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Area Under the Curves question

2022 · 26 Jul · Shift 2 · Q33
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  5. /2022 · 26 Jul · Shift 2 · Q33

Area Under the Curves question

2022 · 26 Jul · Shift 2 · Q33

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area bounded by the curves y=∣x2−1∣y=\left|x^{2}-1\right|y=​x2−1​ and y=1y=1y=1 is
  1. A
    23(2+1)\frac{2}{3}(\sqrt{2}+1)32​(2​+1)
  2. B
    43(2−1)\frac{4}{3}(\sqrt{2}-1)34​(2​−1)
  3. C
    2(2−1)2(\sqrt{2}-1)2(2​−1)
  4. D
    83(2−1)\frac{8}{3}(\sqrt{2}-1)38​(2​−1)
View written solutionFree

Correct answer: D

  1. Understand the curves

We need the area bounded by y=∣x2−1∣andy=1.y=|x^2-1| \quad \text{and} \quad y=1.y=∣x2−1∣andy=1.

First, write ∣x2−1∣|x^2-1|∣x2−1∣ piecewise:

\begin{cases} 1-x^2, & |x|<1,\\ x^2-1, & |x|\ge 1. \end{cases}$$ So the graph is: - for $-1\le x\le 1$, it is the downward parabola $y=1-x^2$, - for $|x|\ge 1$, it is the upward parabola $y=x^2-1$. The line is $y=1$. --- 2. **Find points of intersection** We solve $$|x^2-1|=1.$$ This gives two cases: - Case 1: $x^2-1=1 \Rightarrow x^2=2 \Rightarrow x=\pm\sqrt{2}$ - Case 2: $x^2-1=-1 \Rightarrow x^2=0 \Rightarrow x=0$ Thus the curves meet at: $$( -\sqrt{2},1),\quad (0,1),\quad (\sqrt{2},1).$$ These form two symmetric bounded regions: one on $[-\sqrt{2},0]$ and one on $[0,\sqrt{2}]$. --- 3. **Find the required area using symmetry** Because the graph is symmetric about the $y$-axis, total area is $$2\times \text{(area from }x=0\text{ to }x=\sqrt{2}).$$ From $x=0$ to $x=1$: $$|x^2-1|=1-x^2,$$ so the vertical gap between $y=1$ and the curve is $$1-(1-x^2)=x^2.$$ From $x=1$ to $x=\sqrt{2}$: $$|x^2-1|=x^2-1,$$ so the vertical gap is $$1-(x^2-1)=2-x^2.$$ Hence right-side area is $$\int_0^1 x^2\,dx + \int_1^{\sqrt{2}} (2-x^2)\,dx.$$ Therefore total area is $$A=2\left[\int_0^1 x^2\,dx + \int_1^{\sqrt{2}} (2-x^2)\,dx\right].$$ --- 4. **Evaluate the integrals** First, $$\int_0^1 x^2\,dx=\left[\frac{x^3}{3}\right]_0^1=\frac13.$$ Next, $$\int_1^{\sqrt{2}} (2-x^2)\,dx =\left[2x-\frac{x^3}{3}\right]_1^{\sqrt{2}}.$$ Now evaluate: At $x=\sqrt{2}$, $$2\sqrt{2}-\frac{(\sqrt{2})^3}{3}=2\sqrt{2}-\frac{2\sqrt{2}}{3}=\frac{4\sqrt{2}}{3}.$$ At $x=1$, $$2-\frac13=\frac53.$$ So, $$\int_1^{\sqrt{2}} (2-x^2)\,dx=\frac{4\sqrt{2}}{3}-\frac53=\frac{4\sqrt{2}-5}{3}.$$ Thus right-side area is $$\frac13+\frac{4\sqrt{2}-5}{3}=\frac{4\sqrt{2}-4}{3}=\frac{4}{3}(\sqrt{2}-1).$$ Hence total area is $$A=2\cdot \frac{4}{3}(\sqrt{2}-1)=\frac{8}{3}(\sqrt{2}-1).$$ --- 5. **Match with options** $$\boxed{A=\frac{8}{3}(\sqrt{2}-1)}$$ So the correct option is: **D**. --- 6. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They match.
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