Question

Consider the function $f\left(x\right)=\{\begin{array}{cc}\frac{\tan kx}{x},& x<0\\ 3x+2k^2,& x\ge 0\end{array}$. What is the non-zero value of k for which the function is continuous at $x=0$?

• A. $1/4$
• B. $1/2$
• C. $1$
• D. $2$

Answer 1

The correct option is B $1/2$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{\tan kx}{kx}\times k$

as $x\to 0^{-};kx\to 0^{-}$

$\Rightarrow {lim}_{x\to 0^{-}}1\times k=k$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}(3x+2k^2)$

$=3\left(0\right)+2k^2=2k^2$

Given it is continuous at $x=0$

$\Rightarrow$ LHL$=$RHL

$\Rightarrow k=2k^2\Rightarrow 1=2k$

$k=\frac{1}{2}$