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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.
| 401. |
If `f(x)=|x^nn !2cosxcos(npi)/2 4sinxsin(npi)/2 8|`then findthe value of `(d^n)/(dx^n)([f(x)])_(x=0)dot(n in z)dot` |
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Answer» `(d^(n))/(dx^(n))[f(x)]=|{:((d^(n))/(dx^(n))(x^(n)),n!,2),((d^(n))/(dx^(n))(cos x),cos (npi)/(2),4),((d^(n))/(dx^(n))(sin x), sin (npi)/(2),8):}|` `=|{:(n!,n!,2),(cos(x+(npi)/(2)),cos""(npi)/(2),4),(sin(x+(npi)/(2)),sin""(npi)/(2),8):}|` `"or "(d^(n))/(dx^(n))[f(x)]_(x=0)=0` |
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| 402. |
The derivative of `sec^(-1)((1)/(2x^(2)-1))` with respect to `sqrt(1-x^(2))" at "x=1`, isA. 2B. -2C. non-exisentD. none of these |
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Answer» Correct Answer - C It is evident that `z=sqrt(1-x^(2))` is not differentiable at `x=+-1` So, `(dz)/(dx)` does not exist at `x=+-1.` |
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| 403. |
Select the appropriate option from the given alternatives.Suppose f(x) is the derivative of g(x) and g(x) is the derivative of h(x). If h(x) = a sin x + b cos x + c, then f(x) + h(x) = (A) 0 (B) c (C) -c (D) -2(a sin x + b cos x) |
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Answer» Correct option is:(B) c h(x) = a sin x + b cos x + c Differentiating w.r.t. x, we get h'(x) = a cos x – b sin x = g(x) …..[given] Differentiating w.r.t. x, we get g'(x) = -a sin x – b cos x = f(x) …..[given] ∴ f(x) + h(x) = -a sin x – b cos x + a sin x + b cos x + c ∴ f(x) + h(x) = c |
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| 404. |
The derivative of `cosec^(-1)((1)/(2x^(2)-1))` with respect to `sqrt(1-x^(2))" at "x=(1)/(2)`, isA. -4B. 4C. -1D. none of these |
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Answer» Correct Answer - A Let `y=cosec^(-1)((1)/(2x^(2)-1))` and `z=sqrt(1-x^(2))` We have, `y=cosec^(-1)((1)/(2x^(2)-1))` `y=sin^(-1)(2x^(2)-1)=(pi)/(2)-cos^(-1)(2x^(2)-1)` `y={{:((pi)/(2)-2cos^(-1)x" ,"," if "0lexle1),(-(3pi)/(2)+2cos^(-1)x","," if "-1lexle0.):}` `(dy)/(dx)={{:((2)/(sqrt(1-x^(2)))","," if "0ltxlt1),((2)/(sqrt(1-x^(2)))","," if "-1ltxlt0):}` and, `x=sqrt(1-x^(2))implies(dz)/(dx)=(-x)/(sqrt(1-x^(2)))" for all "x in(-1,1)` `(dy)/(dz)=(dy//dx)/(dz//dx)={{:(-(2)/(x)","," if "0ltxlt1),(-(2)/(x)","," if "-1ltxlt0):}` `implies" "((dy)/(dz))_(x=1//2)=-4` |
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| 405. |
If `f(x)=|{:(x^(n),sinx,cosx),(n!,"sin"(npi)/(2),"cos"(npi)/(2)),(a,a^(2),a^(3)):}|`, then the value of `(d^(n))/(dx^(n))(f(x))" at "x=0" for "n=2m+1` isA. -1B. 0C. 1D. independent of a |
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Answer» Correct Answer - B We have, `:." "f(x)=|{:(x^(n),sinx,cosx),(n!,"sin"(npi)/(2),"cos"(npi)/(2)),(a,a^(2),a^(3)):}|` `implies" "(d^(n))/(dx^(n))(f(x))=|{:(n!,sin((npi)/(2)+x),cos((npi)/(2)-x)),(n!," ""sin"(npi)/(2)," ""cos"(npi)/(2)),(a," "a^(2)," "a^(3)):}|` `implies" "{(d^(n))/(dx^(n))(f(x))}_(x=0)=|{:(n!," ""sin"(npi)/(2)," ""cos"(npi)/(2)),(n!," ""sin"(npi)/(2)," ""cos"(npi)/(2)),(a," "a^(2)," "a^(3)):}|=0` |
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| 406. |
The derivative of `sin^(-1)(3x-4x^(3))` with respect to `sin^(-1)x,` isA. `3," for "|x|lt1`B. `3," for "|x|lt(1)/(2)" and "-3" for "(1)/(2)lt|x|lt1`C. `-3," for "|x|lt12`D. `-3," for "|x|le(1)/(2)" and "3" for "(1)/(2)lt|x|lt1` |
