Maths M.C.Q (1 Marks)

$\because \lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}+x^{4}}-1}{x}\left(\frac{0}{0}\right.$ from $)$

$\lim _{x \rightarrow 0} \frac{\left(1+x^{2}+x^{4}\right)-1}{x\left(\sqrt{1+x^{2}+x^{4}}+1\right.}$

$\lim _{x \rightarrow 0} \frac{x\left(1+x^{2}\right)}{\left(\sqrt{1+x^{2}+x^{4}}+1\right)}=0$

$\lim _{x \rightarrow 0} \frac{e^{\frac{\sqrt{1+x^{2}+x^{4}-1}}{x}-1}}{\left(\frac{\sqrt{1+x^{2}+x^{4}}-1}{x}\right)}=1$

roots are $2 -1$

$\Rightarrow \lim _{x \rightarrow 2^{+}} \frac{\sqrt{1-\cos \left(x^{2}-x-2\right)}}{(x-2)}$

$=\lim _{x \rightarrow 2^{+}} \frac{\sqrt{2 \sin ^{2} \frac{\left(x^{2}-x-2\right)}{2}}}{(x-2)}$

$=\lim _{x \rightarrow 2^{+}} \frac{\sqrt{2} \sin \left(\frac{(x-2)(x+1)}{2}\right)}{(x-2)}$

$=\frac{3}{\sqrt{2}}$

So, $\sqrt {1 + {y^4}}  = 1 + \frac{{{y^4}}}{2}$

$\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt {2 + \frac{{{y^4}}}{2}}  - \sqrt 2 }}{{{y^4}}}$

$ = \frac{{\sqrt 2 \left( {1 + \frac{{{y^4}}}{8} - 1} \right)}}{{{y^4}}} = \frac{{\sqrt 2 }}{8} = \frac{1}{{4\sqrt 2 }}$

$x \to {0^ - }$

$\left. \begin{array}{l}
\left[ x \right] =  - 1\\
\left| x \right| =  - x
\end{array} \right\} \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} \frac{{x\left( { - x - 1} \right)\sin \left( { - 1} \right)}}{{ - x}} =  - \sin 1$

$\mathop {\lim }\limits_{x \to 1 + } \frac{{\left( {1 - x + \sin \left( {1 - x} \right)} \right)\sin \left( { - \frac{\pi }{2}} \right)}}{{\left( {x - 1} \right)\left( { - 1} \right)}}$

$\mathop {\lim }\limits_{x \to 1 + } \frac{{ - \left( {x - 1} \right)\sin \left( {x - 1} \right)}}{{\left( {x - 1} \right)}} =  - 1 + 1 = 0$

$1\left( \pi  \right).{\left( 1 \right)^2} + \mathop {Lt}\limits_{x \to 0} {\left( {1 - \frac{{\sin x\left[ x \right]}}{{x\left[ x \right]}}.\frac{{x\left[ x \right]}}{{\left| x \right|}}} \right)^2}\,\,\,\,\,\,\,\,......\left( i \right)$

$\mathop {Lt}\limits_{x \to {0^ - }} \frac{{x\left[ x \right]}}{{\left| x \right|}} = \frac{x}{{ - x}}\left( { - 1} \right) = 1$

$\mathop {Lt}\limits_{x \to {0^ + }} \frac{{x\left[ x \right]}}{{\left| x \right|}} = \frac{{x\left( 0 \right)}}{x} = 0$

Put in equation $(i)$

$\therefore $ Limit does not exist.

$ = \frac{{4x}}{{\sin 4x}}.\frac{{\cos 4x}}{{{{\cos }^2}2x}}{\cos ^2}x$

$ \Rightarrow 1\,\,\,\,\,as\,\,\,x \to 0$

$\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{3{{\cot }^2}x\left( { - \cos \,e{c^2}x} \right) - {{\sec }^2}x}}{{ - \sin \left( {x + \frac{\pi }{4}} \right)}} = 8$

$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{2\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}x} \right)}}{{\sqrt {1 - x} \left( {\sqrt \pi   + \sqrt {2{{\sin }^{ - 1}}x} } \right)}}$

$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{2{{\cos }^{ - 1}}x}}{{\sqrt {1 - x} }}.\frac{1}{{2\sqrt \pi  }}$

