JEE Advance - Mathematics (2020 - Paper 1 Offline - No. 15)
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 ..........
$$f(x) = (3 - \sin (2\pi x))\sin \left( {\pi x - {\pi \over 4}} \right) - \sin \left( {3\pi x + {\pi \over 4}} \right)$$
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 ..........
Answer
1
Explanation
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 $$4 + \sin (2\pi x) > 0\,\forall x \in R$$.
$$ \Rightarrow \pi x \in \left[ {{\pi \over 4},{{5\pi } \over 4}} \right] \Rightarrow x \in \left[ {{1 \over 4},{5 \over 4}} \right]$$
$$ \therefore $$ $$\alpha = {1 \over 4}$$ and $$\beta = {5 \over 4}$$
Therefore the value of $$(\beta - \alpha ) = 1$$
$$f(x) = (3 - \sin (2\pi x))\sin \left( {\pi x - {\pi \over 4}} \right) - \sin \left( {3\pi x + {\pi \over 4}} \right)$$
$$ = (3 - \sin (2\pi x))\left[ {{{\sin \pi x} \over {\sqrt 2 }} - {{\cos \pi x} \over {\sqrt 2 }}} \right] - \left\{ {{{\sin 3\pi x} \over {\sqrt 2 }} + {{\cos (3\pi x)} \over {\sqrt 2 }}} \right\}$$
$$ = (3 - \sin (2\pi x)){{[\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }} - {1 \over {\sqrt 2 }}[3\sin (\pi x) - 4{\sin ^3}(\pi x) + 4{\cos ^3}(\pi x) - 3\cos (\pi x)]$$
$$ = {{\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }}[3 - \sin (2\pi x) - 3 + 4\{ {\sin ^2}(\pi x) + {\cos ^2}(\pi x) + \sin (\pi x)\cos (\pi x)\} ]$$
$$ = {{\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }}[4 + \sin (2\pi x)]$$
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 $$4 + \sin (2\pi x) > 0\,\forall x \in R$$.
$$ \Rightarrow \pi x \in \left[ {{\pi \over 4},{{5\pi } \over 4}} \right] \Rightarrow x \in \left[ {{1 \over 4},{5 \over 4}} \right]$$
$$ \therefore $$ $$\alpha = {1 \over 4}$$ and $$\beta = {5 \over 4}$$
Therefore the value of $$(\beta - \alpha ) = 1$$
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