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Answer» Correct Answer - B We have, `y=sin^(-1)(3x-4x^(3))={{:(3sin^(-1)x" ,"," if "|x|le(1)/(2)),(pi-3sin^(-1)x" ,"," if "(1)/(2)ltxle1),(-pi-3sin^(-1)x","," if "-1lexlt-(1)/(2)):}` Let`z=sin^(-1)x.` Then, `y={{:(3z" ,"," if "|x|le(1)/(2)),(pi-3z" ,"," if "(1)/(2)ltxle1),(-pi-3z","," if "-1lexlt-(1)/(2)):}` `implies" "(dy)/(dz)={{:(3" ,"," if "|x|lt(1)/(2)),(-3","," if "(1)/(2)lt|x|lt1):}` |
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| 407. |
If `x^m y^n=(x+y)^(m+n),p rov et h a t(dy)/(dx)=y/xdot` |
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Answer» `"We have "x^(m)y^(n)=(x+y)^(m+n).` Taking log on both sides, we get m log x + n log y = (m+n) log (x+y) Differentiating both sides w.r.t.x, we get `m(1)/(x)+n(1)/(y)(dy)/(dx)=(m+n d)/(x+y dx)(x+y)` `"or "(m)/(x)+(n)/(y)(dy)/(dx)=(m+n)/(x+y)(1+(dy)/(dx))` `"or "{(n)/(y)-(m+n)/(x+y)}(dy)/(dx)=(m+n)/(x+y)-(m)/(x)` `"or "{(nx+ny-my-ny)/(y(x+y))}(dy)/(dx)={(mx+nx-mx-my)/((x+y))}` `"or "(nx-my)/(y(x+y))(dy)/(dx)=(nx-my)/((x+y)x)` `"or "(dy)/(dx)=(y)/(x)` |
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| 408. |
If `x=a^sqrt[sin^-1t]` and `y=a^sqrt[cos^-1t]`, then show that `[dy]/[dx]=-y/x` |
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Answer» We have `x^(2)=a^(sin^(-1)t) and y^(2)=a^(cos^(-1)t)` `rArr2x(dx)/(dt)=a^(sin^(-1)t).(1)/(sqrt(1-t^(2)))and2y(dy)/(dt)=a^(cos^(-1)t).((-1))/(sqrt(1-t^(2)))` `rArr(2y)/(2x).((dy//dt))/((dx//dt))=(-a^(cos^(-1)t))/(sqrt(1=t^(2)))xx(sqrt(1-t^(2)))/(a^(sin^(-1)t))` `rArr(y)/(x).(dy)/(dx)=(a^(cos^(-1)t))/(a^(sin^(-1)t))=-(y^(2))/(x^(2))` `rArr(dy)/(dx)=(-y)/(x)" "["on dividing both sides by "(y)/(x)].` Hence,`(dy)/(dx)=(-y)/(x).` |
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| 409. |
If `y=asinx+bcosx ,t h e ny^2+((dy)/(dx))^2`is afunction of `x`(b) function of`y`function of `xa n dy`(d) constantA. function of xB. function of yC. function of x and yD. constant |
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Answer» y=a sin x + b cos x Differentiating with respect to x, we get `(dy)/(dx)=a cos x - b sin x` `"Now, "((dy)/(dx))^(2)=(a cos x - b sin x)^(2)` `=a^(2)cos^(2)x+b^(2)sin^(2)x-2ab sin x cos x` `"and "y^(2)=(a sin x + b cos x)^(2)` `a^(2)sin^(2) x +b^(2)cos^(2) x +2ab sin x cos x` `"So, "((dy)/(dx))^(2)+y^(2)=a^(2)(sin^(2)x + cos^(2)x)+b^(2)(sin^(2)x+ cos^(2)x)` `=(a^(2)+b^(2))=` constant |
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| 410. |
`y=tan^(-1)((acosx-bsinx)/(bcosx+asinx)),w h e r e-pi/2-1` |
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Answer» `"Let "y=tan^(-1)((a cos x- b sin x)/(b cos x + a sin x)),` `=tan^(-1)(((a)/(b)-tan x)/(1+(a)/(b) tan x))` `=tan^(-1)((a)/(b))-tan^(-1)(tan x)` `=tan^(-1)((a)/(b))-x" "[because-(pi)/(2)ltxlt(pi)/(2)` `therefore" "(dy)/(dx)=0-1=-1` |
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| 411. |
If `y=(1/2)^(n-1)cos(ncos^(-1)x),`then prove that `y`satisfies the differential equation `(1-x^2)``(d^2y)/(dx^2)-x(dy)/(dx)+n^2y=0` |
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Answer» `y=(1//2^(n-1))cos (n cos^(-1)x)` `therefore" "(dy)/(dx)=-(1)/(2^(n-1))sin(n cos^(-1)x)[(-n)/sqrt((1-x^(2)))]` `"or "(1-x^(2))((dy)/(dx))^(2)=(n^(2))/(2^(2n-2))sin^(2)(n cos^(-1)x)` `=(n^(2))/(2^(2n-2))[1-cos^(2)(n cos^(-1)x)]` `=n^(2)[(1)/(2^(2n-2))-y^(2)]` Differentiating both sides w.r.t x, we get `(1-x^(2))2(dy)/(dx)(d^(2)y)/(dx^(2))-2x((dy)/dx)^(2)=-2n^(2)y(dy)/(dx)` `"or "(1-x^(2))(d^(2)y)/(dx^(2))-x(dy)/(dx)+n^(2)y=0` |
