Limit Continuity and Differentiability 7 Question 6
6. Let $f(x)=\begin{array}{cc}-1, & -2 \leq x<0 \ x^{2}-1, & 0 \leq x \leq 2\end{array}$ and $g(x)=|f(x)|+f(|x|)$. Then, in the interval $(-2,2), g$ is
(2019 Main, 11 Jan I)
(a) not differentiable at one point
(b) not differentiable at two points
(c) differentiable at all points
(d) not continuous
Show Answer
Answer:
Correct Answer: 6. (a)
Solution:
- Given equation is
$$ (2 x)^{2 y}=4 \cdot e^{2 x-2 y} $$
On applying ’ $\log _e$ ’ both sides, we get
$$ \begin{aligned} \log _e(2 x)^{2 y} & =\log _e 4+\log _e e^{2 x-2 y} \\ 2 y \log _e(2 x) & =\log _e(2)^{2}+(2 x-2 y) \end{aligned} $$
$\left[\because \log _e n^{m}=m \log _e n\right.$ and $\left.\log _e e^{f(x)}=f(x)\right]$
$\Rightarrow \quad\left(2 \log _e(2 x)+2\right) y=2 x+2 \log _e(2)$
$$ \Rightarrow \quad y=\frac{x+\log _e 2}{1+\log _e(2 x)} $$
On differentiating ’ $y$ ’ w.r.t. ’ $x$ ‘, we get
$$ \begin{aligned} \frac{d y}{d x} & =\frac{\left(1+\log _e(2 x)\right) 1-\left(x+\log _e 2\right) \frac{2}{2 x}}{\left(1+\log _e(2 x)\right)^{2}} \\ & =\frac{1+\log _e(2 x)-1-\frac{1}{x} \log _e 2}{\left(1+\log _e(2 x)\right)^{2}} \end{aligned} $$
So, $\left(1+\log _e(2 x)\right)^{2} \frac{d y}{d x}=\frac{x \log _e(2 x)-\log _e 2}{x}$
