# Determine the value of K so that the function. $f(x) = \left\{ \begin{array}{l l} Kx^2 & \quad if \quad x \leq 2 \\ 3 & \quad if \quad x > 2 \end{array} \right.$ is continuous .

$\begin{array}{1 1}(A)\;2 \\(B)\;\frac{1}{4} \\(C)\;2 \\(D)\;\frac{3}{4} \end{array}$

## 1 Answer

Since a polynomial function is continuous and a constant function is continuous,the given function is continuous for all $x > 2$ and for all $x > 2$
So Consider the points $x=2$
We have $f(2) =4k$
$\lim \limits_{x \to {2+} } f(x)= \lim \limits_{h \to 0} f(2 +h)$
$\lim _{h \to 0} 3=3$
$f(x)_{x \to 2-}$$=\lim \limits_{h \to 0} f(2-h)$
$\qquad=\lim \limits_{h \to 0} k(2-h)^2$
$\qquad=4k$
$\therefore$ for continuity we must have
$4K=3$
$k= \large\frac{3}{4}$
Hence D is the correct answer.
answered Apr 22, 2014 by

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