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| 412. |
`cot^(-1)((cosx-sinx)/(cosx+sinx))` |
| Answer» Correct Answer - 1 | |
| 413. |
Find `(dy)/(dx)`, when: `y=x^(sinx)+(sinx)^(cosx)` |
| Answer» Correct Answer - `x^(sinx).{(sinx)/(x)+cosx.logx}+(sinx)^(cosx).{cosxcot x-sin x log(sinx)}` | |
| 414. |
Find `(dy)/(dx)`, when: `y=(sinx)^(cosx)` |
| Answer» Correct Answer - `(sinx)^(cosx).{cotxcostx-sinxlog(sinx)}` | |
| 415. |
Find the differential equation of the family of curves `y=A e^(2x)+B e^(-2x)`, where A and B are arbitrary constants. |
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Answer» `y=A tan^(-1)(B tan""(x)/(2)),` `"where "A=(2)/(sqrt(a^(2)-b^(2))),B=sqrt((a-b)/(a+b))` `Ab=(2)/(sqrt((a-b)(a+b)))sqrt((a-b)/(a+b))=(2)/(a+b)` `(dy)/(dx)=(ABsec^(2)""(x)/(2)xx(1)/(2))/(1+B^(2)tan^(2)""(x)/(2))` `=(1)/(a+b).(a+b)/((a+b)cos^(2)""(x)/(2)+(a-b)sin^(2)""(x)/(2))` `=(1)/(a+b cos x)" (1)"` `therefore" " (d^(2)y)/(dx^(2))=(b sin x)/((a+b cos x)^(2))` |
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| 416. |
If x=a (cos t + t sin t) and y=a ( sin t- t cos t), find `(d^(2)y)/(dx^(2))` |
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Answer» It is given that x =a (cos t + t sin t) and y=a (sin t - t cos t). Therefore, `(dx)/(dt)=a[-sin t+ sin t + t cos t]= at cos t` `(dy)/(dt)=a [ cos t -{cos t - t sin t} ] = at sin t` `therefore" "(dy)/(dx)=(((dy)/(dt)))/(((dx)/(dt)))=(at sin t)/(at cos t)= tan t` `"Then, "(d^(2)y)/(dx^(2))=(d)/(dx)((dy)/(dx))=(d)/(dx)(tan t)` `=(d)/(dt) (tan t)(dt)/(dx)` `=sec^(2) t. (dt)/(dx)` `sec^(2)t. (1)/(at cos t)` `(sec^(3) t)/(at)` |
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| 417. |
Given that `cosx/2.cosx/4.cosx/8....=sinx/x` Then find the sum `1/2^2sec^2x/2+1/2^4sec^2x/4+...` |
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Answer» `"We have "cos""(x)/(2)cdotcos""(x)/(4)cdotcos""(x)/(8)...=(sin x)/(x)." Then find the sum "(1)/(2^(2))sec^(2)""(x)/(2)+(1)/(2^(4))sec^(2)""(x)/(4)+...` Taking log on both sides, we get `log cos""(x)/(2)+log cos""(x)/(4)+log cos""(x)/(8)...+...=log sin x-log x` Differentiating both sides with respect to x, we get `-(1)/(2)(sin""(x)/(2))/(cos""(x)/(2))-(1)/(4)(sin""(x)/(4))/(cos""(x)/(4))-(1)/(8)(sin""(x)/(8))/(cos""(x)/(8))...=(cos x)/(sin x)-(1)/(x)` `"or "-(1)/(2)tan""(x)/(2)-(1)/(4)tan""(x)/(4)-(1)/(8)tan"(x)/(8)-...=cot x -(1)/(x)` Differntiating both sides with respect to x, we get `-(1)/(2^(2))sec^(2)""(x)/(2)-(1)/(4^(2))sec^(2)""(x)/(4)-(1)/(8^(2))sec^(2)""(x)/(8)-...=-cosec^(2)x+(1)/(x^(2))` `"or "(1)/(2^(2))sec^(2)""(x)/(2)+(1)/(4^(2))sec^(2)""(x)/(4)+(1)/(8^(2))sec^(2)""(x)/(8)+...=cosec^(2)x-(1)/(x^(2))`. |
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| 418. |
If `x=cos e ctheta-sinthetaa n dy=cos e c^ntheta-sin^ntheta,`then show that`(x^2+4)((dy)/(dx))^2=n^2(y^2+4)dot` |
