Let fx-sin-1sin x- - -x2 - Then number of solutions of equation fx-0 in x- - - is

F(x)=|sin^ 1(sin x)| ( π x2 ) Draw curve of y=|sin^ 1(sin x)| and y=π x2 Now find number of intersection points of curves. So, there are two solutions of f( ...

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Byju's Answer

Standard XI

Mathematics

Properties of Conjugate of a Complex Number

Let fx=|sin-1...

Question

Let $f (x) = ∣ ∣ {sin}^{- 1} (sin x) ∣ ∣ - (\frac{π - x}{2})$ . Then number of solutions of equation $f (x) = 0$ in $x \in [- π, π]$ is

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Solution

The correct option is A $2$
$f (x) = | {sin}^{- 1} (sin x) | - (\frac{π - x}{2})$
Draw curve of $y = | {sin}^{- 1} (sin x) |$ and $y = \frac{π - x}{2}$
Now find number of intersection points of curves.

So, there are $2$ solutions of $f (x) = 0$

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Let f(x)=∣∣sin−1(sinx)∣∣−(π−x2). Then number of solutions of equation f(x)=0 in x∈[−π,π] is

The correct option is A 2f(x)=|sin−1(sin x)|−(π−x2) Draw curve of y=|sin−1(sin x)| and y=π−x2 Now find number of intersection points of curves. So, there are 2 solutions of f(x)=0

Let $f (x) = ∣ ∣ {sin}^{- 1} (sin x) ∣ ∣ - (\frac{π - x}{2})$ . Then number of solutions of equation $f (x) = 0$ in $x \in [- π, π]$ is

The correct option is A $2$
$f (x) = | {sin}^{- 1} (sin x) | - (\frac{π - x}{2})$
Draw curve of $y = | {sin}^{- 1} (sin x) |$ and $y = \frac{π - x}{2}$
Now find number of intersection points of curves.

So, there are $2$ solutions of $f (x) = 0$