# If the function $f(x) = \left\{ \begin{array}{l l} 3ax+b & \quad for \quad x > 1 \\ 11 & \quad for \quad x = 1 \\ 5ax-2b & \quad for \quad x < 1 \end{array} \right.$ is continuous at $x=1$.find the values of a and b .

$\begin{array}{1 1}(A)\;a=4\;b=2 \\(B)\;a=3\;b=2 \\(C)\;a=0\;b=1 \\(D)\;a=3\;b=6 \end{array}$

We have $f(1)=11$
$\lim \limits_{x \to 1+} f(x) =\lim \limits_{h \to 0} f(1+h)$
$\qquad= \lim \limits_{h \to 0} \{ 3a(1+h) +b\}$
$\qquad= \lim \limits_{h \to 0} \{ (3a+b) +3ah \}$
$\qquad= (3a+b)$
$\lim \limits_{x \to 1-} f(x) =\lim \limits_{h \to 0} f(1-h)$
$\qquad= \lim \limits_{h \to 0} \{ 5a(1-h) -2b \}$
$\qquad= \lim \limits_{h \to 0} \{ (5a-2b) -5ah \}$
$\qquad= (5a-2b)$
Since $f(x)$ is continuous at $x=1$ we have
$\lim \limits_{x \to {1+}} f(x)=\lim \limits_{x \to {1-}}=f(1)$
$3a+b=5a-2b$
$3a+b=11$
$5a-2b=11$
$6a+2b=22$
$5a-2b=11$
______________
$\qquad 11a=33$
$\qquad a=3$
$3 \times 3 +b =11$
$9 +b=11$
$b=2$
Hence B is the correct answer.