QJEEAdv20Ma4fc0: JEE Advanced 2020 Mathematics - Class 11 JEEAdv20Ma4fc0 Solution: Let f : [0, 2] R be the function defined by If are such that ,…

Question

Verified Solution • 2025–26 Session moderate

Let f : [0, 2]

\to

R be the function defined by

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

If

\alpha ,\,\beta \in [0,2]

are such that

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

, then the value of

\beta - \alpha

is ..........

Accepted Answer

The given function f : [0, 2]

\to

R defined by

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

= (3 - \sin (2\pi x))\left[ {{{\sin \pi x} \over {\sqrt 2 }} - {{\cos \pi x} \over {\sqrt 2 }}} \right] - \left\{ {{{\sin 3\pi x} \over {\sqrt 2 }} + {{\cos (3\pi x)} \over {\sqrt 2 }}} \right\}

= (3 - \sin (2\pi x)){{[\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }} - {1 \over {\sqrt 2 }}[3\sin (\pi x) - 4{\sin ^3}(\pi x) + 4{\cos ^3}(\pi x) - 3\cos (\pi x)]

= {{\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }}[3 - \sin (2\pi x) - 3 + 4\{ {\sin ^2}(\pi x) + {\cos ^2}(\pi x) + \sin (\pi x)\cos (\pi x)\} ]

= {{\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }}[4 + \sin (2\pi x)]

As,

f(x) \ge 0\forall \in [\alpha ,\beta ]

, where

\alpha ,\beta \in [0,2]

, so

\sin (\pi x) - \cos (\pi x) \ge 0

as

4 + \sin (2\pi x) &gt; 0\,\forall x \in R

.

\Rightarrow \pi x \in \left[ {{\pi \over 4},{{5\pi } \over 4}} \right] \Rightarrow x \in \left[ {{1 \over 4},{5 \over 4}} \right]

\therefore

\alpha = {1 \over 4}

and

\beta = {5 \over 4}

Therefore the value of

(\beta - \alpha ) = 1

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