Mathematics M.C.Q (1 Marks)

$= {\tan ^{ - 1}}(\sqrt x ) - {\tan ^{ - 1}}(x)$

==>$y' = \frac{1}{{1 + x}}.\frac{1}{{2\sqrt x }} - \frac{1}{{1 + {x^2}}} $

$\Rightarrow y'(1) = \frac{1}{2}.\frac{1}{2} - \frac{1}{2} = \frac{{ - 1}}{4}$.

Differentiate with respect to  $x,$

$1 = \frac{{dy}}{{dx}}\sqrt {1 - {y^2}} + y.\frac{1}{{2\sqrt {1 - {y^2}} }}\,.\,( - 2y)\,.\,\frac{{dy}}{{dx}}$

==> $1 = \frac{{dy}}{{dx}}\sqrt {1 - {y^2}} - \frac{{{y^2}}}{{\sqrt {1 - {y^2}} }}\,.\,\frac{{dy}}{{dx}}$

==> $1 = \frac{{dy}}{{dx}}\left[ {\frac{{1 - {y^2} - {y^2}}}{{\sqrt {1 - {y^2}} }}} \right]$

==> $1 = \frac{{dy}}{{dx}}\left[ {\frac{{1 - 2{y^2}}}{{\sqrt {1 - {y^2}} }}} \right]$

$\frac{{dy}}{{dx}} = \frac{{\sqrt {1 - {y^2}} }}{{1 - 2{y^2}}}$.

==> $y' = \frac{{\frac{3}{2}b{x^{7/4}} - \frac{5}{4}(a + b{x^{3/2}})\,{x^{1/4}}}}{{{x^{5/2}}}}$

$\because$ $y' = 0$ at $x = 5$

$\therefore$ $\frac{3}{2}b{x^{7/4}} - \frac{5}{4}(a + b{x^{3/2}}){x^{1/4}} = 0$ at $x = 5$

==> $6b{x^{3/2}} - 5(a + b{x^{3/2}}) = 0$ at $x = 5$

==> $b{x^{3/2}} = 5a$ at $x = 5$ ==> $b{(5)^{3/2}} = 5a$

==> $\frac{a}{b} = \frac{{{5^{3/2}}}}{5}$ 

==> $a:b = \sqrt 5 :1$.

$ \Rightarrow \,\,y = x - 1,$ $x > 0$ and $0 \le x - 1 \le \pi $

$\therefore $ $y = x - 1$, $1 \le x \le \pi + 1$

we have, $1 < \frac{{5\pi }}{4} < \pi + 1$

$\therefore \,\,\,\,y = x - 1,$ $1 \le x \le \pi + 1$ and $\,\,\frac{{5\pi }}{4} \in \,[\,1,\,\,\pi + 1\,]$

${\left. {\frac{{dy}}{{dx}}} \right|_{x = \frac{{5\pi }}{4}}} = {\left. \begin{array}{l}1\\\end{array} \right|_{x = \frac{{5\pi }}{4}}} = 1$.

Then $f'(x) = 2ax + b$ also, $f(1) = f( - 1)$

$a + b + c = a - b + c$ ==>$ b = 0$

$\therefore $ $f'(x) = 2ax$; 

$\therefore $ $f'({a_1}) = 2a{a_1}$

$f'(a{ _2}) = 2a{a_2}$, $f'({a_3}) = 2a{a_3}$

As ${a_1},{a_2},{a_3}$ are in $ A.P.$ 

$f'({a_1}),\,f'({a_2}),\,f'({a_3})$ are in $A.P.$

Put $\sqrt x = \tan \theta \Rightarrow \theta = {\tan ^{ - 1}}\sqrt x $

$\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}\frac{{(\tan \theta (3 - {{\tan }^2}\theta )}}{{1 - 3{{\tan }^2}\theta }}} \right)$

$= \frac{d}{{dx}}\left( {{{\tan }^{ - 1}}\frac{{(3\tan \theta - {{\tan }^3}\theta )}}{{1 - 3{{\tan }^2}\theta }}} \right)$

$= \frac{d}{{dx}}({\tan ^{ - 1}}(\tan 3\theta ) = \frac{d}{{dx}}(3\theta )$

$= \frac{d}{{dx}}(3.{\tan ^{ - 1}}\sqrt x ) = \frac{3}{{2\sqrt x (1 + x)}}$.

