Mathematics Hothouse

Derivatives: part 7: IITJEE tutorial problems practice

January 7, 2021 – 8:39 pm

Problem 1: Differential coefficient of $\sec{\arctan{x}}$ is

(a) $\frac{x}{1+x^{2}}$ (b) $x\sqrt{1+x^{2}}$ (c) $\frac{1}{\sqrt{1+x^{2}}}$ (d) $\frac{x}{\sqrt{1+x^{2}}}$

Problem 2: If $\sin{(x+y)} = \log{(x+y)}$ , then $\frac{dy}{dx}$ is equal to :

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

Problem 3: If $y = \arcsin{\sqrt{x-ax}-\sqrt{a-ax}}$ , then $\frac{dy}{dx}$ is equal to:

(a) $\frac{1}{2\sqrt{x}\sqrt{1-x}}$ (b) $\sin{(\sqrt{x})} \times \sin{(\sqrt{a})}$

Problem 4: For the differentiable function f, the value of : $\lim_{h \rightarrow 0} \frac{(f(x+h))^{2}-(f(x))^{2}}{2h}$ is equal to:

(a) $(f^{'}(x))^{2}$ (b) $\frac{1}{2}(f(x))^{2}$ (c) $f(x)f^{'}(x)$ (d) zero

Problem 5: The derivative of $\arctan{\frac{\sqrt{1+x^{2}}-1}{x}}$ w.r.t. $\arctan{(\frac{2x\sqrt{1-x^{2}}}{1-2x^{2}})}$ at $x=0$ is :

(a) $\frac{1}{8}$ (b) $\frac{1}{4}$ (c) $\frac{1}{2}$ (d) 1

Problem 6: If $x = e^{y+e^{y+e^{y+e^{y+ \ldots}}}}$ then $\frac{dy}{dx}$ is

(a) $\frac{x}{1+x}$ (b) $\frac{1}{x}$ (c) $\frac{1-x}{x}$ (d) $\frac{-1}{x^{2}}$

Problem 7: Consider the following statements:

(1) $(\frac{f}{g})^{'} = \frac{f^{'}}{g^{'}}$ (2) $\frac{(fg)^{'}}{fg} = \frac{f^{'}}{f} + \frac{g^{'}}{g}$

(3) $\frac{(f+g)^{'}}{f+g} = \frac{f^{'}}{f} + \frac{g^{'}}{g}$ (4) $\frac{(f/g)^{'}}{f/g} = \frac{f^{'}}{f} + \frac{g^{'}}{g}$

Which of the following statements are true?

(a) 1 and 2 (b) 2 and 3 (c) 2 and 4 (d) 3 and 4

Problem 8: If $y=e^{x+3\log{x}}$ then $\frac{dy}{dx} =$

(a) $e^{x+3\log{x}}$ (b) $e^{x}.x^{2}(x+3)$ (c) $e^{x}. e^{3\log{x}}$ (d) $3x^{2}e^{x}$

Problem 9: If $y=\sin^{2}(x \deg)$ , then find the value of $\frac{dy}{dx}$ is:

(a) $\frac{\pi}{360}\sin{(2 x \deg)}$ (b) $\frac{\pi}{2}\sin{(2x\deg)}$ (c) $180 \sin {(2x\deg)}$ (d) $\frac{\pi}{180}\sin{(2x\deg)}$

Problem 10: If $y=\log_{a}{x} + \log_{x}{a} + \log_{x}{x}+ \log_{a}{a}$ then the value of $\frac{dy}{dx}$ is:

(a) $\frac{1}{x}+x\log{a}$ (b) $\frac{\log{a}}{x} + \frac{x}{\log{a}}$ (c) $\frac{1}{x \log{a}}+ x \log{a}$ (d) $\frac{1}{x\log{x}} - \frac{\log{a}}{x(\log{x})^{2}}$

Cheers,

Nalin Pithwa.

