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← JEE Main 2022 · 24 Jun · Shift 1
MathematicsInteger+4 / -1Question 26 of 90

JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 26

Let f(θ)=sin⁡θ+∫−π/2π/2(sin⁡θ+tcos⁡θ)f(t) dtf(\theta)=\sin\theta+\displaystyle\int_{-\pi/2}^{\pi/2}(\sin\theta+t\cos\theta)f(t)\,dt. Then the value of ∣∫0π/2f(θ) dθ∣\left|\displaystyle\int_0^{\pi/2}f(\theta)\,d\theta\right| is ______.

Answer and worked solution

Official answer: 1

Form of ff. f(θ)=(1+A)sin⁡θ+Bcos⁡θf(\theta)=(1+A)\sin\theta+B\cos\theta, where A=∫−π/2π/2f(t) dtA=\int_{-\pi/2}^{\pi/2}f(t)\,dt and B=∫−π/2π/2tf(t) dtB=\int_{-\pi/2}^{\pi/2}tf(t)\,dt.

Equations. A=(1+A)⋅0+B⋅2=2BA=(1+A)\cdot0+B\cdot2=2B. B=(1+A)∫tsin⁡t dt+B∫tcos⁡t dt=2(1+A)+0B=(1+A)\int t\sin t\,dt+B\int t\cos t\,dt=2(1+A)+0.

Solve. B=2+4BB=2+4B, so B=−23B=-\frac23, A=−43A=-\frac43, and f(θ)=−13sin⁡θ−23cos⁡θf(\theta)=-\frac13\sin\theta-\frac23\cos\theta.

Integral. ∫0π/2f=−13−23=−1\int_0^{\pi/2}f=-\frac13-\frac23=-1; absolute value 1.

Answer: 1

Topics:Properties of definite integrals
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