If the domain of the function f(x)=loge(4x2+11x+6)+sin−1(4x+3)+cos−1(10x+63) is (α,β] . Then 8|α+β| is equal to

f(x)=loge(4x2+11x+6)+sin−1(4x+3)+cos−1(10x+63)loge(4x2+11x+6)⇒4x2+11x+6>04x2+8x+3x+6>0(4x+3)(x+2)>0x∈(−∞,−2)∪−34,∞sin−1(4x+3)⇒ -1≤4x+3≤1 ⇒-4≤4x≤2⇒x∈[−1,−1/2]cos−1(10x+63)⇒-1≤10x+63≤1 ⇒-9≤10x≤-3⇒x∈−910,−310⇒x∈−34,−12α=−34,β=−12α+β=−54

If the domain of the function f(x)=loge(4x2+11x+6)+sin−1(4x+3)+cos−1(10x+63) is (α,β] . Then 8|α+β| is equal to

f(x)=loge(4x2+11x+6)+sin−1(4x+3)+cos−1(10x+63)loge(4x2+11x+6)⇒4x2+11x+6>04x2+8x+3x+6>0(4x+3)(x+2)>0x∈(−∞,−2)∪−34,∞sin−1(4x+3)⇒ -1≤4x+3≤1 ⇒-4≤4x≤2⇒x∈[−1,−1/2]cos−1(10x+63)⇒-1≤10x+63≤1 ⇒-9≤10x≤-3⇒x∈−910,−310⇒x∈−34,−12α=−34,β=−12α+β=−54

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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 $8 | α + β |$ is equal to

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Detailed Solution

$f (x) = \log_{e} (4 x^{2} + 11 x + 6) + \sin^{- 1} (4 x + 3) + \cos^{- 1} (\frac{10 x + 6}{3})$
$\log_{e} (4 x^{2} + 11 x + 6)$
$\begin{array}{l} ⇒ 4 x^{2} + 11 x + 6 > 0 \\ 4 x^{2} + 8 x + 3 x + 6 > 0 \\ (4 x + 3) (x + 2) > 0 \\ x \in (- \infty, - 2) \cup (- \frac{3}{4}, \infty) \end{array}$
$\sin^{- 1} (4 x + 3)$
$\begin{array}{l} ⇒ -1≤ 4 x + 3 \leq 1 \Rightarrow - 4 \leq 4 x \leq 2 \\ \Rightarrow x \in [- 1, - 1 / 2] \end{array}$
$\cos^{- 1} (\frac{10 x + 6}{3})$
$\begin{array}{l} ⇒-1≤ \frac{10 x + 6}{3} ≤1 \Rightarrow - 9 \leq 10 x \leq - 3 \\ \Rightarrow x \in [- \frac{9}{10}, - \frac{3}{10}] \\ \Rightarrow x \in (- \frac{3}{4}, - \frac{1}{2}] \\ α = - \frac{3}{4}, β = - \frac{1}{2} \\ α + β = - \frac{5}{4} \end{array}$