By Nalin Pithwa | Posted in calculus, IITJEE Mains | Comments (0)

Derivatives: part 6: IITJEE tutorial practice problems

January 6, 2021 – 1:41 am

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

By Nalin Pithwa | Posted in calculus, IITJEE Mains | Comments (0)

Derivatives: part 5: IITJEE maths tutorial problems for practice

November 14, 2020 – 5:09 pm

Problem 1:

The derivative of $arcsec (\frac{1}{1-2x^{2}})$ w.r.t. $\sqrt{1-x^{2}}$ at $x=\frac{1}{2}$ is

(a) 2 (b) -4 (c) 1 (d) -2

Problem 2:

If $y = \sin{\sin{x}}$ and $\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} \tan{x} + f(x)=0$ , then $f(x) =$

(a) $\sin^{2}{x} \sin{(\cos{x})}$ (b) $\cos^{2}{x}\sin{\cos{x}}$ (c) $\sin^{2}{x} \cos{\sin{x}}$ (d) $\cos^{2}{x} \sin{\sin{x}}$

Problem 3:

If $f(x) = \log_{a}{\log_{a}{x}}$ , then $f^{'}(x)$ is

(a) $\frac{\log_{a}{e}}{x \log_{e}{x}}$ (b) $\frac{\log_{e}{a}}{x}$ (c) $\frac{\log_{e}{a}}{x\log_{a}{x}}$ (d) $\frac{x}{\log_{e}{a}}$

Problem 4:

If $y=\log {\tan{\frac{x}{2}}} + \arcsin{\cos{x}}$ , then $\frac{dy}{dx}$ is

(a) $cosec (x) -1$ (b) $cosec (x) +1$ (c) $cosec (x)$ (d) x

Problem 5:

If $y^{x}=x^{y}$ , then $\frac{dy}{dx}$ is

(a) $\frac{y}{x}$ (b) $\frac{x}{y}$ (c) $\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$ (d) $\frac{x \log{y}}{y \log{x}}$

Problem 6:

Let f, g, h and k be differentiable in $(a,b)$ , if F is defined as $F(x) = \left | \begin{array}{cc} f(x) & g(x) \\ h(x) & k(x) \end{array} \right |$ for all a, b, then $F^{'}$ is given by:

(i) $\left | \begin{array}{cc} f & g \\ h & k \end{array} \right| + \left | \begin{array}{cc}f & g \\ h^{'} & k \end{array} \right |$

(ii) $\left | \begin{array}{cc}f & g^{'} \\ h & k^{'} \end{array}\right | + \left | \begin{array}{cc} f^{'} & g \\ h & k^{'} \end{array} \right |$

(iii) $\left | \begin{array}{cc}f^{'} & g^{'} \\ h & k \end{array} \right | + \left | \begin{array}{cc}f & g \\ h^{'} & h^{'} \end{array} \right |$

(iv) $\left | \begin{array}{cc}f & g \\ h^{'} & k^{'} \end{array} \right | + \left | \begin{array}{cc}f^{'} & g \\h & k \end{array} \right |$

Problem 7:

If $pv=81$ , then $\frac{dp}{dv}$ at $v=9$ is equal to:

(i) 1 (ii) -1 (iii) 2 (iv) 3

Problem 8:

If $x^{2}+y^{2}=1$ , then

(i) $yy^{''}-2(y^{'})^{2}+1=0$ (ii) $yy^{''} - (y^{'})^{2}-1=0$ (iii) $yy^{''} + (y^{'})^{2} + 1 = 0$ (iv) $yy^{''} - 2(y^{'})^{2}-1=0$

Problem 9:

If $y = \arctan{\frac{\sqrt{x}-1}{\sqrt{x}+1}} + \arctan{\frac{\sqrt{x}+1}{\sqrt{x}-1}}$ , then the value of $\frac{dy}{dx}$ will be

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

Problem 10:

