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Prove that f(x) = sinx + 3 cosx has maximum value at x = π6

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प्रश्न

Prove that f(x) = sinx + `sqrt(3)` cosx has maximum value at x = `pi/6`

योग
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उत्तर

We have: f (x) = sinx + `sqrt(3)` cosx 

= `2(1/2 sin x + sqrt(3)/2  cos x)`

= `2(cos  pi/3  sin x + sin  pi/3  cos x)`

= `2 sin (x + pi/3)`

f'(x) = `2cos(x + pi/3)`

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

`"f''"(x)_(x = pi/6) = - 2 sin (pi/6 + pi/3)`

= `- 2 sin  pi/2`

= – 2.1

= – 2< 0  ....(Maxima)

= `- 2 xx sqrt(3)/2`

= `- sqrt(3) < 0`  .....(Maxima)

Maximum value of the function at x = `pi/6` is 

`sin  pi/6 + sqrt(3) cos  pi/6 = 1/2 + sqrt(3) * sqrt(3)/2` = 2

Hence, the given function has maximum value at x = `pi/6` and the maximum value is 2.

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अध्याय 6: Application Of Derivatives - Exercise [पृष्ठ १३७]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 6 Application Of Derivatives
Exercise | Q 24 | पृष्ठ १३७

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