JEE Mathematics 2025 Question with Solution

Given: $f(x) = \frac{1}{\sqrt{10 + 3x - x^2}}$

Find: If the domain is $(a,b)$, find $(1+a)^2 + b$.

For the function to be defined, the expression inside the square root in the denominator must be strictly positive:

$$10 + 3x - x^2 > 0$$

Rearrange:

$$x^2 - 3x - 10 < 0$$

Factorizing,

$$(x-5)(x+2) < 0$$

So the inequality holds between the roots:

$$-2 < x < 5$$

Hence,

$$a = -2, \quad b = 5$$

Now compute:

$$(1+a)^2 + b = (1-2)^2 + 5$$ $$= (-1)^2 + 5$$ $$= 1 + 5 = 6$$

From the given options and the solution, there is a discrepancy: the solution discusses a different function and evaluates $(1+a)^2 + b^2$ instead, giving $26$. Based on the recorded correct option, the marked answer is A.