Start free

Mathematics · Area under the Curves

JEE Advanced 2023 — Paper 1 — Question 8

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

f(x)={n(1−2nx) if 0≤x≤12n2n(2nx−1) if 12n≤x≤34n4n(1−nx) if 34n≤x≤1nnn−1(nx−1) if 1n≤x≤1f(x)=\left\{\begin{array}{ccc} 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{array}\right.

If n is such that the area of the region bounded by the curves x=0,x=1,y=0\mathrm{x}=0, \mathrm{x}=1, \mathrm{y}=0 and y=f(x)\mathrm{y}=\mathrm{f}(\mathrm{x}) is 4 , then the

maximum value of the function ff is

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

The function f(x)f(x) consists of four linear segments. For 0≤x≤12n0\le x\le \frac{1}{2n}, f(x)=n(1−2nx)f(x)=n(1-2nx): line from (0,n)(0,n) to (12n,0)(\frac{1}{2n},0).

Area =12⋅12n⋅n=14= \frac12\cdot\frac{1}{2n}\cdot n = \frac14. For 12n≤x≤34n\frac{1}{2n}\le x\le \frac{3}{4n}, f(x)=2n(2nx−1)f(x)=2n(2nx-1): line from (12n,0)(\frac{1}{2n},0) to (34n,n)(\frac{3}{4n},n).

Area =12⋅14n⋅n=18= \frac12\cdot\frac{1}{4n}\cdot n = \frac18. For 34n≤x≤1n\frac{3}{4n}\le x\le \frac{1}{n}, f(x)=4n(1−nx)f(x)=4n(1-nx): line from (34n,n)(\frac{3}{4n},n) to (1n,0)(\frac{1}{n},0).

Area =12⋅14n⋅n=18= \frac12\cdot\frac{1}{4n}\cdot n = \frac18. For 1n≤x≤1\frac{1}{n}\le x\le 1, f(x)=nn−1(nx−1)f(x)=\frac{n}{n-1}(nx-1): line from (1n,0)(\frac{1}{n},0) to (1,n)(1,n).

Area =12⋅(1−1n)⋅n=n−12= \frac12\cdot\left(1-\frac1n\right)\cdot n = \frac{n-1}{2}. Total area A=14+18+18+n−12=12+n−12=n2A = \frac14+\frac18+\frac18+\frac{n-1}{2} = \frac12+\frac{n-1}{2} = \frac{n}{2}. Given A=4A=4, we have n2=4⇒n=8\frac{n}{2}=4 \Rightarrow n=8. The maximum of ff on [0,1][0,1] occurs at x=0,34n,1x=0,\frac{3}{4n},1, all giving f=nf=n. Hence the maximum value is n=8n=8.

Solution figure

Answer key and solution verified before publishing.

Practise Area under the Curves

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Advanced 2023
Paper
Paper 1
Subject
Mathematics
Chapter
Area under the Curves
Topic
Area under the Curves

More Area under the Curves questions from this paper