JEE Mathematics 2024 Question with Solution

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

Find: If the domain is $(\alpha, \beta]$, find $5\beta - 4\alpha$.

For the logarithmic part, the argument must be positive:

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

Factor the denominator:

$$4x^2 + x - 3 = (4x - 3)(x + 1)$$

So,

$$\frac{2x + 3}{(4x - 3)(x + 1)} > 0$$

The critical points are $x = -\frac{3}{2}, -1, \frac{3}{4}$. From sign analysis,

$$D_1 = \left(-\frac{3}{2}, -1\right) \cup \left(\frac{3}{4}, \infty\right)$$

For the inverse cosine part, its argument must lie in $[-1,1]$:

$$-1 \leq \frac{2x - 1}{x + 2} \leq 1$$

First,

$$\frac{2x - 1}{x + 2} \geq -1$$

which gives

$$\frac{3x + 1}{x + 2} \geq 0$$

So,

$$x \in (-\infty, -2) \cup \left[-\frac{1}{3}, \infty\right)$$

Next,

$$\frac{2x - 1}{x + 2} \leq 1$$

which gives

$$\frac{x - 3}{x + 2} \leq 0$$

So,

$$x \in (-2, 3]$$

Hence,

$$D_2 = \left(( -\infty, -2) \cup \left[-\frac{1}{3}, \infty\right)\right) \cap (-2, 3] = \left[-\frac{1}{3}, 3\right]$$

Now take the intersection:

$$\text{Domain}(f) = D_1 \cap D_2 = \left(\left(-\frac{3}{2}, -1\right) \cup \left(\frac{3}{4}, \infty\right)\right) \cap \left[-\frac{1}{3}, 3\right] = \left(\frac{3}{4}, 3\right]$$

Thus,

$$\alpha = \frac{3}{4}, \qquad \beta = 3$$

Now compute:

$$5\beta - 4\alpha = 5(3) - 4\left(\frac{3}{4}\right) = 15 - 3 = 12$$

Therefore, the value of $5\beta - 4\alpha$ is $12$. The correct option is B.

The solution says "The Correct Option is A", but the actual working concludes the value is $12$, which matches option B.

Given: The function is the sum of a logarithmic function and an inverse cosine function.

Find: The interval $(\alpha, \beta]$ and then $5\beta - 4\alpha$.

Use domain rules directly:

• For $\log_e\left(\frac{2x + 3}{4x^2 + x - 3}\right)$, require

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

which gives

$$\left(-\frac{3}{2}, -1\right) \cup \left(\frac{3}{4}, \infty\right)$$

• For $\cos^{-1}\left(\frac{2x - 1}{x + 2}\right)$, require

$$-1 \leq \frac{2x - 1}{x + 2} \leq 1$$

which gives

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

Their intersection is immediately

$$\left(\frac{3}{4}, 3\right]$$

So,

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

and

$$5\beta - 4\alpha = 15 - 3 = 12$$

Therefore, the correct option is B.