MHTCET 18 Apr 2026 Shift 2 Paper

Concept:For a real square root, the expression inside it must be non-negative, and the denominator must never be zero.Explanation:We require $\frac{x}{1+x} \geq 0$ and $1+x \neq 0$.The critical points are $x = -1$ and $x = 0$.For $x < -1$, both $x$ and $1+x$ are negative, so $\frac{x}{1+x} > 0$. Hence $(-\infty, -1)$ is included.At $x = -1$, the denominator becomes zero, so this point is excluded.For $-1 < x < 0$, the fraction is negative, so this interval is rejected.At $x = 0$, $\frac{0}{1+0} = 0 \geq 0$, so $0$ is included.For $x > 0$, both are positive, so $\frac{x}{1+x} > 0$ and $(0, \infty)$ is included.Combining the valid intervals, the domain is $(-\infty, -1) \cup [0, \infty)$.Answer:$$(-\infty, -1) \cup [0, \infty)$$Hence, the correct option is A.