Question

If $f(x)= \begin{cases} \frac{8^{x}-4^{x}-2^{x}+1}{x^{2}}, & \text{if $x>\,0$ } \\[2ex] e ^{x}\,sin\,x+\pi x +\lambda {\text{ln}} 4, & \text{if $x \leq 0$ } \end{cases}$
is continuous at $x = 0$. Then, the value of $\lambda$ is

• A. $4 \log _{e} 2$
• B. $2 \log _{e} 2$
• C. $\log _{e} 2$
• D. None of these

Answer 1

Answer: (C)

$f(0) =0+0+\lambda \ln 4=\lambda \ln 4\, \dots(1)$
$ R.H.L. =\displaystyle\lim _{x \rightarrow 0+} f(x)=\displaystyle\lim _{h \rightarrow 0} f(0+h) $
$=\displaystyle\lim _{h \rightarrow 0} \frac{8^{h}-4^{h}-2^{h}+1^{h}}{h^{2}}=\displaystyle\lim _{h \rightarrow 0} \frac{\left(4^{h}-1\right)\left(2^{h}-1\right)}{h \cdot h}$
$=\displaystyle\lim _{h \rightarrow 0}\left(\frac{4^{k}-1}{h}\right) \displaystyle\lim _{h \rightarrow 0}\left(\frac{2^{h}-1}{h}\right)=\ln 4 \cdot \ln 2\, dots(2)$
$\therefore f (0) =R.H.L.$
$\Rightarrow \lambda=\ln \, 2$