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Functions question

2020 · Shift 1 · Q33
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Functions question

2020 · Shift 1 · Q33

JEE AdvancedMathematicsFunctionsNumerical+4 / −1
Let f : [0, 2] →\to→ R be the function defined by f(x)=(3−sin⁡(2πx))sin⁡(πx−π4)−sin⁡(3πx+π4)f(x) = (3 - \sin (2\pi x))\sin \left( {\pi x - {\pi \over 4}} \right) - \sin \left( {3\pi x + {\pi \over 4}} \right)f(x)=(3−sin(2πx))sin(πx−4π​)−sin(3πx+4π​) If α, β∈[0,2]\alpha ,\,\beta \in [0,2]α,β∈[0,2] are such that {x∈[0,2]:f(x)≥0}=[α,β]\{ x \in [0,2]:f(x) \ge 0\} = [\alpha ,\beta ]{x∈[0,2]:f(x)≥0}=[α,β], then the value of β−α\beta - \alphaβ−α is ..........
Numerical answer
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Correct answer: 1

  1. Given function

We need to analyze

f(x)=(3−sin⁡2πx)sin⁡(πx−π4)−sin⁡(3πx+π4).f(x)=(3-\sin 2\pi x)\sin\left(\pi x-\frac{\pi}{4}\right)-\sin\left(3\pi x+\frac{\pi}{4}\right).f(x)=(3−sin2πx)sin(πx−4π​)−sin(3πx+4π​).

We are told that

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

and we must find β−α\beta-\alphaβ−α.

  1. Simplify the expression

Let

A=πx−π4.A=\pi x-\frac{\pi}{4}.A=πx−4π​.

Then

3πx+π4=3(πx−π4)+π=3A+π.3\pi x+\frac{\pi}{4}=3\left(\pi x-\frac{\pi}{4}\right)+\pi=3A+\pi.3πx+4π​=3(πx−4π​)+π=3A+π.

Hence

sin⁡(3πx+π4)=sin⁡(3A+π)=−sin⁡3A.\sin\left(3\pi x+\frac{\pi}{4}\right)=\sin(3A+\pi)=-\sin 3A.sin(3πx+4π​)=sin(3A+π)=−sin3A.

So

f(x)=(3−sin⁡2πx)sin⁡A+sin⁡3A.f(x)=(3-\sin 2\pi x)\sin A+\sin 3A.f(x)=(3−sin2πx)sinA+sin3A.

Now use

sin⁡2πx=sin⁡(2A+π2)=cos⁡2A.\sin 2\pi x=\sin\bigl(2A+\tfrac{\pi}{2}\bigr)=\cos 2A.sin2πx=sin(2A+2π​)=cos2A.

Thus

f(x)=(3−cos⁡2A)sin⁡A+sin⁡3A.f(x)=(3-\cos 2A)\sin A+\sin 3A.f(x)=(3−cos2A)sinA+sin3A.

Using

cos⁡2A=1−2sin⁡2A,\cos 2A=1-2\sin^2 A,cos2A=1−2sin2A,

we get

3−cos⁡2A=3−(1−2sin⁡2A)=2+2sin⁡2A=2(1+sin⁡2A).3-\cos 2A=3-(1-2\sin^2 A)=2+2\sin^2 A=2(1+\sin^2 A).3−cos2A=3−(1−2sin2A)=2+2sin2A=2(1+sin2A).

Also,

sin⁡3A=3sin⁡A−4sin⁡3A.\sin 3A=3\sin A-4\sin^3 A.sin3A=3sinA−4sin3A.

Therefore

f(x)=2(1+sin⁡2A)sin⁡A+(3sin⁡A−4sin⁡3A).f(x)=2(1+\sin^2 A)\sin A+(3\sin A-4\sin^3 A).f(x)=2(1+sin2A)sinA+(3sinA−4sin3A).

Simplify:

f(x)=2sin⁡A+2sin⁡3A+3sin⁡A−4sin⁡3Af(x)=2\sin A+2\sin^3 A+3\sin A-4\sin^3 Af(x)=2sinA+2sin3A+3sinA−4sin3A =5sin⁡A−2sin⁡3A=5\sin A-2\sin^3 A=5sinA−2sin3A =sin⁡A (5−2sin⁡2A).=\sin A\,(5-2\sin^2 A).=sinA(5−2sin2A).

So

f(x)=sin⁡(πx−π4)(5−2sin⁡2(πx−π4)).f(x)=\sin\left(\pi x-\frac{\pi}{4}\right)\left(5-2\sin^2\left(\pi x-\frac{\pi}{4}\right)\right).f(x)=sin(πx−4π​)(5−2sin2(πx−4π​)).
  1. Determine the sign of f(x)f(x)f(x)

Since

0≤sin⁡2A≤1,0\le \sin^2 A\le 1,0≤sin2A≤1,

we have

5−2sin⁡2A≥5−2=3>0.5-2\sin^2 A\ge 5-2=3>0.5−2sin2A≥5−2=3>0.

Thus the factor 5−2sin⁡2A5-2\sin^2 A5−2sin2A is always positive.

Hence the sign of f(x)f(x)f(x) is the sign of

sin⁡A=sin⁡(πx−π4).\sin A=\sin\left(\pi x-\frac{\pi}{4}\right).sinA=sin(πx−4π​).

Therefore,

f(x)≥0  ⟺  sin⁡(πx−π4)≥0.f(x)\ge 0 \iff \sin\left(\pi x-\frac{\pi}{4}\right)\ge 0.f(x)≥0⟺sin(πx−4π​)≥0.
  1. Solve the inequality

We need

sin⁡(πx−π4)≥0.\sin\left(\pi x-\frac{\pi}{4}\right)\ge 0.sin(πx−4π​)≥0.

Now sin⁡θ≥0\sin\theta\ge 0sinθ≥0 when

θ∈[2nπ,(2n+1)π],n∈Z.\theta\in[2n\pi,(2n+1)\pi],\quad n\in\mathbb Z.θ∈[2nπ,(2n+1)π],n∈Z.

So

πx−π4∈[2nπ,(2n+1)π].\pi x-\frac{\pi}{4}\in[2n\pi,(2n+1)\pi].πx−4π​∈[2nπ,(2n+1)π].

Add π4\frac{\pi}{4}4π​ and divide by π\piπ:

x∈[2n+14,2n+54].x\in\left[2n+\frac14,2n+\frac54\right].x∈[2n+41​,2n+45​].

Now intersect with [0,2][0,2][0,2].

For n=0n=0n=0:

x∈[14,54].x\in\left[\frac14,\frac54\right].x∈[41​,45​].

For all other integers nnn, the interval lies outside [0,2][0,2][0,2].

Hence

{x∈[0,2]:f(x)≥0}=[14,54].\{x\in[0,2]: f(x)\ge 0\}=\left[\frac14,\frac54\right].{x∈[0,2]:f(x)≥0}=[41​,45​].

So

α=14,β=54.\alpha=\frac14,\qquad \beta=\frac54.α=41​,β=45​.

Therefore

β−α=54−14=1.\beta-\alpha=\frac54-\frac14=1.β−α=45​−41​=1.
  1. Compare with stored answer

Derived answer is 111, which matches the stored correct answer.

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