$ \Rightarrow \,\,\frac{{dr}}{{d\phi }} = \frac{1}{2}{\left[ {2\phi + {{\cos }^2}\left( {2\phi + \frac{\pi }{4}} \right)} \right]^{ - 1/2}}$

$\left[ {2 - 2 \times 2\sin \,\left( {2\phi + \frac{\pi }{4}} \right) \times \cos \left( {2\phi  + \frac{\pi }{4}} \right)} \right]$

${\left( {\frac{{dr}}{{d\phi }}} \right)_{r = \frac{\pi }{4}}} = \frac{1}{2}{\left[ {\frac{\pi }{2} + {{\cos }^2}\frac{{3\pi }}{4}} \right]^{ - 1/2}} \times 2\left[ {\left( {1 - \sin \left( {\pi + \frac{\pi }{2}} \right)} \right)} \right]$

${\left( {\frac{{dr}}{{d\phi }}} \right)_{r = \frac{\pi }{4}}} = \frac{1}{2}{\left( {\frac{\pi }{2} + \frac{1}{2}} \right)^{ - 1/2}} \times 2\,(1 + 1) = 2 \times {\left( {\frac{2}{{\pi + 1}}} \right)^{1/2}}$.

Let $a = r\sin \theta $ and $b = r\cos \theta $

$\therefore$ $y = {\tan ^{ - 1}}\left[ {\frac{{r\sin (\theta - x)}}{{r\cos (\theta - x)}}} \right]$

$\Rightarrow$ $y = \theta - x$ $\Rightarrow$ $y = {\tan ^{ - 1}}\left( {\frac{a}{b}} \right) - x$

$\Rightarrow$ $\frac{{dy}}{{dx}} = - 1$.

$ \Rightarrow $ $f[f\{ f(x)\} ] = \frac{{ - x}}{{ - x - 1 + x}} = x$

$\therefore $ Derivative of $f[f\{ f(x)\} ] = 1$.

==>$y = \sqrt {\frac{{(1 + x)(1 + x)}}{{(1 - x)(1 + x)}}} = \sqrt {\frac{{{{(1 + x)}^2}}}{{1 - {x^2}}}} $

Differentiating with respect to $x,$ we get

$\frac{{dy}}{{dx}} = \frac{{{{(1 - x)}^{1/2}}\frac{d}{{dx}}{{(1 + x)}^{1/2}} - {{(1 + x)}^{1/2}}\frac{d}{{dx}}{{(1 - x)}^{1/2}}}}{{(1 - x)}}$

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

$x = \frac{2}{3}y$.

Trick : $\log y = \frac{1}{2}\log (1 + x) - \frac{1}{2}\log (1 - x)$

==> $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{2(1 + x)}} + \frac{1}{{2(1 - x)}}$

==> $\frac{{dy}}{{dx}} = \frac{1}{{(1 + x)(1 - x)}} \times \sqrt {\frac{{1 + x}}{{1 - x}}} $

$ = \frac{1}{{{{(1 + x)}^{1/2}}{{(1 - x)}^{3/2}}}}$.

$ = \frac{d}{{dx}}{\tan ^{ - 1}}\left[ {\tan \left( {\frac{\pi }{4} - x} \right)} \right] = - 1$.

==> $ - \sin (x + y)\left( {1 + \frac{{dy}}{{dx}}} \right) = y\cos x + \sin x\frac{{dy}}{{dx}}$

==> $\frac{{dy}}{{dx}} = - \frac{{y\cos x + \sin (x + y)}}{{\sin (x + y) + \sin x}}$.

==> $1 = \frac{{\cos y.\frac{{dy}}{{dx}}.\sin (a + y) - \sin y\cos (a + y)\frac{{dy}}{{dx}}}}{{{{\sin }^2}(a + y)}}$

$ = \frac{{\frac{{dy}}{{dx}}.\sin (a + y - y)}}{{{{\sin }^2}(a + y)}} $

$\Rightarrow \frac{{dy}}{{dx}} = \frac{{{{\sin }^2}(a + y)}}{{\sin a}}$.

Differentiating w.r.t. $x$ of $y,$ we get

==> ${\sec ^2}(x + y)\left( {1 + \frac{{dy}}{{dx}}} \right) + {\sec ^2}(x - y)\left( {1 - \frac{{dy}}{{dx}}} \right) = 0$

==> $\frac{{dy}}{{dx}} = \frac{{{{\sec }^2}(x + y) + {{\sec }^2}(x - y)}}{{{{\sec }^2}(x - y) - {{\sec }^2}(x + y)}}$.

