JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Application of Derivatives

Write $f(x)=1+\frac{4x}{2x+\sin x}$. Expanding near $0$ with $\sin x\approx x-\frac{x^3}{6}$ gives $f(x)\approx \frac{7}{3}+\frac{2}{27}{x}^2$, so $f$ has a local minimum at $x=0$. Away from $0$, $f^{\prime }(x)=\frac{4(\sin x-x\cos x)}{(2x+\sin x{)}^2}$, which vanishes where $\tan x=x$; these roots occur once in each period, near $4.49,7.73,10.90,14.07,17.28,\dots$. Testing the sign of $\sin x-x\cos x$ across each root shows the maxima fall at $4.49,10.90,17.28,\dots$ and the minima at $7.73,14.07,\dots$. Thus in $[\pi ,6\pi ]$ there are three maxima ($4.49,10.90,17.28$), and in $[2\pi ,4\pi ]$ there is one minimum ($7.73$). So the minimum at $0$, the count of three maxima, and the count of one minimum all hold.