetacosx+sin thetasin x))}` `=tan^(-1){(sin(theta-x))/(cos(theta-x))}=tan^(-1){tan(theta-x)}` `=theta-x=(tan^(-1).(a)/(b)-x)[because (a)/(b)=tan theta rArr theta= tan^(-1).(1)/(b)].` `therefore(dy)/(dx)=(d)/(dx)(tan^(-1).(a)/(b)-x)` `=(d)/(dx)(tan^(-1).(a)/(b))-(d)/(dx)(x)=-1" "[because tan^(-1).(a)/(b)="constant"].` |
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| 356. |
Find the derivative of `sec^(-1)((1)/(2x^(2)-1))" w.r.t. "sqrt(1-x^(2))" at "x=(1)/(2).` |
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Answer» Correct Answer - 4 `"Let "y=sec^(-1)((1)/(2x^(2)-1))and z= sqrt(1-x^(2))` Put `x= cos theta. ` Therefore, `y=sec^(-1) (sec 2theta)=2theta and z=sqrt(1-cos^(2)theta)=sin theta` `therefore" "(dy)/(d""theta)=2, (dz)/(d""theta)=c os theta or (dy)/(dz)=(2)/(cos theta)=(2)/(x)` `"At "x=(1)/(2), (dy)/(dz)=4` |
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| 357. |
If y sin (sin x) and `(d^(2)y)/(dx^(2))+(dy)/(dx)` tan x + f(x) = 0, then find (x). |
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Answer» Correct Answer - `cos^(2)x sin (sin x)` `(dy)/(dx)=cos (sin x) cos x` `(d^(2)y)/(dx^(2))=-cos ( sin x) sin x + cos x [-sin (sin x)] cos x` `therefore" "(d^(2)y)/(dx^(2))+(dy)/(dx)tan x= -c os ( sin x) sin x-cos^(2) x sin (sin x)+cos (sin x )cos x tan x` `=-cos^(2)x sin (sin x)` `therefore" "(d^(2)y)/(dx^(2))+(dy)/(dx)tan x + cos^(2)x sin (sin x) =0` `therefore" "f(x)=cos^(2)x sin (sin x)` |
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| 358. |
If `x=a(1-costheta)`, `y=a(theta+sintheta)`, prove that `(d^2y)/(dx^2)=-1/a`at `theta=pi/2`. |
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Answer» We have `x=a(theta+sin theta)and y=a(1-cos theta)` `rArr (dx)/(d theta)=a(1+cos theta) and (dy)/(d theta)=a sin theta` `rArr(dy)/(dx)=((dy//d theta))/((dx//d theta))` `=(asin theta)/(a(1+costheta))=(2sin(theta//2)cos(theta//2))/(2cos^(2)(theta//2))=tan.(theta)/(2)` `rArr(d^(2)y)/(dx^(2))=(d)/(dx)(tan.(theta)/(2))=(1)/(2)sec^(2).(theta)/(2).(d theta)/(dx)=((1)/(2)sec^(2).(theta)/(2))xx(1)/(a(1+cos theta))` `rArr((d^(2)y)/(dx^(2)))_(theta=(pi)/(2))=(1)/(2)sec^(2).(pi)/(4).(1)/(a(1+cos.(pi)/(2)))=(1)/(a).` |
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| 359. |
If `y={x+sqrt(x^2+1)}^m`, show that `(x^2+1)y_2+x y_1-m^2 y=0`A. `m^(2)y`B. `my^(2)`C. `m^(2)y^(2)`D. none of these |
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Answer» Correct Answer - A We have, `y^(1//m)={x+sqrt(1+x^(2))}` `impliesy={x+sqrt(1+x^(2))}^(m)` `implies(dy)/(dx)=m{x+sqrt(1+x^(2))}^(m-1){1+(x)/(sqrt(x^(2)+1))}=m({xsqrt(1+x^(2))}^(m))/(sqrt(1+x^(2)))` `implies" "(dy)/(dx)=(my)/(sqrt(1+x^(2)))` `impliesy_(1)""^(2)(1+x^(2))=m^(2)y^(2)` `implies2y_(1)y_(2)(1+x^(2))+2xy_(1)""^(2)=2m^(2)yy_(1)` `impliesy_(2)(1+x^(2))+y_(1)=m^(2)y` |
