## Monthly Archives: November 2020

### Derivatives: part 5: IITJEE maths tutorial problems for practice

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

### Derivatives: part 4: IITJEE maths tutorial problems for practice

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

### How to find square root of a binomial quadratic surd

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

### IITJEE Mains or Advanced Maths Doubt Solving Tutorials

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

Contact Nalin Pithwa