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

$$ \begin{aligned} & e^{1} \frac{d^{2} y}{d x^{2}}+e^{1}-\frac{1}{e}+0 \frac{d^{2} y}{d x^{2}}+-\frac{1}{e}+-\frac{1}{e}=0 \\ & \left.\Rightarrow \quad \frac{d^{2} y}{d x^{2}}\right| _{(0,1)}=\frac{1}{e^{2}} \\ & \text { So, } \quad \frac{d y}{d x}, \frac{d^{2} y}{d x^{2}} \text { at }(0,1) \text { is }-\frac{1}{e}, \frac{1}{e^{2}} \end{aligned} $$



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