Derivatives: part 6: IITJEE tutorial practice problems « Mathematics Hothouse

Derivatives: part 6: IITJEE tutorial practice problems

Problem 1:

If $\sec {(\frac{x+y}{x-y})}=a$ , then $\frac{dy}{dx}$ is (i) $\frac{x}{y}$ (ii) $\frac{y}{x}$ (iii) y (iv) $x$

Problem 2:

If $f(x) = x+ 2$ , when $-1<x<1$ ;

$f(x)=5$ , when $x=3$ ;

$f(x) = 8-x$ , when $x>3$ ; then, at $x=3$ , the value of $f^{'}(x)$ is

(a) 1 (b) -1 (c) 0 (d) does not exist.

Problem 3:

If $y = x \tan{y}$ , then $\frac{dy}{dx}$ is equal to

(i) $\frac{\tan{y}}{x-x^{2}-y^{2}}$ (ii) $\frac{\tan{y}}{y-x}$

(iii) $\frac{y}{x-x^{2}-y^{2}}$ (iv) $\frac{\tan{x}}{x-y^{2}}$

Problem 4:

If g is the inverse function of f and $f^{'}(x) = \frac{1}{1+x^{n}}$ , then $g^{'}(x)$ is equal to

(i) $1 + (g(x))^{n}$ (ii) $1+g(x)$ (iii) $1-g(x)$ (iv) $1-(g(x))^{n}$

Problem 5:

If $f(x) = \log_{x^{2}}(\log{x})$ then $f(x)$ at $x=c$ is :

(i) 0 (ii) 1 (iii) $\frac{1}{e}$ (iv) $\frac{1}{2e}$

Problem 6:

If $y = (\sin{x})^{\tan{x}}$ then $\frac{dy}{dx}$ is equal to :

(i) $(\sin{x})^{\tan{x}}(1+ \sec^{2}{x} \log{\sin{x}})$

(ii) $\tan{x}. (\sin{x})^{\tan{x}-1} \times \cos{x}$

(iii) $(\sin{x})^{\tan{x}}\sec^{2}{x} \times \log{\sin{x}}$

(iv) $\tan{x} (\sin{x})^{\tan{x}-1}$

Problem 7:

If $y = \sqrt{\sin{x}+y}$ , then $\frac{dy}{dx}$ equals:

(i) $\frac{\sin{x}}{2y-1}$ (ii) $\frac{\sin{x}}{1-2y}$ (iii) $\frac{\cos{x}}{1-2y}$

(iv) $\frac{\cos{x}}{2y-1}$

Problem 8:

If $x = \sqrt{\frac{1-t^{2}}{1+t^{2}}}$ and $y = \sqrt{\frac{\sqrt{1+t^{2}}-sqrt{1-t^{2}}}{\sqrt{1+t^{2}}+\sqrt{1-t^{2}}}}$

then the value of $\frac{d^{2}y}{dx^{2}}$ at $t=0$ is given by:

(a) 0 (b) 1/2 (c) 1 (d) -1

Problem 9:

If $x = a \cos^{3}{\theta}$ , $y = a \sin^{3}{\theta}$ , then $\sqrt{1 + (\frac{dy}{dx})^{2}}$ is equal to:

(i) $\sec^{2}{\theta}$ (ii) $\tan^{2}{\theta}$ (iii) $\sec{\theta}$ (iv) $|\sec{\theta}|$

Problem 10:

If $y = \arcsin{\sqrt{1-x}} + \arccos{\sqrt{x}}$ , then $\frac{dy}{dx}$ equals:

(a) $\frac{1}{\sqrt{x(1-x)}}$ (b) $\frac{1}{x(1+x)}$ (c) $\frac{-1}{\sqrt{x(1-x)}}$ (d) none

Regards,

Nalin Pithwa

This entry was written by Nalin Pithwa, posted on January 6, 2021 at 1:41 am, filed under calculus, IITJEE Mains. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL.

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