Limit Continuity And Differentiability Ques 43
- $f(x)=\left\{\begin{array}{cc}b \sin ^{-1}\left(\frac{x+c}{2}\right), & -\frac{1}{2}<x<0 \\ \frac{1}{2}, & x=0 \\ \frac{e^{a x^{\prime 2}}-1}{x}, & 0<x<\frac{1}{2}\end{array}\right.$
If $f(x)$ is differentiable at $x=0$ and $|c|<\frac{1}{2}$, then find the value of $a$ and prove that $64 b^2=\left(4-c^2\right)$.
(2004, 4M)
Show Answer
Answer:
Correct Answer: 43.$(a=1)$
Solution: Since, $f(x)$ is differentiable at $x=0$.
$\Rightarrow \quad$ It is continuous at $x=0$.
i.e. $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0)$
Here, $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} \frac{e^{a h / 2}-1}{h}=\lim _{h \rightarrow 0} \frac{e^{a / / 2}-1}{a \frac{h}{2}} \cdot \frac{a}{2}=\frac{a}{2}$
Also, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} b \sin ^{-1}\left(\frac{c-h}{2}\right)=b \sin ^{-1} \frac{c}{2}$
$\therefore \quad b \sin ^{-1} \frac{c}{2}=\frac{a}{2}=\frac{1}{2}$
$\Rightarrow \quad a=1$
Also, it is differentiable at $x=0$
$R f^{\prime}\left(0^{+}\right)=L f^{\prime}\left(0^{-}\right)$
$R f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{\frac{e^{h / 2}-1}{h}-\frac{1}{2}}{h}$ $\quad [\because a=1]$
$ =\lim _{h \rightarrow 0} \frac{2 e^{H / 2}-2-h}{2 h^2}=\frac{1}{8} $
and $L f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{b \sin ^{-1}\left(\frac{c-h}{2}\right)-\frac{1}{2}}{-h}=\frac{b / 2}{\sqrt{1-\frac{c^2}{4}}}$
$\therefore \quad \frac{b}{\sqrt{4-c^2}}=\frac{1}{8}$
$\Rightarrow \quad 64 b^2=\left(4-c^2\right)$
$\Rightarrow \quad a=1 \quad$ and $\quad 64 b^2=\left(4-c^2\right)$
BETA