Assuming $x = \cos \theta $

$\mathop {\lim }\limits_{\theta  \to {0^ + }} \frac{{2\theta }}{{\sqrt 2 \sin \left( {\frac{\theta }{2}} \right)}}.\frac{1}{{2\sqrt \pi  }} = \sqrt {\frac{2}{\pi }} $

$ = \frac{{{{\left( 1 \right)}^2}.\left( {2\sqrt 2 } \right)}}{{\frac{1}{2}}} = 4\sqrt 2 $

$k = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{f\left( {3 + x} \right) - f\left( {2x} \right) - f\left( 3 \right)\left( {f\left( 2 \right)} \right)}}{{x\left( {1 + f\left( {2 - x} \right) - f\left( 2 \right)} \right)}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{f'\left( {3 + x} \right) + f'\left( {2 - x} \right)}}{{\left( {1 + f\left( {2 - x} \right) - f\left( 2 \right)} \right) - xf'\left( {2 - x} \right)}}$

$ = 0$

$ \Rightarrow {e^k} = 1$

$\frac{0}{0}$ from, so we use $L-$ Hopital Rule

$ = \mathop {\lim }\limits_{x \to 2} \frac{{f'\left( x \right).2f\left( x \right)}}{1}$

$ = f'\left( 2 \right).2f\left( 2 \right)$

$ = 12f'\left( 2 \right)$

$\mathop {\lim }\limits_{x \to k} \frac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}} = \frac{{{k^2} + {k^2} + {k^2}}}{{2k}}.........\left( 2 \right)$

$(1)=(2)$

$ \Rightarrow k = \frac{8}{3}$

$ = \int\limits_0^1 {{{\left( {1 + x} \right)}^{1/3}}} dx = \frac{3}{4}\left( {{2^{4/3}} - 1} \right)$

$1 - a + b = 0\,\,\,\,\,\,\,.......\left( i \right)$

$2 - a = 5\,\,\,\,\,\,\,.....\left( {ii} \right)$

$ \Rightarrow a + b =  - 7$

$f\left( x \right)$ attains maximum value when $\left| {x - 2} \right| = 0 \Rightarrow x = 2 = \alpha $

$g\left( x \right) = \left| {x + 1} \right|$

$g\left( x \right)$ attins minimum value of $x =  - 1 = \beta $

$\mathop {\lim }\limits_{x \to  - \alpha \beta } \frac{{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)}}{{{x^2} - 6x + 8}}$

$ = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 2} \right)\left( {x - 4} \right)}}$

$ = \frac{{\left( {2 - 1} \right)\left( {2 - 3} \right)}}{{\left( {2 - 4} \right)}} = \frac{1}{2}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{x + 2\sin x}}{{{x^2} + 2\sin x + 1 - {{\sin }^2}x - x + 1}}$ $\left( {\sqrt {{x^2} + 2\sin x + 1}  + \sqrt {{{\sin }^2}x - x + 1} } \right)$

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{x + 2\sin x}}{{{x^2} + 2\sin x - {{\sin }^2}x + x}}.\left( 2 \right)$

Applying $L'H$ Rule

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{2.\left( {1 + 2\cos x} \right)}}{{2x + 2\cos x - 2\sin x\cos x + 1}}$

$ = \frac{{2\left( 3 \right)}}{{2 + 1}} = 2$

Hence the correct answer is option $(A)$.

$ = \mathop {\lim }\limits_{x \to {0^ + }} x\left( {\frac{{1 + 2 + 3 + .... + 15}}{x}} \right) - \left( {\left\{ {\frac{1}{x}} \right\} + \left\{ {\frac{2}{x}} \right\} + .. + \left\{ {\frac{{15}}{x}} \right\}} \right)$

$\because $ $0 \le \left\{ {\frac{r}{x}} \right\} < 1\,\,\,\,\,\,\,\,\, \Rightarrow 0 \le x\left\{ {\frac{r}{x}} \right\} < x$

$\therefore \mathop {\lim }\limits_{x \to {0^ + }} x\left( {\frac{{1 + 2 + 3 + .... + 15}}{x}} \right) = \frac{{15 \times 16}}{2} = 120$

$L = \,\,\,\mathop {\lim }\limits_{x \to 0} \frac{{\left( {x\tan 2x - 2x\tan \,x} \right)}}{{{{\left( {1 - \cos \,2x} \right)}^2}}} = \mathop {\lim }\limits_{x \to 0} K$ (say)

