MHTCET 20 Apr 2026 Shift 1 Paper

Concept:For a square root function $\sqrt{g(x)}$ to be defined, the expression inside the root must be non-negative: $g(x) \geq 0$. The domain of a sum of functions is the intersection of the individual domains of the functions.Explanation:Given $f(x) = \sqrt{x-1} + \sqrt{6-x}$, both square root terms must be defined for the same value of $x$.For $\sqrt{x-1}$, we require $x-1 \geq 0$.This simplifies to $x \geq 1$.For $\sqrt{6-x}$, we require $6-x \geq 0$.This simplifies to $x \leq 6$.Both conditions must hold together, so $x$ must satisfy $x \geq 1$ and $x \leq 6$.Combining these inequalities gives $1 \leq x \leq 6$.This means the permissible values of $x$ form the closed interval from 1 to 6.Answer:The domain of $f(x)$ is $[1,6]$.Therefore, the correct option is Option B: $[1,6]$.