The domain of the function f(x)= sin−1[log4(x4)]+17x−x2−16 is

f(x)=sin−1(logx(x4))+17x−x2−16 is defied iff −1≤log4(x4)≤1,and17x−x2−16≥0⇒ 4−1 ≤x4≤4, ⇒ x2−17x+16≤0⇒ 14 ≤x4≤ 4, ⇒ x2−16x−x+16≤0⇒ 1≤ x≤16, ⇒ (x−16)(x−1) ≤0∴x∈ [1, 6]

The domain of the function f(x)= sin−1[log4(x4)]+17x−x2−16 is

f(x)=sin−1(logx(x4))+17x−x2−16 is defied iff −1≤log4(x4)≤1,and17x−x2−16≥0⇒ 4−1 ≤x4≤4, ⇒ x2−17x+16≤0⇒ 14 ≤x4≤ 4, ⇒ x2−16x−x+16≤0⇒ 1≤ x≤16, ⇒ (x−16)(x−1) ≤0∴x∈ [1, 6]

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$T h e d o m a i n o f t h e f u n c t i o n f (x) = \sin^{- 1} [\log_{4} (\frac{x}{4})] + \sqrt{17 x - x^{2} - 16} i s$

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$[- 1, 1]$

$[1, 4]$

$(0, 16]$

$[1, 16]$

answer is D.

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Detailed Solution

$\begin{array}{l} f (x) = \sin^{- 1} (\log_{x} (\frac{x}{4})) + \sqrt{17 x - x^{2} - 16} i s d e f i e d \\ i f f - 1 \leq \log_{4} (\frac{x}{4}) \leq 1, a n d 17 x - x^{2} - 16 \geq 0 \\ \Rightarrow 4^{- 1} \leq \frac{x}{4} \leq 4, \Rightarrow x^{2} - 17 x + 16 \leq 0 \\ \Rightarrow \frac{1}{4} \leq \frac{x}{4} \leq 4, \Rightarrow x^{2} - 16 x - x + 16 \leq 0 \\ \Rightarrow 1 \leq x \leq 16, \Rightarrow (x - 16) (x - 1) \leq 0 \\ ∴ x \in [1, 6] \end{array}$