Test Index

VITEEE 2017 Paper

Section: Mathematics

Question : 99 of 125

Marks: +1, -0

\left[0, \frac{3\pi}{2}\right]

is -

Solution:

f(x) = 2sinx + sin2x

f ’ (x) = 2 cos x + 2 cos 2x = 2 (cos x + cos 2x)

∴ f'(x) = 0 ⇒ 2cos2x + cosx - 1 = 0

cos x =

\frac{-1\pm 3}{4}

= - 1 ,

\frac{1}{2}

∴ x =

\pi, \frac{\pi}{3}

Now , f (0) = 0 , f

\left(\frac{3\pi}{2}\right)

= - 2

f (π) = 0 , f

\left(\frac{\pi}{3}\right)

= 2

\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}

\frac{3\sqrt{3}}{2}

∴ difference between greatest value and least value

\frac{3\sqrt{3}}{2}

+ 2

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