Let the domains of the functions f(x)=log4 log3 log7(8-log2(x^2+4x+5)) and g(x)=sin^-1((7x+10)/(x-2)) be (alpha, beta) and [gamma, delta], respectively. Then alpha^2+beta^2+gamma^2+delta^2 is equal to :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q1 – JEE-MAIN 2025

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Home Maths Relations and FunctionsQuestion #381

Let the domains of the functions

f (x) = \log_{4} \log_{3} \log_{7} (8 - \log_{2} (x^{2} + 4 x + 5))

and

g (x) = \sin^{- 1} (\frac{7 x + 10}{x - 2})

(α, β)

and

[γ, δ]

, respectively. Then

α^{2} + β^{2} + γ^{2} + δ^{2}

is equal to :

Clue 1 of 3: Domain of Logarithmic and Inverse Trigonometric Functions

Recall the conditions for the domain of $\log_{a} x$ and $\sin^{- 1} x$ .

🧠 Full Solution Path

Step 1: Determine the domain of

f (x)

✦ Active

The function is $f (x) = \log_{4} \log_{3} \log_{7} (8 - \log_{2} (x^{2} + 4 x + 5))$ . For $f (x)$ to be defined, we must satisfy the following conditions:

1. x^{2} + 4 x + 5 > 0 ⟹ (x + 2)^{2} + 1 > 0 (True for all x \in R)

2. 8 - \log_{2} (x^{2} + 4 x + 5) > 0 ⟹ \log_{2} (x^{2} + 4 x + 5) < 8 ⟹ x^{2} + 4 x + 5 < 2^{8} = 256 ⟹ x^{2} + 4 x - 251 < 0

3. \log_{7} (8 - \log_{2} (x^{2} + 4 x + 5)) > 0 ⟹ 8 - \log_{2} (x^{2} + 4 x + 5) > 7^{0} = 1 ⟹ \log_{2} (x^{2} + 4 x + 5) < 7 ⟹ x^{2} + 4 x + 5 < 2^{7} = 128 ⟹ x^{2} + 4 x - 123 < 0

4. \log_{3} \log_{7} (8 - \log_{2} (x^{2} + 4 x + 5)) > 0 ⟹ \log_{7} (8 - \log_{2} (x^{2} + 4 x + 5)) > 3^{0} = 1 ⟹ 8 - \log_{2} (x^{2} + 4 x + 5) > 7^{1} = 7 ⟹ \log_{2} (x^{2} + 4 x + 5) < 1 ⟹ x^{2} + 4 x + 5 < 2^{1} = 2 ⟹ x^{2} + 4 x + 3 < 0

The most restrictive condition is $x^{2} + 4 x + 3 < 0$ . Factoring gives $(x + 1) (x + 3) < 0$ , which implies $x \in (- 3, - 1)$ . Thus, $(α, β) = (- 3, - 1)$ , so $α = - 3$ and $β = - 1$ .

Step 2: Determine the domain of

g (x)

○ Expand

The function is $g (x) = \sin^{- 1} (\frac{7 x + 10}{x - 2})$ . For $g (x)$ to be defined, the argument of $\sin^{- 1}$ must be in $[- 1, 1]$ , and the denominator cannot be zero. So, $- 1 \leq \frac{7 x + 10}{x - 2} \leq 1$ and $x \neq 2$ . This can be split into two inequalities:

1. \frac{7 x + 10}{x - 2} \leq 1 ⟹ \frac{7 x + 10 - (x - 2)}{x - 2} \leq 0 ⟹ \frac{6 x + 12}{x - 2} \leq 0 ⟹ \frac{6 (x + 2)}{x - 2} \leq 0

This inequality holds for x \in [- 2, 2) .

2. \frac{7 x + 10}{x - 2} \geq - 1 ⟹ \frac{7 x + 10 + (x - 2)}{x - 2} \geq 0 ⟹ \frac{8 x + 8}{x - 2} \geq 0 ⟹ \frac{8 (x + 1)}{x - 2} \geq 0

This inequality holds for x \in (- \infty, - 1] \cup (2, \infty) .

The intersection of these two intervals is $[- 2, - 1]$ . Thus, $[γ, δ] = [- 2, - 1]$ , so $γ = - 2$ and $δ = - 1$ .

Step 3: Calculate

α^{2} + β^{2} + γ^{2} + δ^{2}

○ Expand

We have $α = - 3$ , $β = - 1$ , $γ = - 2$ , $δ = - 1$ . The required sum is:

α^{2} + β^{2} + γ^{2} + δ^{2} = (- 3)^{2} + (- 1)^{2} + (- 2)^{2} + (- 1)^{2} = 9 + 1 + 4 + 1 = 15

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Maths Question 1 – JEE-MAIN 2025

Clue 1 of 3: Domain of Logarithmic and Inverse Trigonometric Functions

🧠 Full Solution Path