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QJEEAdv23Ma8c33: JEE Advanced 2023 Mathematics - Class 12 JEEAdv23Ma8c33 Solution: Let be a natural number and be the function defined by If is…

Let n≥2n \geq 2 be a natural number and f:[0,1]→Rf:[0,1] \rightarrow \mathbb{R} be the function defined by

f(x)={n(1−2nx)amp; if 0≤x≤12n2n(2nx−1)amp; if 12n≤x≤34n4n(1−nx)amp; if 34n≤x≤1nnn−1(nx−1)amp; if 1n≤x≤1f(x)= \begin{cases}n(1-2 n x) & \text { if } 0 \leq x \leq \frac{1}{2 n} \\\\ 2 n(2 n x-1) & \text { if } \frac{1}{2 n} \leq x \leq \frac{3}{4 n} \\\\ 4 n(1-n x) & \text { if } \frac{3}{4 n} \leq x \leq \frac{1}{n} \\\\ \frac{n}{n-1}(n x-1) & \text { if } \frac{1}{n} \leq x \leq 1\end{cases}

If nn is such that the area of the region bounded by the curves x=0,x=1,y=0x=0, x=1, y=0 and y=f(x)y=f(x) is 4 , then the maximum value of the function ff is :

Step-by-Step Verified Solution

f(x)f(x) is decreasing in [0,12n]\left[0, \frac{1}{2 n}\right]

increasing in [12n,34n]\left[\frac{1}{2 n}, \frac{3}{4 n}\right]

decreasing in [34n,1n]\left[\frac{3}{4 n}, \frac{1}{n}\right]

increasing in [1n,1]\left[\frac{1}{n}, 1\right]

JEE Advanced 2023 Paper 1 Online Mathematics - Application of Integration Question 3 English Explanation

 Area amp;=12×12n×n+12×12n×n+12×(1−1n)×n4amp;=14+14+n−124amp;=12+n−124amp;=n2namp;=8\begin{aligned} \text { Area } & =\frac{1}{2} \times \frac{1}{2 n} \times n+\frac{1}{2} \times \frac{1}{2 n} \times n+\frac{1}{2} \times\left(1-\frac{1}{n}\right) \times n \\\\ 4 & =\frac{1}{4}+\frac{1}{4}+\frac{n-1}{2} \\\\ 4 & =\frac{1}{2}+\frac{n-1}{2} \\\\ 4 & =\frac{n}{2} \\\\ n & =8\end{aligned}

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