Limit Continuity and Differentiability 7 Question 36
37. Let $h(x)=\min {x, x^{2} }$ for every real number of $x$, then
(a) $h$ is continuous for all $x$
(1998, 2M)
(b) $h$ is differentiable for all $x$
(c) $h^{\prime}(x)=1, \forall x>1$
(d) $h$ is not differentiable at two values of $x$
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Answer:
Correct Answer: 37. $(1,2)$
Solution:
- Here, $\lim _{x \rightarrow 1} \frac{F(x)}{G(x)}=\frac{1}{14}$
$\Rightarrow \quad \lim _{x \rightarrow 1} \frac{F^{\prime}(x)}{G^{\prime}(x)}=\frac{1}{14} \quad$ [using L’Hospital’s rule]…(i)
As $\quad F(x)=\int _{-1}^{x} f(t) d t \Rightarrow F^{\prime}(x)=f(x)$
and $\quad G(x)=\int _{-1}^{x} t|f{f(t)}| d t$
$\Rightarrow \quad G^{\prime}(x)=x|f{f(x)}|$
$\therefore \quad \lim _{x \rightarrow 1} \frac{F(x)}{G(x)}=\lim _{x \rightarrow 1} \frac{F^{\prime}(x)}{G^{\prime}(x)}=\lim _{x \rightarrow 1} \frac{f(x)}{x|f{f(x)}|}$
$=\frac{f(1)}{1|f{f(1)}|}=\frac{1 / 2}{|f(1 / 2)|}$
Given, $\quad \lim _{x \rightarrow 1} \frac{F(x)}{G(x)}=\frac{1}{14}$
$\therefore \quad \frac{\frac{1}{2}}{\left|f \frac{1}{2}\right|}=\frac{1}{14} \Rightarrow\left|f \frac{1}{2}\right|=7$ Download Chapter Test http://tinyurl.com/y6dl84lx
