If 2 f sin x - f cos x - x - x - then range of f x isA- -3- -3-B- -2 -3- -3-C- -2 -3- -6-D -6- -6-

The correct option is B Put x = sin^ 1x 2f(x)+f(√(1 x^2))=sin^ 1 x → (1) x = cos ^ 1x ⇒ 2f (√(1 x^2))+f(x)=cos^ 1→ (2) (1)× (2) ⇒ 4f(x)+2f(√(1 x^2))=2sin^ ...

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Standard XII

Mathematics

Integration as Antiderivative

if 2 f sin x ...

Question

$i f 2 f (s i n x) + f (c o s x) = x \forall x ϵ t h e n r a n g e o f f (x) i s$

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Solution

The correct option is B
$P u t x = s i n^{- 1} x 2 f (x) + f (\sqrt{1 - x^{2}}) = s i n^{- 1} x \to (1) x = c o s^{- 1} x \Rightarrow 2 f (\sqrt{1 - x^{2}}) + f (x) = c o s^{- 1} \to (2) (1) \times (2) \Rightarrow 4 f (x) + 2 f (\sqrt{1 - x^{2}}) = 2 s i n^{- 1} x \frac{f (x) + 2 f (\sqrt{1 - x^{2}}) = c o s^{- 1} x}{3 f (x) = 2 s i n^{- 1} x - c o s^{- 1} x} f (x) = \frac{2}{3} s i n^{- 1} x - \frac{1}{3} (\frac{π}{3} - s i n^{- 1} x) = s i n^{- 1} x - \frac{π}{6} f_{m a x} = \frac{π}{2} - \frac{π}{6} = \frac{π}{3}, f_{m i n} = - \frac{π}{2} - \frac{π}{6} = \frac{- 4 π}{6} = \frac{- 2 π}{3} = [\frac{- 2 π}{3}, \frac{π}{3}]$

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if 2f(sinx)+f(cosx)=x∀xϵ then range of f(x) is

$i f 2 f (s i n x) + f (c o s x) = x \forall x ϵ t h e n r a n g e o f f (x) i s$