Let $f(x) = \left | \begin{array}{ccc} x^{3} & \sin{x} & \cos{x} \\ 0 & -1 & 0 \\ p & p^{2} & p^{3} \end{array} \right |$ , where p is a constant, then $\frac{d^{3}}{dx^{3}}(f(x))$ at $x=0$ is

(a) p (b) $p+p^{2}$ (c) $p+p^{3}$ (d) independent of p

Regards,

Nalin Pithwa

By Nalin Pithwa | Posted in calculus, IITJEE Mains | Comments (0)

Derivatives: part 4: IITJEE maths tutorial problems for practice

November 13, 2020 – 11:16 pm

Problem 1:

Given $x=x(t)$ , $y=y(t)$ , then $\frac{d^{2}y}{dx^{2}}$ is equal to

(a) $\frac{\frac{d^{2}y}{dt^{2}}}{\frac{d^{2}x}{dt^{2}}}$

(b) $\frac{\frac{d^{2}y}{dt^{2}} \times \frac{dx}{dt} - \frac{dy}{dt} \times \frac{d^{2}x}{dt^{2}}}{(\frac{dx}{dt})^{3}}$

(c) $\frac{\frac{dx}{dt} \times \frac{d^{2}y}{dt^{2}} - \frac{d^{2}x}{dt^{2}} \times \frac{dy}{dt}}{(\frac{dx}{dt})^{2}}$

(d) $\frac{1}{\frac{d^{2}x}{dy^{2}}}$

Problem 2:

$\frac{d}{dx}(\arctan{\sec{x}+ \tan{x}})$ is equal to

(a) 0 (b) $\sec{x}-\tan{x}$ (c) $\frac{1}{2}$ (d) 2

Problem 3:

If $y= \sqrt{x + \sqrt{x + \sqrt{x} + \ldots}}$ , then $\frac{dy}{dx}$ is equal to :

(a) 1 (b) \ $\frac{1}{xy}$ (c) $\frac{1}{2y-x}$ (d) $\frac{1}{2y-1}$

Problem 4:

If $f(x) = \left| \begin{array}{ccc} x & x^{2} & x^{3} \\ 1 & 2x & 3x^{2} \\ 0 & 2 & 6x \end{array} \right|$ , then $f^{'}(x) =$

(a) 12 (b) $6x^{2}$ (c) $6x$ (d) $12x^{2}$

Problem 5:

If $y = (\frac{x^{a}}{x^{b}}) ^{a+b} \times (\frac{x^{b}}{x^{c}})^{b+c} \times (\frac{x^{c}}{x^{a}})^{c+a}$ , then $\frac{dy}{dx}=$

(a) 0 (b) 1 (c) $a+b+c$ (d) abc

Problem 6:

If $y = \arctan{\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}}$ , then $\frac{dy}{dx}$ is equal to

(a) $\frac{1}{1-x^{2}}$ (b) $\frac{1}{\sqrt{1-x^{2}}}$ (c) $\frac{1}{1+x^{2}}$ (d) $\frac{1}{\sqrt{1+x^{2}}}$

Problem 7:

If $x=at^{2}$ , $y=2at$ , then $\frac{d^{2}y}{dx^{2}}=$

(a) $\frac{1}{t^{2}}$ (b) $\frac{1}{2at^{3}}$ (c) $\frac{1}{t^{3}}$ (d) $\frac{-1}{2at^{3}}$

Problem 8:

If $y=ax^{n+1} +bx^{-n}$ , then $x^{2}\frac{d^{2}y}{dx^{2}}=$

(a) $n(n-1)y$ (b) $ny$ (c) $n(n+1)y$ (d) $n^{2}y$

Problem 9:

If $x=t^{2}$ , $y=t^{3}$ , then $\frac{d^{2}y}{dx^{2}}=$

(a) $\frac{3}{2}$ (b) $\frac{3}{4t}$ (c) $\frac{3}{2t}$ (d) 0

Problem 10:

