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Question
Let [x] denote the greatest integer less than or equal to x. If \[f\left( x \right) = \sin^{- 1} x, g\left( x \right) = \left[ x^2 \right]\text{ and } h\left( x \right) = 2x, \frac{1}{2} \leq x \leq \frac{1}{\sqrt{2}}\]
Options
\[\text{fogoh}\left( x \right) = \frac{\pi}{2}\]
fogoh (x) = π
\[\text{ho f og = hogo f}\]
\[\text{ho f og ≠ hogo f}\]
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Solution
(c) \[\text{ho fog = hogo f}\]
\[\text{We have}, \]
\[g\left( x \right) = \left[ x^2 \right] \]
\[ = 0 \left(As\frac{1}{2} \leq x \leq \frac{1}{\sqrt{2}}\therefore \frac{1}{4} \leq x^2 \leq \frac{1}{2} \right)\]
\[\text{fog}\left( x \right) = f\left( g\left( x \right) \right) = \sin^{- 1} \left( 0 \right)\]
\[ = 0\]
\[\text{hofog}\left( x \right) = h\left( f\left( g\left( x \right) \right) \right) = 2 \times 0 = 0\]
\[\text{And}\]
\[f\left( x \right) = \sin^{- 1} x\]
\[Now, \]
\[for, x \in \left[ \frac{1}{2}, \frac{1}{\sqrt{2}} \right]\]
\[f\left( x \right) \in \left[ \frac{\pi}{6}, \frac{\pi}{4} \right]\]
\[f\left( x \right) \in \left[ 0 . 52, 0 . 78 \right]\]
\[gof\left( x \right) = 0 \left( As, f\left( x \right) \in \left[ 0 . 52, 0 . 78 \right] \right)\]
\[ = 0\]
\[\text{hogof}\left( x \right) = h\left( g\left( f\left( x \right) \right) \right) = 2 \times 0 = 0\]
\[\therefore \text{hofog = hogof} = 0\]
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