Question

Let $f\left(x\right)=\{\begin{array}{cc}ax+1,& x<1\\ 3,& x=1\\ b{x}^2+1,& x>1\end{array}$. The value of ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ for which $f\left(x\right)$ is continuous at $x=1$, is given by

• A. $2,2$
• B. $1,2$
• C. $3,2$
• D. $1,1$

Answer 1

The correct option is A $2,2$

$f\left(x\right)=\{\begin{array}{c}ax+1x<1\\ 3x=1\\ b{x}^2+1x>1\end{array}$

To Find: Value of 'a' and 'b' for which $f\left(x\right)$ is continuous at $x=0$

Sol: $\underset{x\to 1^{-}}{lim}$$f\left(x\right)$$\underset{x\to 1}{lim}$ $ax+1=(a+1)$

$f\left(1\right)=3$

$\underset{x\to 1^{+}}{lim}$ $f\left(x\right)$ $=\underset{x\to 1}{lim}$ $(b{x}^2+1)$ = $(b+1)$

LHL $=f\left(1\right)$ $=RHL$

$\Rightarrow a+1=3=b+1\Rightarrow a=b=2$

Hence, Correct answer is $a=2$ & $b=2.$