Question

Let $f(x)=x^3-x^2-3x-1,g(x)=(x+1)a$ 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) $\underset{x\to \infty }{\lim }h(x)=\infty$ and
(iii) $\underset{x\to -1}{\lim }h(x)=\frac{1}{2}$.
The value of $h(1)$ is

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

Answer 1

Answer: (C)

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