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| 360. |
Prove that `(d)/(dx)("cosec"^(-1)x)=(-1)/(|x|sqrt(x^(2)-1))`, where `x in R-[-1,1]`. |
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Answer» Let `y="cosec"^(-1)x`, where `x in R-[-1,1] and y in [-(pi)/(2),(pi)/(2)]-{0}.` Then, `x="cosec y"` `rArr (dx)/(dx)=-"cosec y cot y, where cosec y cot y " gt 0` `rArr (dy)/(dx)=(-1)/("cosec y cot y")=(-1)/(("cosec y")sqrt("cosec"^(2)y-1))=(-1)/(|x|sqrt(x^(2)-1)).` Hence, `(d)/(dx)("cosec"^(-1)x)=(-1)/(|x|sqrt(x^(2)-1)).` |
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| 361. |
Differentiate `(sinx)^(logx)` w.r.t. x. |
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Answer» Let `y=(sinx)^(logx)`. Taking logarithm on both sides of (i), we get `logy=(logx)(log sin x).` On differentiating both sides of (ii) w.r.t. x, we get `(1)/(y).(dy)/(dx)=(logx).(d)/(dx)(log sinx)+(logsinx).(d)/(dx)(logx)` `=log(x).(1)/(sinx).cos x+(log sin x).(1)/(x)` `=(logx)cotx+((log sin x))/(x)` `rArr" "(dy)/(dx)=y.[(logx)cotx+((log sin x))/(x)]` `rArr" "(dy)/(dx)=(sinx)^(logx).[(logx)cotx+((log sinx))/(x)].` |
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| 362. |
`cos(sinsqrt(ax+b))` |
| Answer» Correct Answer - `(-acossqrt(ax+b))/(2sqrt(ax+b)).sin{sin sqrt(ax+b)}` | |
| 363. |
Find `(dy)/(dx)fory=log(x+sqrt(a^2+x^2))dot` |
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Answer» `y=log (x+sqrt(a^(2)+x^(2)))` `"Then "(dy)/(dx)=(d)/(x){log (x+sqrt(a^(2)+x^(2)))}` `=(1)/(x+sqrt(a^(2)+x^(2)))(d)/(dx)(x+sqrt(a^(2)+x^(2)))` `=(1)/(x+sqrt(a^(2)+x^(2)))xx{1+(1)/(2)(a^(2)+x^(2))^(-1//2)(d)/(dx)(a^(2)+x^(2))}` `=(1)/(x+sqrt(a^(2)+x^(2))){1+(1)/(2sqrt(a^(2)+x^(2)))xx2x}` `(1)/(x+sqrt(a^(2)+x^(2)))xx(sqrt(a^(2)+x^(2))+x)/(sqrt(a^(2)+x^(2)))` `=(1)/(sqrt(a^(2)+x^(2)))` |
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| 364. |
Differentiate `(x)/(sinx)` w.r.t . sinx. |
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Answer» Correct Answer - `(tan x-x)/(sin^(2)x)` `"Let "u=(x)/(sin x ) and v = sin x` `therefore" "(du)/(dx)=(sin x.(d)/(dx)x -x. (d)/(dx) sin x)/((sin x)^(2))` `=(sin x - x cos x)/(sin^(2)x)" ....(i)"` `"And "(dv)/(dx)=(d)/(dx)( sin x )= cos x" ...(ii)"` `therefore" "(du)/(dv)=(du//dx)/(dv//dx)=((sin x- x cos x)/sin^(2)x)/(cos x )` `=(tan x-x)/(sin^(2)-x)` |
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| 365. |
Find `(dy)/(dx)fory=x^xdot` |