If $y=a+bx^{2}$ , a, b arbitrary constants, then

(a) $\frac{d^{2}}{dx^{2}} = 2xy$ (b) $x \frac{d^{2}y}{dx^{2}} - \frac{dy}{dx} + y=0$ (c) $x \frac{d^{2}y}{dx^{2}} = \frac{dy}{dx}$ (d) $x \frac{d^{2}y}{dx^{2}} = 2xy$

Regards,

Nalin Pithwa

By Nalin Pithwa | Posted in calculus, IITJEE Mains | Comments (0)

How to find square root of a binomial quadratic surd

November 12, 2020 – 10:47 pm

Assume $\sqrt{a+ \sqrt{b} + \sqrt{c} + \sqrt{d}}=\sqrt{x} + \sqrt{y} + \sqrt{z}$ ;

Hence, $a+\sqrt{b} + \sqrt{c} + \sqrt{d} = x+y+z+ 2\sqrt{xy} + 2\sqrt{yz}+ 2\sqrt{zx}$

If then, $2\sqrt{xy}=\sqrt{b}$ , $2\sqrt{yz}=\sqrt{c}$ , $2\sqrt{zx}=\sqrt{d}$ ,

And, if simultaneously, the values of x, y, z thus found satisfy $x+y+z=a$ , we shall have obtained the required root.

Example:

Find the square root of $21-4\sqrt{5}+5\sqrt{3}-4\sqrt{15}$ .

Solution:

Clearly, we can’t have anything like

$21--4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=\sqrt{x} + \sqrt{y} +\sqrt{z}$

We will have to try the following options:

$21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=\sqrt{x} - \sqrt{y} - \sqrt{z}$

$21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=\sqrt{x}-\sqrt{y}+\sqrt{z}$

$21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=\sqrt{x}+\sqrt{y}-\sqrt{z}$ .

Only the last option will work as we now show:

So, once again, assume that $\sqrt{21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}}=\sqrt{x}+\sqrt{y}-\sqrt{z}$

Hence, $21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=z+y+z+2\sqrt{xy}-2\sqrt{yz}+2\sqrt{zx}$

Put $2\sqrt{xy}=8\sqrt{3}$ , $2\sqrt{xz}=4\sqrt{15}$ , $2\sqrt{yz}=4\sqrt{5}$ ;

by multiplication, $xyz = 240$ ; that is $\sqrt{xyz}=4\sqrt{15}$ ; so it follows that : $\sqrt{x}=2\sqrt{3}$ , $\sqrt{y}=2$ , $\sqrt{z}=\sqrt{5}$ .

And, since, these values satisfy the equation $x+y+z=21$ , the required root is $2\sqrt{3}+2-\sqrt{5}$ .

That is all, for now,

Regards,

Nalin Pithwa

By Nalin Pithwa | Posted in algebra | Comments (0)

IITJEE Mains or Advanced Maths Doubt Solving Tutorials

November 9, 2020 – 1:17 pm

Any maths questions from any where or any other branded class problems sets.

Contact Nalin Pithwa

By Nalin Pithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Comments (0)

Derivatives: part 3: IITJEE maths tutorial problems for practice

October 28, 2020 – 2:34 pm

Problem 1:

Differential coefficient of $\log[10]{x}$ w.r.t. $\log[x]{10}$ is

(a) $\frac{(\log{x})^{2}}{(\log{10})^{2}}$ (b) $\frac{(\log[x]{10})^{2}}{(\log{10})^{2}}$ (c) $\frac{(\log[10]{x})^{2}}{(\log{10})^{2}}$ (d) $\frac{(\log{10})^{2}}{(\log{x})^{2}}$

Problem 2:

The derivative of an even function is always:

(a) an odd function (b) does not exist (c) an even function (d) can be either even or odd.

