NTA Abhyas 2020
Exam NTA Abhyas
Year 2020
Subject Mathematics
Chapter Differentiation NTA Abhyas 2020 · Mathematics · Differentiation
If f ( x ) = co s ^ - 1 ( x (4co s ^ 2 x - 1)) , then (1)/(π) f ^ '...
If
f ( x ) = c o s − 1 f ( x ) = co{s}^{- 1} f ( x ) = co s − 1 \left(\sin x
\left(4co{s}^{2}x - 1\right)\right) , then
1 π f ′ ( π 3 ) ⋅ f ( π 10 ) \dfrac{1}{\pi}{f}^{'}\left(\dfrac{\pi}{3}\right) ⋅ f\left(\dfrac{\pi}{10}\right) π 1 f ′ ( 3 π ) ⋅ f ( 10 π ) is
Step-by-step solution
f ( x ) = c o s − 1 { s i n x ( 4 ( 1 − s i n 2 x ) − 1 ) } f\left(x\right) = co{s}^{- 1}\left\{sin x\left(4\left(1 - si{n}^{2}x\right) - 1\right)\right\} f ( x ) = co s − 1 { s in x ( 4 ( 1 − s i n 2 x ) − 1 ) } f ( x ) = c o s − 1 { s i n x ( 3 − 4 s i n 2 x ) } = c o s − 1 { 3 s i n x − 4 s i n 3 x } f\left(x\right) = co{s}^{- 1}\left\{sin x\left(3 - 4si{n}^{2}x\right)\right\} = co{s}^{- 1}\left\{3sin x - 4si{n}^{3}x\right\} f ( x ) = co s − 1 { s in x ( 3 − 4 s i n 2 x ) } = co s − 1 { 3 s in x − 4 s i n 3 x } f ( x ) = c o s − 1 ( sin 3 x ) = π 2 − s i n − 1 ( sin 3 x ) f\left(x\right) = co{s}^{- 1}\left(\sin 3x\right) = \dfrac{\pi}{2} - si{n}^{- 1}\left(\sin 3x\right) f ( x ) = co s − 1 ( sin 3 x ) = 2 π − s i n − 1 ( sin 3 x ) In the neighbourhood of
x = π 3 ⇒ f ( x ) = π 2 − ( π − 3 x ) x = \dfrac{\pi}{3} ⇒ f\left(x\right) = \dfrac{\pi}{2} - \left(\pi - 3x\right) x = 3 π ⇒ f ( x ) = 2 π − ( π − 3 x ) f ( x ) = 3 x − π 2 f\left(x\right) = 3x - \dfrac{\pi}{2} f ( x ) = 3 x − 2 π ∴ f ′ ( π 3 ) = 3 ∴ {f}^{'}\left(\dfrac{\pi}{3}\right) = 3 ∴ f ′ ( 3 π ) = 3 In the neighbourhood of
x = π 10 ⇒ f ( x ) = π 2 − ( 3 x ) x = \dfrac{\pi}{10} ⇒ f\left(x\right) = \dfrac{\pi}{2} - \left(3x\right) x = 10 π ⇒ f ( x ) = 2 π − ( 3 x ) f ( π 10 ) = π 2 − 3 π 10 = π 5 f\left(\dfrac{\pi}{10}\right) = \dfrac{\pi}{2} - \dfrac{3\pi}{10} = \dfrac{\pi}{5} f ( 10 π ) = 2 π − 10 3 π = 5 π
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