Let n ≥ 2 be a natural number and f: [0, 1] → ℝ be the function defined by f(x) = \ cases n(1 - 2nx) & if & 0 ≤ x ≤ (1)/(2n) 2n(2nx - 1) & if & (1)/(2n) ≤ x ≤ (3)/(4n) 4n(1 - nx) & if & (3)/(4n) ≤ x ≤ (1)/(n) (n)/(n

Correct answer: 8 Given, f(x) = \ array ll n(1 - 2nx), 0 ≤ x) Here, x ∈ [0, 1] Now f(x) = n - 2 n ^ 2 x is decreasing in [0, (1)/(2n)] And f(x) = 4 n ^ 2 x - 4n increasing in [(1)/(2n), (3)/(4n)] And f(x) = 4n - 4 n ^ 2 x decreasing in [(3)/(4n), (1)/(n)] And f(x) = n ^ 2 x - n n - 1 increasing in [(1)/(n), 1] Now plotting the graph of above function we get, Also, f(x) ∈ [0, n] Now from above diagram area will be, A = (1)/(4) + (1)/(8) + (1)/(8) + (n - 1)/(2) = 4 \ asn ≥ 2\ ⇒ (1)/(2) + (n - 1)/(2) = 4 ⇒ (n)/(2) = 4 ⇒ n = 8 So, f (x) _ = n = 8

JEE Advanced 2023Paper 1

Exam: JEE Advanced
Year: 2023
Paper: Paper 1
Subject: Mathematics
Chapter: Area Under Curves

JEE Advanced 2023 · Paper 1 · Mathematics · Area Under Curves

Let n ≥ 2 be a natural number and f: [0, 1] → ℝ be the function...

Let $n \ge 2 be a natural number and$ f: \left[0, 1\right] \to \mathbb{R} $be the function defined by$ f\left(x\right) = \left\{ $\begin{cases}n\left(1 - 2nx\right) & if & 0 \le x \le \dfrac{1}{2n} \\ 2n\left(2nx - 1\right) & if & \dfrac{1}{2n} \le x \le \dfrac{3}{4n} \\ 4n\left(1 - nx\right) & if & \dfrac{3}{4n} \le x \le \dfrac{1}{n} \\ \dfrac{n}{n - 1}\left(nx - 1\right) & if & \dfrac{1}{n} \le x \le 1\end{cases}$ \right)

If $n$ is such that the area of the region bounded by the curves $x = 0, x = 1, y = 0$ and $y = f\left(x\right)$ is $4$ , then the maximum value of the function $f$ is

$Let n ≥ 2 be a natural number and f: [0, 1] → ℝ be the function defined by f(x) = \ cases n(1 - 2nx) & if & 0 ≤ x ≤$

Correct answer

Step-by-step solution

Given,
$f\left(x\right) = \left\{\begin{array}{ll}n\left(1 - 2nx\right), 0 \le x\right)Here,$ x \in \left[0, 1\right] $Now$ f\left(x\right) = n - 2{n}^{2}x $is decreasing in \left[0, \dfrac{1}{2n}\right]$
And $f\left(x\right) = 4{n}^{2}x - 4n$ increasing in \left[\dfrac{1}{2n}, \dfrac{3}{4n}\right] $And$ f\left(x\right) = 4n - 4{n}^{2}x $decreasing in \left[\dfrac{3}{4n}, \dfrac{1}{n}\right]$
And $f\left(x\right) = \dfrac{{n}^{2}x - n}{n - 1}$ increasing in \left[\dfrac{1}{n}, 1\right] $Now plotting the graph of above function we get,<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/3bd70984-de3a-458c-a400-1700a8803b3b.png" style="width: 300px; height: 169px;" />Also,$ f\left(x\right) \in \left[0, n\right] $Now from above diagram area will be,$ A = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{n - 1}{2} = 4 $\left\{\text{as}n \ge 2\right\}$
$⇒ \dfrac{1}{2} + \dfrac{n - 1}{2} = 4$
$⇒ \dfrac{n}{2} = 4$
$⇒ n = 8$
So, f{\left(x\right)}_\max = n = 8

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