$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad k \quad, \quad x=0$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1} ,\,\,\, x>0$
is continuous at $x=0$, then $\frac{1}{a}+\frac{1}{b}+\frac{4}{k}$ is equal to :
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1} \cdot \frac{\sqrt{x^{2}+1}+1}{\sqrt{x^{2}+1}+1}$ (Rationalisation) $\lim _{x \rightarrow 0^{+}}-\frac{2 \sin ^{2} x}{x^{2}} \cdot\left(\sqrt{x^{2}+1}+1\right)=-4$
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{1}{x} \ell n\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right)$
$\lim _{x \rightarrow 0^{-}} \frac{\ln \left(1+\frac{x}{a}\right)}{\left(\frac{x}{a}\right) \cdot a}+\frac{\ell n\left(1-\frac{x}{b}\right)}{\left(-\frac{x}{b}\right) \cdot b}$
$=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}$
So $\frac{1}{a}+\frac{1}{b}=-4=k$
$\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{4}{k}=-4-1=-5$
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$(A)$ $P(E)=\frac{4}{5}, P(F)=\frac{3}{5}$
$(B)$ $P(E)=\frac{1}{5}, P(F)=\frac{2}{5}$
$(C)$ $P(E)=\frac{2}{5}, P(F)=\frac{1}{5}$
$(D)$ $P(E)=\frac{3}{5}, P(F)=\frac{4}{5}$
Statement $-1 :$ $f'\left( 4 \right) = 0$
Statement $-2 :$ $ f $ is continuous in $ [2,5] $ , differentiable in $ (2,5) $ and $f(2)=f(5).$