$ \Rightarrow K = \frac{{x\left[ {\frac{{2\tan \,x}}{{1 - {{\left( {\tan \,x} \right)}^2}}}} \right] - 2x\,\tan x}}{{{{\left( {1 - \left( {1 - 2{{\sin }^2}x} \right)} \right)}^2}}}$

$ = \frac{{2x\,\tan x - \left[ {2x\,\tan x - 2x\,{{\tan }^3}x} \right]}}{{4{{\sin }^4}\,x \times \left( {1 - {{\tan }^2}x} \right)}}$

$ = \frac{{2x\,{{\tan }^3}x}}{{4{{\sin }^4}\,x \times \left( {1 - {{\tan }^2}x} \right)}}$

$ = \frac{{2x\,{{\tan }^3}x}}{{4{{\sin }^4}\,x \times \left( {\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}$

$ = \frac{{2x\,\frac{{{{\sin }^3}x}}{{{{\cos }^3}x}}}}{{4{{\sin }^4}\,x \times \left( {\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}$

$ \Rightarrow K = \frac{x}{{2\sin \,x \times \left( {{{\cos }^2}x - {{\sin }^2}x} \right)\cos x}}$

$L = \mathop {\lim }\limits_{x \to 0} \frac{x}{{2\sin \,x}} \times \mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{x}{{2\sin \,x}} \times \mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos 0\left( {{{\cos }^2}0 - {{\sin }^2}0} \right)}}$

$ = \frac{1}{2}$

Here $'L'$ is in the indeterminate from i.e.,$\frac{0}{0}$

$\therefore $ usinh the $L'$ Hosoital rule we get:

$L = \,\,\,\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{3}{{\left( {27 + x} \right)}^{\frac{{ - 2}}{3}}}}}{{ - \frac{2}{3}{{\left( {27 + x} \right)}^{\frac{{ - 1}}{3}}}}} = \frac{{\frac{1}{3} \times {{\left( {27} \right)}^{\frac{{ - 2}}{3}}}}}{{\frac{{ - 2}}{3} \times {{\left( {27} \right)}^{\frac{{ - 1}}{3}}}}} =  - \frac{1}{6}$

Put $\frac{\pi }{2} - x = t \Rightarrow $ as $x \to \frac{\pi }{2} \Rightarrow t \to 0$

$ = \mathop {\lim }\limits_{t \to 0} \frac{{\cot \left( {\frac{\pi }{2} - t} \right)\left( {1 - \sin \left( {\frac{\pi }{2} - t} \right)} \right)}}{{8{t^3}}}$

$ = \mathop {\lim }\limits_{t \to 0} \frac{{\tan \,t\left( {1 - \cos \,t} \right)}}{{8{t^3}}}$

$ = \mathop {\lim }\limits_{t \to 0} \frac{{\tan \,t}}{{8t}}.\frac{{1 - \cos t}}{{{t^2}}}$

$ = \frac{1}{8}.1.\frac{1}{2} = \frac{1}{{16}}$

Rationalise 

$ \Rightarrow A = \,\,\,\mathop {\lim }\limits_{x \to 3} \frac{{\left( {3x - 9} \right) \times \left( {2x - 4 + \sqrt 2 } \right)}}{{\left\{ {\left( {2x - 4 - 2} \right)} \right\} \times \left( {\sqrt {3x}  + 3} \right)}}$

        $ = \,\,\,\mathop {\lim }\limits_{x \to 3} \frac{{3\left( {x - 3} \right)}}{{2\left( {x - 3} \right)}} \times \frac{{\sqrt {2x - 4}  + \sqrt 2 }}{{\left( {\sqrt {3x}  + 3} \right)}}$

         $ = \frac{3}{2} \times \frac{{2\sqrt 2 }}{6} = \frac{1}{{\sqrt 2 }}$

$\,\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {\frac{1}{n}} \right)}^a} + {{\left( {\frac{2}{n}} \right)}^a} + ..... + {{\left( {\frac{n}{n}} \right)}^a}}}{{{{\left( {n + 1} \right)}^{a - 1}}\left[ {{n^a} + \frac{{n\left( {n + 1} \right)}}{2}} \right]}}.$