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Answer» As x = cosec theta - sin theta, we have `x^(2)+4=(cosec theta- sin theta)^(2)+4=(cosec theta + sin theta)^(2)" (1)"` `"and "y^(2)+4=(cosec^(n)theta - sin^(n)theta)^(2)+4 = (cosec^(n)theta+ sin^(n)theta)^(2)" (2)"` Now, `(dy)/(dx)=(((dy)/(d""theta)))/(((dx)/(d""theta)))=(n(cosec^(n-1)theta)(-cosec theta cot theta)-n sin^(n-1)theta cos theta)/(-cosec theta cot theta- cos theta)` `=(n(cosec^(n)thetacot theta +sin^(n-1)theta cos theta))/((cosec theta cot theta +cos theta))` `(ncot theta (cosec^(n)theta + sin^(n)theta))/(cot theta (cosec theta +sin theta))` `=(n(cosec^(n)theta+sin^(n)theta))/((cosec theta +sin theta))=(nsqrt(y^(2)+4))/(sqrt(x^(2)+4))" [from (1) and (2) ]"` `"Squaring both sides, we get "((dy)/(dx))^(2)=(n^(2)(y^(2)+4))/((x^(2)+4))` `"or "(x^(2)+4)((dy)/(dx))^(2)=n^(2)(y^(2)+4)` |
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| 419. |
If `x=a{costheta+logtan""(theta)/(2)}" and "y=asintheta,` then `(dy)/(dx)` is equal toA. `cot theta`B. `tantheta`C. `sintheta`D. `costheta` |
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Answer» Correct Answer - B We have, `x=a{costheta+logtan""(theta)/(2)}" and "y=asintheta` `implies(dx)/(d theta)=a{-sintheta+("sec"^(2)(theta)/(2))/(2"tan"(theta)/(2))}" and "(dy)/(d theta)=acostheta` `implies(dx)/(d theta)=a{-sintheta+(1)/(sintheta)}" and "(dy)/(d theta)=acostheta` `implies" "(dx)/(d theta)=a(cos^(2)theta)/(sintheta)" and",(dy)/(d theta)=acostheta` `(dy)/(dx)=((dy)/(d theta))/((dx)/(d theta))=(acostheta)/(a(cos^(2)theta)/(sintheta))=tantheta` |
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| 420. |
If `P_n`is the sum of a `GdotPdot`upto `n`terms `(ngeq3),`then prove that `(1-r)(d P_n)/(d r)=(1-n)P_n+n P_(n-1),`where `r`is the common ratio of `GdotPdot` |
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Answer» Let the first term of G.P. be `alpha`. Then `P_(n)=alpha[(1-r^(n))/(1-r)]` `(dp_(n))/(dr)=alpha[((1-r)(-nr^(n-1))+(1-r^(n)))/((1-r^())^(2))]` `therefore" "(1-r)(dP_(n))/(dr)=alpha((-nr^(n-1)+nr^(n))/(1-r))+((1-r^(n))/(1-r))alpha` `=alphan.((1.r^(n-1)-1+r^(n))/(1-r))+P_(n)` `=ncdotP_(n-1)-nP_(n)+P_(n)` `=(1-n)P_(n)+nP_(n-1)` |
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| 421. |
Given that `cos(x/2).cos(x/4).cos(x/8)..... = sinx/x`Prove that `(1/2^2)sec^2(x/2) +(1/2^4)sec^2(x/4) +..... = cosec^2x - 1/(x^2)` |
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Answer» `"We have "cos""(x)/(2)cdotcos""(x)/(4)cdotcos""(x)/(8)...=(sin x)/(x)." Then find the sum "(1)/(2^(2))sec^(2)""(x)/(2)+(1)/(2^(4))sec^(2)""(x)/(4)+...` Taking log on both sides, we get `log cos""(x)/(2)+log cos""(x)/(4)+log cos""(x)/(8)...+...=log sin x-log x` Differentiating both sides with respect to x, we get `-(1)/(2)(sin""(x)/(2))/(cos""(x)/(2))-(1)/(4)(sin""(x)/(4))/(cos""(x)/(4))-(1)/(8)(sin""(x)/(8))/(cos""(x)/(8))...=(cos x)/(sin x)-(1)/(x)` `"or "-(1)/(2)tan""(x)/(2)-(1)/(4)tan""(x)/(4)-(1)/(8)tan"(x)/(8)-...=cot x -(1)/(x)` Differntiating both sides with respect to x, we get `-(1)/(2^(2))sec^(2)""(x)/(2)-(1)/(4^(2))sec^(2)""(x)/(4)-(1)/(8^(2))sec^(2)""(x)/(8)-...=-cosec^(2)x+(1)/(x^(2))` `"or "(1)/(2^(2))sec^(2)""(x)/(2)+(1)/(4^(2))sec^(2)""(x)/(4)+(1)/(8^(2))sec^(2)""(x)/(8)+...=cosec^(2)x-(1)/(x^(2))`. |
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| 422. |
If `x=a((1+t^2)/(1-t^2))`and `y=(2t)/(1-t^2)`, find `(dy)/(dx)` |
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Answer» We have `x=a[-1+(2)/((1-t^(2)))]=1[-1+2(1-t^(2))^(-1)]` `rArr (dx)/(dt)=a[0+2(-1)(1-t^(2))^(-2)(-2t)]=axx(4t)/((1-t^(2))^(2))=(4at)/((1-t^(2))^(2)).` And, `y=(2t)/((1-t^(2)))` `rArr(dy)/(dt)=((1-t^(2)).2-2t(-2t))/((1-t^(2))^(2))(2(1+t^(2)))/((1-t^(2))^(2))` `therefore(dy)/(dx)=((dy//dt))/((dx//dt))={(2(1+t^(2)))/((1-t^(2))^(2))xx((1-t^(2))^(2))/(4at)}=((1+t^(2)))/(2at).` |
