If 2fsin x-fcos x-x - x - R then range of fx is

The correct option is B [ 2π/3,π/3 ]Put x=sin^ 1x 2f(x)+f(√(1 x^2))=sin^ 1x→ (1) On Putting x=cos^ 1x ⇒ 2f(√(1 x^2))+f(x)=cos^ 1x→ (2) Eq.(1)× 2⇒ 4f(x)+2f(√(1 ...

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Byju's Answer

Standard X

Mathematics

Using Monotonicity to Find the Range of a Function

If 2fsin x+fc...

Question

If $2 f (s i n x) + f (c o s x) = x \forall x ϵ$ R then range of f(x) is

[\frac{- π}{3}, \frac{π}{3}]

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[\frac{- 2 π}{3}, \frac{π}{3}]

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[\frac{- 2 π}{3}, \frac{π}{6}]

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[\frac{- π}{6}, \frac{π}{6}]

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Solution

The correct option is B $[\frac{- 2 π}{3}, \frac{π}{3}]$
Put $x = s i n^{- 1} x$
$2 f (x) + f (\sqrt{1 - x^{2}}) = s i n^{- 1} x \to (1)$
On Putting $x = c o s^{- 1} x$
$\Rightarrow 2 f (\sqrt{1 - x^{2}}) + f (x) = c o s^{- 1} x \to (2)$
Eq. $(1) \times 2 \Rightarrow 4 f (x) + 2 f (\sqrt{1 - x^{2}}) = 2 s i n^{- 1} x \to (3)$
On subtracting Eq. 2 from Eq. 3 we get -
$3 f (x) = 2 s i n^{- 1} x - c o s^{- 1} x$
$f (x) = \frac{2}{3} s i n^{- 1} x - \frac{1}{3} (\frac{π}{2} - s i n^{- 1} x)$
$= s i n^{- 1} x - \frac{π}{6}$
$f_{m a x} = \frac{π}{2} - \frac{π}{6} = \frac{π}{3}, f_{m i n} = - \frac{π}{2} - \frac{π}{6} = \frac{- 4 π}{6} = \frac{- 2 π}{3}$
$= [\frac{- 2 π}{3}, \frac{π}{3}]$

Suggest Corrections

If 2f(sin x)+f(cos x)=x ∀ x ϵ R then range of f(x) is

If $2 f (s i n x) + f (c o s x) = x \forall x ϵ$ R then range of f(x) is