JEE Mathematics 2023 Question with Solution

Given:

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

Find: The value of $|3\alpha + 10(\beta + \gamma) + 21\delta|$ where the domain is $[\alpha, \beta] \cup (\gamma, \delta)$.

For $\sec^{-1}(t)$ to be defined, we need

$$|t| \ge 1$$

So,

$$\left|\frac{2x}{5x+3}\right| \ge 1$$

This gives two cases:

$$\frac{2x}{5x+3} \ge 1$$

and

$$\frac{2x}{5x+3} \le -1$$

From the first inequality,

$$\frac{2x}{5x+3} \ge 1 \Rightarrow 2x \ge 5x+3 \Rightarrow -3x \ge 3 \Rightarrow x \le -1$$

From the second inequality,

$$\frac{2x}{5x+3} \le -1 \Rightarrow 2x \le -5x-3 \Rightarrow 7x \le -3 \Rightarrow x \le -\frac{3}{7}$$

Thus, the domain is

$$[-1,-\frac{3}{5}] \cup (-\frac{3}{5},-\frac{3}{7}]$$

Hence,

$$\alpha = -1, \quad \beta = -\frac{3}{5}, \quad \gamma = -\frac{3}{5}, \quad \delta = -\frac{3}{7}$$

Now compute:

$$3\alpha + 10(\beta+\gamma) + 21\delta = 3(-1) + 10\left(-\frac{3}{5} + -\frac{3}{5}\right) + 21\left(-\frac{3}{7}\right)$$ $$= -3 + 10\left(-\frac{6}{5}\right) + 21\left(-\frac{3}{7}\right)$$ $$= -3 - 12 - 9 = -24$$

Therefore,

$$|3\alpha + 10(\beta + \gamma) + 21\delta| = 24$$

So the required answer is $24$.

Given: $f(x) = \sec^{-1} \left( \frac{2x}{5x+3} \right)$

Find: $|3\alpha + 10(\beta + \gamma) + 21\delta|$.

The key observation is that the argument of $\sec^{-1}$ must satisfy

$$\left|\frac{2x}{5x+3}\right| \ge 1$$

So split immediately into the two standard cases:

$$\frac{2x}{5x+3} \ge 1 \quad \text{or} \quad \frac{2x}{5x+3} \le -1$$

Using the values extracted from the interval form given in the working,

$$\alpha = -1, \quad \beta = -\frac{3}{5}, \quad \gamma = -\frac{3}{5}, \quad \delta = -\frac{3}{7}$$

Substitute directly:

$$3\alpha + 10(\beta + \gamma) + 21\delta = -3 - 12 - 9 = -24$$

Hence,

$$|3\alpha + 10(\beta + \gamma) + 21\delta| = 24$$

Therefore, the answer is $24$.