$ = \,\frac{{\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^n {{{\left( {\frac{r}{n}} \right)}^a}} }}{{{{\left( {1 + \frac{1}{n}} \right)}^{a - 1}}\left[ {a + \frac{1}{2}\left( {1 + \frac{1}{n}} \right)} \right]}}\, = \frac{1}{{60}}$

$ = \frac{{\int\limits_0^1 {{x^n}dx} }}{{\left( {a + \frac{1}{2}} \right)}} = \frac{1}{{60}} = \frac{{\frac{1}{{a + 1}}}}{{a + \frac{1}{2}}} = \frac{1}{{60}}$

$ \Rightarrow \frac{{\frac{1}{{a + 1}}}}{{\left( {a + \frac{1}{2}} \right)}} = \frac{1}{{60}} \Rightarrow \left( {a + 1} \right)\left( {2a + 1} \right) = 120$

$ \Rightarrow 2{a^2} + 3a - 119 = 0$

$ \Rightarrow 2{a^2} + 17a - 14a - 119 = 0$

$ \Rightarrow \left( {a - 7} \right)\left( {2a + 17} \right) = 0$

                                               $a = 7, - \frac{{17}}{2}$

$\log p = \frac{1}{2}$

$\, = e\left[ {\mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}} - 1} \right)2x} \right]$

        $\, = e\left[ {\mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}} - 1} \right)2x} \right]$

$\therefore 2a = 3 \Rightarrow a = 3/2$

$\, = \,\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {{{\sin }^2}x} \right)}^2}}}{{2x\left( {x + \frac{{{x^3}}}{3} + \frac{{2{x^5}}}{{15}} + ....} \right) - x\left( {2x + \frac{{{2^3}{x^3}}}{3} + \frac{{{2^5}{x^5}}}{{15}} + ....} \right)}}$

             $\, = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\left( {x + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - ....} \right)}^4}}}{{{x^4}\left( {\frac{2}{3} - \frac{8}{5}} \right) + {x^6}\left( {\frac{4}{{15}} - \frac{{64}}{{15}}} \right)}}$

             $ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\left( {1 + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - ....} \right)}^4}}}{{ - 2 + {x^2}\left( { - \frac{{60}}{{15}}} \right) + ......}}$

        (dividing numerator & denominator by ${{x^4}}$)

          $=2$

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{{\sin x}}{e^{{x^2}}} + \frac{1}{2}} \right)\frac{1}{{\cos x}} = 1 + \frac{1}{2} = \frac{3}{2}$

$ = \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi \left( {1 - {{\sin }^2}x} \right)} \right)}}{{{x^2}}}$

$ = \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\sin }^2}x} \right)}}{{{x^2}}}$

$ = \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\sin }^2}x} \right)}}{{{x^2}}}$

$ = \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\sin }^2}x} \right)}}{{\pi {{\sin }^2}x}} \cdot \frac{{\pi {{\sin }^2}x}}{{{x^2}}}$

We know,  $\mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {f\left( x \right)} \right)}}{{f\left( x \right)}} = 1$

So, our limits becomes,

$ = 1 \cdot \pi \left( 1 \right) = \pi $

$\therefore \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}} = \pi $

$\,\mathop {\lim }\limits_{x \to 0} f\left( {\frac{{1 - \cos 3x}}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{x^2}}}{{1 - \cos 3x}}} \right)$

$\, = \,\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{x^2}}}{{2{{\sin }^2}\frac{{3x}}{2}}}} \right)$

$\, = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\frac{9}{4}.{x^2}.\frac{4}{9}}}{{{{\sin }^2}\frac{{3x}}{2}}}} \right)$

$ = \frac{4}{{9 \times 2}}\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{\frac{{{{\sin }^2}\frac{{3x}}{2}}}{{{{\left( {\frac{{3x}}{2}} \right)}^2}}}}}} \right)$

$ = \frac{2}{9}\frac{{\mathop {\lim }\limits_{x \to 0} }}{{\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\frac{{3x}}{2}}}{{{{\left( {\frac{{3x}}{2}} \right)}^2}}}}}$

