MHTCET 25 Apr 2025 Shift 2 Paper

Concept: A function is increasing on an interval if its derivative is positive, i.e., $f'(x) > 0$.Explanation:Given, $f(x) = \log(1+x) - \frac{2x}{2+x}$.The domain of $f$ is $x > -1$, since $\log(1+x)$ requires $1+x > 0$.Differentiate with respect to $x$:$f'(x) = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2}$.Simplify the second term: $\frac{2(2+x) - 2x}{(2+x)^2} = \frac{4}{(2+x)^2}$.So, $f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2}$.Taking the common denominator $(1+x)(2+x)^2$:$f'(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{x^2}{(x+1)(x+2)^2}$.For $f$ to be increasing, $f'(x) > 0$:$\frac{x^2}{(x+1)(x+2)^2} > 0$.Here, $x^2 > 0$ and $(x+2)^2 > 0$ for $x \neq 0, -2$.Thus, the sign depends only on $(x+1)$, so we need $x+1 > 0$, i.e., $x > -1$.Therefore, $f(x)$ is increasing in $(-1, \infty)$.Answer: Option A: $(-1, \infty)$