JEE MAIN - Mathematics (2023 - 6th April Evening Shift - No. 2)
The area bounded by the curves $$y=|x-1|+|x-2|$$ and $$y=3$$ is equal to :
5
4
6
3
Explanation
Given equation of curve $y=|x-1|+|x-2|$ and $y=3$
_6th_April_Evening_Shift_en_2_1.png)
When, $x-1=0 \Rightarrow x=1$, then $y=1$
and when, $x-2=0 \Rightarrow x=2$, then $y=1$
Equation of curve passing through point $(1,1)$ and $(2,1)$ and $(0,3),(3,3)$
$\therefore$ Area bounded by given curves
$$ =\frac{1}{2}(1+3) \times 2(\text { Area of trapezium })=4 $$
When, $x-1=0 \Rightarrow x=1$, then $y=1$
and when, $x-2=0 \Rightarrow x=2$, then $y=1$
Equation of curve passing through point $(1,1)$ and $(2,1)$ and $(0,3),(3,3)$
$\therefore$ Area bounded by given curves
$$ =\frac{1}{2}(1+3) \times 2(\text { Area of trapezium })=4 $$
Comments (0)
