Let the domain of the function f(x) = cos^{-1 | vedclass

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(B) For $f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)$,we require $-1 \leq \frac{4x+5}{3x-7} \leq 1$.Case $1$: $\frac{4x+5}{3x-7} \geq -1$ $\Rightar

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(B) For $f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)$,we require $-1 \leq \frac{4x+5}{3x-7} \leq 1$.Case $1$: $\frac{4x+5}{3x-7} \geq -1$ $\Rightarrow \frac{4x+5+3x-7}{3x-7} \geq 0$ $\Rightarrow \frac{7x-2}{3x-7} \geq 0$. This gives $x \in (-\infty, 2/7] \cup (7/3, \infty)$.Case $2$: $\frac{4x+5}{3x-7} \leq 1$ $\Rightarrow \frac{4x+5-3x+7}{3x-7} \leq 0$ $\Rightarrow \frac{x+12}{3x-7} \leq 0$. This gives $x \in [-12, 7/3)$.Taking the intersection,the domain of $f(x)$ is $[-12, 2/7]$. Thus,$\alpha = -12$ and $\beta = 2/7$.For $g(x) = \log_2(2-6\log_{27}(2x+5))$,we require $2-6\log_{27}(2x+5) > 0$ and $2x+5 > 0$.$6\log_{27}(2x+5) Also,$2x+5 > 0 \Rightarrow x > -5/2$.So,the domain of $g(x)$ is $(-5/2, -1)$. Thus,$\gamma = -5/2$ and $\delta = -1$.Now,$|7(\alpha+\beta) + 4(\gamma+\delta)| = |7(-12 + 2/7) + 4(-5/2 - 1)| = |-84 + 2 - 10 - 4| = |-96| = 96$.

Let the domain of the function $f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)$ be $[\alpha, \beta]$ and the domain of $g(x) = \log_2\left(2-6\log_{27}(2x+5)\right)$ be $(\gamma, \delta)$. Then $|7(\alpha+\beta)+4(\gamma+\delta)|$ is equal to . . . . . . .

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