$ \Rightarrow \sec x\frac{{dy}}{{dx}} + y\sec x\tan x + {\sec ^2}x + 2xy + {x^2}\frac{{dy}}{{dx}} = 0$

==> $\frac{{dy}}{{dx}} = - \frac{{2xy + {{\sec }^2}x + y\sec x\tan x}}{{{x^2} + \sec x}}$.

Differentiating both sides,

$\cos (xy)\frac{d}{{dx}}(xy) + x\left\{ { - \frac{1}{{{y^2}}}} \right\}\frac{{dy}}{{dx}} + \frac{1}{y} = 2x - \frac{{dy}}{{dx}}$

==> $[x\cos (xy) - \frac{x}{{{y^2}}} + 1]\frac{{dy}}{{dx}} = 2x - \frac{1}{y} - y\cos (xy)$

==> $\frac{{dy}}{{dx}} = \left[ {\frac{{2x{y^2} - y - {y^3}\cos (xy)}}{{x{y^2}\cos (xy) - x + {y^2}}}} \right]$.

Find $\frac{{dx}}{{dy}}$ and then $\frac{{dy}}{{dx}}$.

$\frac{{dy}}{{dx}} = - \frac{{\partial f/\partial x}}{{\partial f/\partial y}} $

$= - \frac{{\cos (x + y) - \frac{1}{{x + y}}}}{{\cos (x + y) - \frac{1}{{x + y}}}} = - 1$.

$ = {2^x}{.2^y}\frac{{dy}}{{dx}}.\log 2 + {2^y}{.2^x}\log 2$

==> ${2^x} + {2^y}\frac{{dy}}{{dx}} = {2^{x + y}}\frac{{dy}}{{dx}} + {2^{x + y}}$

==> $\frac{{dy}}{{dx}} = \frac{{{2^{x + y}} - {2^x}}}{{{2^y} - {2^{x + y}}}} = {2^{x - y}}\frac{{{2^y} - 1}}{{1 - {2^x}}}$.

${2^x}(\log 2) + {2^y}(\log 2)\frac{{dy}}{{dx}}$ $= {2^{(x + y)}}.(\log 2)\,\left( {1 + \frac{{dy}}{{dx}}} \right)$

==> ${2^x} + {2^y}\frac{{dy}}{{dx}} = {2^{x + y}} + {2^{x + y}}\left( {\frac{{dy}}{{dx}}} \right)$

==> $\frac{{dy}}{{dx}}({2^y} - {2^{x + y}}) = {2^{x + y}} - {2^x}$

==>$\frac{{dy}}{{dx}} = \frac{{{2^{x + y}} - {2^x}}}{{{2^y} - {2^{x + y}}}}$.

$\therefore $ ${\left( {\frac{{dy}}{{dx}}} \right)_{x = y = 1}} = \frac{{{2^2} - 2}}{{2 - {2^2}}} = \frac{2}{{ - 2}} = - 1.$

$\Rightarrow m\log x + n\log y = \log 2 + (m + n)\log (x + y)$

Differentiating both sides w.r.t. $x,$

$\frac{m}{x} + \frac{n}{y}\,\frac{{dy}}{{dx}} = \frac{{m + n}}{{x + y}}\left[ {1 + \frac{{dy}}{{dx}}} \right]$

==> $\frac{{dy}}{{dx}} = \frac{y}{x}$.

$ = {\log _{10}}x + \frac{{{{\log }_e}10}}{{\log x}} + 1 + 1$

$ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{x}{\log _{10}}e - \frac{{{{\log }_e}10}}{{x{{({{\log }_e}x)}^2}}}$.

Differentiating w.r.t.  $ x $ of  $y,$  we get

$\frac{{dy}}{{dx}} = {\left( {\frac{{1 - x}}{{1 + x}}} \right)^{1/4}}\frac{1}{4}{\left( {\frac{{1 + x}}{{1 - x}}} \right)^{ - 3/4}}\left[ {\frac{{(1 - x) + (1 + x)}}{{{{(1 - x)}^2}}}} \right] - \frac{1}{2}.\frac{1}{{1 + {x^2}}}$

$ = \frac{1}{2}\left( {\frac{{1 - x}}{{1 + x}}} \right)\frac{1}{{{{(1 - x)}^2}}} - \frac{1}{{2(1 + {x^2})}}$

$ = \frac{1}{2}.\frac{1}{{(1 - {x^2})}} - \frac{1}{2}\frac{1}{{(1 + {x^2})}} = \frac{{{x^2}}}{{1 - {x^4}}}$.

