JEE Advanced 2023 — Paper 1 — MATH-III solution

JEE Advanced2023Paper 1MATH-III

Let $n \geq 2$ be a natural number and $f : [0, 1] \to R$ be the function defined by

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

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

Your answer

Solution

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

Compute the three triangle areas:

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

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

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 $x = 1$ ).

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