$\left\{ {\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1} \right\}$

$ = \frac{2}{9}\left[ {\frac{1}{1}} \right] = \frac{2}{9}$

$\, \Rightarrow \,\mathop {\lim }\limits_{x \to 2} \frac{{\tan \left( {x - 2} \right)\left\{ {{x^2} + kx - 2x - 2k} \right\}}}{{{{\left( {x - 2} \right)}^2}}} = 5$

$\, \Rightarrow \mathop {\lim }\limits_{x \to 2} \frac{{\tan \left( {x - 2} \right)\left\{ {x\left( {x - 2} \right) + k\left( {x - 2} \right)} \right\}}}{{\left( {x - 2} \right) \times \left( {x - 2} \right)}} = 5$

$\, \Rightarrow \mathop {\lim }\limits_{x \to 2} \left( {\frac{{\tan \left( {x - 2} \right)}}{{\left( {x - 2} \right)}}} \right) \times \mathop {\lim }\limits_{x \to 2} \left( {\frac{{\left( {k + 2} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)}}} \right) = 5$

$ \Rightarrow 1 \times \mathop {\lim }\limits_{x \to 2} \left( {k + x} \right) = 5$

{$\because $ $\mathop {\lim }\limits_{h \to 0} \frac{{\tan \,h}}{h} = 1$}

or $k + 2 = 5$

$ \Rightarrow \boxed{k = 3}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x\left( {3 + \cos x} \right)}}{{x \times \frac{{\tan 4x}}{{4x}} \times 4x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x}}{{{x^2}}} \times \mathop {\lim }\limits_{x \to 0} \frac{{\left( {3 + \cos x} \right)}}{4} \times \frac{1}{{\mathop {\lim }\limits_{x \to 0} \frac{{\tan 4x}}{{4x}}}}$

$ = 2 \times \frac{4}{4} \times 1$              ($\because $ $\mathop {\lim }\limits_{\theta  \to 0} \frac{{\sin \theta }}{\theta } = 1$ and $\mathop {\lim }\limits_{\theta  \to 0} \frac{{\tan \theta }}{\theta } = 1$)

$=2$

$\, = \,\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x}} \right)\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{2x + 1}}} \right) - {{\tan }^{ - 1}}\left( 1 \right)} \right]$

$ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x}} \right).{\tan ^{ - 1}}\left( {\frac{{\frac{{x + 1}}{{2x + 1}} - 1}}{{1 + \frac{{x + 1}}{{2x + 1}}}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \frac{1}{x}.{\tan ^{ - 1}}\left( {\frac{{ - x}}{{3x + 2}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{{\tan }^{ - 1}}\left( {\frac{x}{{3x + 2}}} \right)}}{{\frac{x}{{3x + 2}}}} \times \frac{1}{{3x + 2}}} \right] =  - \frac{1}{2}$

$\lim _{x \rightarrow \infty}\left(\frac{\log _0 x}{\log _0(x+1)}\right)^\beta \cdot \frac{\left(\log _0 x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0$

$\lim _{x \rightarrow \infty} \frac{\left(\log _8 x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0 \quad \text { Put } \log _e x=t$

$\lim _{t \rightarrow \infty} \frac{t^{\alpha-\beta}}{\left(e^t\right)^{\alpha \beta+2}}=0$

As we know $\lim _{ x \rightarrow \infty} \frac{ x }{ e ^{ x }}=0$

$\alpha \beta+2>0 \Rightarrow \alpha \beta>-2$

$\because \int_{x^2}^x \sqrt{\frac{1-t}{t}} d t=\sin ^{-1} \sqrt{x}+\sqrt{x} \sqrt{1-x}-\sin ^{-1} x-x \sqrt{1-x^2}$

$\Rightarrow \lim _{x \rightarrow 0}\left(\frac{\sin ^{-1} \sqrt{x}+\sqrt{x} \sqrt{1-x}-\sin ^{-1} x-x \sqrt{1-x^2}}{\sqrt{x}} \leq f(x) g(x) \leq \frac{2 \sqrt{x}}{\sqrt{x}}\right)$

$\Rightarrow 2 \leq \lim _{x \rightarrow 0} f(x) g(x) \leq 2$

$\Rightarrow \lim _{x \rightarrow 0} f(x) g(x)=2$

use expansion

$\beta=\lim _{x \rightarrow 0} \frac{\left(1+x^3\right)-\left(1-\frac{x^3}{3}\right)}{x^3}+\lim _{x \rightarrow 0} \frac{\left(\left(1-\frac{x^2}{2}\right)-1\right)}{x^2} \frac{\sin x}{x}$

