Question

If $f\left(x\right)=\{\begin{array}{cc}x+1& ,x\le 1\\ 3-a{x}^2& ,x>1\end{array}$ is continuous at $x=1$, then the value of $a$ is.

• A. $-1$
• B. $2$
• C. $-3$
• D. $1$

Answer 1

Answer: (D)

$1$

$f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}x+1& ,x\le 1\\ 3-a{x}^2& ,x>1\end{array}\end{array}$ is continuous at $x=1$

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

$\underset{x\to 1^{-}}{lim}(x+1)=\underset{x\to 1^{+}}{lim}(3-a{x}^2)=(x+1){|}_{x=1}$

$\underset{x\to 1^{+}}{lim}(3-a{x}^2)=1+1$

$3-a{\left(1\right)}^2=2$

$-a=-1$

$a=1$