Q82 / 170 · Manipal (MET) 2018 · Mathematics

If $f(x)=\left\{\begin{array}{cc}x+1, & x \leq 1 \\ 3-a x^2, & x\gt1\end{array}\right.$ is continuous at $x=1$, then the value of $a$ is

  • Correct answer
    A
    1
  • B
    2
  • C
    -3
  • D
    -2

Solution

Since, $f(x)$ is continuous at $x=1$. $\begin{array}{ll} \therefore & \lim _{h \rightarrow 0^{+}} f(1+h)=f(1) \\ \Rightarrow & \lim _{h \rightarrow 0} 3-a(1+h)^2=2 \end{array}$ $\begin{aligned} \Rightarrow & 3-a(1+0)^2 & =2 \\ \Rightarrow & a & =1\end{aligned}$

Hence, the correct answer is option: 'A'.

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