Problem 3:

The derivative of $\arcsin{x}$ w.r.t. $\arccos{\sqrt{1-x^{2}}}$ is

(a) $\frac{1}{\sqrt{1-x^{2}}}$ (b) $\arccos{x}$ (c) $1$ (d) $\arctan{(\frac{1}{\sqrt{1-x^{2}}})}$

Problem 4:

If $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$ , then $\frac{dy}{dx}$ is

(a) $\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$ (b) $\sqrt{1-x^{2}}$ (c) $\frac{\sqrt{1-x^{2}}}{\sqrt{1-y^{2}}}$ (d) $\sqrt{1-y^{2}}$

Problem 5:

$\frac{d}{dx} \arcsin{2x\sqrt{1-x^{2}}}$ is equal to

(a) $\frac{2}{\sqrt{1-x^{2}}}$ (b) $\cos{2x}$ (c) $\frac{1}{2\sqrt{1-x^{2}}}$ (d) $\frac{1}{\sqrt{1-x^{2}}}$

Problem 6:

If $y=\arctan{\frac{x}{2}}-\arccos{\frac{x}{2}}$ , then $\frac{dy}{dx}$ is

(a) $\frac{2}{1+x^{2}}$ (b) $\frac{2}{4+x^{2}}$ (c) $\frac{4}{4+x^{2}}$ (d) $0$

Problem 7:

If $y=\arccos{(\frac{\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}})}$ , then $\frac{dy}{dx}$ is equal to:

(a) $\frac{1}{2}$ (b) $\frac{2}{3}$ (c) $3$ (d) $\frac{3}{2}$

Problem 8:

If $y = \arctan{\frac{4x}{1+5x^{2}}} + \arctan{\frac{2+3x}{3-2x}}$ , then $\frac{dy}{dx}$ is

(a) $\frac{1}{1+x^{2}}$ (b) $\frac{5}{1+25x^{2}}$ (c) $1$ (d) $\frac{3}{1+9x^{2}}$

Problem 9:

If $2^{x}+2^{y}=2^{x+y}$ , then $\frac{dy}{dx}$ is equal to

(a) $\frac{2^{x}+2^{y}}{2^{x}-2^{y}}$ (b) $2^{x-y} \times \frac{2^{y}-1}{1-2^{x}}$ (c) $\frac{2^{x}+2^{y}}{1+2^{x+y}}$ (d) $\frac{2^{x+y}-2^{x}}{2^{y}}$

Problem 10:

If $y^{2}=p(x)$ , a polynomial of degree 3, then $2\frac{d}{dx}(y^{3}\frac{d^{2}y}{dx^{2}})$ is equal to

(a) $p^{'''}(x)+p^{'}(x)$ (b) $p^{''}(x).p^{'''}(x)$ (c) $p^{'''}(x).p(x)$ (d) a constant.

Regards,

Nalin Pithwa.

By Nalin Pithwa | Posted in calculus, IITJEE Mains | Comments (0)

Derivatives : part 2: IITJEE Maths : Tutorial problems for practice

October 26, 2020 – 4:22 am

Problem 1:

If $f(a)=2$ , $f^{'}(a)=1$ , $g(a)=-1$ , $g^{'}(a)=2$ , then the value of $\lim_{x \rightarrow a}\frac{g(x)f(a)-g(a)f(x)}{x-a}$ is

(a) -5 (b) $\frac{1}{5}$ (c) 5 (d) 0

Problem 2:

Let $y = \arcsin{(\frac{2x}{1+x^{2}})}$ , $0 < x <1$ and $0 < y < \frac{\pi}{2}$ , then $\frac{dy}{dx}$ is equal to :

(a) $\frac{2}{1+x^{2}}$ (b) $\frac{2x}{1+x^{2}}$ (c) $\frac{-2}{1+x^{2}}$ (d) none

Problem 3:

Let $f(x) = ax^{2}+1$ for $x \leq 1$

and $f(x)= x+a$ for $x \leq 1$ then f is derivable at $x=1$ , if

(a) $a=0$ (b) $a = \frac{1}{2}$ (c) $a=1$ (d) $a=2$

Problem 4:

