KCET 2026 Mathematics Solved Paper

Concept:For a square root function to be defined, the radicand must be non-negative, and the denominator must not be zero.Explanation:We need $\sqrt{\frac{x-7}{9-x}}$ to be defined.So the expression inside the square root must satisfy $\frac{x-7}{9-x} \geq 0$.Also, the denominator must not be zero, so $9 - x \neq 0$, which gives $x \neq 9$.The critical points of the expression are $x = 7$ and $x = 9$.At $x = 7$, the value is $\frac{7-7}{9-7} = 0$, and $\sqrt{0}$ is defined, so $x = 7$ is allowed.For any $x$ between $7$ and $9$, for example $x = 8$, we get $\frac{8-7}{9-8} = 1 > 0$, so the radicand is positive and the square root is defined.At $x = 9$, the denominator becomes zero, so $x = 9$ is not allowed.For $x < 7$ or $x > 9$, the fraction becomes negative, which is not allowed under the square root.Therefore, the valid values of $x$ are $7 \leq x < 9$.Thus the domain is $[7,9)$.Answer:Option B: $[7,9)$