If fx-sin-1cos xcos-1sin x - x - -0-2- then the number of points of non differentiability is

Given : nbsp;f(x)=sin^ 1(cos x)cos^ 1(sin x) ⇒ f(x)= ( π2 cos^ 1(cos x) ) ( π2 sin^ 1(sin x)) ⇒ f(x) = (π2 x ) ( π2 x ) ; 0≤ x≤π2 ( π2 x ) ( π2 (π x) ...

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Standard XII

Mathematics

Graphical Interpretation of Differentiability

If fx=sin-1co...

Question

If $f (x) = {sin}^{- 1} (cos x) {cos}^{- 1} (sin x) \forall x \in [0, 2 π],$ then the number of point(s) of non differentiability is

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Solution

The correct option is B $1$
Given : $f (x) = {sin}^{- 1} (cos x) {cos}^{- 1} (sin x)$
$\Rightarrow f (x) = (\frac{π}{2} - {cos}^{- 1} (cos x)) (\frac{π}{2} - {sin}^{- 1} (sin x)) \Rightarrow f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} (\frac{π}{2} - x) (\frac{π}{2} - x); 0 \leq x \leq \frac{π}{2} (\frac{π}{2} - x) (\frac{π}{2} - (π - x)); \frac{π}{2} \leq x \leq π (\frac{π}{2} - (2 π - x)) (\frac{π}{2} - (π - x)); π \leq x < \frac{3 π}{2} (\frac{π}{2} - (2 π - x)) (\frac{π}{2} - (x - 2 π)); \frac{3 π}{2} \leq x \leq 2 π \end{matrix}$

$\Rightarrow f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} {(\frac{π}{2} - x)}^{2}; 0 \leq x \leq \frac{π}{2} - {(\frac{π}{2} - x)}^{2}; \frac{π}{2} \leq x \leq π (π - x)^{2} - \frac{π^{2}}{4}; π \leq x < \frac{3 π}{2} \frac{π^{2}}{4} - (2 π - x)^{2}; \frac{3 π}{2} \leq x \leq 2 π \end{matrix}$

$f (x)$ is non differentiable at $x = π$

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If f(x)=sin−1(cosx)cos−1(sinx) ∀x∈[0,2π], then the number of point(s) of non differentiability is

If $f (x) = {sin}^{- 1} (cos x) {cos}^{- 1} (sin x) \forall x \in [0, 2 π],$ then the number of point(s) of non differentiability is