The area bounded by the curves y=|x-1|+|x-2| and y=3 is equal to :

Area Under the Curves PYQ — JEE Main 2023 · 6 Apr · Shift 2 · Q25 with Solution

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Correct answer: B

We have $y=|x-1|+|x-2|.$ The critical points are $x=1$ and $x=2$ .

So we consider three regions:

So the curve is:

3-2x, & x<1,\\ 1, & 1\le x\le 2,\\ 2x-3, & x>2. \end{cases}$$ 2. **Find points of intersection with $y=3$** We solve $$|x-1|+|x-2|=3.$$ Using the piecewise form: - For $x<1$: $$3-2x=3 \implies x=0.$$ - For $1\le x\le 2$: $$1=3$$ no solution. - For $x>2$: $$2x-3=3 \implies 2x=6 \implies x=3.$$ Thus the two curves meet at $x=0$ and $x=3$. 3. **Set up the bounded area** Between $x=0$ and $x=3$, the line $y=3$ lies above the curve $y=|x-1|+|x-2|$. Therefore, $$\text{Area}=\int_0^3 \left[3-\bigl(|x-1|+|x-2|\bigr)\right]dx.$$ Now evaluate piecewise. 4. **Compute the area on each interval** ### From $x=0$ to $x=1$ Here $y=3-2x$, so $$3-(3-2x)=2x.$$ Thus, $$A_1=\int_0^1 2x\,dx=\left[x^2\right]_0^1=1.$$ ### From $x=1$ to $x=2$ Here $y=1$, so $$3-1=2.$$ Thus, $$A_2=\int_1^2 2\,dx=2(2-1)=2.$$ ### From $x=2$ to $x=3$ Here $y=2x-3$, so $$3-(2x-3)=6-2x.$$ Thus, $$A_3=\int_2^3 (6-2x)\,dx=\left[6x-x^2\right]_2^3=(18-9)-(12-4)=9-8=1.$$ 5. **Add all parts** $$\text{Total Area}=A_1+A_2+A_3=1+2+1=4.$$ 6. **Compare with the stored answer** Our derived answer is **4**, which corresponds to **Option B**.

View written solutionFree

Correct answer: B

We have $y=|x-1|+|x-2|.$ The critical points are $x=1$ and $x=2$ .

So we consider three regions:

So the curve is:

3-2x, & x<1,\\ 1, & 1\le x\le 2,\\ 2x-3, & x>2. \end{cases}$$ 2. **Find points of intersection with $y=3$** We solve $$|x-1|+|x-2|=3.$$ Using the piecewise form: - For $x<1$: $$3-2x=3 \implies x=0.$$ - For $1\le x\le 2$: $$1=3$$ no solution. - For $x>2$: $$2x-3=3 \implies 2x=6 \implies x=3.$$ Thus the two curves meet at $x=0$ and $x=3$. 3. **Set up the bounded area** Between $x=0$ and $x=3$, the line $y=3$ lies above the curve $y=|x-1|+|x-2|$. Therefore, $$\text{Area}=\int_0^3 \left[3-\bigl(|x-1|+|x-2|\bigr)\right]dx.$$ Now evaluate piecewise. 4. **Compute the area on each interval** ### From $x=0$ to $x=1$ Here $y=3-2x$, so $$3-(3-2x)=2x.$$ Thus, $$A_1=\int_0^1 2x\,dx=\left[x^2\right]_0^1=1.$$ ### From $x=1$ to $x=2$ Here $y=1$, so $$3-1=2.$$ Thus, $$A_2=\int_1^2 2\,dx=2(2-1)=2.$$ ### From $x=2$ to $x=3$ Here $y=2x-3$, so $$3-(2x-3)=6-2x.$$ Thus, $$A_3=\int_2^3 (6-2x)\,dx=\left[6x-x^2\right]_2^3=(18-9)-(12-4)=9-8=1.$$ 5. **Add all parts** $$\text{Total Area}=A_1+A_2+A_3=1+2+1=4.$$ 6. **Compare with the stored answer** Our derived answer is **4**, which corresponds to **Option B**.