$\beta=\lim _{x \rightarrow 0} \frac{4 x^3}{3 x^3}+\lim _{x \rightarrow 0} \frac{-x^2}{2 x^2}$

$\beta=\frac{4}{3}-\frac{1}{2}=\frac{5}{6}$

$6 \beta=5$

$\therefore$ Using Lopital rule,

$=2 \lim _{x \rightarrow a^{+}} \frac{\left(\frac{1}{\sqrt{x}-\sqrt{\alpha}}\right) \cdot \frac{1}{2 \sqrt{x}}}{\left(\frac{1}{e^{\sqrt{x}}-e^{\sqrt{a}}}\right) \cdot e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}}$

$=\frac{2}{e^{\sqrt{a}}} \lim _{x \rightarrow a^{+}} \frac{\left(e^{\sqrt{x}}-e^{\sqrt{a}}\right)}{(\sqrt{x}-\sqrt{\alpha})}\left(\frac{0}{0}\right)$

$=\frac{2}{e^{\sqrt{a}}} \lim _{x \rightarrow a^{+}} \frac{\left(e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}-0\right)}{\left(\frac{1}{2 \sqrt{x}}-0\right)}=2$

$\lim _{x \rightarrow a^{+}} f(g(x))=\lim _{x \rightarrow a^{+}} f(2)$

$=f(2)=\sin \frac{\pi}{6}=\frac{1}{2}$

$=0.50$

Now $\int_0^{\pi / 3} g(x) d x=\int_0^{\pi / 3} f(x) d x-3 \int_0^{\pi / 3} \cos 3 x d x=0$

Hence $g ( x )=0$ has a root in $\left(0, \frac{\pi}{3}\right)$

$(B)$ Let $h ( x )=f( x )-3 \sin 3 x +\frac{6}{\pi}$

Now $\int_0^{\pi / 3} h(x) d x=\int_0^{\pi / 3} f(x) d x-3 \int_0^{\pi / 3} \sin 3 x d x+\int_0^{\pi / 3} \frac{6}{\pi} d x$ $=0-2+2=0$

Hence $h(x)=0$ has a root in $\left(0, \frac{\pi}{3}\right)$

$(C)$ $\lim _{x \rightarrow 0} \frac{x \int_0^x f(t) d t}{1- e ^{x^2}}=\lim _{x \rightarrow 0} \underbrace{\left.\frac{x^2}{1-e^{x^2}}\right)}_{-1} \underbrace{\frac{\int_0^x f(t) d t}{x}}_{\text {Apply }}$

$=-1 \lim _{x \rightarrow 0} \frac{f( x )}{1}=-1$

$(D)$ $\lim _{ x \rightarrow 0} \frac{(\sin x ) \int_0 f( t ) dt }{ x ^2}$

$=\lim _{x \rightarrow 0} \underbrace{\left(\frac{\sin x}{x}\right)}_1\underbrace{\frac{\int_0^x f(t) d t}{x}}_{\text {Apply }}$

$=1 \lim _{x \rightarrow 0} \frac{f(x)}{1}=1$

Ans. $A,B,C$

$\frac{e^{\left(\frac{\ln (1-x)}{x}\right)}-\frac{1}{e}}{x^2}$

$=\lim _{x-0^{+}} \frac{1}{e} \frac{e^{\left(1+\frac{(n(1-x)}{x}\right)}-1}{x^{ a }}$

$=\frac{1}{e} \lim _{ x -0^{+}} \frac{1+\frac{\ell \operatorname{n}(1-x)}{ x }}{ x ^{ a }}$

$=\frac{1}{e} \lim _{x-0^{+}} \frac{\ln (1-x)+x}{x^{(a+1)}}$

$=\frac{1}{e} \lim _{x \rightarrow 0^{+}} \frac{\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\ldots \ldots . .\right)+x}{x^{a+1}}$

Thus, $a=1$

$=\frac{1}{ n } \sum_{ r =1}^{ n } \log \left(1+\frac{ r }{ n }\right)$

$\lim _{ n \rightarrow \infty} \ell n y _{ n }=\lim _{ n \rightarrow \infty} \frac{1}{n}$ $\sum_{ r =1}^{ n } \log \left(1+\frac{ r }{ n }\right)$

