Q.
Let g : $R → R$ be a differentiable function with $g(0) = 0, g'(0) = 0$ and $g'(1) ≠ 0$. Let
$f\left(x\right) = \begin{pmatrix}\frac{x}{\left|x\right|}g\left(x\right),&x \ne 0\\ 0,&x = 0\end{pmatrix}$and $h\left(x\right) = e^{|x|}$ for all $x ∈ R$. Let $\left(f \,o \,h\right) \left(x\right) $ denote $f\left(h\left(x\right)\right)$ and $\left(h\, o\, f\right)\left(x\right)$ denote $h\left(f\left(x\right)\right)$. Then which of the following is (are) true ?
JEE AdvancedJEE Advanced 2015
Solution:
$g(0) = 0, \quad g′(0) = 0, \quad g′(1) ≠ 0$
$f(x) = \begin{cases} g(x) & \quad \text{ } x > 0 \\ 0 & \quad \text{ } x = 0 \\ - g(x) & \quad \text{ } x < 0 \end{cases} $
$f'(x) = \begin{cases} g'(x) & \quad \text{ } x > 0 \\ 0 & \quad \text{ } x = 0 \\ - g'(x) & \quad \text{ } x < 0 \end{cases} $
f(x) is diff. at $x = 0$, as g′(0) = 0
So option (A) is true.
$h(x) = e^{|x|}$
not diff. at $x = 0$
So option (B) is not true
$f \,o\, h(x) = f(h(x))\, as\, h(x) > 0$
So $f(h(x)) = g(h(x))$
$\frac{df\left(h\left(x\right)\right)}{dx} = g′\left(h\left(x\right)\right) h′\left(x\right)$ since $g′\left(1\right) \ne 0$
at $x = 0, f(h(x))$ is not differentiable
So option (C) is not true
$h \,o\, f (x) = h(f(x))$
$= e^{|f(x)|}$
Differentiable at $x = 0$
So option (D) is true.