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Answer» Correct Answer - `x^(x)(1+ log x)` `"Let "y=x^(x). Then, y= e^(xlog x).` Differentiating both sides w.r.t. x, we get `(dy)/(dx)=e^(x log x)(d)/(dx)(x log x)` `=x^(x)(log x+ x(1)/(x))" "[becausee^(x log x)=x^(x)]` `=x^(x)(1+ log x)` |
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| 366. |
If `y=(tan^(-1)x)^2`, then prove that `(1+x^2)^2 y_2+2x (1+x^2)y_1=2`. |
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Answer» Given: `y=(tan^(-1)x)^(2)." …(i)"` On differentiating both sides of (i) w.r.t. x, we get `y_(1)=2 tan^(-1)x.(1)/((1+x^(2))` `rArr (1+x^(2))y_(1)=2 tan^(-1)x` `rArr (1+x^(2))^(2)y_(1)^(2)=4(tan^(-1)x)^(2)" [on squaring both sides]"` `rArr (1+x^(2))^(2)y_(1)^(2)-4y=0." ...(ii)"` On differentiating both sides of (ii) w.r.t. x, we get `(1+x^(2))^(2).2y_(1)y_(2)+y_(1)^(2).2(1+x^(2)).2x-4y_(1)=0` `rArr (1+x^(2))^(2)y_(2)+2x(1+x^(2))y_(1)-2=0.` Hence, `(1+x^(2))^(2)y_(2)+2x(1+x^(2))y_(1)=2.` |
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| 367. |
`sqrt("cosec"(x^(3)+1))` |
| Answer» Correct Answer - `(-acossqrt(ax+b))/(2sqrt(ax+b)).sin{sin sqrt(ax+b)}` | |
| 368. |
Differentiate `sin3xcos5x` w.r.t .`x`. |
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Answer» Let `y=sin 3x cos 5x=(1)/(2){2 cos 5x sin 3x}` `=(1)/(2){sin(5x+3x)-sin (5x-3x)}` `=(1)/(2).(sin 8x-sin 2x)=(1)/(2).(d)/(dx)(sin 2x)` `=((1)/(2).8 cos 8x-(1)/(2)xx2 cos2x)=(4 cos 8x -cos2x)`. |
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| 369. |
Find `(dy)/(dx)fory=xsinxlogxdot` |
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Answer» We have `(d)/(dx) (x sin x log x)` `={(d)/(dx)(x)}sin x log x+x{(d)/(dx)(sin x)} log x + x sin x{(d)/(dx) (log x)}` `=1xxsinxxlog x + x xx cos x xx log x + x xx sin x xx(1)/(x)` `=sin x log x + x cos x log x + sin x` |
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| 370. |
Differentiate `y=(e^x)/(1+sinx)` |
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Answer» Using quotient rule, we have `(dy)/(dx)=(d)/(dx)((e^(x))/(1+ sin x))` `=((1+ sin x)cdot(d)/(dx)(e^(x))-e^(x)cdot(d)/(dx)(1+ sin x))/(1+ sin x )^(2)` `=((1+sin x)cdote^(x)-e^(x)cdot(0 + cos x))/((1+ sin x)^(2))=(e^(x)(1+ sin x - cos x))/((1+ sin x )^(2))` |
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| 371. |
If `y=sin^(-1)x`, prove that `(1-x^2)(d^2y)/(dx^2)-x(dy)/(dx)=0` |
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Answer» Given: `y=sin^(-1)x." …(i)"` On differentiating both sides of (i) w.r.t. x, we get `y_(1)=(1)/(sqrt(1-x^(2))` `rArr y_(1)^(2)=(1)/((1-x^(2)))` `rArr(1-x^(2))y_(1)^(2)=1." …(ii)"` On differentiating both sides of (ii) w.r.t. x, we get `(1-x^(2)).2y_(1)y_(2)+y_(1)^(2)(-2x)=0` `rArr (1-x^(2))y_(2)-xy_(1)=0.` Hence, `(1-x^(2))(d^(2)y)/(dx^(2))-x(dy)/(dx)=0.` |
