QJEEAdv16Ma55e0: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Ma55e0 Solution: Let f: R and g : R R be twice differentiable functions such…

Let f : R $\to$ (0, $\infty$ ) and g : R $\to$ R.

f(x) > 0 $\forall$ x $\in$ R

It is given that f'(2) = 0, g(2) = 0, f''(2) $\ne$ 0 and g'(2) $\ne$ 0.

It is also given that

$\mathop {\lim }\limits_{x \to 2} {{f(x)g(x)} \over {f'(x)g'(x)}} = 1$ $\left( {{0 \over 0}} \right)$

Applying L' Hospital rule, we get

$\mathop {\lim }\limits_{x \to 2} {{f'(x)g(x) + g'(x)f(x)} \over {f''(x)g'(x) + f'(x)g''(x)}} = 1$

For finite limit, we get

${{f'(2)g(2) + g'(2)f(2)} \over {f''(2)g'(2) + f'(2)g''(2)}} = 1$

${{g'(2)f(2)} \over {f''(2)g'(2)}} = 1$

{{f(2)} \over {f''(2)}} = 1 \Rightarrow f''(2) = f(2) > 0 and f'(2) = 0 which means that f(x) has local minima at x = 2.

Hence, option (A) is correct.

$f(2) - f''(2) = 0$

Therefore, we can say that $f(x) - f''(x) = 0$ has at least one solution in x $\in$ R.

Hence, option (D) is correct.