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| 423. |
If `x=a(cost+tsint)`and `y=a(sint-tcost),fin d(d^2y)/(dx^2)dot` |
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Answer» It is given that x =a (cos t + t sin t) and y=a (sin t - t cos t). Therefore, `(dx)/(dt)=a[-sin t+ sin t + t cos t]= at cos t` `(dy)/(dt)=a [ cos t -{cos t - t sin t} ] = at sin t` `therefore" "(dy)/(dx)=(((dy)/(dt)))/(((dx)/(dt)))=(at sin t)/(at cos t)= tan t` `"Then, "(d^(2)y)/(dx^(2))=(d)/(dx)((dy)/(dx))=(d)/(dx)(tan t)` `=(d)/(dt) (tan t)(dt)/(dx)` `=sec^(2) t. (dt)/(dx)` `sec^(2)t. (1)/(at cos t)` `(sec^(3) t)/(at)` |
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| 424. |
Differentiate `tan^(-1){(sqrt(1+x^2)-sqrt(1-x^2))/(sqrt(1+x^2)+sqrt(1-x^2))}`with respect to `cos^(-1)x^2` |
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Answer» Let `u=tan^(-1){(sqrt(1+x^(2))-sqrt(1-x^(2)))/(sqrt(1+x^(2))+sqrt(1-x^(2)))} and v = cos^(-1)x^(2).` Then, `cos^(-1)x^(2)=v rArr x^(2)=cos v`. Putting `x^(2)=cos v`, we get `u=tan^(-1){(sqrt(1+cosv)-sqrt(1-cosv))/(sqrt(1+cosv)+sqrt(1-cosv))}` `rArrtan^(-1){(sqrt(2cos^(2)(v//2))-sqrt(2sin^(2)(v//2)))/(sqrt(2cos^(2)(v//2))+sqrt(2sin^(2)(v//2)))}` `=tan^(-1){(cos(v//2)-sin(v//2))/(cos(v//2)+sin(v//2))}=tan^(-1){(1-tan(v//2))/(1+tan(v//2))}` `" [dividing num. and denom. by cos (v/2)]"` `=tan^(-1){tan((pi)/(4)-(v)/(2))}=((pi)/(4)-(v)/(2)).` `thereforeu=((pi)/(4)-(v)/(2))rArr (du)/(dv)=(-1)/(2).` |
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| 425. |
If `x = a sin 2t(1 + cos 2t) and y = b cos 2 t(1 – cos 2 t),` show that `((d y)/(d x))_(at t =pi/4)=b/a.` |
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Answer» We have `x=a sin 2t(1+cos 2t)` `rArr(dx)/(dt)=a.[sin2t(-2 sin 2t)+(1+cos 2t)(2cos 2t)]` `=(2a).[-sin^(2)2t+cos 2t+cos^(2)2t]` `=(2a)[cos4t+cos2t]" "{because" "(cos^(2)2t-sin^(2)2t)=cos4t}.` And, `y=b cos 2t (1-cos2t)` `rArr (dy)/(dx)=b[cos2t(2sin 2t)+(1-cos2t)(-2sin2t)]` `=(2b)[sin 2t cos 2t-sin 2t +sin 2t cos 2t)]` `(2b)[2sin 2t cos 2t -sin 2t]=(2b)[sin 4t-sin 2t].` `therefore(dy)/(dx)=((dy//dt))/((dx//dt))=((2b)(sin 4t -sin 2t))/((2a)(cos 4t +cos2t))` `rArr((dy)/(dx))_((t=(pi)/4))=(b)/(a).({sin(4xx(pi)/(4))-sin(2xx(pi)/(4))})/({cos(4xx(pi)/(4))+cos(2xx(pi)/(4))})` `=(b)/(a).((sin pi-sin.(pi)/(2)))/((cos pi+cos.(pi)/(2)))=(b)/(a).((0-1))/((-1+0))=(b)/(a).` |
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| 426. |
Find `(dy)/(dx)`, when `x=a(t+sint) and y=a(1-cost).` |
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Answer» We have `x=a(t+sint)rArr (dx)/(dt)=a(1+cost)`, `y=a(1-cost)rArr (dy)/(dt)=a sin t.` `therefore(dy)/(dx)=((dy)/(dt)xx(dt)/(dx))=(asint)/(a(1+cost))=(2a sin(t//2)cos(t//2))/(2a cos^(2)(t//2))=tan.(t)/(2).` |
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| 427. |
Find `(dy)/(dx)`, when `x=costheta+cos2 theta,y=sin theta+sin2theta` |
| Answer» Correct Answer - `-((costheta+2cos2theta)/(sin theta+2sin 2theta))` | |
| 428. |
If `f(x)=log{(u(x))/(v(x))}, u(1)=v(1)`and `u^(prime)(1)=v^(prime)(1)=2`, then find the valueof `f^(prime)(1)`. |
| Answer» Correct Answer - A | |
| 429. |
If `x=a(cos theta+log tan(theta/2))` and `y=a sin theta` then find `(dy)/(dx)` at `theta=pi/3` and `theta=pi/4` |
