Question

Let $f(x)=x^{3}-x^{2}-3 x-1, g(x)=(x+1) a$ and $h(x)=\frac{f(x)}{g(x)}$
where $h$ is a rational function such that:
(i) it is continuous everywhere except when $x=-1$,
(ii) $\displaystyle\lim _{x \rightarrow \infty} h(x)=\infty$ and
(iii) $\displaystyle\lim _{x \rightarrow-1} h(x)=\frac{1}{2}$.
The value of $h(1)$ is

• A. 1 / 2
• B. 1 / 4
• C. -1 / 2
• D. 1

Answer 1

Answer: (C)

$h(x)=\frac{f(x)}{g(x)}$
$\Rightarrow h(x)=\frac{x^{3}-x^{2}-3 x-1}{(x+1) a}\, \dots(i)$
Now, $\displaystyle\lim _{x \rightarrow-1} h(x)=\frac{1}{2}$
$\Rightarrow \displaystyle\lim _{x \rightarrow-1} \frac{x^{3}-x^{2}-3 x-1}{(x+1) a}=\frac{1}{2}$
$\Rightarrow \frac{2}{a}=\frac{1}{2} $
$\Rightarrow a=4$
$\therefore h(x)=\frac{x^{3}-x^{2}-3 x-1}{4(x+1)} $
$\therefore h(1)=-1 / 2$