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