JEE Mathematics 2025 Question with Solution

Given:

• $f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)$
• $g(x) = \log_2\left(2 - 6\log_{27}(2x+5)\right)$

Find:

• $|7(\alpha+\beta) + 4(\gamma+\delta)|$

For $f(x)$ to be defined, the argument of $\cos^{-1}$ must satisfy

$$-1 \leq \frac{4x+5}{3x-7} \leq 1$$

This gives two inequalities:

$$\frac{4x+5}{3x-7} - 1 \leq 0$$ $$\frac{x+12}{3x-7} \leq 0$$

So,

$$-12 \leq x < \frac{7}{3}$$

Also,

$$\frac{4x+5}{3x-7} + 1 \geq 0$$ $$\frac{7x-2}{3x-7} \geq 0$$

So,

$$x \leq \frac{2}{7} \text{ or } x > \frac{7}{3}$$

Taking intersection,

$$x \in \left[-12, \frac{2}{7}\right]$$

Hence,

$$\alpha = -12, \quad \beta = \frac{2}{7}$$

For $g(x)$ to be defined, both logarithmic conditions must hold. First,

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

Next,

$$2 - 6\log_{27}(2x+5) > 0$$ $$\log_{27}(2x+5) < \frac{1}{3}$$ $$2x+5 < 27^{1/3} = 3$$ $$x < -1$$

Therefore,

$$x \in \left(-\frac{5}{2}, -1\right)$$

So,

$$\gamma = -\frac{5}{2}, \quad \delta = -1$$

Now compute:

$$7(\alpha+\beta) + 4(\gamma+\delta)$$ $$= 7\left(-12 + \frac{2}{7}\right) + 4\left(-\frac{5}{2} - 1\right)$$ $$= 7\left(-\frac{82}{7}\right) + 4\left(-\frac{7}{2}\right)$$ $$= -82 - 14 = -96$$

Hence,

$$\left|7(\alpha+\beta) + 4(\gamma+\delta)\right| = 96$$

Therefore, the required value is $96$.

Note: The question text writes $\log_2$ in the inner logarithm of $g(x)$, but the extracted solution works with $\log_{27}$. The final answer has been derived from the solution working, which is consistent with $96$.

Given:

• Domain of $f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)$ is $[\alpha,\beta]$
• Domain of $g(x) = \log_2\left(2 - 6\log_{27}(2x+5)\right)$ is $(\gamma,\delta)$

Find:

• $|7(\alpha+\beta) + 4(\gamma+\delta)|$

For $f(x)$,

$$\frac{x+12}{3x-7} \leq 0$$

Critical points are $x=-12$ and $x=\frac{7}{3}$. Sign analysis gives the valid interval

$$[-12, \tfrac{7}{3})$$

For the second inequality,

$$\frac{7x-2}{3x-7} \geq 0$$

Critical points are $x=\frac{2}{7}$ and $x=\frac{7}{3}$. Sign analysis gives

$$(-\infty, \tfrac{2}{7}] \cup (\tfrac{7}{3}, \infty)$$

Intersecting these two sets,

$$[-12, \tfrac{2}{7}]$$

Thus,

$$\alpha = -12, \quad \beta = \frac{2}{7}$$

For $g(x)$,

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

and

$$2 - 6\log_{27}(2x+5) > 0$$ $$\log_{27}(2x+5) < \frac{1}{3}$$

Since the base $27>1$,

$$2x+5 < 27^{1/3} = 3$$ $$x < -1$$

Therefore,

$$-\frac{5}{2} < x < -1$$

So,

$$\gamma = -\frac{5}{2}, \quad \delta = -1$$

Now,

$$\alpha+\beta = -12 + \frac{2}{7} = -\frac{82}{7}$$ $$\gamma+\delta = -\frac{5}{2} - 1 = -\frac{7}{2}$$

Hence,

$$7(\alpha+\beta) + 4(\gamma+\delta) = 7\left(-\frac{82}{7}\right) + 4\left(-\frac{7}{2}\right) = -82 - 14 = -96$$

Therefore,

$$|7(\alpha+\beta) + 4(\gamma+\delta)| = 96$$

So the required answer is $96$.