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| 372. |
Differentiate each of the following w.r.t. x: (i)`(ax+b)^(m)" "(ii)(2x+3)^(5)" "(iii)sqrt(ax^(2)+2bx+c)` |
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Answer» (i) Let `y=(ax+b)^(m)`. Putting `(ax+b)=t`, we get `y=t^(m) and t=(ax+b)` `rArr" "(dy)/(dt)=mt^(m-1)and (dt)/(dx)=a` `rArr" "(dy)/(dx)=((dy)/(dt)xx(dt)/(dx))=(mt^(m-1)xxa)=mat^(m-1)=ma(ax+b)^(m-1)` `therefore" "(d)/(dx)(ax+b)^(m)=ma(ax+b)^(m-1)`. Let `y=(2x+3)^(5)`. Putting `(2x+3)=t`, we get `y=t^(5)and t=2x+3` `rArr" "(dy)/(dt)=5t^(4)and (dt)/(dx)=2` `rArr" "(dy)/(dx)=((dy)/(dt)xx(dt)/(dx))=10t^(4)=10(2x+3)^(4)`. (iii) Let `y=sqrt(ax^(2)+2bx+c).` Putting `(ax^(2)+2bx+c)=t`, we get `rArr" "(dy)/(dx)=(1)/(2)t^(-1//2)=(1)/(2sqrtt)and (dt)/(dx)=(2ax+2b)=2(ax+b)` `rArr" "(dy)/(dx)=((dy)/(dt)xx(dt)/(dx))=(1)/(2sqrtt)xx2(ax+b)` `=((ax+b))/(sqrtt)=((ax+b))/(sqrt(ax^(2)+2bx+c)).` |
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| 373. |
`"Evaluate "lim_(hrarr0) ((a+h)^(2). sin ^(-1)(a+h)-a^(2) sin^(-1) a)/(h).` |
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Answer» `underset(hrarr0)lim((a+h)^(2). sin ^(-1)(a+h)-a^(2) sin^(-1) a)/(h).` `=(d)/(da)(a^(2) sin^(-1) a)` `=(sin^(-1)a) (d)/(da)(a^(2)) +(a^(2))(d)/(da)(sin^(-1)a)` `=(a^(2))/(sqrt(1-a^(2)))+2a sin^(-1)a` |
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| 374. |
`"Evaluate "lim_(hrarr0)((a+h)^(2). sin ^(-1)(a+h)-a^(2) sin^(-1) a)/(h).` |
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Answer» `underset(hrarr0)lim((a+h)^(2). sin ^(-1)(a+h)-a^(2) sin^(-1) a)/(h).` `=(d)/(da)(a^(2) sin^(-1) a)` `=(sin^(-1)a) (d)/(da)(a^(2)) +(a^(2))(d)/(da)(sin^(-1)a)` `=(a^(2))/(sqrt(1-a^(2)))+2a sin^(-1)a` |
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| 375. |
If `y=3 e^(2x)+2 e^(3x)`, prove that `(d^2y)/(dx^2)-5(dy)/(dx)+6y=0`. |
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Answer» We have `y=3e^(2x)+2e^(3x)" …(i)"` `rArr(dy)/(dx)=3.(d)/(dx)(e^(2x))+2.(d)/(dx)(e^(3x))" [on differentiating (i) w.r.t. x]"` `=(3xx2e^(2x))+(2xx3e^(3x))=(6x^(2x)+6e^(3x))` `rArr (dy)/(dx)=6(e^(2x)+e^(3x))." ...(ii)"` On differentiating (ii) w.r.t. x, we get `(d^(2)y)/(dx^(2))=6.{(d)/(dx)(e^(2x))+(d)/(dx)(e^(3x))}` `=6.(2e^(2x)+3e^(3x)).` `therefore((d^(2)y)/(dx^(2))-5(dy)/(dx)+6y)` `=6(2e^(2x)+3e^(3x))-30(e^(2x)+e^(3x))+(18e^(2x)+12e^(2x))` `=(12-30+18)e^(2x)+(18-30+12)e^(3x)=0.` Hence, `((d^(2)y)/(dx^(2))-5(dy)/(dx)+6y)=0.` |
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| 376. |
`sin5x cos 3x` |
