The area bounded by the lines y = || x - 1 | - 2 | is \underline\hspace2cm .

Area Under the Curves PYQ — JEE Main 2021 · 26 Feb · Shift 1 · Q40 with Solution

View written solutionFree

Correct answer: 4

Interpret the curve

We need the area bounded by $y=\bigl||x-1|-2\bigr|.$

Let $f(x)=\bigl||x-1|-2\bigr|.$

Since this is an absolute-value graph, the bounded region is the finite region enclosed by the graph and the $x$ -axis.

Find where the graph meets the $x$ -axis

For intersections with the $x$ -axis, set $y=0$ : $\bigl||x-1|-2\bigr|=0.$

This gives $|x-1|-2=0$ so $|x-1|=2.$

Hence, $x-1=\pm 2 \implies x=3 \text{ or } x=-1.$

So the graph touches the $x$ -axis at $(-1,0)$ and $(3,0)$ .

Write the function piecewise

First analyze $|x-1|$ :

If $x\ge 1$ , then $|x-1|=x-1$ .
If $x<1$ , then $|x-1|=1-x$ .

Now handle the outer absolute value.

For $x\ge 1$

$f(x)=|(x-1)-2|=|x-3|.$ Thus,

for $1\le x\le 3$ , $f(x)=3-x$ ,
for $x\ge 3$ , $f(x)=x-3$ .

For $x<1$

$f(x)=|(1-x)-2|=|-x-1|=|x+1|.$ Thus,

for $-1\le x<1$ , $f(x)=x+1$ ,
for $x<-1$ , $f(x)=-(x+1)=-x-1$ .

Therefore, in the bounded interval $[-1,3]$ ,

\begin{cases} x+1, & -1\le x\le 1,\\ 3-x, & 1\le x\le 3. \end{cases}$$ 4. **Identify the bounded region** Between $x=-1$ and $x=3$, the graph lies above the $x$-axis and forms a triangle-like shape with vertex at $x=1$: $$f(1)=\bigl||1-1|-2\bigr|=2.$$ So the enclosed region has base from $x=-1$ to $x=3$, i.e. length $4$, and height $2$. Hence area should be $$\frac12\times 4\times 2=4.$$ Let us also confirm by integration. 5. **Compute the area by integration** $$A=\int_{-1}^{1}(x+1)\,dx+\int_{1}^{3}(3-x)\,dx.$$ First integral: $$\int_{-1}^{1}(x+1)\,dx=\left[\frac{x^2}{2}+x\right]_{-1}^{1} =\left(\frac12+1\right)-\left(\frac12-1\right)=2.$$ Second integral: $$\int_{1}^{3}(3-x)\,dx=\left[3x-\frac{x^2}{2}\right]_{1}^{3} =\left(9-\frac92\right)-\left(3-\frac12\right)=2.$$ Thus, $$A=2+2=4.$$ 6. **Compare with stored answer** Our derived area is $$\boxed{4}.$$ The stored correct answer is $8$, which does **not** match. The likely mistake is counting both the upper and lower symmetric parts, but the graph $y=\bigl||x-1|-2\bigr|$ is always nonnegative, so only one bounded region with the $x$-axis exists, having area $4$.

More from Area Under the Curves

View written solutionFree

Correct answer: 4

Interpret the curve

We need the area bounded by $y=\bigl||x-1|-2\bigr|.$

Let $f(x)=\bigl||x-1|-2\bigr|.$

Since this is an absolute-value graph, the bounded region is the finite region enclosed by the graph and the $x$ -axis.

Find where the graph meets the $x$ -axis

For intersections with the $x$ -axis, set $y=0$ : $\bigl||x-1|-2\bigr|=0.$

This gives $|x-1|-2=0$ so $|x-1|=2.$

Hence, $x-1=\pm 2 \implies x=3 \text{ or } x=-1.$

So the graph touches the $x$ -axis at $(-1,0)$ and $(3,0)$ .

Write the function piecewise

First analyze $|x-1|$ :

If $x\ge 1$ , then $|x-1|=x-1$ .
If $x<1$ , then $|x-1|=1-x$ .

Now handle the outer absolute value.

For $x\ge 1$

$f(x)=|(x-1)-2|=|x-3|.$ Thus,

for $1\le x\le 3$ , $f(x)=3-x$ ,
for $x\ge 3$ , $f(x)=x-3$ .

For $x<1$

$f(x)=|(1-x)-2|=|-x-1|=|x+1|.$ Thus,

for $-1\le x<1$ , $f(x)=x+1$ ,
for $x<-1$ , $f(x)=-(x+1)=-x-1$ .

Therefore, in the bounded interval $[-1,3]$ ,

\begin{cases} x+1, & -1\le x\le 1,\\ 3-x, & 1\le x\le 3. \end{cases}$$ 4. **Identify the bounded region** Between $x=-1$ and $x=3$, the graph lies above the $x$-axis and forms a triangle-like shape with vertex at $x=1$: $$f(1)=\bigl||1-1|-2\bigr|=2.$$ So the enclosed region has base from $x=-1$ to $x=3$, i.e. length $4$, and height $2$. Hence area should be $$\frac12\times 4\times 2=4.$$ Let us also confirm by integration. 5. **Compute the area by integration** $$A=\int_{-1}^{1}(x+1)\,dx+\int_{1}^{3}(3-x)\,dx.$$ First integral: $$\int_{-1}^{1}(x+1)\,dx=\left[\frac{x^2}{2}+x\right]_{-1}^{1} =\left(\frac12+1\right)-\left(\frac12-1\right)=2.$$ Second integral: $$\int_{1}^{3}(3-x)\,dx=\left[3x-\frac{x^2}{2}\right]_{1}^{3} =\left(9-\frac92\right)-\left(3-\frac12\right)=2.$$ Thus, $$A=2+2=4.$$ 6. **Compare with stored answer** Our derived area is $$\boxed{4}.$$ The stored correct answer is $8$, which does **not** match. The likely mistake is counting both the upper and lower symmetric parts, but the graph $y=\bigl||x-1|-2\bigr|$ is always nonnegative, so only one bounded region with the $x$-axis exists, having area $4$.

Area Under the Curves question

Area Under the Curves question

For $x\ge 1$

For $x<1$

More from Area Under the Curves

Area Under the Curves question

Area Under the Curves question

For $x\ge 1$

For $x<1$

More from Area Under the Curves

For x≥1x\ge 1x≥1

For x<1x<1x<1

More from Area Under the Curves

For x≥1x\ge 1x≥1

For x<1x<1x<1

More from Area Under the Curves

For $x\ge 1$

For $x<1$

For $x\ge 1$

For $x<1$