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Let f(x)=sin^(-1)x+|sin^(-1)x|+sin^(-1)|...

Let `f(x)=sin^(-1)x+|sin^(-1)x|+sin^(-1)|x|` The range of f(x) is (a) `[0,(pi)/(2)]` (b) `[0,(3pi)/(2)]` (c) `[0,(pi)/(4)]` (d) `[0,pi]`

A

`[0,(pi)/(2)]`

B

`[0,(3pi)/(2)]`

C

`[0,(pi)/(4)]`

D

`[0,pi]`

Text Solution

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To find the range of the function \( f(x) = \sin^{-1} x + |\sin^{-1} x| + \sin^{-1} |x| \), we will analyze the function for two cases: when \( x \geq 0 \) and when \( x < 0 \). ### Step 1: Case 1 - \( x \geq 0 \) For \( x \geq 0 \): - \( \sin^{-1} x \) is defined and non-negative since \( x \) is in the interval \([0, 1]\). - Thus, \( |\sin^{-1} x| = \sin^{-1} x \). - Also, \( \sin^{-1} |x| = \sin^{-1} x \) since \( |x| = x \). Therefore, in this case: \[ f(x) = \sin^{-1} x + \sin^{-1} x + \sin^{-1} x = 3 \sin^{-1} x \] ### Step 2: Determine the range for \( x \in [0, 1] \) The function \( \sin^{-1} x \) ranges from \( 0 \) to \( \frac{\pi}{2} \) as \( x \) goes from \( 0 \) to \( 1 \). Therefore: - When \( x = 0 \), \( f(0) = 3 \cdot \sin^{-1}(0) = 0 \). - When \( x = 1 \), \( f(1) = 3 \cdot \sin^{-1}(1) = 3 \cdot \frac{\pi}{2} = \frac{3\pi}{2} \). Thus, for \( x \geq 0 \), the range of \( f(x) \) is: \[ [0, \frac{3\pi}{2}] \] ### Step 3: Case 2 - \( x < 0 \) For \( x < 0 \): - \( \sin^{-1} x \) is negative, so \( |\sin^{-1} x| = -\sin^{-1} x \). - Since \( x \) is negative, \( |x| = -x \), thus \( \sin^{-1} |x| = \sin^{-1} (-x) = -\sin^{-1} x \). Therefore, in this case: \[ f(x) = \sin^{-1} x - \sin^{-1} x - \sin^{-1} x = -\sin^{-1} x \] ### Step 4: Determine the range for \( x \in [-1, 0) \) The function \( \sin^{-1} x \) ranges from \( -\frac{\pi}{2} \) to \( 0 \) as \( x \) goes from \( -1 \) to \( 0 \). Therefore: - When \( x = -1 \), \( f(-1) = -\sin^{-1}(-1) = -(-\frac{\pi}{2}) = \frac{\pi}{2} \). - When \( x \) approaches \( 0 \), \( f(x) \) approaches \( -\sin^{-1}(0) = 0 \). Thus, for \( x < 0 \), the range of \( f(x) \) is: \[ (0, \frac{\pi}{2}] \] ### Step 5: Combine the ranges Now we combine the ranges from both cases: - From Case 1 (for \( x \geq 0 \)): \( [0, \frac{3\pi}{2}] \) - From Case 2 (for \( x < 0 \)): \( (0, \frac{\pi}{2}] \) The overall range of \( f(x) \) is: \[ [0, \frac{3\pi}{2}] \] ### Conclusion Thus, the range of \( f(x) \) is \( [0, \frac{3\pi}{2}] \), which corresponds to option (b). ---

To find the range of the function \( f(x) = \sin^{-1} x + |\sin^{-1} x| + \sin^{-1} |x| \), we will analyze the function for two cases: when \( x \geq 0 \) and when \( x < 0 \). ### Step 1: Case 1 - \( x \geq 0 \) For \( x \geq 0 \): - \( \sin^{-1} x \) is defined and non-negative since \( x \) is in the interval \([0, 1]\). - Thus, \( |\sin^{-1} x| = \sin^{-1} x \). - Also, \( \sin^{-1} |x| = \sin^{-1} x \) since \( |x| = x \). ...
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