| Answer» Correct Answer - `(4cos8x+cos2x)` | |
| 377. |
`sin^(-1){sqrt((1-cosx)/(2))}` |
| Answer» Correct Answer - `(1)/(2)` | |
| 378. |
`sin 2x sinx` |
| Answer» Correct Answer - `((3)/(2)sin3x-(1)/(2)sinx)` | |
| 379. |
If `y=e^x(sinx+cosx)`prove that `(d^2y)/(dx^2)-2(dy)/(dx)+2y=0`. |
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Answer» We have `y=e^(x)(sin x +cosx)` `rArr(dy)/(dx)=e^(x).(d)/(dx)(sinx+cosx)+(sinx+cosx).(d)/(dx)(e^(x))` `=e^(x)(cosx-sinx)+(sinx+cosx).e^(x)=2e^(x)cosx` `rArr (d^(2)y)/(dx^(2))=2.(d)/(dx)(e^(x)cosx)` `=.{e^(x).(d)/(dx)(cosx)+cosx.(d)/(dx)(e^(x))}` `=2.{e^(x)(-sinx)+(cosx)e^(x)}=2e^(x)(cosx-sinx).` `therefore((d^(2)y)/(dx^(2))-2(dy)/(dx)+2y)` `=2e^(x)(cosx-sinx)-4e^(x)cosx+2e^(x)(sinx+cosx)=0.` Hence, `(d^(2)y)/(dx^(2))-2(dy)/(dx)+2y=0.` |
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| 380. |
Evaluate `int e^(2x) sin 3x dx`. |
| Answer» `e^(2x)(3cos3x+2 sin 3x)` | |
| 381. |
Find `(dy)/(dx)`, when : `y=sin((1+x^(2))/(1-x^(2)))` |
| Answer» Correct Answer - `(4x)/((1-x^(2))^(2)).cos((1+x^(2))/(1-x^(2)))` | |
| 382. |
If `y=Acosn x+Bsinn x ,`show that `(d^2y)/(dx^2)+n^2 y=0`. |
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Answer» We have `y=A cos nx +B sin nx` `rArr (dy)/(dx)=(d)/(dx)(A cos nx)+(d)/(dx)(B sin nx)` `=-An sin nx+Bn cos nx` `=n(B cos nx-A sin nx)` `rArr (d^(2)y)/(dx^(2))=n.(d)/(dx)(B cos nx-A sin nx)` `=n.{B.(d)/(dx)(cosnx)-A.(d)/(dx)(sinnx)}` `=n.{-Bn sin nx-An cosnx}` `=-n^(2)(A cos nx+B sin nx)=-n^(2)y` `rArr(d^(2)y)/(dx^(2))+n^(2)y=0.` |
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| 383. |
`int cos 4x. cos 2x dx` |
| Answer» Correct Answer - `-(3sin 6x+sin2x)` | |
| 384. |
`(3-4x)^(5)` |
| Answer» Correct Answer - `-20(3-4x)^(4)` | |
| 385. |
If `y=e^(4x) sin 3x`, find `(d^(2)y)/(dx^(2)).` |
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Answer» Let `y=e^(4x)sin 3x.` Then, `(dy)/(dx)=3e^(4x)cos3x+4e^(4x)sin 3x=e^(4x)(3cos 3x+4sin3x).` `therefore (d^(2)y)/(dx^(2))=(d)/(dx)((dy)/(dx)){e^(4x)(3cos 3x+4sin3x)}` `=e^(4x)(-9sin 3x+12 3x)+4e^(4x)(3cos 3x+4sin 3x)` `=e^(4x)(7sin 3x+24 cos 3x).` |
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| 386. |
If `e^y(x+1)=1,` show that `(d^2y)/(dx^2)=((dy)/(dx))^2.` |
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Answer» We have `e^(y)(x+1)=1 rArre^(y)=(1)/((x+1))" ...(i)"` `rArr y=log{(1)/((x+1))}=log1-log(x+1)` `rArr y=-log (x+1)." ...(ii)"` `therefore(dy)/(dx)=(-1)/((x+1))` `rArr(d^(2)y)/(dx^(2))=(1)/((x+1)^(2))=((dy)/(dx))^(2).` Hence, `(d^(2)y)/(dx^(2))=((dy)/(dx))^(2).` |
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| 387. |
