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Find the difference between the greatest and least values of the function f(x) = sin2x – x, on [-π2,π2]

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प्रश्न

Find the difference between the greatest and least values of the function f(x) = sin2x – x, on `[- pi/2, pi/2]`

योग
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उत्तर

f(x) = sin2x – x

⇒ f′(x) = 2 cos2x – 1

Therefore, f′(x) = 0

⇒ cos2x = `1/2`

⇒ 2x is `(-pi)/3` or `pi/3`

⇒ x = `- pi/6` or `pi/6`

`"f"(- pi/2) = sin(- pi) + pi/2 = pi/2`

`"f"(- pi/6) = sin(-(2pi)/6) + pi/6 - - sqrt(3)/2 + pi/6`

`"f"(pi/6) = sin((2pi)/6) - pi/6 = sqrt(3)/2 - pi/6`

`"f"(pi/2) = sin(pi) - pi/2 = - pi/2`

Clearly, `pi/2` is the greatest value and `- pi/2` is the least.

Therefore, difference = `pi/2 + pi/2` = π

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अध्याय 6: Application Of Derivatives - Solved Examples [पृष्ठ १३०]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 6 Application Of Derivatives
Solved Examples | Q 17 | पृष्ठ १३०

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