If $f(x) = ax^{2}+b$ for $x \leq 1$

if $f(x)=b x^{2}+ax+c$ for $x>1$ , where $b \neq 0$ , then $f(x)$ is continuous and differentiable at $x=1$ , if

(a) $c=0, a=2b$ (b) $a=2b, c \in \Re$ (c) $a=b, c=0$ (d) $a=2b, c \neq 0$

Problem 5:

$\lim_{h \rightarrow 0} \frac{\cos^{2}(x+h)- \cos^{2}(x)}{h}$ is equal to

(a) $\cos^{2}(x)$ (b) $-\sin{2x}$ (c) $\sin{x} \cos{x}$ (d) $2\sin{x}$

Problem 6:

$\lim_{h \rightarrow 0} \frac{\sin{\sqrt{x+h}-\sin{\sqrt{x}}}}{h}$ is equal to

(a) $\cos {\sqrt{x}}$ (b) $\frac{1}{2\sin{\sqrt{x}}}$ (c) $\frac{\cos{\sqrt{x}}}{2\sqrt{x}}$ (d) $\sin{\sqrt{x}}$

Problem 7:

$(\arccos{x})^{'}= \frac{-1}{\sqrt{1-x^{2}}}$ where

(a) $-1 < x <1$ (b) $-1 \leq x \leq 1$ (c) $-1 \leq x < 1$ (d) $-1 < x \leq 1$

Problem 8:

$\frac{d}{dx}(\arctan{(\frac{3x-x^{2}}{1-3x^{2}})})$ is equal to

(a) $\frac{3}{1+x^{2}}$ (b) $\frac{3}{1+9x^{2}}$ (c) $\sec^{2}{x}$ (d) $\frac{1}{9+x^{2}}$

Problem 9:

If $x=a\cos^{3}(t)$ and $y=a\sin^{3}(t)$ , then $\frac{dy}{dx}$ is equal to

(a) $\cos{t}$ (b) $\cot{t}$ (c) $cosec{(t)}$ (d) $-\tan{t}$

Problem 10:

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

(a) -1 (b) $\cos{t}$ (c) $cosec{(t)}$ (d) $-\tan{t}$

Regards,

Nalin Pithwa

By Nalin Pithwa | Posted in calculus, IITJEE Mains | Comments (0)

Derivatives: part 1: IITJEE Maths Tutorial Problems Practice

October 26, 2020 – 12:36 am

Problem 1:

If $y=x^{x}$ , $x>0$ , then find $\frac{dy}{dx}$ .

Problem 2:

If $y= x^{x^{x^{\ldots}}}$ , then find the value of $x\frac{dy}{dx}$ .

Problem 3:

Find the derivative of $e^{\ln{x}}$ w.r.t. x.

Problem 4:

Let $f(x) = \log{(x+\sqrt{x^{2}+1})}$ , then find the value of $f^{'}(x)$ .

Problem 5:

If $y= \arctan{\frac{\sqrt{1+x^{2}}-1}{x}}$ , then find the value of $y^{'}(0)$ .

Problem 6:

If $y=t^{2}+t-1$ , then find the value of $\frac{dy}{dx}$ .

Problem 7:

If $x=a(t-\sin{t})$ , $y=a(1+\cos{t})$ , then evaluate $\frac{dy}{dx}$ .

Problem 8:

If $x^{y}=e^{x-y}$ , then evaluate $\frac{dy}{dx}$ .

Problem 9:

If $y= \sec^{-1}{(\frac{x+1}{x-1})} + \arcsin{(\frac{x-1}{x+1})}$ , then evaluate $\frac{dy}{dx}$ .

Problem 10:

If $y = \arctan{(\frac{\sin{x}+\cos{x}}{\cos{x}-\sin{x}})}$ , then find $\frac{dy}{dx}$

Problem 11:

If $\sqrt{x}+\sqrt{y}=4$ , then evaluate $\frac{dy}{dx}$ at $y=1$ .

