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Functions

Section: Mathematics

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Question : 14 of 30

Marks: +1, -0

Let

f:[0,2]\rightarrow R

be the function defined by

f(x)=(3-\sin(2\pi x))\sin\left(\pi x-\frac{\pi}{4}\right)

-\sin\left(3\pi x+\frac{\pi}{4}\right)

\alpha, \beta

∈ [0,2] are such that

\{x\in[0,2]: f(x)\ge 0\}=[\alpha, \beta],

then the value of

\beta-\alpha

[JEE Adv 2020 P1]

Your Answer:

Solution:

Let

\pi x=\theta

f(x)=(3-\sin 2\theta)\sin\left(\theta-\frac{\pi}{4}\right)

-\sin\left(3\theta+\frac{\pi}{4}\right)

=(3-\sin 2\theta)\frac{\sin\theta-\cos\theta}{\sqrt{2}}

-\left(\frac{\sin 3\theta}{\sqrt{2}}+\frac{\cos 3\theta}{\sqrt{2}}\right)

=\frac{1}{\sqrt{2}}\left[(3-\sin 2\theta)(\sin\theta-\cos\theta)\right.

-\left(3\sin\theta-3\cos\theta-4\sin^{3}\theta+4\cos^{3}\theta\right)]

=\frac{1}{\sqrt{2}}\left[(3-\sin 2\theta)(\sin\theta-\cos\theta)\right.

-\left\{3(\sin\theta-\cos\theta)-4(\sin\theta-\cos\theta)\right.

\left(1+\frac{\sin 2\theta}{2}\right)\}]

=\frac{(\sin\theta-\cos\theta)}{\sqrt{2}}(3-\sin 2\theta

+1+2\sin 2\theta)

\therefore

For

f(x)\ge 0

\therefore[\alpha, \beta]=\left[\frac{1}{4}, \frac{5}{4}\right]

\Rightarrow[\beta-\alpha]=\frac{5}{4}-\frac{1}{4}=1

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