Question

$f\left(x\right)=\{\begin{array}{cc}\frac{\sin ax}{x}+2,& -2\le x<0\\ 3x+5,& 0<x\le 0\\ \sqrt{x^2+8}-b,& 1<x\le 2\end{array}$ is continuous on $[-2,2]$. Then the value of $a$ and $b$ are

• A. $3,5$
• B. $-3,5$
• C. $3,-5$
• D. $-3,-5$

Answer 1

The correct option is C $3,-5$

$f\left(x\right)=\{\begin{array}{c}\frac{\sin ax}{x}+2\\ 3x+5\\ \sqrt{x^2+8}-b\end{array}\begin{array}{c}2\le x<0\\ 0<x\le 1\\ 1<x\le 1\end{array}$

At $x=0$ left named limit $\underset{x\to 0}{lim}\frac{\sin ax}{x}+2$

$\underset{x\to 0}{lim}a\left(\frac{\sin ax}{ax}\right)+2=a\left(a1\right)+2=a+2$

Right hand limit $\underset{x\to 0}{lim}3x+5=5$

$\Rightarrow a+2=5\Rightarrow a=3$ (for $f$ to be continuous $L.H.L=R.H.L$)

At $x=1,L.H.L,\underset{x\to 1}{lim}3x+5=8$

$R.H.L=\underset{x\to 1}{lim}\sqrt{x^2+8}-b=\sqrt{9}-b=3-b$

$\Rightarrow 3-b=8\Rightarrow b=-5\Rightarrow \left(C\right)$