Problem 12:

If $f(x) = \frac{x-4}{2\sqrt{x}}$ , then evaluate $f^{'}(0)$ .

Problem 13:

If $f^{'}(x) = \sin{\log{x}}$ and $y=f(\frac{2x+3}{3-2x})$ , find $\frac{dy}{dx}$ . One of the given choices is correct:

(a) $\frac{12\cos{(\log{x})}}{x(3-2x)^{2}}$

(b) $\frac{12\sin{\log{(\frac{2x+3}{3-2x})}}}{(3-2x)^{2}}$

(d) none of these

Problem 14:

If $f(0)=0=g(0)$ and $f^{'}(0)=6=g^{'}(0)$ , then $\lim_{x \rightarrow 0} \frac{f(x)}{g(x)}$ is given by:

(a) 1 (b) 0 (c) 12 (d) -1

Regards,

Nalin Pithwa

By Nalin Pithwa | Posted in calculus, IITJEE Mains | Comments (0)

Limits and Continuity: part 11: IITJEE Maths tutorial problems for practice

October 18, 2020 – 8:00 am

Problem 1:

A function $f(x)$ is defined as follows:

$f(x) = \frac{(e^{2x}-1)(1-\cos{x})}{\tan^{2}{(x)}\log{(1+2x)}}$ when $x \neq 0$

$f(0) = \log{a}$ is continuous at $x=0$ .

The value of a should be

(i) $\frac{e}{2}$ (b) $\frac{1}{2e}$ (c) 2 (d) none

Problem 2:

If $f(x) = \frac{e^{(2x)} + e^{(-2x)} -2}{1-\cos{(4x)}}$ when $x \neq 0$ is continuous at $x=0$ , then what is the value of $f(0)$

Problem 3:

Given $f(x) = x + a$ , when $-1 \leq x \leq 0$

and $f(x) = x + b$ when $0 < x \leq 1$

and $f(x) = c -x$ when $1 < x \leq 2$

if f is continuous at $x=0$ and $x=1$ and $f(2)=1$ , then the value of $3a+b-2c=$

(i) 0 (ii) 1 (iii) 2 (iv) 3

Problem 4:

If the function $f(x)$ is continuous on its domain where

$f(x) = x^{2} + ax + b$ for $0 \leq x < 2$

$f(x)=4x-1$ for $2 \leq x < 4$

$f(x)=ax^{2+17b}$ for $4 \leq x \leq 6$

then the quadratic equation whose roots are 2a and 2b is:

(i) $x^{2}+2x-8$ (b) $x^{2}-2x-8=0$ (c) $x^{2}+2x+8$ (d) $x^{2}-2x+8=0$

Problem 5:

The value of c for which the function

$f(x) = \frac{\sin{(x)} + \sin{((a+1)x)}}{x}$ when $x<0$

$f(x) = c$ when $x=0$

$f(x) = \frac{(x+bx^{2})^{\frac{1}{2}}-x^{\frac{1}{2}}}{bx^{\frac{3}{2}}}$

is continuous at $x=0$ is

(i) 1/2 (ii) -1/2 (iii) 2 (iv) -2

Problem 6:

If $f(x) = \frac{\sin{x\pi}}{x-1}+a$ when $x<1$

$f(x) = 2x$ , when $x=1$

$f(x)= \frac{1+\cos{x\pi}}{\pi (1-x)^{2}} + b$ when $x>1$

is continuous at $x=1$ , then a and b have the values:

(i) $3\pi, 3\frac{\pi}{2}$ (ii) $3\pi, \frac{\pi}{2}$ (iii) $\pi, \frac{\pi}{2}$ (iv) $\pi, 3\frac{\pi}{2}$

Problem 7:

If $f(x) = \frac{(\sin{x} - \cos{x})^{2}}{\sqrt{2}-\sin{x}-\cos{x}}$ , when $x \neq \frac{\pi}{4}$ is continuous at $x=\frac{\pi}{4}$ then $f(\frac{\pi}{4})=$