JEE Mathematics 2025 Question with Solution

Given:

$$f(x) = \log_e \left( \frac{2x - 3}{5 + 4x} \right) + \sin^{-1} \left( \frac{4 + 3x}{2 - x} \right)$$

Find: The value of $\alpha^2 + 4\beta$ when the domain is $[\alpha, \beta]$.

For the domain of $f(x)$, both parts must be defined.

For the logarithmic part, the argument must be strictly positive:

$$\frac{2x - 3}{5 + 4x} > 0$$

So,

$$x \in \left( -\infty, -\frac{5}{4} \right) \cup \left( \frac{3}{2}, \infty \right)$$

For the inverse sine part, the argument must satisfy:

$$-1 \le \frac{4 + 3x}{2 - x} \le 1$$

Break this into two inequalities:

$$\frac{4 + 3x}{2 - x} \ge -1$$

and

$$\frac{4 + 3x}{2 - x} \le 1$$

From the first inequality:

$$\frac{4 + 3x}{2 - x} + 1 \ge 0$$ $$\frac{6 + 2x}{2 - x} \ge 0$$

From the second inequality:

$$\frac{4 + 3x}{2 - x} - 1 \le 0$$ $$\frac{2 + 4x}{2 - x} \le 0$$

Combining both gives:

$$x \in \left[ -3, -\frac{1}{2} \right]$$

Now intersect the two domain conditions:

$$\left[ -3, -\frac{1}{2} \right] \cap \left[ \left( -\infty, -\frac{5}{4} \right) \cup \left( \frac{3}{2}, \infty \right) \right]$$ $$= \left[ -3, -\frac{5}{4} \right)$$

So,

\alpha = -3, \quad \beta = -\frac{5}{4} $$](streamdown:incomplete-link)

Now compute:

$$\alpha^2 + 4\beta = (-3)^2 + 4\left(-\frac{5}{4}\right) = 9 - 5 = 4$$

Therefore, the correct option is B.

Given:

$$f(x) = \log_e \left( \frac{2x-3}{5+4x} \right) + \sin^{-1} \left( \frac{4+3x}{2-x} \right)$$

Find: The value of $\alpha^2 + 4\beta$.

For

$$\log_e \left( \frac{2x-3}{5+4x} \right)$$

the argument must be positive. This happens when numerator and denominator have the same sign.

Case 1:

$$2x - 3 > 0, \quad 5 + 4x > 0$$

which gives

$$x > \frac{3}{2}$$

Case 2:

$$2x - 3 < 0, \quad 5 + 4x < 0$$

which gives

$$x < -\frac{5}{4}$$

Hence, the logarithmic domain is

$$(-\infty, -\frac{5}{4}) \cup (\frac{3}{2}, \infty)$$

For

$$\sin^{-1} \left( \frac{4+3x}{2-x} \right)$$

we need

$$-1 \le \frac{4+3x}{2-x} \le 1$$

Using the inequalities stated in the solution, we obtain:

$$x \ge -3$$

and

$$x \le -\frac{1}{2}$$

Therefore,

$$x \in \left[ -3, -\frac{1}{2} \right]$$

Now take the overlap:

$$\left[ -3, -\frac{1}{2} \right] \cap \left[ (-\infty, -\frac{5}{4}) \cup (\frac{3}{2}, \infty) \right] = \left[ -3, -\frac{5}{4} \right)$$

the solution later writes the domain as $[-3, -\frac{5}{4}]$, but the logarithmic condition excludes $x = -\frac{5}{4}$ because the denominator becomes zero there. This does not affect the final value of $\beta$ used in the solution.](streamdown:incomplete-link)

Thus,

$$\alpha = -3, \quad \beta = -\frac{5}{4}$$

and

$$\alpha^2 + 4\beta = 9 - 5 = 4$$

Therefore, the answer is $4$, so the correct option is B.