Question

If the area of the figures which are defined on the coordinate plane by the following inequalities $|y|+2|x|⩽x^2+1$ is k. Find 6k

Answer 1

Given:

$|y|+2|x|\le x^2+1$

So we arrange it

$|y|=\le x^2+1-2|x|$

$\because x^2=(|x|{)}^2$

$|y|\le (|x|{)}^2+1-2|x|$

$|y|\le (|x|-1{)}^2$

Square of any quantity is always positive

$|y|\le (|x|-1{)}^2$

Now $y=(x-1{)}^2$ we draw graph

Ref. image

So our required answer

$A=4{\int }_0^1(x-1{)}^2$

$=4\left(\frac{(x-1{)}^3}{3}\right){|}_0^1$

$K=4\times \frac{1}{3}=\frac{4}{3}$

Now we have to find value of $6K$

$=6\times \frac{4}{3}$

$=2\times 4$

$=8$

So our Answer is $8$.