Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{8^x-4^x-2^x+1^x}{x^2},x>0& \\ e^x\sin x+\pi x+\lambda \ln 4,x\le 0& \end{array}$is continuous at $x=0$, then $\lambda$ is a

• A. rational number
• B. irrational number
• C. natural number
• D. complex number

Answer 1

The correct option is B irrational number

$\underset{x\to -h}{lim}f\left(x\right)=\lambda \ln 4$
$\underset{x\to 0+h}{lim}f\left(x\right)=\frac{2^{3x}-2^{2x}-2^x+1^x}{x^2}=\lambda \ln 4$

$\Rightarrow \underset{h\to 0}{lim}\frac{2^{3h}-2^{2h}-2^h+1^h}{h^2}=\lambda \ln 4$

$\Rightarrow \underset{h\to 0}{lim}\frac{\left(2^{2h}\right(2^h-1)-1(2^h-1\left)\right)}{h^2}=\lambda \ln 4$

$\Rightarrow \underset{h\to 0}{lim}\frac{(2^h+1){(2^h-1)}^2}{h^2}=\lambda \ln 4$

$\Rightarrow \underset{h\to 0}{lim}\frac{2{(2^h-1)}^2}{h^2}=\lambda \ln 4$

Applying L'hospital's rule

$\Rightarrow \underset{h\to 0}{lim}(2^h-1)\frac{\ln 4}{h}=\lambda \ln 4$

$\Rightarrow \ln 2\ln 4=\lambda \ln 4$

$\Rightarrow \ln 2=\lambda$

Hence, option 'B' is correct.