Let the domain of the function f(x) = cos^-1((4x+5)/(3x-7)) be [alpha, beta] and the domain of g(x) = log_2(2 - 6log_27(2x+5)) be (gamma, delta). Then |7(alpha + beta) + 4(gamma + delta)| is equal to _______

The correct answer is Option 96. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q21 – JEE-MAIN 2025

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Home Maths Relations and FunctionsQuestion #384

Let the domain of the function

f (x) = \cos^{- 1} (\frac{4 x + 5}{3 x - 7})

[α, β]

and the domain of

g (x) = \log_{2} (2 - 6 \log_{27} (2 x + 5))

(γ, δ)

. Then

| 7 (α + β) + 4 (γ + δ) |

is equal to _______

Enter Calculated Numerical Value:

Clue 1 of 3: Domain of Inverse Trigonometric Functions

For an inverse cosine function $\cos^{- 1} (u)$ to be defined, its argument $u$ must satisfy the condition $- 1 \leq u \leq 1$ .

🧠 Full Solution Path

Step 1: Determine the domain of

f (x)

✦ Active

The function $f (x) = \cos^{- 1} (\frac{4 x + 5}{3 x - 7})$ is defined when $- 1 \leq \frac{4 x + 5}{3 x - 7} \leq 1$ and $3 x - 7 e 0$ . We solve the two inequalities separately:

\frac{4 x + 5}{3 x - 7} \leq 1 ⟹ \frac{4 x + 5 - (3 x - 7)}{3 x - 7} \leq 0 ⟹ \frac{x + 12}{3 x - 7} \leq 0

This inequality holds for $x \in [- 12, \frac{7}{3})$ .

\frac{4 x + 5}{3 x - 7} \geq - 1 ⟹ \frac{4 x + 5 + (3 x - 7)}{3 x - 7} \geq 0 ⟹ \frac{7 x - 2}{3 x - 7} \geq 0

This inequality holds for $x \in (- \infty, \frac{2}{7}] \cup (\frac{7}{3}, \infty)$ . The intersection of these two solution sets gives the domain of $f (x)$ as $x \in [- 12, \frac{2}{7}]$ . Therefore, $α = - 12$ and $β = \frac{2}{7}$ . So, $α + β = - 12 + \frac{2}{7} = - \frac{82}{7}$ .

Step 2: Determine the domain of

g (x)

○ Expand

The function $g (x) = \log_{2} (2 - 6 \log_{27} (2 x + 5))$ is defined when its arguments are positive. First, the argument of the outer logarithm must be positive:

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

Since the base $27 > 1$ , we can write $2 x + 5 < 27^{1 / 3} ⟹ 2 x + 5 < 3 ⟹ 2 x < - 2 ⟹ x < - 1$ . Second, the argument of the inner logarithm must be positive:

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

Combining these conditions, the domain of $g (x)$ is $x \in (- \frac{5}{2}, - 1)$ . Therefore, $γ = - \frac{5}{2}$ and $δ = - 1$ . So, $γ + δ = - \frac{5}{2} - 1 = - \frac{7}{2}$ .

Step 3: Calculate the final expression○ Expand

Substitute the calculated values of $(α + β)$ and $(γ + δ)$ into the given expression:

| 7 (α + β) + 4 (γ + δ) | = | 7 (- \frac{82}{7}) + 4 (- \frac{7}{2}) |

= | - 82 + 2 (- 7) |

= | - 82 - 14 |

= | - 96 |

= 96

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Maths Question 21 – JEE-MAIN 2025

Clue 1 of 3: Domain of Inverse Trigonometric Functions

🧠 Full Solution Path