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JEE Advanced 2020 · Paper 1, Mathematics question 15

Let f:[0,2]→Rf:[0,2]\to\mathbb R be the function defined by

f(x)=(3−sin⁡(2πx))sin⁡(πx−π4)−sin⁡(3πx+π4).f(x) = \left(3 - \sin(2\pi x)\right)\sin\left(\pi x - \frac\pi4\right) - \sin\left(3\pi x + \frac\pi4\right).

If α,β∈[0,2]\alpha, \beta \in [0,2] are such that {x∈[0,2]:f(x)≥0}=[α,β]\{x\in[0,2] : f(x)\ge 0\} = [\alpha,\beta], then the value of β−α\beta - \alpha is ______

Answer and worked solution

Official answer: 1

Substitute t=πx−π4t=\pi x-\frac\pi4. Then:

  • sin⁡2πx=sin⁡(2t+π2)=cos⁡2t\sin2\pi x=\sin\left(2t+\frac\pi2\right)=\cos2t;
  • sin⁡(3πx+π4)=sin⁡(3t+π)=−sin⁡3t\sin\left(3\pi x+\frac\pi4\right)=\sin(3t+\pi)=-\sin3t.

Simplify, with s=sin⁡ts=\sin t:
f=(3−cos⁡2t)s+sin⁡3t=(2+2s2)s+3s−4s3=s (5−2s2).f=(3-\cos2t)s+\sin3t=(2+2s^2)s+3s-4s^3=s\,(5-2s^2).
Since 5−2s2>05-2s^2>0, f≥0  ⟺  sin⁡t≥0f\ge0\iff\sin t\ge0.

Solve. For x∈[0,2]x\in[0,2], t∈[−π4,7π4]t\in[-\frac\pi4,\frac{7\pi}4], and sin⁡t≥0\sin t\ge0 means t∈[0,π]t\in[0,\pi]. So x∈[14,54]x\in[\frac14,\frac54] and
β−α=1.\beta-\alpha=1.
Answer: 1

Topics:Trigonometric equationsTrigonometric ratios and identities
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