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Answer» We have `x=a{cos theta +log tan(theta//2)}` `rArr(dx)/(d theta)=a{-sin theta+(sec^(2)(theta//2))/(2tan(theta//2))}=a{-sin theta+(1)/(2sin (theta//2)cos(theta//2))}` `=a{-sintheta+(1)/(sin theta)}=(a(1-sin^(2)theta))/(sintheta)=(acos^(2)theta)/(sin theta).` And, `y=a sin theta rArr (dy)/(dx)=a cos theta.` `therefore(dy)/(dx)=((dy)/(d theta)xx(d theta)/(dx))=(a cos theta.(sin theta)/(a cos^(2)theta))=tan theta.` `therefore[(dy)/(dx)]_(theta=pi//2)=tan.(pi)/(4)=1.` |
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| 430. |
If `x=costheta,y=sin50," then "(1-x^(2))(d^(2)y)/(dx^(2))-x(dy)/(dx)=`A. `-5y`B. 5yC. 25yD. `-25y` |
| Answer» Correct Answer - D | |
| 431. |
if `2x^2-3xy+y^2+x+2y-8=0` then `(dy)/(dx)`A. `(3y-4x-1)/(2y-3x+2)`B. `(3y+4x+1)/(2y+3x+2)`C. `(3y-4x+1)/(2y-3x-2)`D. `(3y-4x+1)/(2y+3x+2)` |
| Answer» Correct Answer - A | |
| 432. |
Find `(dy)/(dx)`for the function:`y=(log)_esqrt((1+sinx)/(1-s ingx)),w h e r ex=pi/3` |
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Answer» Correct Answer - 2 `y=log sqrt((1+ sin x)/(1- sin x))=(1)/(2)[log_(e)(1+ sin x) -log _(e) (1- sin x)]` `therefore" "(dy)/(dx)=(1)/(2){(1)/(1+ sin x dx)(d)/(dx)(1+ sin x)-(1)/(1- sin x )(d)/(dx)(1- sin x )}` `=(1)/(2){(cos x)/(1+ sin x)+(cos x )/(1-sin x)}` `=(1)/(2)cos x ((2)/(1- sin^(2)x))=(cos x)/(cos^(2)x)=sec x` `"or "(dy)/(dx):|_(x=(pi)/(3))=sec""(pi)/(3)=2` |
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| 433. |
If`ysqrt(x^2+1)=log(sqrt(x^2+)1-x),s howt h a t(x^2+1)(dy)/(dx)+x y+1=0` |
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Answer» `ysqrt(x^(2)+1)=log{(sqrt(x^(2)+1)-x}` Differentiating both sides w.r.t. x, we get `(dy)/(dx)sqrt(x^(2)+1)+y(1)/(2sqrt(x^(2)+1))2x=(1)/(sqrt(x^(2)+1)-x)xx{(1)/(2)(2x)/(sqrt(x^(2)+1))-1}` `"or "(x^(2)+1)(dy)/(dx)+xy=sqrt(x^(2)+1)(-1)/(sqrt(x^(2))+1)` `"or "(x^(2)+1)(dy)/(dx)+xy+1=0` |
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| 434. |
If `y=log{((1+x)/(1-x))^(1//4)}-(1)/(2)tan^(-1)x," then "(dy)/(dx)=`A. `(x)/(1-x^(2))`B. `(x^(2))/(1-x^(4))`C. `(x)/(1+x^(4))`D. `(x)/(1-x^(4))` |
| Answer» Correct Answer - B | |
| 435. |
Differentiate `sec^-1""(1)/(2x^2-1)` with respect to `sqrt(1-x^2)`A. -4B. 4C. 2D. -2 |
| Answer» Correct Answer - B | |
| 436. |
Differentiate `sin^(-1)(1-2x^2), 0 |
| Answer» Correct Answer - `(-2)/(sqrt(1-x^(2)))` | |
| 437. |
Differentiate:\(\frac{d}{d\theta}(cos\theta-sin\theta)^2\) |
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Answer» \(\frac{d}{d\theta}(cos\theta-sin\theta)^2\) = 2(cos θ - sin θ) \(\frac{d}{d\theta}(cos\theta-sin\theta)\) = 2(cos θ - sin θ) (-sinθ - cos θ) = -2(cos θ - sin θ) (cos θ - sin θ) = -2(cos θ - sin θ) (cos θ - sin θ) = -2 (cos2θ - sin2θ) = -2 cos 2θ |
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| 438. |
`sqrt(tanx)` |
| Answer» Correct Answer - `(sec^(2)x)/(2sqrt(tanx))` | |
| 439. |
Differentiate w.r.t. x: `(i)tan^(-1)((cosx)/(1+sinx))" "(ii)tan^(-1)(secx+tanx)` |
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Answer» Let `y=tan^(-1)((cos+1)/(1+sinx))=tan^(-1){(sin((pi)/(2)-x))/(1+cos((pi)/(2)-x))}` `=tan^(-1){(2sin((pi)/(4)-(x)/(2))cos((pi)/(4)-(x)/(2)))/(2cos^(2)((pi)/(4)-(x)/(2))}}` `tan^(-1){tan((pi)/(4)-(x)/(2))}=((pi)/(4)-(x)/(2)).` `thereforey=((pi)/(4)-(x)/(2)).` Hence,`(dy)/(dx)=(d)/(dx)((pi)/(4)-(x)/(2))=(d)/(dx)((pi)/(4))-(d)/(dx)((x)/(2))=(0-(1)/(2))=(-1)/(2).` (ii) Let `y=tan^(-1)(sec x+tanx)` `=tan^(-1)((1)/(cosx)+(sinx)/(cosx))=tan^(-1)((1+sinx)/(cosx))` `=tan^(-1){(1-cos((pi)/(2)+x))/(sin((pi)/(2)+x))}` `{because cos ((pi)/(2)+x)=-sinx,sin((pi)/(2)+x)=cosx}` `=tan^(-1){(2sin^(2)((pi)/(4)+(x)/(2)))/(2sin((pi)/(4)+(x)/(2))cos((pi)/(4)+(x)/(2)))}` `=tan^(-1){tan((pi)/(4)+(x)/(2))}=((pi)/(4)+(x)/(2)).` `therefore y=((pi)/(4)+(x)/(2)).` Hence, `(dy)/(dx)=(d)/(dx)((pi)/(4)+(x)/(2))=(d)/(dx)((pi)/(4))+(d)/(dx)((x)/(2))=(0+(1)/(2))=(1)/(2).` |