The differential coefficient of `f(logx)` with respect to x, where `f(x)=logx,` isA. `(x)/(logx)`B. `(xlogx)^(-1)`C. `(logx)/(x)`D. `xlogx` |
| Answer» Correct Answer - B | |
| 388. |
The derivative of `tan^(-1)((sqrt(1+x^2)-1)/x)`with respect to `tan^(-1)((2xsqrt(1-x^2))/(1-2x^2))`at `x=0`is`1/8`(b) `1/4`(c) `1/2`(d) 1A. `1//8`B. `1//4`C. `1//2`D. 1 |
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Answer» `"Let "y=tan^(-1)((sqrt(1+x^(2))+1)/(x)) and z = tan^(-1)((2xsqrt(1-x^(2)))/(1-2x^(2)))` Putting x = tan `theta` in y, we get `y=tan^(-1)((sectheta-1)/(tan theta))=tan^(-1)(tan""(theta)/(2))=(1)/(2)tan^(-1)x` `therefore" "(dy)/(dx)=(1)/(2(1+x^(2)))` Putting `x= sin theta` in z, we get `z=tan^(-1)""((2 sin theta cos theta)/(cos 2theta))=tan^(-1)(tan 2theta)=2theta=2sin^(-1)x` `therefore" "(dz)/(dx)=(2)/(sqrt(1-x^(2)))` `"Thus, "(dy)/(dx)=((dy)/(dx))/((dz)/(dx))=(1)/(4(1+x^(2)))sqrt(1-x^(2))or ((dy)/(dz))_(x=0)=(1)/(4)` |
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| 389. |
The derivative of `sin^(-1)((sqrt(1+x)+sqrt(1-x))/(2))` with respect to x isA. `-(1)/(2sqrt(1-x^(2)))`B. `(1)/(2sqrt(1-x^(2)))`C. `(2)/(sqrt(1-x^(2)))`D. `(-2)/(sqrt(1-x^(2)))` |
| Answer» Correct Answer - A | |
| 390. |
Find `(dy)/(dx)`for the functions:`y=x^3e^xsinx` |
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Answer» Correct Answer - `x^(2)e^(x)(3 sin x + x sin x + x cos x )` `y=x^(3)e^(x) sin x` `therefore" " (dy)/(dx)=(d)/(dx)(x^(3)e^(x)sin x)` `={(d)/(dx)(x^(3))}e^(x) sin x +x^(3){(d)/(dx)(e^(x))}sin +x^(3)e^(x){(d)/(dx)(sin x)}` `=3x^(2)e^(x) sin x +x^(3)e^(x) sin x +x^(3) e^(x) cos x` `=x^(2)e^(x)(3 sin x + x sin x + x cos x)` |
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| 391. |
`cot^(-1)((sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx)))=(x)/(2), x in (0,(pi)/(4))` |
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Answer» We have `(1+sinx)={cos^(2)(x//2)+sin^(2)(x//2)+2sin(x//2)cos(x//2)}` `={cos(x//2)+sin(x//2)}^(2).` `(1-sinx)={cos^(2)(x//2)+sin^(2)(x//2)-2sin(x//2)cos(x//2)}` `={cos(x//2)-sin(x//2)}^(2).` `thereforesqrt(1+sinx)=sqrt({cos(x//2)+sin(x//2)}^(2))=cos((x)/(2))+sin((x)/(2))` `andsqrt(1-sinx)=sqrt({cos(x//2)-sin(x//2)}^(2))=cos((x)/(2))-sin((x)/(2)).` `thereforey=cot^(-1){([cos(x//2)+sin(x//2)]+[cos(x//2)-sin(x//2)])/([cos(x//2)+sin(x//2)]-[cos(x//2)-sin(x//2)])}` `=cot^(-1){(2cos(x//2))/(2sin(x//2))}=cot^(-1){cot(x//2)}=(x)/(2)` `rArr(dy)/(dx)=(d)/(dx)((x)/(2))=(1)/(2).` |
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| 392. |
If `y="tan"^(-1)((sqrt(1+sinx)+sqrt(1-sinx)))/((sqrt(1+sinx)-sqrt(1-sinx)))," find "(dy)/(dx).` |
