Let f(x)=(π2−cos−1(cosx))(π2−sin−1(sinx)), then

f(x)=π2−cos−1⁡(cos⁡x)π2−sin−1⁡(sin⁡x)f(x)=π2−x2x∈0,π2π2−x2x∈π2,πx−3π2x−π2=x2−2πx+3π24x∈π,3π2f(x)=x−3π25π2−x=−x2+4πx−15π24x∈3π2,2π; Number of points of non differentiability =3 Range ∈−π24,π24

Let f(x)=(π2−cos−1(cosx))(π2−sin−1(sinx)), then

f(x)=π2−cos−1⁡(cos⁡x)π2−sin−1⁡(sin⁡x)f(x)=π2−x2x∈0,π2π2−x2x∈π2,πx−3π2x−π2=x2−2πx+3π24x∈π,3π2f(x)=x−3π25π2−x=−x2+4πx−15π24x∈3π2,2π; Number of points of non differentiability =3 Range ∈−π24,π24

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Let $f (x) = (\frac{π}{2} - \cos^{- 1} (\cos x)) (\frac{π}{2} - \sin^{- 1} (\sin x)),$ then

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Range of $f (x)$ is $[- \frac{π^{2}}{4}, \frac{π^{2}}{4}]$

Range of $f (x)$ is $[- \frac{π^{2}}{2}, \frac{π^{2}}{2}]$

Number of points in the interval $(0, 4 π)$ where $f (x)$ is non-differentiable is 4

Number of points in the interval $(0, 4 π)$ where $f (x)$ is non-differentiable is 3

answer is A, D.

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Detailed Solution

$f (x) = (\frac{π}{2} - \cos^{- 1} (\cos x)) (\frac{π}{2} - \sin^{- 1} (\sin x))$

$f (x) = \{\begin{cases} {(\frac{π}{2} - x)}^{2} & x \in (0, \frac{π}{2}) \\ {(\frac{π}{2} - x)}^{2} & x \in (\frac{π}{2}, π) \\ (x - \frac{3 π}{2}) (x - \frac{π}{2}) = x^{2} - 2 π x + \frac{3 π^{2}}{4} & x \in (π, \frac{3 π}{2}) \\ f (x) = (x - \frac{3 π}{2}) (\frac{5 π}{2} - x) = - x^{2} + 4 π x - \frac{15 π^{2}}{4} & x \in (\frac{3 π}{2}, 2 π) \end{cases};$