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| 440. |
Differentiate the following functions with respect to `x :``(2^2cotx)/(sqrt(x))`(ii) `e^xlogsqrt(x)tanx` |
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Answer» i) `y= (2^2 cot x)/(sqrtx)` `= (4cot x)/sqrtx` `dy/dx = (4[sqrtx d/dx(cot x) - cot x d/dx(sqrtx)])/(sqrtx)^2` `= 4[sqrtx(-cosec^2 x) - cot x xx 1/(2 sqrtx) ]/x` `dy/dx = -(4cosec^2 x)/sqrtx - cot x/(2x sqrtx) ` ii) `y= e^x log sqrtx tan x` `y = e^x 1/2 log x tan x` `y= 1/2[ e^x log x tan x]` `dy/dx = 1/2[e^x log x * tan x + e^x 1/x tan x + e^x log x * sec^2 x]` `= e^x/2[ log x*tan x + tanx/x + log x*sec^2 x]` Answer |
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| 441. |
Differentiate `{x^(tanx)+sqrt((x^(2)+1)/(x))}` w.r.t. x. |
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Answer» Let `y=u+v`, where `y=x^(tanx) and v=sqrt((x^(2)+1)/(x)).` Now, `u=x^(tanx)` `rArr logu=(tanx)(logx)` `rArr(1)/(u).(du)/(dx)=(tanx).(d)/(dx)(logx)+(logx).(d)/(dx)(tanx)` `" [differentiating w.r.t. x]"` `=(tanx).(1)/(x)+(logx)sec^(2)x` `rArr(du)/(dx)=u.[(tanx)/(x)+(logx)sec^(2)x]` `rArr(du)/(dx)=x^(tanx).{(tanx)/(x)+(logx)sec^(2)x}." ...(i)"` And, `v=sqrt((x^(2)+1)/(x))` `rArr log v=(1)/(2).{log (x^(2)+1)-logx}` `rArr(1)/(v).(dv)/(dx)=(1)/(2).{(2x)/((x^(2)+1))-(1)/(x)}" [differentiating w.r.t. x]"` `rArr(dv)/(dx)=(v)/(2).{(2x^(2)-(x^(2)+1))/(x(x^(2)+1))}` `rArr(dv)/(dx)=(1)/(2)sqrt((x^(2)+1)/(x)).{(x^(2)-1)/(x(x^(2)+1))}." ...(ii)"` `therefore y=u+v` `rArr(dy)/(dx)=(du)/(dx)+(dv)/(dx)` `rArr(dy)/(dx)=x^(tanx).{(tanx)/(x)+(logx)sec^(2)x}+(1)/(2).sqrt((x^(2)+1)/(x)).{((x^(2)-1))/(x(x^(2)+1))}.` |
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| 442. |
Differentiate `tan^(-1)((sqrt(1+x^(2))-1)/(x))` w.r.t. `tan^(-1)x.` |
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Answer» Let `u=tan^(-1)((sqrt(1+x^(2))-1)/(x)) and v=tan^(-1)x` Now, `v=tan^(-1)x rArr x = tanv.` Putting `x=tan v,` we get `u=tan^(-1){(sqrt(1+tan^(2)v)-1)/(tanv)}=tan^(-1)((secv-1)/(tanv))` `=tan^(-1)((1-cosv)/(sinv))=tan^(-1){(2sin^(2)(v//2))/(2sin(v//2)cos(v//2))}` `=tan^(-1){tan.(v)/(2)}=(v)/(2).` `thereforeu=(v)/(2)rArr(du)/(dv)=(1)/(2).` |
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| 443. |
Differentiate `(e^(x^(2))tan^(-1)x)/(sqrt(1+x^(2)))` w.r.t. x. |
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Answer» Let `y=(e^(x^(2))tan^(-1)x)/(sqrt(1+x^(2)))." ...(i)"` Taking logarithm on both sides of (i), we get `logy=x^(2)+log(tan^(-1)x)-(1)/(2)log (1+x^(2))." …(ii)"` On differentiating both sides of (ii) w.r.t. x, we get `(1)/(y).(dy)/(dx)=2x+(1)/(tan^(-1)x).(1)/((1+x^(2)))-(1)/(2).(2x)/((1+x^(2))).` `rArr(dy)/(dx)=y[2x+(1)/((1+x^(2))tan^(-1)x)-(x)/((1+x^(2)))]` `=(e^(x^(2))tan^-1)/(sqrt(1+x^(2))).[2x+(1)/((1+x^(2))tan^(-1)x)-(x)/((1+x^(2)))].` |
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| 444. |
Differentiate `tan^(-1)((sqrt(1-x^(2)))/(x))` w.r.t. `cos^(-1)(2xsqrt(1-x^(2))),` when `x ne0.`A. `-(1)/(2)`B.C.D. |
