Let $\mathrm{n} \geq 2$ be a natural number and $\mathrm{f}:[0,1] \rightarrow \mathbb{R}$ be the function defined by $$\mathrm{f}(\mathrm{x})= \begin{cases}\mathrm{n}(1-2 \mathrm{nx}) & \text { if } 0 \leq \mathrm{x} \leq \frac{1}{2 \mathrm{n}} \\ 2 \mathrm{n}(2 \mathrm{nx}-1) & \text { if } \frac{1}{2 \mathrm{n}} \leq \mathrm{x} \leq \frac{3}{4 \mathrm{n}} \\ 4 \mathrm{n}(1-\mathrm{nx}) & \text { if } \frac{3}{4 \mathrm{n}} \leq \mathrm{x} \leq \frac{1}{\mathrm{n}} \\ \frac{\mathrm{n}}{\mathrm{n}-1}(\mathrm{nx}-1) & \text { if } \frac{1}{\mathrm{n}} \leq \mathrm{x} \leq 1\end{cases}$$ If n is such that the area of the region bounded by the curves $\mathrm{x}=0, \mathrm{x}=1, \mathrm{y}=0$ and $\mathrm{y}=\mathrm{f}(\mathrm{x})$ is 4, then the maximum value of the function f is

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Area Under Curves

Let

n \geq 2

be a natural number and

f : [0, 1] \to R

be the function defined by

f (x) = ⎩ ⎨ ⎧ n (1 - 2 nx) 2 n (2 nx - 1) 4 n (1 - nx) \frac{n}{n - 1} (nx - 1) if 0 \leq x \leq \frac{1}{2 n} if \frac{1}{2 n} \leq x \leq \frac{3}{4 n} if \frac{3}{4 n} \leq x \leq \frac{1}{n} if \frac{1}{n} \leq x \leq 1

If n is such that the area of the region bounded by the curves

x = 0, x = 1, y = 0

and

y = f (x)

is 4, then the maximum value of the function f is

Answer: 8

Each piece of

f

is a straight segment, so the region under the graph splits into triangles. The first piece rises from

0

n

over width

\frac{1}{2 n}

(area

\frac{1}{4}

); the second and third pieces form two triangles each of base

\frac{1}{4 n}

and height

n

(area

\frac{1}{8}

each); and the last piece is a triangle of base

\frac{n - 1}{n}

and height

n

(area

\frac{n - 1}{2}

). The total area is

\frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{n - 1}{2} = \frac{n}{2}

. Setting this equal to

4

gives

n = 8

. Since the graph attains its greatest value

n

at the peaks, the maximum value of

f

8

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Integer answer