Question
The largest value of the non-negative integer a for which $\lim _{x \rightarrow 1}\left\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}^{\frac{1-x}{1 \sqrt{x}}}=\frac{1}{4}$ is
$\Rightarrow \quad \lim _{x \rightarrow 1}\left\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}^{x+\sqrt{x}}=\frac{1}{4}$
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
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$f(x)=\sin x-e^{x} \,\,\,\, \text { if } x \leq 0$
$\quad\quad\quad a+[-x] \,\,\,\, \text { if } 0\,<\,x\,<\,1$
$\quad\quad\quad 2 x-b \,\,\,\,\,\,\,\, \text { if } \geq 1$
where $[\mathrm{x}]$ is the greatest integer less than or equal to $\mathrm{x}$. If $\mathrm{f}$ is continuous on $\mathrm{R}$, then $(\mathrm{a}+\mathrm{b})$ is equal to: