Question

If $f(x)=\{\begin{array}{ll}\frac{8^x-4^x-2^x+1}{x^2},& \text{if x>0 }\\ e^{x}sinx+\pi x+\lambda \text{ln}4,& \text{if x≤0 }\end{array}$
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)

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