${1 \over {x{{\log }_e}x}} \cdot \frac{1}{{{\log }_{e}}7}$  = $\frac{{{\log }_{7}}e}{x{{\log }_{ex}}}$

Differentiating w.r.t.  $x$  of $y$ ,we get

$\frac{{dy}}{{dx}} = \frac{{1 - \sqrt x }}{{1 + \sqrt x }}\left[ {\frac{{(1 - \sqrt x )\frac{1}{{2\sqrt x }} + (1 + \sqrt x )\frac{1}{{2\sqrt x }}}}{{{{(1 - \sqrt x )}^2}}}} \right]$

$ = \frac{1}{{2(1 - x)\sqrt x }}[1 - \sqrt x + 1 + \sqrt x ] = \frac{1}{{\sqrt x (1 - x)}}$.

$ = \frac{1}{2}.\frac{1}{{\sin \left( {\frac{\pi }{4} + \frac{x}{2}} \right)\cos \left( {\frac{\pi }{4} + \frac{x}{2}} \right)}} = \frac{1}{{\sin \left( {\frac{\pi }{2} + x} \right)}} = \frac{1}{{\cos x}} = \sec x$.

==> $\log y = 2x + \log \cos x - \log x - \log \sin x$

$\frac{1}{y}\frac{{dy}}{{dx}} = 2 + \left( {\frac{{ - \sin x}}{{\cos x}}} \right) - \frac{1}{x} - \frac{{\cos x}}{{\sin x}}$

==> $\frac{{dy}}{{dx}} = {e^{2x}}\left[ {\frac{2}{x}\cot x - \frac{1}{x} - \frac{1}{{{x^2}}}\cot x - \frac{{{{\cot }^2}x}}{x}} \right]$

$ = \frac{{{e^{2x}}}}{{{x^2}}}[(2x - 1)\cot x - x\,{\rm{cose}}{{\rm{c}}^2}x]$.

$2y\frac{{dy}}{{dx}} = \frac{{(1 - {e^x}){e^x} + (1 + {e^x}){e^x}}}{{{{(1 - {e^x})}^2}}} = \frac{{2{e^x}}}{{{{(1 - {e^x})}^2}}}$

$\therefore \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{{(1 - {e^x})}^2}}}\sqrt {\left[ {\frac{{1 - {e^x}}}{{1 + {e^x}}}} \right]\left[ {\frac{{1 - {e^x}}}{{1 - {e^x}}}} \right]} $

$ = \frac{{{e^x}}}{{(1 - {e^x})\sqrt {1 - {e^{2x}}} }}$.

==> $\frac{{dy}}{{dx}} = \frac{{(\log \sin x)\left( {\frac{{{{\sec }^2}x}}{{\tan x}}} \right) - (\log \tan x)(\cot x)}}{{{{(\log \sin x)}^2}}}$

==> ${\left( {\frac{{dy}}{{dx}}} \right)_{\pi /4}} = \frac{{ - 4}}{{\log 2}}$      (On simplification).

$ = [{\log _e}{\log _e}x + {\log _e}({\log _2}e)]{\log _2}e$

$\therefore \frac{{dy}}{{dx}} = {\log _2}e.\frac{1}{{x{{\log }_e}x}}$.

Differentiating both sides w.r.t. $x,$  we get

$\frac{{dy}}{{dx}}\sqrt {{x^2} + 1} + y.\frac{1}{{2\sqrt {{x^2} + 1} }}.2x = \frac{1}{{\sqrt {{x^2} + 1} - x}} \times \left\{ {\frac{1}{2}\frac{{2x}}{{\sqrt {{x^2} + 1} }} - 1} \right\}$

==> $({x^2} + 1)\frac{{dy}}{{dx}} + xy = \sqrt {{x^2} + 1} .\frac{{ - 1}}{{\sqrt {{x^2} + 1} }}$

==> $({x^2} + 1)\frac{{dy}}{{dx}} + xy + 1 = 0$.

$ = {({\log _{\cot x}}{(\cot x)^{ - 1}})^2} \Rightarrow f'(x) = 0$ .

Differentiating w.r.t.  $x,$ we get $f'(x) = 6x{e^{{x^2}}}$;

$\therefore f(0) = 3$ and $f'(0) = 0$

==>$f'(x) - 2xf(x) + \frac{1}{3}f(0) - f'(0)$

$ = 6x{e^{{x^2}}} - 6x{e^{{x^2}}} + \frac{1}{3}(3) - 0 = 1$ .

