NIMCET 2015 Question Paper with solutions

Question : 43 of 120

Marks: +1, -0

A function f : (0, π) → R defined by f(x) = 2 sin x + cos 2x has

Solution:

We have,

f(x)=2\sin x+\cos 2x

f'(x)=2\cos x-2\sin 2x

f"(x)=-2\sin x-4\cos2x

For maxima or minima,

f'(x)=0

⇒2\cos x-2\sin2x=0

⇒\cos x-2\sin x \cos x=0

⇒\cos x(1-2\sin x)=0

\cos x=0

1-2\sin x=0

Now,

f''({π}/{2})=-2\sin{π}/{2}-4\cos({2π}/{2})

=-2+4=2>0

f''({π}/{6})=-2\sin{π}/{6}-4\cos({2π}/{6})

=-1-2=-3<0

f''({5π}/{6})=-2\sin{5π}/{6}-4\cos({10π}/{6})

=-1-2=-3<0

Hence, fix) is minimum at

π/2

and maximum at

π/6

∴ f(x) has both local minimum and local maximum.

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