Tag: maths

Questions Related to maths

If $f\left( x \right) =\sqrt { { x }^{ 2 }-2x+1 } $, then

  1. $f^{ ' }\left( x \right) =1,\forall x$

  2. $f^{ ' }\left( x \right) =1, \forall x\ge 1$

  3. $f^{ ' }\left( x \right) =1, \forall x\le 1$

  4. $f^{ ' }\left( x \right) =1,if\quad x>1\quad and\quad f^{ ' }\left( x \right) =-1\quad if\quad x<1$


Correct Option: D

Derivative of $(\sin x)^x + \sin^{-1} \sqrt{x}$ with respect to $x$ is

  1. $(x \cot x + \log \sin x) + \dfrac{1}{2\sqrt{x - x^2}}$

  2. $(x \cot x + \log \sin x) + \dfrac{1}{\sqrt{x - x^2}}$

  3. $(\sin x)^x (x \cot x + \log \,x) + \dfrac{1}{\sqrt{x - x^2}}$

  4. $(\sin x)^x (x \cot x + \log \sin x) + \dfrac{1}{2\sqrt{x - x^2}}$


Correct Option: D
Explanation:

Let $y=(\sin x)^x$

$\Rightarrow \log y=x \log (\sin x)$
Now differentiating both sides with respect to $x$ 
$\dfrac{1}{y}\dfrac{dy}{dx}=x\cot x+\log \sin x$
or, $\dfrac{dy}{dx}=(\sin x )^x{x\cot x+\log \sin x}$........(1).
Again let $z=(\sin x)^x+\sin^{-1}\sqrt{x}$
Now differentiating both sides with respect to $x$.
$\dfrac{dz}{dx}=\dfrac{dy}{dx}+\dfrac{1}{\sqrt{1-x}}.\dfrac{1}{2\sqrt{x}}$
$\dfrac{dz}{dx}=(\sin x)^x{x\cot x+\log \sin x}+\dfrac{1}{2\sqrt{x-x^2}}$ [Using (1)]

Let f(x) be a differentiable function satisfying $f(x+y)=f(x)+f(y)\forall x, y \in R$ and $f(0)=1$ then $\displaystyle\lim _{x\rightarrow 0}\dfrac{2^{f(\tan^2x)}-2^{f(\sin^2x)}}{x^3f(\sin x)}$ equals to?

  1. $\dfrac{1}{2} ln2$

  2. $ln 2$

  3. $\dfrac{1}{4}ln 2$

  4. $\dfrac{1}{8} ln2$


Correct Option: A

If $t={ \sin {  }  }^{ -1 }{ 2 }^{ s }$ Then $\dfrac { ds }{ dt }$ is equal to

  1. $\dfrac { \log { 2 } }{ \sqrt { 1-t^{ 2 } } }$

  2. $\dfrac { \sin { t } }{ \log { 2 } }$

  3. $\dfrac { \cot { t } }{ \log { 2 } }$

  4. None of these


Correct Option: D

If $u=e^{x}(xcosy-ysiny)$ then $\frac{d^{2}y}{dx^{2}}+\frac{d^{2}u}{dy^{2}}=0$.

  1. True

  2. False


Correct Option: A

If $U=tan^{-1}(\dfrac{x^3+y^3}{x+y})$ , then $x\dfrac{du}{dx}+y\dfrac{du}{dy}=sinu$.

  1. True

  2. False


Correct Option: B

If $y=\sqrt{x}-\dfrac{1}{\sqrt{x}}$, then $2x\dfrac{dy}{dx}+y$=

  1. $\sqrt{x}$

  2. $2\sqrt{x}$

  3. $3\sqrt{x}$

  4. $\dfrac{\sqrt{x}}{2}$


Correct Option: B
Explanation:

$y=\sqrt x-\dfrac{1}{\sqrt x}\ \dfrac{dy}{dx}=\dfrac{d}{dx}(x)^{\dfrac{1}{2}}-\dfrac{d}{dx}(\dfrac{1}{\sqrt x})=\dfrac{1}{2\sqrt x}-(-\dfrac{1}{2})x^{-\dfrac{3}{2}}=\dfrac{1}{2\sqrt x}+\dfrac{x^{-\dfrac{3}{2}}}{2}\ \dfrac{2x dy}{dx}+y=2x[\dfrac{1}{2\sqrt x}+\dfrac{x^{-\dfrac{3}{2}}}{2}]+\sqrt x-\dfrac{1}{\sqrt x}\ \quad =\sqrt x+\dfrac{1}{\sqrt x}+\sqrt x-\dfrac{1}{\sqrt x}\ \quad=2\sqrt x$


If $x\sqrt {1+y}+y\sqrt {1+x}=0$ then  $\dfrac {dy}{dx}=\dfrac {1}{(1+x)^{2}}$

  1. True

  2. False


Correct Option: A

Let $f(x)$ be a function continuous on $[1, 2]$ and differentiable on $(1, 2)$ satisfying $f(1)=2, f(2)=3$ and $f'(x)\ge 1\forall x\in (1, 2)$. Define $g(x)=\displaystyle \int _{1}^{x}{f(t)dt}\forall x\in [1, 2]$ then the greatest value of  $g(x)$ on $[1, 2]$ is-

  1. $3$

  2. $5$

  3. $\dfrac{5}{2}$

  4. $\dfrac{3}{2}$


Correct Option: C