Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{\tan kx}{x},& x<0\\ 3x+2k^2,& x\ge 0\end{array}$. If $f\left(x\right)$ is continuous at $x=0$, then number of values of $k$ is

• A. 1
• B. 2
• C. more than 2
• D. none

Answer 1

The correct option is B 2

$(LHL\text{at}x=0)=\underset{x\to 0^{-}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(0-h)$
$=\underset{h\to 0}{lim}\frac{\tan k(-h)}{-h}$
$=\underset{h\to 0}{lim}\frac{\tan kh}{kh}\times k=k$
$(RHL\text{at}x=0)=\underset{x\to 0^{+}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(0+h)$
$=\underset{h\to 0}{lim}(3h+2k^2)$
$=2k^2$
As $f\left(x\right)$ is continuous at $x=0$.
$\therefore (LHL\text{at}x=0)=(RHL\text{at}x=0)$
$\Rightarrow k=2k^2\Rightarrow 2k^2-k=0\Rightarrow k=0\text{or}\frac{1}{2}$