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JEE Advanced 2025 Paper 2 · Mathematics · Indefinite and Definite Integrals · hard

Let f:[0,∞)→mathbb R be continuous and satisfy f(x)=cos x+2∫0^xsin (x-t) f(t) dt. Then f(ln 3)= ______.

Numerical answer

Let f:[0,∞)→Rf:[0,\infty)\to\mathbb R be continuous and satisfy f(x)=cos⁡x+2∫0xsin⁡(x−t) f(t) dtf(x)=\cos x+2\displaystyle\int_0^x\sin(x-t)\,f(t)\,dt. Then f(ln⁡3)=f(\ln3)= ______. Give the answer correct to two decimal places.
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Answer: 1.67

Write g(x)=∫0xsin⁡(x−t)f(t) dtg(x)=\int_0^x\sin(x-t)f(t)\,dt, so f=cos⁡x+2gf=\cos x+2g. By Leibniz' rule g′(x)=∫0xcos⁡(x−t)f(t)dtg'(x)=\int_0^x\cos(x-t)f(t)dt, g′(0)=0g'(0)=0, and g′′(x)=f(x)−g(x)g''(x)=f(x)-g(x). Then f′′=−cos⁡x+2g′′=−cos⁡x+2f−2g=−cos⁡x+2f−(f−cos⁡x)=ff''=-\cos x+2g''=-\cos x+2f-2g=-\cos x+2f-(f-\cos x)=f. Initial conditions: f(0)=1f(0)=1, f′(0)=0f'(0)=0. So f′′=ff''=f gives f(x)=cosh⁡xf(x)=\cosh x. Hence f(ln⁡3)=3+1/32=53≈1.67f(\ln3)=\frac{3+1/3}{2}=\frac53\approx1.67. A numerical solution of the ODE system reproduces 1.66671.6667.
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