$\ell n\left(\lim _{n \rightarrow \infty} y_n\right)=\int_0^1 \log (1+x) d x$

$\ell n ( L )=[\log (1+ x ) \cdot x ]_0^1-\int_0^1 \frac{ x }{1+ x } dx$

$=\log 2-\left[\left. x \right|_0 ^1-\ell n \mid 1+ x \|_0^1\right]$

$=\log 2-1+[\ell n 2-0]$

$=\ell n \left(\frac{4}{ e }\right)$

$L =\frac{4}{ e }$

${[ L ]=1}$

$\Rightarrow \lim _{t \rightarrow x} \frac{f(x) \cos (t)-f^{\prime}(t) \sin (x)}{1}=\sin ^2(x)$

$\Rightarrow f(x) \cos x-f^{\prime}(x) \sin (x)=\sin ^2(x)$

$\Rightarrow \frac{d}{d x}\left(\frac{f(x)}{\sin x}\right)=-1$

$\Rightarrow f(x)=-x \sin (x) \quad(\because C=0)$

For $f(x)=-x \sin (x)$, option $(B), (C), (D)$ are correct.

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} \frac{-h(2+h)}{h} \cos \frac{1}{h} \text { does not exist }$

$\Rightarrow g(x)=\left.f(t) \operatorname{cosec}(t)\right|_x ^{\pi / 2}=3-\frac{f(x)}{\sin x}$

$\Rightarrow \lim _{x \rightarrow 0} g(x)=\lim _{x \rightarrow 0}\left(3-\frac{f(x)}{\sin x}\right)$

$=3-f^{\prime}(0)=2$

$\lim _{x \rightarrow 0} \frac{x^2\left(\beta x-\frac{\beta^3 x^3}{3!}+\ldots\right)}{a x-\left(x-\frac{x^3}{3!}+\ldots\right)}=1$

$\lim _{x \rightarrow 0} \frac{x^3\left(\beta-\frac{\beta^3 x^2}{3!}+\ldots\right)}{(\alpha-1) x+\frac{x^3}{3!}-\ldots}=1$

$\lim _{x \rightarrow 0} \frac{x^2\left(\beta-\frac{\beta^3 x^2}{3!}+\ldots\right)}{(\alpha-1)+\frac{x^2}{3!}-\ldots}=1$

$\Rightarrow a-1=0 \Rightarrow a=1$

$\text { At } a=1$

$\lim _{x \rightarrow 0} \frac{x^2\left(\beta-\frac{\beta^3 x^2}{3!}+\cdots\right)}{\frac{x^2}{3!} \ldots}=1$

$\Rightarrow \beta \times 3!=1 \Rightarrow \beta=\frac{1}{6}$

$\therefore 6(\alpha+\beta)=6\left(1+\frac{1}{6}\right)=7$

$f(x)=\lim _{n \rightarrow \infty}\left\{\frac{n^n(x+n)\left(x+\frac{n}{2}\right) \ldots\left(x+\frac{n}{2}\right)}{n!\left(x^2+n^2\right)\left(x^2+\frac{n^2}{4}\right) \ldots\left(x^2+\frac{n^2}{n^2}\right)}\right\}$

$\Rightarrow \log f(x)=\lim _{n \rightarrow \infty} \frac{x}{n} \log \left\{\prod_{r=1}^n \frac{\left(x+\frac{n}{r}\right)}{\left(x^2+\frac{n^2}{r^2}\right)} \cdot \frac{n}{r}\right\}$

$\Rightarrow \log f(x)=\lim _{n \rightarrow \infty} \frac{x}{2} \sum_{r=1}^n \log \left\{\left(\frac{x+\frac{n}{r}}{x^2+\frac{n^2}{r^2}}\right) \cdot \frac{1}{\frac{r}{n}}\right\}$

$\Rightarrow \log f(x)=\lim _{n \rightarrow \infty} \frac{x}{n} \sum_{r=1}^n \log \left\{\frac{1+\frac{r}{n} x}{1+\left(\frac{r}{n} x\right)^2}\right\}$

$\Rightarrow \log f(x)=x \lim _{n \rightarrow \infty} \sum_{r=1}^n \log \left\{\frac{1+\frac{r}{n} x}{1+\left(\frac{r}{n} x\right)^2}\right\} \frac{1}{n}$

$\Rightarrow \log f(x)=x \int_0^1 \log \left\{\frac{1+t x}{1+(t x)^2}\right\} d t$

$\Rightarrow \log f(x)=\int_0^x \log \left(\frac{1+u}{1+u^2}\right) d u \text {, where } u=t tx$

$\Rightarrow \frac{f^{\prime}(x)}{f(x)}=\log \left(\frac{1+x}{1+x^2}\right)$

We observe that $f(x)>0$ for all $a>0$.

