Limit Continuity and Differentiability 7 Question 32
33. Let $f:[a, b] \rightarrow[1, \infty)$ be a continuous function and
$$ g: R \rightarrow R \text { be defined as } g(x)=\begin{array}{ccc} 0, & \text { if } & x<a \\ \int _a^{x} f(t) d t, & \text { if } & a \leq x \leq b . \\ \int _a^{b} f(t) d t, & \text { if } & x>b \end{array} $$
Then,
(2013)
(a) $g(x)$ is continuous but not differentiable at $a$
(b) $g(x)$ is differentiable on $R$
(c) $g(x)$ is continuous but not differentiable at $b$
(d) $g(x)$ is continuous and differentiable at either $a$ or $b$ but not both
$$ -x-\frac{\pi}{2}, \quad x \leq-\frac{\pi}{2} $$
Show Answer
Answer:
Correct Answer: 33. $x=0$
Solution:
- Given, $(a+b x) e^{y / x}=x \Rightarrow y=x \log \frac{x}{a+b x}$
$\Rightarrow \quad y=x[\log (x)-\log (a+b x)]$
On differentiating both sides, we get
$$ \begin{aligned} \frac{d y}{d x} & =x \frac{1}{x}-\frac{b}{a+b x}+1[\log (x)-\log (a+b x)] \\ \Rightarrow x \frac{d y}{d x} & =x^{2} \frac{a}{x(a+b x)}+y \\ \Rightarrow x y _1 & =\frac{a x}{a+b x}+y \end{aligned} $$
Again, differentiating both sides, we get
$$ \begin{array}{rlrl} \Rightarrow & & x y _2+y _1 & =a \frac{(a+b x) \cdot 1-x \cdot b}{(a+b x)^{2}}+y _1 \\ \Rightarrow & x^{3} y _2 & =\frac{a^{2} x^{2}}{(a+b x)^{2}} \\ \Rightarrow & & x^{3} y _2 & =\frac{a x}{(a+b x)} \\ \Rightarrow & x^{3} y _2 & =\left(x y _1-y\right)^{2} \\ \Rightarrow & x^{3} \frac{d^{2} y}{d x^{2}} & =x \frac{d y}{d x}-y \end{array} $$
[from Eq. (ii)]
