Question

If the function $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}1+\sin \frac{\pi }{2}x,for& -\infty <x\le 1\end{array}\\ \begin{array}{cc}ax+b,for& 1<x<3\end{array}\\ \begin{array}{cc}6\tan \frac{\pi x}{12},for& 3\le x<6\end{array}\end{array}$ is continous in the interval $(-\infty ,6)$ then the value of $a$ and $b$ are?

• A. $0,2$
• B. $1,1$
• C. $2,0$
• D. $2,1$

Answer 1

The correct option is C $2,0$

Function $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}1+\sin \frac{\pi }{2}x,& -\infty <x\le 1\end{array}\\ \begin{array}{cc}ax+b,& 1<x<3\end{array}\\ \begin{array}{cc}6\tan \frac{\pi x}{12},& 3\le x\le 6\end{array}\end{array}$

is continuous in the interval $(-\infty ,6)$.

We know that $f\left(x\right)$ will also be continuous at $x=1$ and $x=3$.
Therefore, for continuity at $x=1,f\left(1\right)=\underset{x\to 1}{lim}f\left(x\right)$

$\Rightarrow 1+\sin \frac{\pi }{2}=\underset{x\to 1}{lim}(ax+b)\Rightarrow a+b=2$ ...(1)

Similarly, for continuity at $x=3,f\left(3\right)=\underset{x\to 3}{lim}f\left(x\right)$

$\Rightarrow 6\tan \frac{3\pi }{12}=\underset{x\to 3}{lim}(ax+b)\Rightarrow 3a+b=6$ ...(2)

Solving equation (1) and (2), we get $a=2$ and $b=0$