QJEEAdv08Ma4227: JEE Advanced 2008 Mathematics - Class 12 JEEAdv08Ma4227 Solution: Let , and let be the left hand derivative of at . If , then

Given, g(x) = {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}},0 < x < 2,m \ne 0,n > 0,m,n are integer.

Left hand derivative (L.H.D.) of $|x - 1|$ at $x = 1$ is P.

As $$|x-1|=\{\begin{array}{cc}x-1,& x\ge 1\\ -(1-x),& x<1\end{array}$$

$\therefore$ L.H.D. at $x = 1$ will be $-1$ .

[ $\therefore$ L.H.D. at $x = 1$ is $\mathop {\lim }\limits_{h \to 0} {{f(1 - h) - f(1)} \over { - h}}$ ]

$p = - 1$

Also, $\mathop {\lim }\limits_{x \to {1^ + }} g(x) = p = - 1$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} {{{{(1 + h - 1)}^n}} \over {\log {{\cos }^m}(1 + h - 1)}} = - 1$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} {{{h^n}} \over {\log {{\cos }^m}h}} = - 1$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} {{{h^n}} \over {m\log \cos \,h}} = - 1$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} {{n\,.\,{h^{n - 1}}} \over {m\log \cos \,h}} = - 1$ [using L. Hospital rule]

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {{{ - n} \over m}} \right)\,.\,{{{h^{n - 2}}} \over {\left( {{{\tan \,h} \over h}} \right)}} = - 1$

$\Rightarrow \left( {{n \over m}} \right)\mathop {\lim }\limits_{h \to 0} {{{h^{n - 2}}} \over {\left( {{{\tan \,h} \over h}} \right)}} = 1$

$n - 2 = 0$ and ${n \over m} = 1 \Rightarrow m = n = 2$

or

[f(x) = |x - 1| = - x + 1,x &lt; 1f'(x) = - 1

So, L.H.D. at $x = - 1$ is $-1$

$\therefore$ $P = - 1$

Now, $\mathop {\lim }\limits_{x \to {1^ + }} {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}} = p$ .... (i)

Let $x - 1 = t$ in L.H.S. of eq. (i)

$\mathop {\lim }\limits_{t \to } {{{t^n}} \over {m\log \cos \,t}},\left( {{0 \over 0}\,\mathrm{form}} \right)$ [using L' Hospital rule]

We get, $\mathop {\lim }\limits_{t \to 0} {{n{t^{n - 1}}} \over {m\tan \,t}} = {{ - n} \over m}\mathop {\lim }\limits_{t \to 0} {{{t^{n - 1}}} \over {\tan \,t}}\left( {{0 \over 0}\,\mathrm{form}} \right)$

Again using L' Hospital rule,

${{ - n} \over m}\mathop {\lim }\limits_{t \to 0} {{(n - 1){t^{n - 2}}} \over {{{\sec }^2}t}} = P$

$\Rightarrow {{ + n} \over m}\mathop {\lim }\limits_{t \to 0} {{(n - 1){t^{n - 2}}} \over {{{\sec }^2}t}} = - 1$

For non-zero answer

$n - 2 = 0 \Rightarrow n = 2$ Also $m = n = 2$ .