If a real valued function $f(x)=\left\{\begin{array}{cl}\frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6} & , \text { for…
If a real valued function
$f(x)=\left\{\begin{array}{cl}\frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6} & , \text { for } x \neq 3 \\ l & , \text { for } x=3\end{array}\right.$
$x=3$ and $l$ is a finite value, then $l-k=$
$\frac{31}{11}$
$\frac{124}{11}$
$24$
$32$
Solution
Since, $f(x)$ is continuous at $x=3 \Rightarrow \lim _{x \rightarrow 3} f(x)=f(3)$ Now, $\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3} \frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6}$
For this limit to exist, we must have
$\begin{aligned}
& \lim _{x \rightarrow 3} 2 x^2+(k+2) x+9=0 \\
& \Rightarrow 18+3 k+6+9=0 \Rightarrow k=-11
\end{aligned}$ So, $\lim _{x \rightarrow 3} \frac{2 x^2-9 x+9}{3 x^2-7 x-6}=\lim _{x \rightarrow 3} \frac{4 x-9}{6 x-7}=\frac{3}{11}=l$
Thus, $l-k=\frac{3}{11}+11=\frac{124}{11}$