Question

Let $f:R\to R$ be defined as:

$f\left(x\right)=\left\{\begin{array}{l}2\sin \left(\frac{-\pi x}{2}\right)ifx<-1\\ \left|a{x}^2+x+b\right|if-1\le x\le 1\\ \sin \left(\pi x\right)ifx>1\end{array}\right.$

If $f\left(x\right)$ is continuous on $R$, then $a+b$equals to:

• A. $3$
• B. $-1$
• C. $-3$
• D. $1$

Answer 1

The correct option is B

$-1$

Explanation for correct option:

Determine the value of $a+b$.

$f\left(x\right)$ is Continuous on $R$

Case I: For $x=1$, we have: $f\left(1^{-}\right)=f\left(1\right)=f\left(1^{+}\right)$

$\left|a+1+b\right|=\underset{x\to 1}{\lim }\sin \left(\mathrm{\pi x}\right)$

Putting the limit we get,

$$\begin{gathered} \Rightarrow \left|a+1+b\right|=0 \\ \Rightarrow a+b=-1...\left(i\right) \end{gathered}$$

Case II: For $x=-1$, we have: $f(-1^{-})=f(-1)=f(-1^{+})$

$\Rightarrow \underset{x\to -1}{\lim }2\sin \left(\frac{-\mathrm{\pi x}}{2}\right)=\left|a-1+b\right|$

Putting the limits we have,

$\Rightarrow \left|a-1+b\right|=2\mathbf{}\left[\because sin\left(\frac{\pi }{2}\right)=1\right]$

Thus we have, $a-1+b=2\mathrm{or}\mathrm{a}-1+\mathrm{b}=-2$

solving we get,

$$\begin{gathered} a+b=3...\left(ii\right) \\ \mathrm{a}+\mathrm{b}=-1...\left(iii\right) \end{gathered}$$

Solving simultaneously $\left(i\right)\mathrm{and}\left(ii\right)$

$$\begin{gathered} LHS: \\ 0 \\ RHS: \\ 4 \\ LHS\ne RHS \end{gathered}$$

Hence, rejected.

Similarly, Solving simultaneously $\left(i\right)\mathrm{and}\left(iii\right)$

$$\begin{gathered} LHS: \\ 0 \\ RHS: \\ 0 \\ LHS=RHS \end{gathered}$$

Therefore, $a+b=-1$

Hence, the correct answer is option (B).