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1.

The derivative of `cos(2tan^(-1)sqrt((1-x)/(1+x)))-2cos^(-1)sqrt((1-x)/(2))` w.r.t. x isA. `1-(1)/(sqrt(1-x^(2)))`B. `1-(1)/(sqrt(1+x^(2)))`C. `2-(1)/(sqrt(1-x^(2)))`D. `2-(1)/(sqrt(1+x^(2)))`

Answer» Correct Answer - A
Let `x=cos theta, therefore theta=cos^(-1)x`
`therefore" "y=cos (2 tan^(-1)sqrt((1-x)/(1+x)))-2cos^(-1).sqrt((1-x)/(2))`
`=cos(2tan^(-1)(tan.(theta)/(2)))-2 cso^(-1)(sin.(theta)/(2))`
`=cos theta-2((pi)/(2)-sin^(-1)(sin.(theta)/(2)))`
`=cos theta-pi+theta`
`therefore" "y=x-pi+cos^(-1)x`
`therefore" "(dy)/(dx)=1-(1)/(sqrt(1-x^(2)))`
Discussion
2.

`(d)/(dx)[cos^(-1)(xsqrtx-sqrt((1-x)(1-x^(2))))]=`A. `(1)/(sqrt(1-x^(2)))-(1)/(2sqrt(x-x^(2)))`B. `(-1)/(sqrt(1-x^(2)))-(1)/(2sqrt(x-x^(2)))`C. `(1)/(sqrt(1-x^(2)))+(1)/(2sqrt(x-x^(2)))`D. `(1)/(sqrt(1-x^(2)))`

Answer» Correct Answer - B
`y=cos^(-1)[xsqrtx-sqrt(1-(sqrtx)^(2))sqrt(1-x^(2))]`
`=cos^(-1)x+cos^(-1)sqrtx`
`therefore" "(dy)/(dx)=(-1)/(sqrt(1-x^(2)))-(1)/(2sqrt(x-x^(2)))`
Discussion
3.

Let `y=x^3-8x+7a n dx=f(t)dot`If `(dy)/(dt)=2`and `x=3`at `t=0,`then `(dx)/(dt)`at `t=0`is given by1 (b)`(19)/2`(c) `2/(19)`(d) none of theseA. 1B. `(19)/(2)`C. `(2)/(19)`D. None of these

Answer» Correct Answer - C
`y=x^(3)-8x+7 and x=f(t)`
`(dy)/(dx)=2 and x = 3" at "t = 0`
`because" "(dy)/(dx)=(dy//dt)/(dx//dt)`
`rArr" "(dx)/(dx)=(dy//dt)/(dy//dx)`
`rArr" "(dx)/(dt=(2)/(3x^(2)-8)`
`rArr" "because " at "t=0, x=3`
`therefore" "(dx)/(dt)("at "t=0)=(2)/(19)`
Discussion
4.

if `x=(1+t)/t^3 ,y=3/(2t^2)+2/t` satisfies `f(x)*{(dy)/(dx)}^3=1+(dy)/(dx)` then `f(x)` is:A. `x`B. `(x^(2))/(1+X^(2))`C. `x+x+(1)/(x)`D. `x-(1)/(x)`

Answer» Correct Answer - A
`(dy)/(dx)=(dy//dt)/(dx//dt)=(-((3+2r))/(t^(3)))/(-((3+2t))/(t^(4)))=t`
Since `f(x)((dy)/(dx))^(3)=1+(dy)/(dx)`
`rArr" "f(x)t^(3)=1+t`
`rArr" "f(x)=(1+t)/(t^(3))=x`
Discussion
5.

If `x=sectheta-costheta` and `y=sec^n theta- cos^n theta` then show that `(x^2+4)((dy)/(dx))^2=n^2(y^2+4)`A. 8B. 16C. 64D. 49

Answer» Correct Answer - C
`(dy)/(dx)=(((dy)/(d theta)))/(((dx)/(d theta)))=(8sec^(8) theta tan theta+8cos^(7) thetasin theta)/(sec thetatan theta+sin theta)`
`=(8 tan theta(sec^(8)theta+cos^(8)theta)^(2))/(tan theta (sec theta+ cos theta))`
`therefore" "((dy)/(dx))^(2)=(64(sec^(8)theta+cos^(8)theta)^(2))/((sec theta+cos theta)^(2))`
`=(64[(sec^(8)theta-cos^(8)theta)^(2)+4])/([(sec theta-cos theta)^(2)+4])`
`=(64(y^(2)+4))/((x^(2)+4))`
`therefore" "((x^(2)+4)/(y^(2)+4))((dy)/(dx))^(2)=64`
Discussion
6.

A curve parametrically given by `x=t+t^(3)" and "y=t^(2)," where "t in R." For what vlaue(s) of t is "(dy)/(dx)=(1)/(2)`?A. `(1)/(3)`B. 2C. 3D. 1

Answer» Correct Answer - A::D
`(dy)/(dx)=(2t)/(1+3t^(2))=(1)/(2)`
`rArr" "3t^(2)-4t+1=0`
Discussion
7.

If `y^3-y=2x ,t h e n(x^2-1/(27))(d^2y)/(dx^2)+x(dy)/d=``y`b. `y/3`c. `y/9`d. `y/(27)`A. yB. `(y)/(3)`C. `(y)/(9)`D. `(y)/(27)`

Answer» Correct Answer - C
We have
`y^(3)-y=2x`
Differentiating both sides w.r.t. x, we get
`(3y^(2)-1)(dy)/(dx)=2 rArr (dy)/(dx)=(2)/(3y^(2)-1)" (1)"`
Again, differentiating both sides w.r.t. x, we get
`(d^(2)y)/(dx^(2))=(-2.6y((dy)/(dx)))/((3y^(2)-1))`
Using (1), we get
`(d^(2)y)/(dx^(2))=(-2y)/((3y^(2)-1)^(3))" (2)"`
Now,
`(x^(2)-(1)/(27))(d^(2)y)/(dx^(2))+(xdy)/(dx)`
`=(x^(2)-(1)/(27))((-24y)/((3y^(2)-1)^(2)))+(2x)/((3y^(2)-1))`
`=(y^(2)((y^(2)-1)^(2))/(4)-(1)/(27))((-24y)/((3y^(2)-1)^(3)))+(y(y^(2)-1))/(3y^(2)-1)`
`" "(because y^(3)-y=2x)`
`=({27y^(2)(y^(2)-1)^(2)-4})/(108)((-24y))/((3y^(2)-1))+(y(y^(2)-1))/(3y^(2)-1)`
`=(y)/(9){(-54y^(2)(y^(2)-1)^(2)+8)/((3y^(2)-1)^(3))+(9(y^(2)-1))/(3y^(2)-1)}`
`=(y)/(9){(-2(1+alpha)(alpha-2)^(2)+8)/(alpha^(3))}+(3(alpha-2))/(alpha)`
`" where "alpha=3y^(2)-1`
`=(y)/(9)`
Discussion