MCQ
If $f(x)=\left\{\begin{array}{ll}k x^2 & \text { if } x \leq 2 \\ 3 & \text { if } x>2\end{array}\right.$ is continuous at $x=2,$ then the value of $k$ is
  • A
    3
  • B
    4
  • $\frac{3}{4}$
  • D
    $\frac{4}{3}$

Answer

Correct option: C.
$\frac{3}{4}$
(C)
$f(2)=k(2)^2=4 k$
$\lim _{x \rightarrow 2^{+}} f (x)=\lim _{x \rightarrow 2^{+}} 3=3$
Since the function is continous at $x=2$,
$\lim _{x \rightarrow 2^{+}} f (x)= f (2)$
$\Rightarrow 4 k =3$
$\Rightarrow k =\frac{3}{4}$

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