JEE Advanced 2023 · Paper 1 · Mathematics Q9

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MathematicsInteger+4Question 9 of 51

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

$f(x) = \begin{cases} n(1 - 2nx) & \text{if } 0 \le x \le \dfrac{1}{2n}\\[2mm] 2n(2nx - 1) & \text{if } \dfrac{1}{2n} \le x \le \dfrac{3}{4n}\\[2mm] 4n(1 - nx) & \text{if } \dfrac{3}{4n} \le x \le \dfrac1n\\[2mm] \dfrac{n}{n-1}(nx - 1) & \text{if } \dfrac1n \le x \le 1 \end{cases}$

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

Answer and worked solution

Official answer: 8

Each piece of $f$ is linear, so each area is a triangle.

Piece	Width	Peak height	Area
$0\le x\le\frac1{2n}$	$\frac1{2n}$	$n$	$\frac14$
$\frac1{2n}\le x\le\frac3{4n}$	$\frac1{4n}$	$n$	$\frac18$
$\frac3{4n}\le x\le\frac1n$	$\frac1{4n}$	$n$	$\frac18$
$\frac1n\le x\le1$	$1-\frac1n$	$n$	$\frac{n-1}{2}$

Total area $=\frac12+\frac{n-1}2=\frac n2=4$ , so $n=8$ .

All four pieces peak at the value $n$ , so the maximum of $f$ is $\mathbf 8$ .

Topics:Area under a curve Graphs of functions

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JEE Advanced 2023 · Paper 1, Mathematics question 9