Also,

$\log \left(\frac{1+x}{1+x^2}\right) > 0 \Leftrightarrow \frac{1+x}{1+x^2} > 1 \Leftrightarrow x>x^2 \Leftrightarrow 0$

$ < x < 1$

$\therefore \frac{f^{\prime}(x)}{f(x)}>0 \text { for } 0

$ < 0 \text { for } x > 1$

$\Rightarrow f(x)$ is increasing on $(0,1)$ and decreasing on $(1, \infty)$.

$\Rightarrow f\left(\frac{1}{2}\right) \leq f(1) \text { and } f\left(\frac{1}{3}\right) \leq f\left(\frac{2}{3}\right)$

So option$(b)$ is correct and option $(a)$ is incorrect.

From$(ii)$, we obtain

$f^{\prime}(2) < 0$. So,option $(c)$ is correct.

From$(i)$, we obtain

$\frac{f^{\prime}(3)}{f(3)}-\frac{f^{\prime}(2)}{f(2)}=\frac{\log (4)}{10}-\frac{\log (3)}{5}=\log \left(\frac{2}{3}\right) < 0$

$\Rightarrow \frac{f^{\prime}(3)}{f(3)} < \frac{f^{\prime}(2)}{f(2)} \text {. So, option (d) is not correct. }$

$f(x)=\sin \left(\frac{1}{3} g(g(x))\right)$

$\Rightarrow g(g(x)) \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \forall x \in R$

Also, $g(g(g(x))) \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \forall x \in R$

Hence, $f(x)$ and $f(g(x)) \in\left[-\frac{1}{2}, \frac{1}{2}\right]$

$\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim _{x \rightarrow 0} \frac{\sin \left(\frac{1}{3} g(g(x))\right)}{\frac{1}{3} g(g(x))} \cdot \frac{\frac{1}{3} g(g(x))}{g(x)}$

$\lim _{\alpha \rightarrow 0} \frac{e\left(e^{\left(\cos (\alpha)^n-1\right)}-1\right)\left(\cos \alpha^n-1\right)}{\left(\cos \left(\alpha^n\right)-1\right) \alpha^m \alpha^{2 n}} \alpha^{2 n}=-\frac{e}{2} \text { if and only if } 2 n-m=0$

$f(-x)=-f(x)$

$\text { Given } f(1)=\frac{1}{2}$

$\lim _{x \rightarrow 1} \frac{F(x)}{G(x)}=\lim _{x \rightarrow 1} \frac{\frac{F(x)-F(1)}{x-1}}{\frac{G(x)-G(1)}{x-1}}=\frac{f(1)}{|f(f(1))|}=\frac{1}{14}$

$\Rightarrow \frac{1 / 2}{|f(1 / 2)|}=\frac{1}{14}$

$\Rightarrow f\left(\frac{1}{2}\right)=7 .$

$\Rightarrow \quad \lim _{x \rightarrow 1}\left\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}^{x+\sqrt{x}}=\frac{1}{4}$

Hence $\lim _{x \rightarrow 1}\left\{\frac{-a x+\sin (x-1)+a}{(x-1)+\sin (x-1)}\right\}^{1+\sqrt{x}}=\frac{1}{4}$

put $x=1+h$,

$\lim _{h \rightarrow 0}\left\{\frac{-a h+\sin h}{h+\sinh }\right\}^{1+\sqrt{1+h}}=\frac{1}{4}$

or $\quad \frac{- a +1}{2}=\frac{1}{2} \quad$ or $-\frac{1}{2} \Rightarrow \quad a =0$ or $2$

But at $a = 2 , \frac{-a h+\sinh }{h+\sinh }$ tends to negative value

So correct Answer is $a =0$

However $a = 2$ may be accepted if this is not considered