Consider the functions f, g: \mathbbR arrow \mathbbR defined by f(x)=x^2+\frac512 \quad and \quad g(x)= \begincases2(1-\frac4|x|3), & |x| \leq \frac34 0, & |x|\gt \frac34\endcases If \alpha is the area of the region \(x, y) \in \mathbbR \times \mathbbR:|x| \leq \frac34, 0 \leq y \leq \min \f(x), g(x) , then the value of 9 \alpha is

Application of Integration PYQ — JEE Advanced 2022 · Shift 2 · Q26 with Solution

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Correct answer: 6

Given functions

$f(x)=x^2+\frac{5}{12}$

and

$g(x)=\begin{cases}2\left(1-\frac{4|x|}{3}\right), & |x|\le \frac34\\[4pt]0, & |x|>\frac34\end{cases}$

We need the area

$\alpha=\int_{-3/4}^{3/4} \min\{f(x),g(x)\}\,dx$

because the region is

$\{(x,y): |x|\le 3/4,\ 0\le y\le \min(f(x),g(x))\}.$

Use symmetry

Both $f(x)$ and $g(x)$ are even functions, so $\min\{f(x),g(x)\}$ is also even. Hence,

$\alpha=2\int_0^{3/4} \min\{f(x),g(x)\}\,dx.$

For $x\in[0,3/4]$ ,

$g(x)=2\left(1-\frac{4x}{3}\right)=2-\frac{8x}{3}.$

So we compare

$f(x)=x^2+\frac{5}{12}, \qquad g(x)=2-\frac{8x}{3}.$

Find intersection points

Solve

$x^2+\frac{5}{12}=2-\frac{8x}{3}.$

Multiply by $12$ :

$12x^2+5=24-32x$ $12x^2+32x-19=0.$

Now,

=\frac{-32\pm\sqrt{1024+912}}{24} =\frac{-32\pm 44}{24}.$$ Thus, $$x=\frac{12}{24}=\frac12, \qquad x=\frac{-76}{24}=-\frac{19}{6}.$$ In the interval $[0,3/4]$, the relevant intersection is $$x=\frac12.$$ 4. **Determine which function is smaller** Check at $x=0$: $$f(0)=\frac{5}{12}, \qquad g(0)=2,$$ so $f(x)<g(x)$ near $0$. Check at $x=3/4$: $$f\left(\frac34\right)=\frac{9}{16}+\frac{5}{12}>0, \qquad g\left(\frac34\right)=0,$$ so $g(x)<f(x)$ near $3/4$. Hence, $$\min\{f(x),g(x)\}=\begin{cases}f(x), & 0\le x\le \frac12,\\[4pt]g(x), & \frac12\le x\le \frac34. \end{cases}$$ Therefore, $$\alpha=2\left(\int_0^{1/2} \left(x^2+\frac{5}{12}\right)dx+\int_{1/2}^{3/4} \left(2-\frac{8x}{3}\right)dx\right).$$ 5. **Evaluate the first integral** $$\int_0^{1/2} \left(x^2+\frac{5}{12}\right)dx =\left[\frac{x^3}{3}+\frac{5x}{12}\right]_0^{1/2}.$$ At $x=\frac12$, $$\frac{(1/2)^3}{3}+\frac{5}{12}\cdot\frac12 =\frac{1}{24}+\frac{5}{24}=\frac14.$$ So, $$\int_0^{1/2} \left(x^2+\frac{5}{12}\right)dx=\frac14.$$ 6. **Evaluate the second integral** $$\int_{1/2}^{3/4} \left(2-\frac{8x}{3}\right)dx =\left[2x-\frac{4x^2}{3}\right]_{1/2}^{3/4}.$$ At $x=\frac34$, $$2\cdot\frac34-\frac{4}{3}\cdot\frac{9}{16} =\frac32-\frac34=\frac34.$$ At $x=\frac12$, $$2\cdot\frac12-\frac{4}{3}\cdot\frac14 =1-\frac13=\frac23.$$ Thus, $$\int_{1/2}^{3/4} \left(2-\frac{8x}{3}\right)dx=\frac34-\frac23=\frac{1}{12}.$$ 7. **Compute total area** $$\alpha=2\left(\frac14+\frac{1}{12}\right)=2\left(\frac{3+1}{12}\right)=2\cdot\frac13=\frac23.$$ Therefore, $$9\alpha=9\cdot\frac23=6.$$ 8. **Comparison with stored answer** Our derived answer is $6$, which matches the stored correct answer.

