If x in [0, 1],then the number of solution(s) | vedclass

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(B) Given the equation: $2[\cos^{-1}x] + 6[	ext{sgn}(\sin x)] = 3$ for $x \in [0, 1]$.Step $1$: Analyze the range of $\cos^{-1}x$ for $x \in [0, 1]$.

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$0$

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(B) Given the equation: $2[\cos^{-1}x] + 6[	ext{sgn}(\sin x)] = 3$ for $x \in [0, 1]$.Step $1$: Analyze the range of $\cos^{-1}x$ for $x \in [0, 1]$.Since $x \in [0, 1]$,$\cos^{-1}x \in [0, \pi/2]$.Thus,$[\cos^{-1}x]$ can take values $0$ or $1$ (since $\pi/2 \approx 1.57$).Step $2$: Analyze the range of $	ext{sgn}(\sin x)$ for $x \in [0, 1]$.For $x \in (0, 1]$,$\sin x > 0$,so $	ext{sgn}(\sin x) = 1$.For $x = 0$,$\sin 0 = 0$,so $	ext{sgn}(0) = 0$.Step $3$: Test the cases.Case $1$: If $x = 0$,then $[\cos^{-1}0] = [\pi/2] = 1$ and $	ext{sgn}(\sin 0) = 0$.The equation becomes $2(1) + 6(0) = 2 
eq 3$.Case $2$: If $x \in (0, 1]$,then $	ext{sgn}(\sin x) = 1$,so $[	ext{sgn}(\sin x)] = 1$.The equation becomes $2[\cos^{-1}x] + 6(1) = 3$,which simplifies to $2[\cos^{-1}x] = -3$,or $[\cos^{-1}x] = -1.5$.Since the greatest integer function must result in an integer,there is no solution for this case.Conclusion: There are no values of $x$ that satisfy the equation. Therefore,the number of solutions is $0$.

If $x \in [0, 1]$,then the number of solution$(s)$ of the equation $2[\cos^{-1}x] + 6[\text{sgn}(\sin x)] = 3$ is (where $[.]$ denotes the greatest integer function and $\text{sgn}(x)$ denotes the signum function of $x$)-

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