If the domain of the function f ( x ) = log e ( 4 x 2 + 11 x + 6 ) + sin - 1 ( 4 x + 3 ) + cos - 1 ( 10 x + 6 3 ) is then 36 is equal to [2023]

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Solution

Q.
If the domain of the function $f (x) = \log_{e} (4 x^{2} + 11 x + 6) + \sin^{- 1} (4 x + 3) + \cos^{- 1} (\frac{10 x + 6}{3})$ is $(α, β],$ then $36$ $| α + β |$ is equal to [2023]

1 45

2 63

3 72

4 54

Ans.
(1)

We have,

$f (x) = \log_{e} (4 x^{2} + 11 x + 6) + \sin^{- 1} (4 x + 3) + \cos^{- 1} (\frac{10 x + 6}{3})$

Let $f_{1} (x) = \log_{e} (4 x^{2} + 11 x + 6)$

$f_{2} (x) = \sin^{- 1} (4 x + 3)$

and $f_{3} (x) = \cos^{- 1} (\frac{10 x + 6}{3})$

Now, we will find the domain of $f_{1}, f_{2}$ and $f_{3}$

Consider, $f_{1} (x) = \log (4 x^{2} + 11 x + 6)$

Now, $4 x^{2} + 11 x + 6 > 0$

$\Rightarrow x > \frac{- 11 + \sqrt{121 - 96}}{8} and x < \frac{- 11 - \sqrt{121 - 96}}{8}$

$\Rightarrow x > \frac{- 3}{4} and x < - 2$

Consider, $f_{2} (x) = \sin^{- 1} (4 x + 3)$

So, $- 1 \leq 4 x + 3 \leq 1$

$\Rightarrow - 4 \leq 4 x \leq - 2 \Rightarrow - 1 \leq x \leq - \frac{1}{2}$

$∴$ $D (f_{2} (x)) is$ $[- 1, - \frac{1}{2}]$

Consider, $f_{3} (x) = \cos^{- 1} (\frac{10 x + 6}{3})$

So, $- 1 \leq \frac{10 x + 6}{3} \leq 1$

$\Rightarrow - 3 \leq 10 x + 6 \leq 3 \Rightarrow \frac{- 9}{10} \leq x \leq \frac{- 3}{10}$

$\Rightarrow D (f_{3} (x)) is [\frac{- 9}{10}, \frac{- 3}{10}]$

Now, $D (f (x))$ is $D (f_{1} (x)) \cap D (f_{2} (x)) \cap D (f_{3} (x))$

$= (- \infty, - 2) \cup (\frac{- 3}{4}, \infty) \cap [- 1, \frac{- 1}{2}] \cap [\frac{- 9}{10}, \frac{- 3}{10}] = (\frac{- 3}{4}, \frac{- 1}{2}]$

So, $α = - \frac{3}{4} and β = \frac{- 1}{2}$

So, $36$ $| α + β |$ $= 36$ $| - \frac{3}{4} - \frac{1}{2} |$ $= 36 \times \frac{5}{4} = 45$

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