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View written solutionFree

Correct answer: 6

Given functions

$f(x)=x^2+\frac{5}{12}$

and

$g(x)=\begin{cases}2\left(1-\frac{4|x|}{3}\right), & |x|\le \frac34\\[4pt]0, & |x|>\frac34\end{cases}$

We need the area

$\alpha=\int_{-3/4}^{3/4} \min\{f(x),g(x)\}\,dx$

because the region is

$\{(x,y): |x|\le 3/4,\ 0\le y\le \min(f(x),g(x))\}.$

Use symmetry

Both $f(x)$ and $g(x)$ are even functions, so $\min\{f(x),g(x)\}$ is also even. Hence,

$\alpha=2\int_0^{3/4} \min\{f(x),g(x)\}\,dx.$

For $x\in[0,3/4]$ ,

$g(x)=2\left(1-\frac{4x}{3}\right)=2-\frac{8x}{3}.$

So we compare

$f(x)=x^2+\frac{5}{12}, \qquad g(x)=2-\frac{8x}{3}.$

Find intersection points

Solve

$x^2+\frac{5}{12}=2-\frac{8x}{3}.$

Multiply by $12$ :

$12x^2+5=24-32x$ $12x^2+32x-19=0.$

Now,

=\frac{-32\pm\sqrt{1024+912}}{24} =\frac{-32\pm 44}{24}.$$ Thus, $$x=\frac{12}{24}=\frac12, \qquad x=\frac{-76}{24}=-\frac{19}{6}.$$ In the interval $[0,3/4]$, the relevant intersection is $$x=\frac12.$$ 4. **Determine which function is smaller** Check at $x=0$: $$f(0)=\frac{5}{12}, \qquad g(0)=2,$$ so $f(x)<g(x)$ near $0$. Check at $x=3/4$: $$f\left(\frac34\right)=\frac{9}{16}+\frac{5}{12}>0, \qquad g\left(\frac34\right)=0,$$ so $g(x)<f(x)$ near $3/4$. Hence, $$\min\{f(x),g(x)\}=\begin{cases}f(x), & 0\le x\le \frac12,\\[4pt]g(x), & \frac12\le x\le \frac34. \end{cases}$$ Therefore, $$\alpha=2\left(\int_0^{1/2} \left(x^2+\frac{5}{12}\right)dx+\int_{1/2}^{3/4} \left(2-\frac{8x}{3}\right)dx\right).$$ 5. **Evaluate the first integral** $$\int_0^{1/2} \left(x^2+\frac{5}{12}\right)dx =\left[\frac{x^3}{3}+\frac{5x}{12}\right]_0^{1/2}.$$ At $x=\frac12$, $$\frac{(1/2)^3}{3}+\frac{5}{12}\cdot\frac12 =\frac{1}{24}+\frac{5}{24}=\frac14.$$ So, $$\int_0^{1/2} \left(x^2+\frac{5}{12}\right)dx=\frac14.$$ 6. **Evaluate the second integral** $$\int_{1/2}^{3/4} \left(2-\frac{8x}{3}\right)dx =\left[2x-\frac{4x^2}{3}\right]_{1/2}^{3/4}.$$ At $x=\frac34$, $$2\cdot\frac34-\frac{4}{3}\cdot\frac{9}{16} =\frac32-\frac34=\frac34.$$ At $x=\frac12$, $$2\cdot\frac12-\frac{4}{3}\cdot\frac14 =1-\frac13=\frac23.$$ Thus, $$\int_{1/2}^{3/4} \left(2-\frac{8x}{3}\right)dx=\frac34-\frac23=\frac{1}{12}.$$ 7. **Compute total area** $$\alpha=2\left(\frac14+\frac{1}{12}\right)=2\left(\frac{3+1}{12}\right)=2\cdot\frac13=\frac23.$$ Therefore, $$9\alpha=9\cdot\frac23=6.$$ 8. **Comparison with stored answer** Our derived answer is $6$, which matches the stored correct answer.

Application of Integration question

Application of Integration question

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