Q93 / 170 · Manipal (MET) 2019 · Mathematics
A function $f(x)=\frac{x^2-3 x+2}{x^2+2 x-3}$ is
- Amaximum at $x=-3$
- Bmaximum at $x=-3$ and maximum at $x=1$
- C$\operatorname{maximum}$ at $x=1$
- Correct answerDfunction is increasing in its domain
Solution
$f(x)=\frac{x^2-3 x+2}{x^2+2 x-3}$
On differentiating with respect to $x$, we get $f^{\prime}(x)=\frac{\left(x^2+2 x-3\right)(2 x-3)-\left(x^2-3 x+2\right)}{(2 x+2)}$ $\left(2 x^3+4 x^2-6 x-3 x^2-6 x+9\right)$ $\begin{aligned} & =\frac{-\left(2 x^3-6 x^2+4 x+2 x^2-6 x+4\right)}{\left(x^2+2 x-3\right)^2} \\ & =\frac{5 x^2-10 x+5}{\left(x^2+2 x-3\right)^2}\gt0\end{aligned}$ $=\frac{5(x-1)^2}{(x+3)^2(x-1)^2}\gt0=\frac{5}{(x+3)^2}\gt0$ $\therefore f(x)$ is increasing in its domain.
On differentiating with respect to $x$, we get $f^{\prime}(x)=\frac{\left(x^2+2 x-3\right)(2 x-3)-\left(x^2-3 x+2\right)}{(2 x+2)}$ $\left(2 x^3+4 x^2-6 x-3 x^2-6 x+9\right)$ $\begin{aligned} & =\frac{-\left(2 x^3-6 x^2+4 x+2 x^2-6 x+4\right)}{\left(x^2+2 x-3\right)^2} \\ & =\frac{5 x^2-10 x+5}{\left(x^2+2 x-3\right)^2}\gt0\end{aligned}$ $=\frac{5(x-1)^2}{(x+3)^2(x-1)^2}\gt0=\frac{5}{(x+3)^2}\gt0$ $\therefore f(x)$ is increasing in its domain.
Hence, the correct answer is option: 'D'.