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Answer» We have `y=tan^(-1){(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))xx(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)+sqrt(1-sinx))}` `=tan^(-1){((1+sinx)+(1-sinx)+2sqrt(1-sin^(2)x))/((1+sinx)-(1-sinx))}=tan^(-1)((1+cosx)/(sinx))` `=tan^(-1){(2cos^(2)(x//2))/(2sin(x//2)cos(x//2))}=tan^(-1){"cot"(x)/(2)}=tan^(-1){tan((pi)/(2)-(x)/(2))}` `=((pi)/(2)-(x)/(2)).` `therefore(dy)/(dx)=(d)/(dx)((pi)/(2)-(x)/(2))=(d)/(dx)((pi)/(2))-(d)/(dx)((x)/(2))=(0-(1)/(2))=-(1)/(2).` |
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| 393. |
Differentiate w.r.t. x : `(i)cot^(-1)((1)/(x))" "(ii)tan^(-1)((2x)/(1-x^(2)))" "(iii)cot^(-1)((1-x)/(1+x))` |
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Answer» (i) Let `y=cot^(-1)((1)/(x)).` Putting `x= tan theta`, we get `y=cot^(-1)((1)/(tan theta))=cot^(-1)(cot theta)=theta=tan^(-1)x.` `therefore(dy)/(dx)=(1)/((1+x^(2))).` Hence, `(d)/(dx){cot^(-1)((1)/(2))}=(1)/((1+x^(2))).` (ii) Let `y=tan^(-1)((2x)/(1-x^(2))).` Putting `x=tan theta`, we get `y=tan^(-1)((2tan theta)/(1-tan^(2)theta))=tan^(-1)(tan2 theta)=2theta=2 tan^(-1)x.` `therefore(dy)/(dx)=(2)/((1+x^(2))).` Hence, `(d)/(dx){tan^(-1)((2x)/(1-x^(2)))}=(2)/((1+x^(2))).` (iii) Let `y=cot^(-1)((1-x)/(1+x)).` Putting `x=tan theta,` we get `y=cot^(-1)((1-tan theta)/(1+tan theta))=cot^(-1){tan((pi)/(4)-theta)}` `=cot^(-1)[cot{(pi)/(2)-((pi)/(4)-theta)}]=((pi)/(4)+theta)=(pi)/(4)+tan^(-1)x.` `therefore (dy)/(dx)=(1)/((1+x^(2)))` Hence, `(d)/(dx){cot^(-1)((1-x)/(1+x))}=(1)/((1+x^(2))).` |
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| 394. |
Differentiate w.r.t. x: `(i)cos^(-1)((1-x^(2))/(1+x^(2)))" "(ii)sin^(-1)((2x)/(1+x^(2)))" "(iii)sec^(-1)((1)/(2x^(2)-1))` |
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Answer» (i) Let `y=cos^(-1)((1-x^(2))/(1+x^(2)))` Putting `x=tan theta,` we get `y=cos^(-1)((1-tan^(2)theta)/(1+tan^(2)theta))=cos^(-1)(cos2 theta)=2theta=2tan^(-1)x`. `therefore(dy)/(dx)=(2)/((1+x^(2))).` Hence, `(d)/(dx){cos^(-1)((1-x^(2))/(1+x^(2)))}=(2)/((1+x^(2))).` (ii) Let `y=sin^(-1)((2x)/(1+x^(2)))`. Putting `x=tan theta,` we get `y=sin^(-1)((2tan theta)/(1+tan^(2)theta))=sin^(-1)(sin 2 theta)=2 theta=2tan^(-1)x`. `therefore(dy)/(dx)=(2)/((1+x^(2)))`. Hence, `(d)/(dx){sin^(-1)((2x)/(1+x^(2)))}=(2)/((1+x^(2))).` (iii) Let `y=sec^(-1)((1)/(2x^(2)-1))`. Putting `x=cos theta,` we get `y=sec^(-1)((1)/(2cos^(2)theta-1))=sec^(-1)((1)/(cos 2 theta))` `=sec^(-1)(sec 2 theta)=2 theta=2cos^(-1)x`. `therefore y=2cos^(-1)x.` Hence, `(dy)/(dx)=2(d)/(dx)(cos^(-1)x)=(-2)/(sqrt(1-x^(2))).` |
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| 395. |
