$f(x)=\left\{\begin{array}{cc}\frac{\left(2 x^2-a x+1\right)-\left(a x^2+3 b x+2\right)}{x+1} & ; \text { if } x \neq-1 \\ k & , \text { if } x=-1\end{array}\right.$
is a real valued function. If $a, b, k \in \mathrm{R}$ and $f$ is continuous on $\mathbf{R}$ then $k=$
$-\frac{1}{3}$
$6$
$a-2$
$a-3$
Solution
Given the function
$f(x)=\left\{\begin{array}{cc}\frac{\left(2 x^2-a x+1\right)-\left(a x^2+3 b x+2\right)}{x+1} & \text { if } x \neq-1 \\ k & \text { if } x=1\end{array}\right.$
Now, $\lim _{x \rightarrow-1} \frac{(2-a) x^2-(a+3 b) x-1}{x+1}$
For existence of limit
$2-a+a+3 b-1=0 \Rightarrow b=-\frac{1}{3}$
Now, $\lim _{x \rightarrow-1} \frac{(2-a) x^2-(a-1) x-1}{x+1}$
$\begin{aligned} & =\lim _{x \rightarrow-1} \frac{2 x(2-a)-(a-1)}{1} \\ & =\frac{-2(2-a)-a+1}{1}=-4+a+1=a-3\end{aligned}$
Since, $f(x)$ is continuous at $x=-1$
So, $k=a-3$