मराठी

F(x) = sinx – sin2x in [0, π] - Mathematics

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प्रश्न

f(x) = sinx – sin2x in [0, π]

बेरीज

उत्तर

We have, f(x) = sinx – sin2x in [0, π]

We know that all trigonometric functions are continuous and differentiable their domain, given function is also continuous and differentiable

So, condition of mean value theorem are satisfied.

Hence, there exists atleast one c ∈ (0, π) such that,

f'(c) = `("f"(pi) - "f"(0))/(pi - 0)`

⇒ cos c – 2 cos 2c =  `(sin pi - sin 2pi - sin 0 + sin 0)/(pi - 0)`

⇒ 2 cos 2c – cos c = 0

⇒ 2(2 cos2c – 1) – cos c = 0

⇒ 4cos2c – cos c – 2 = 0

⇒ cos c = `(1 +- sqrt(1 + 32))/8`

= `(1 +- sqrt(33))/8`

⇒ c = `cos^-1 ((1 +- sqrt(33))/8) ∈ (0, π)`

Hence, mean value theorem has been verified.

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पाठ 5: Continuity And Differentiability - Exercise [पृष्ठ ११२]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 5 Continuity And Differentiability
Exercise | Q 75 | पृष्ठ ११२

व्हिडिओ ट्यूटोरियलVIEW ALL [3]

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