Differentiate w.r.t. x: `(i)cos^(-1)(4x^(3)-3x)" "(ii)sin^(-1)((1-x^(2))/(1+x^(2)))" "(iii)sec^(-1)((x^(2)+1)/(x^(2)-1))` |
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Answer» (i) Let `y=cos^(-1)(4x^(3)-3x).` Putting `x=cos theta`, we get `y=cos^(-1)(4cos^(3) theta-3 cos theta)=cos^(-1)(cos 3 theta)=3 theta.` `therefore y=3 theta rArr y= 3 cos^(-1)x` `rArr (dy)/(dx)=((-3)/(sqrt(1-x^(2))))`. Let `y=sin^(-1)((1-x^(2))/(1+x^(2)))`. Putting `x=tan theta,` we get `y=sin^(-1)((1-tan^(2)theta)/(1+tan^(2)theta))=sin^(-1)(cos2 theta)=sin^(-1)[sin((pi)/(2)-2 theta)]` `=((pi)/(2)-2 theta)=((pi)/(2)-2 tan^(-1)x).` `therefore y=((pi)/(2)-2 tan^(-1)x).` Hence, `(dy)/(dx)=(d)/(dx)((pi)/(2)-2 tan^(-1)x)=(d)/(dx)((pi)/(2))-2.(d)/(dx)(tan^(-1)x)` `={0-(2)/((1+x^(2)))}=(-2)/((1+x^(2))).` Let `y=sec^(-1)((x^(2)+1)/(x^(2)-1)).` Putting `x=cot theta`, we get Putting `x=cot theta`, we get `y=sec^(-1)((cot^(2)theta+1)/(cot^(2) theta-1))=sec^(-1)((1+tan^(2)theta)/(1-tan^(2)theta))` `=sec^(-1)(sec 2 theta)=2 theta=2 cot^(-1)x.` `therefore (dy)/(dx)=(-2)/((1+x^(2)))` Hence, `(d)/(dx){sec^(-1)((x^(2)+1)/(x^(2)-1))}=(-2)/((1+x^(2))).` |
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| 396. |
`sin^(-1)((1-tan^(2)x)/(1+tan^(2)x))` |
| Answer» Correct Answer - `-2` | |
| 397. |
`"cosec"^(-1)((1+tan^(2)x)/(2tanx))` |
| Answer» Correct Answer - 2 | |
| 398. |
Differentiate the following functions:(i) 1/x (ii) 1/√x (iii) 1/x1/3 |
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Answer» We know, d/dx (xm) = m(xm-1) (i) d/dx (1/x ) = x-1 = -x– 2 = -1/x2 (ii) d/dx (1/√x ) = x-1/2 = -1/2* x– 3/2 (iii) d/dx (1/x1/3) = x-1/3 = -1/3 * x– 4/3 |
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| 399. |
Differentiate the following functions:(i) x-3 (ii) x1/3 |
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Answer» We know, d/dx (xm) = m(xm-1) (i) d/dx (x-3 ) = -3x-4 (ii) d/dx (x1/3) = 1/3 x-2/3 |
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| 400. |
Let `f(x)=(x^2-x)/(x^2+2x)` then `d(f^(-1)x)/(dx)` is equal toA. `(-3)/((1-x)^(2))`B. `(3)/((1-x)^(2))`C. `(1)/((1-x)^(2))`D. none of these |
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Answer» Correct Answer - D We have, `f(x)=(x^(2)-x)/(x^(2)+2x)` Clearly, f(x) is not derfined at `x=0, -2` So, Domian `(f)=R-{-2,0}.` For all `x in ` domain (f), we have `f(x)=(x^(2)-x)/(x^(2)+2x)=(x-1)/(x+2)=1-(3)/(x+2)` Now, `fof^(-1)(x)=x` `implies" "f(f^(-1)(x))=x` `implies" "1-(3)/(f^(-1)(x)+2)=x` `implies" "1-x=(3)/(f^(-1)(x)+2)` `implies" "f^(-1)(x)=(3)/(1-x)-2implies(d)/(dx){f^(-1)(x)}=(3)/((1-x)^(2))` |
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