| Answer» Correct Answer - Put `x=cos theta.` | |
| 445. |
Differentiate `tan^(-1){(sqrt(1+x^(2))-1)/(x)}` w.r.t. x. |
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Answer» Let `y=tan^(-1){(sqrt(1+x^(2))-1)/(x)}` Putting `x=tan theta`, we get `y=tan^(-1){(sqrt(1+tan^(2)theta-1))/(tan theta)}=tan^(-1){(sec theta-1)/(tan theta)}` `=tan^(-1){(((1)/(cos theta)-1))/(sin theta).cos theta}=tan^(=1)((1-cos theta)/(sin theta))` `=tan^(-1){(2sin^(2)(theta//2))/(2sin(theta//2)cos(theta//2))}=tan^(-1){"tan"(theta)/(2)}` `=(theta)/(2)=(1)/(2)tan^(-1)x.` `therefore y=(1)/(2)tan^(-1)x` `rArr(dy)/(dx)=(1)/(2).(d)/(dx)(tan^(-1)x)=(1)/(2(1+x^(2))).` |
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| 446. |
Suppose `f(x)=e^(ax) + e^(bx)`, where `a!=b`, and that `fprimeprime(x)-2fprime(x)-15f(x)=0` for all `x`. Then the value of `ab` is equal to:A. 25B. 9C. -15D. -9 |
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Answer» `(a^(2)-2a-15)e^(ax)+(b^(2)-2b-15)e^(bx)=0` `"or "(a^(2)-2a-15)=0 and b^(2)-2b-15=0` `"or "(a-5)(a+3)=0 and (b-5)(b+3)=0` `i.e., a=5 or -3 and b=5 or -3` `therefore" "aneb` Hence, `a=5 and b=-3 or a =-3 and b=5` `"or "ab=-15` |
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| 447. |
Differentiate `sqrt(cot^(-1)sqrtx)`, w.r.t. x. |
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Answer» Let `y=sqrt(cot^(-1)sqrtx)`. Putting `sqrtx=t and cot^(-1)sqrtx=cot^(-1)sqrtx=u`, we get `y=sqrtu,` where `u=cot^(-1)t and t=sqrtx`. Now, `y=sqrtu rArr (dy)/(du)=(1)/(2)u^(-1//2)=(1)/(2sqrtu),` `u=cot^(-1)t rArr (du)/(dt)=(-1)/((1+t^(2)))`. And, `t=sqrtx rArr (dt)/(dx)=(1)/(2)x^(-1//2)=(1)/(2sqrtx)`. `therefore(dy)/(dx)=((dy)/(du)xx(du)/(dt)xx(dt)/(dx))=(-1)/(4sqrtu(1+t^(2))sqrtx)` `" "=(-1)/(4(sqrt(cot^(-1)t))(1+t^(2))sqrtx)" "[because u=cot^(-1)t]` `" "=(-1)/(4(sqrt(cot^(-1)sqrtx))(1+x)sqrtx)" "[because t=sqrtx].` |
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| 448. |
Differentiate `tan^(-1)((sqrt(1+x^(2))+1)/(x))` w.r.t. x. |
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Answer» Let `y=tan^(-1)((sqrt(1+x^(2))+1)/(x)).` Putting `x=tan theta,` we get `y=tan^(-1)((sec theta+1)/(tan theta))=tan^(-1)((1+costheta)/(sin theta))` `=tan^(-1){(2cos^(2)(theta//2))/(2sin(theta//2)cos(theta//2))}=tan^(-1){cot.(theta)/(2)}` `=tan^(-1){tan((pi)/(2)-(theta)/(2))}=((pi)/(2)-(theta)/(2))=(pi)/(2)-(1)/(2)tan^(-1)x.` `therefore(dy)/(dx)=(1)/(2(1+x^(2))).` Hence, `(d)/(dx){tan^(-1)((sqrt(1+x^(2))+1)/(x))}=(-1)/(2(1+x^(2)))` |
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| 449. |
Observe the following statements: `"I. If "f(x)=ax^(41)+bx^(-40)," then "(f'(x))/(f(x))=1640x^(2)` `"II. "(d)/(dx){tan^(-1)((2x)/(1-x^(2)))}=(1)/(1+x^(2))` Which of the following is correct ?A. I is true, but II is falseB. Both I and II trueC. Neither I nor II is trueD. I is false, but II is true |
| Answer» Correct Answer - A | |
| 450. |
`d/(dx)[sin^2cot^(- 1)sqrt((1-x)/(1+x))]` isA. -1B. `(1)/(2)`C. `-(1)/(2)`D. 1 |
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Answer» `"Let "y=sin^(2)cot^(-1){sqrt((1-x)/(1+x))}` `"Put "x=cos theta or theta = cos^(-1)x` `therefore" "y=sin^(2)cot^(-1){sqrt((1-cos theta)/(1+cos theta))}= sin ^(2) cot^(-1) (tan""(theta)/(2))` `=sin^(2)((pi)/(2)-(theta)/(2))` `=cos^(2)((theta)/(2))` `=(1+ cos theta)/(2)` `=(1+x)/(2)` `therefore" "(dy)/(dx)=(1)/(2)` |
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