$\therefore \frac{{dy}}{{dx}} = \frac{{\cot x.\log \cos x + (\log \sin x)\tan x}}{{{{(\log \cos x)}^2}}}$.

==> $y = x + \frac{3}{4}\left[ {\log (x + 2) - \log (x - 2)} \right]$

$\frac{{dy}}{{dx}} = 1 + \frac{3}{4}\left[ {\frac{1}{{x + 2}} - \frac{1}{{x - 2}}} \right] = 1 - \frac{3}{{{x^2} - 4}}$

$\frac{{dy}}{{dx}} = \frac{{{x^2} - 7}}{{{x^2} - 4}}$.

$ = 2{\tan ^{ - 1}}(\log x)$

==> $f'(x) = 2.\frac{1}{{1 + {{(\log x)}^2}}}.\frac{1}{x}.$

${\rm{\,Therefore\,\, }}f'(e) = \frac{1}{e}$.

$= \log {e^x} + \log {\left( {\frac{{x - 2}}{{x + 2}}} \right)^{3/4}}$

==> $y = x + \frac{3}{4}\,[\log (x - 2) - \log (x + 2)]$

==> $\frac{{dy}}{{dx}} = 1 + \frac{3}{4}\,\left[ {\frac{1}{{x - 2}} - \frac{1}{{x + 2}}} \right] $

$= 1 + \frac{3}{{({x^2} - 4)}}$

==> $\frac{{dy}}{{dx}} = \frac{{{x^2} - 1}}{{{x^2} - 4}}$.

==> $\cos y\frac{{dy}}{{dx}} + {e^{ - x\cos y}}\left\{ {( - x)\,\left( { - \sin y\frac{{dy}}{{dx}}} \right) + \cos y( - 1)} \right\}\, = 0$

==> $\cos y\frac{{dy}}{{dx}} + x\sin y\,\,{e^{ - x\cos y}}\frac{{dy}}{{dx}} - \cos y{e^{ - x\cos y}} = 0$

==> $\frac{{dy}}{{dx}} = \frac{{\cos y\,\,{e^{ - x\cos y}}}}{{\cos y + x\sin y\,\,{e^{ - x\cos y}}}}$

==> ${\left( {\frac{{dy}}{{dx}}} \right)_{(1,\,\pi )}} = \frac{{\cos \pi \,\,{e^{ - \cos \pi }}}}{{\cos \pi + \sin \pi \,\,{e^{ - \cos \pi }}}}$

$= \frac{{( - 1)e}}{{ - 1 + 0}} = e$.

==> $f(x) = {\log _5}\left( {\frac{{{{\log }_e}x}}{{{{\log }_e}7}}} \right)$

==> $f(x) = {\log _5}{\log _e}x - {\log _5}{\log _e}7$

==> $f(x) = \frac{{{{\log }_e}{{\log }_e}x}}{{{{\log }_e}5}} - {\log _5}{\log _e}7$

Now, $f'(x) = \frac{1}{{x{{\log }_e}x\log 5}} - 0$

==> $f'(x) = \frac{1}{{x{{\log }_e}x\frac{{{{\log }_e}5}}{{{{\log }_e}7}}{{\log }_e}7}}$

$ = \frac{1}{{x(\ln 5)(\ln 7)({{\log }_7}x)}}$.

==> $\frac{1}{y}\frac{{dy}}{{dx}} = 1 + \frac{{dy}}{{dx}}$

==> $ \frac{{dy}}{{dx}} = \frac{y}{1-y} $.

$ \Rightarrow 2y\frac{{dy}}{{dx}} = \frac{1}{x} + \frac{{dy}}{{dx}} $

$\Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{x(2y - 1)}}$.

==> $y = {(\sin x)^y} \Rightarrow {\log _e}y = y\log \sin x$

==> $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{{dy}}{{dx}}[\log \sin x + y\cot x]$

$\therefore \frac{{dy}}{{dx}} = \frac{{{y^2}\cot x}}{{1 - y\log \sin x}}$.

$ \Rightarrow y = \sqrt {\sin x + y} \Rightarrow {y^2} = \sin x + y$

On differentiating both sides, we get

$2y\frac{{dy}}{{dx}} = \cos x + \frac{{dy}}{{dx}} $

$\Rightarrow \frac{{dy}}{{dx}}(2y - 1) = \cos x$.

$ \Rightarrow 2y\frac{{dy}}{{dx}} = y.2x + {x^2}\frac{{dy}}{{dx}}$

==> $\frac{{dy}}{{dx}} = \frac{{2xy}}{{2y - {x^2}}}$.