Question

If $f\left(x\right)={\left(\tan (\frac{\pi }{4}+\ln x)\right)}^{\frac{1}{2}{\log }_{x}e}$ is to be made continuous at $x=0$, then $\left[f\right(1\left)\right]$ should be equal to (where $[.]$denote greatest integer function) :

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (C)

$2$

$\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}{\left(\tan (\frac{\pi }{4}+\ln x)\right)}^{\frac{1}{2}\frac{1}{\log x}}$

which is in $1^{\infty }$ form

$\underset{x\to 1}{lim}f\left(x\right)=e^{\underset{x\to 1}{lim}\frac{1}{2}\frac{\log e}{\log x}(\tan (\frac{\pi }{4}+\ln x)-1)}$

which is in $\frac{0}{0}$ form

$\Rightarrow f\left(x\right)=e^{\frac{1}{2}\log e\underset{x\to 1}{lim}\frac{{\sec }^2(\frac{\pi }{4}+\ln x)}{\frac{1}{x}}}$

$=e^{\frac{1}{2}\log e\times 2}=e^{{\log }_{e}e}=e$

$\Rightarrow \left[f\left(1\right)\right]=\left[e\right]=2(e\approx 2.71)$

Hence, the answer is $2$.