Let f(x) = | x - 2 | and g(x) = f(f(x)), x \in [ 0,4 ] . Then \int\limits_0^3 ( g(x) - f(x) ) dx is equal to:

Definite Integration PYQ — JEE Main 2020 · 4 Sep · Shift 1 · Q35 with Solution

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Correct answer: A

We have $f(x)=|x-2|$ and $g(x)=f(f(x))=\big||x-2|-2\big|,$ for $x\in[0,4]$ .

We need to compute $I=\int_0^3 (g(x)-f(x))\,dx.$

Since $f(x)=|x-2|$ , we write it piecewise:

2-x, & 0\le x\le 2,\\ x-2, & 2\le x\le 4. \end{cases}$$ --- 3. **Find $g(x)=f(f(x))$** Since $$g(x)=\big||x-2|-2\big|,$$ let us simplify on $[0,4]$. Note that for $x\in[0,4]$, we have $$|x-2|\in[0,2].$$ So $|x-2|-2\le 0$, hence $$\big||x-2|-2\big|=2-|x-2|.$$ Thus, $$g(x)=2-|x-2|.$$ Now write this piecewise: $$g(x)=\begin{cases} 2-(2-x)=x, & 0\le x\le 2,\\ 2-(x-2)=4-x, & 2\le x\le 4. \end{cases}$$ --- 4. **Compute $g(x)-f(x)$** For $0\le x\le 2$: $$g(x)-f(x)=x-(2-x)=2x-2.$$ For $2\le x\le 3$: $$g(x)-f(x)=(4-x)-(x-2)=6-2x.$$ So $$I=\int_0^2 (2x-2)\,dx+\int_2^3 (6-2x)\,dx.$$ --- 5. **Evaluate the integrals** First, $$\int_0^2 (2x-2)\,dx=\left[x^2-2x\right]_0^2=(4-4)-0=0.$$ Second, $$\int_2^3 (6-2x)\,dx=\left[6x-x^2\right]_2^3=(18-9)-(12-4)=9-8=1.$$ Therefore, $$I=0+1=1.$$ --- 6. **Check options** The value is $$\int_0^3 (g(x)-f(x))\,dx=1.$$ So the correct option is **A**. --- 7. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** They agree.

View written solutionFree

Correct answer: A

We have $f(x)=|x-2|$ and $g(x)=f(f(x))=\big||x-2|-2\big|,$ for $x\in[0,4]$ .

We need to compute $I=\int_0^3 (g(x)-f(x))\,dx.$

Since $f(x)=|x-2|$ , we write it piecewise:

2-x, & 0\le x\le 2,\\ x-2, & 2\le x\le 4. \end{cases}$$ --- 3. **Find $g(x)=f(f(x))$** Since $$g(x)=\big||x-2|-2\big|,$$ let us simplify on $[0,4]$. Note that for $x\in[0,4]$, we have $$|x-2|\in[0,2].$$ So $|x-2|-2\le 0$, hence $$\big||x-2|-2\big|=2-|x-2|.$$ Thus, $$g(x)=2-|x-2|.$$ Now write this piecewise: $$g(x)=\begin{cases} 2-(2-x)=x, & 0\le x\le 2,\\ 2-(x-2)=4-x, & 2\le x\le 4. \end{cases}$$ --- 4. **Compute $g(x)-f(x)$** For $0\le x\le 2$: $$g(x)-f(x)=x-(2-x)=2x-2.$$ For $2\le x\le 3$: $$g(x)-f(x)=(4-x)-(x-2)=6-2x.$$ So $$I=\int_0^2 (2x-2)\,dx+\int_2^3 (6-2x)\,dx.$$ --- 5. **Evaluate the integrals** First, $$\int_0^2 (2x-2)\,dx=\left[x^2-2x\right]_0^2=(4-4)-0=0.$$ Second, $$\int_2^3 (6-2x)\,dx=\left[6x-x^2\right]_2^3=(18-9)-(12-4)=9-8=1.$$ Therefore, $$I=0+1=1.$$ --- 6. **Check options** The value is $$\int_0^3 (g(x)-f(x))\,dx=1.$$ So the correct option is **A**. --- 7. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** They agree.