JEE Advance - Mathematics (2010 - Paper 2 Offline - No. 11)
Let $$f$$ be a real-valued function defined on the interval $$(-1, 1)$$ such that
$${e^{ - x}}f\left( x \right) = 2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,\,dt,} $$ for all $$x \in \left( { - 1,1} \right)$$,
and let $${f^{ - 1}}$$ be the inverse function of $$f$$. Then $$\left( {{f^{ - 1}}} \right)'\left( 2 \right)$$ is equal to
$${e^{ - x}}f\left( x \right) = 2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,\,dt,} $$ for all $$x \in \left( { - 1,1} \right)$$,
and let $${f^{ - 1}}$$ be the inverse function of $$f$$. Then $$\left( {{f^{ - 1}}} \right)'\left( 2 \right)$$ is equal to
$$1$$
$${{1 \over 3}}$$
$${{1 \over 2}}$$
$${{1 \over e}}$$
Explanation
We have,
$${e^{ - x}}f(x) = 2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,dt\,x \in ( - 1,1)} $$
On differentiating w.r.t. x, we get
$${e^{ - x}}(f'(x) - f(x)) = \sqrt {{x^4} + 1} $$
$$ \Rightarrow f'(x) = f(x) + \sqrt {{x^4} + 1} \,{e^x}$$
$$\because$$ $${f^{ - 1}}$$ is the inverse of f
$$\therefore$$ $${f^{ - 1}}(f(x)) = x$$
$$ \Rightarrow {f^{ - 1'}}(f(x))f'(x) = 1$$
$$ \Rightarrow {f^{ - 1'}}(f(x)) = {1 \over {f'(x)}}$$
$$ \Rightarrow {f^{ - 1'}}(f(x)) = {1 \over {f(x) + \sqrt {{x^4} + 1} \,{e^x}}}$$
At $$x = 0$$, $$f(x) = 2$$
$${f^{ - 1'}}(2) = {1 \over {2 + 1}} = {1 \over 3}$$
Comments (0)
