JEE Advanced 2023 Paper 1, Mathematics Section 3 Q2: Area as Definite Integral

The graph of $f$ on $[0,1]$ consists of triangles. The first three pieces form two small triangles inside $\left[0,\frac{1}{n}\right]$ (one above the $x$-axis between $0$ and $\frac{1}{2n}$ of peak height $n$, another between $\frac{1}{2n}$ and $\frac{1}{n}$ of peak height $n$ at $x=\frac{3}{4n}$). The last piece is a triangle on $\left[\frac{1}{n},1\right]$ rising linearly from $0$ to $f(1)=n$.

Compute the three triangle areas:

First: base $\frac{1}{2n}$, height $n$, area $=\frac{1}{2}\cdot \frac{1}{2n}\cdot n=\frac{1}{4}$.

Second (peak $n$ at $\frac{3}{4n}$ over base $\left[\frac{1}{2n},\frac{1}{n}\right]$ of length $\frac{1}{2n}$): area $=\frac{1}{2}\cdot \frac{1}{2n}\cdot n=\frac{1}{4}$.

Third: base $1-\frac{1}{n}=\frac{n-1}{n}$, height $n$, area $=\frac{1}{2}\cdot \frac{n-1}{n}\cdot n=\frac{n-1}{2}$.

Total area $=\frac{1}{4}+\frac{1}{4}+\frac{n-1}{2}=\frac{1}{2}+\frac{n-1}{2}=\frac{n}{2}$.

Setting $\frac{n}{2}=4$ gives $n=8$. The maximum value of $f$ on $[0,1]$ equals $n=8$ (attained at the triangle peaks and at $x=1$).