Limit Continuity and Differentiability 7 Question 3
3. Let $f(x)=15-|x-10| ; x \in \mathbf{R}$. Then, the set of all values of $x$, at which the function, $g(x)=f(f(x))$ is not differentiable, is
(2019 Main, 9 April I)
(a) ${5,10,15,20}$
(b) ${5,10,15}$
(c) ${10}$
(d) ${10,15}$
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Answer:
Correct Answer: 3. (a)
Solution:
Key Idea Differentiating the given equation twice w.r.t. ’ $x$ ‘.
Given equation is
$$ e^{y}+x y=e $$
On differentiating both sides w.r.t. $x$, we get
$$ \begin{aligned} e^{y} \frac{d y}{d x}+x \frac{d y}{d x}+y & =0 \\ \frac{d y}{d x} & =-\frac{y}{e^{y}+x} \end{aligned} $$
Again differentiating Eq. (ii) w.r.t. ’ $x$ ‘, we get
$$ e^{y} \frac{d^{2} y}{d x^{2}}+e^{y} \frac{d y}{d x}+x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+\frac{d y}{d x}=0 $$
Now, on putting $x=0$ in Eq. (i), we get
$$ \begin{aligned} e^{y} & =e^{1} \\ y & =1 \end{aligned} $$
On putting $x=0, y=1$ in Eq. (iii), we get
$$ \frac{d y}{d x}=-\frac{1}{e+0}=-\frac{1}{e} $$
Now, on putting $x=0, y=1$ and $\frac{d y}{d x}=-\frac{1}{e}$ in
Eq. (iv), we get
$$ \begin{aligned} & e^{1} \frac{d^{2} y}{d x^{2}}+e^{1}-\frac{1}{e}+0 \frac{d^{2} y}{d x^{2}}+-\frac{1}{e}+-\frac{1}{e}=0 \\ & \left.\Rightarrow \quad \frac{d^{2} y}{d x^{2}}\right| _{(0,1)}=\frac{1}{e^{2}} \\ & \text { So, } \quad \frac{d y}{d x}, \frac{d^{2} y}{d x^{2}} \text { at }(0,1) \text { is }-\frac{1}{e}, \frac{1}{e^{2}} \end{aligned} $$
