Mathematics - Area Under Curves Question with Solution | TestHub

Given,

$f\left(x\right)=\left\{\begin{array}{c}\begin{array}{cc}n\left(1-2nx\right),& 0\le x<\frac{1}{2n}\\ 2n\left(2nx-1\right),& \frac{1}{2n}\le x<\frac{3}{4n}\\ 4n\left(1-nx\right),& \frac{3}{4n}\le x\le \frac{1}{n}\\ \frac{n}{n-1}\left(nx-1\right),& \frac{1}{n}\le x\le 1\end{array}\end{array}\right.$

Here, $x\in \left[0, 1\right]$

Now $f\left(x\right)=n-2n^{2}x$ is decreasing in $\left[0, \frac{1}{2n}\right]$ 

And $f\left(x\right)=4n^{2}x-4n$ increasing in $\left[\frac{1}{2n}, \frac{3}{4n}\right]$

And $f\left(x\right)=4n-4n^{2}x$ decreasing in $\left[\frac{3}{4n}, \frac{1}{n}\right]$

And $f\left(x\right)=\frac{n^{2}x-n}{n-1}$ increasing in $\left[\frac{1}{n}, 1\right]$

Now plotting the graph of above function we get,

Also, $f\left(x\right)\in \left[0, n\right]$

Now from above diagram area will be,

$A=\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{n-1}{2}=4$ $\left\{\text{as} n\ge 2\right\}$

$\Rightarrow \frac{1}{2}+\frac{n-1}{2}=4$

$\Rightarrow \frac{n}{2}=4$

$\Rightarrow n=8$

So, $f{\left(x\right)}_{max}=n=8$