Let the domains of the functions log 7 ( 8 – log 2 ( x 2 + 4 x + 5 ) ) and g ( x ) = sin – 1 ( 7 x + 10 x – 2 ) be, respectively. Then 2 + 2 + 2 + 2 is equal to : [2025]

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Solution

Q.
Let the domains of the functions $f (x) = \log_{4} \log_{3}$ $\log_{7} (8 - \log_{2} (x^{2} + 4 x + 5))$ and $g (x) = \sin^{- 1} (\frac{7 x + 10}{x - 2})$ be ( $α, β$ ) and [ $γ, δ$ ], respectively. Then $α^{2} + β^{2} + γ^{2} + δ^{2}$ is equal to : [2025]

1 13

2 16

3 15

4 14

Ans.
(3)

We have,

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

For f(x) to be defined we need,

$\log_{3} \log_{7} (8 - \log_{2} (x^{2} + 4 x + 5)) > 0$

$\Rightarrow \log_{7} (8 - \log_{2} (x^{2} + 4 x + 5)) > 1$

$\Rightarrow 8 - \log_{2} (x^{2} + 4 x + 5) > 7$

$\Rightarrow \log_{2} (x^{2} + 4 x + 5) < 1$

$\Rightarrow x^{2} + 4 x + 5 < 2$

$\Rightarrow x^{2} + 4 x + 3 < 0$

$\Rightarrow (x + 1) (x + 3) < 0$

$\Rightarrow x \in (- 3, - 1) i . e ., α = - 3, β = - 1$ ... (i)

Also, $x^{2} + 4 x + 5 > 0, but D = 16 - 20 = - 4 < 0$

Also, we have $g (x) = \sin^{- 1} (\frac{7 x + 10}{x - 2})$

$\Rightarrow - 1 \leq \frac{7 x + 10}{x - 2} \leq 1 \Rightarrow - 1 \leq \frac{7 (x - 2) + 24}{x - 2} \leq 1$

$\Rightarrow - 8 \leq \frac{24}{x - 2} \leq - 6 \Rightarrow \frac{1}{- 6} \leq \frac{x - 2}{24} \leq \frac{1}{- 8}$

$\Rightarrow - 4 \leq x - 2 \leq - 3 \Rightarrow - 2 \leq x \leq - 1$

$\Rightarrow x \in [- 2, - 1] i . e ., γ = - 2, δ = - 1$

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

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