Let the domain of the function f ( x ) = cos – 1 ( 4 x + 5 3 x – 7 ) be and the domain of g ( x ) = log 2 ( 2 – 6 log 27 ( 2 x + 5 ) ) be . Then | 7 + 4 is equal to __________. [2025]

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Solution

Q.
Let the domain of the function $f (x) = \cos^{- 1} (\frac{4 x + 5}{3 x - 7})$ be $[α, β]$ and the domain of $g (x) = \log_{2} (2 - 6 \log_{27} (2 x + 5))$ be $(γ, δ)$ . Then $| 7 (α + β) + 4 (γ + δ) |$ is equal to __________. [2025]

Ans.
(96)

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

Since, domain of $\cos^{— 1} x$ is [–1, 1].

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

For, $\frac{4 x + 5}{3 x - 7} \geq - 1 \Rightarrow \frac{4 x + 5 + 3 x - 7}{3 x - 7} \geq 0$

$\Rightarrow \frac{7 x - 2}{3 x - 7} \geq 0 \Rightarrow x \in (- \infty, \frac{2}{7}] \cup (\frac{7}{3}, \infty)$ ... (i)

Now, for $\frac{4 x + 5}{3 x - 7} \leq 1 \Rightarrow \frac{4 x + 5 - 3 x + 7}{3 x - 7} \leq 0$

$\Rightarrow \frac{x + 12}{3 x - 7} \leq 0 \Rightarrow x \in [- 12, \frac{7}{3})$ ... (ii)

On combining (i) and (ii), we get domain of f(x) is $[- 12, \frac{2}{7}]$ .

$∴ α = - 12 and β = \frac{2}{7}$

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

For g(x) to be defined, we have

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

$\Rightarrow \log_{27} (2 x + 5) < \frac{1}{3} \Rightarrow (2 x + 5) < {(27)}^{\frac{1}{3}}$

$\Rightarrow 2 x + 5 < 3 \Rightarrow 2 x < - 2 \Rightarrow x < - 1$

Also, 2x + 5 > 0 [For $\log_{27} (2 x + 5)$ to be defined]

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

$∴$ Domain of g(x) is $(\frac{- 5}{2}, - 1)$ $∴ γ = \frac{- 5}{2} and δ = - 1$

$∴$ $| 7 (α + β) + 4 (γ + δ) |$ $=$ $| 7 (- 12 + \frac{2}{7}) + 4 (\frac{- 5}{2} - 1) |$

$=$ $| - 82 - 14 |$ $= 96$

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