The area bounded by the curve y = |x2 - 9| and the line y = 3 is :

Area Under the Curves PYQ — JEE Main 2022 · 26 Jun · Shift 1 · Q28 with Solution

View written solutionFree

Correct answer: NONE OF THE GIVEN OPTIONS, $32\SQRT{3}+8\SQRT{6}-54$

Given curves

We need the area bounded by $y=|x^2-9|$ and the line $y=3.$

Write the modulus function piecewise

Since

\begin{cases} 9-x^2, & |x|\le 3,\\ x^2-9, & |x|\ge 3. \end{cases}$$ So the curve is: - downward parabola $y=9-x^2$ between $x=-3$ and $x=3$, - upward branches $y=x^2-9$ for $|x|\ge 3$. 3. **Find points of intersection with $y=3$** We solve $$|x^2-9|=3.$$ This gives two cases: - $x^2-9=3 \Rightarrow x^2=12 \Rightarrow x=\pm 2\sqrt3$, - $x^2-9=-3 \Rightarrow x^2=6 \Rightarrow x=\pm \sqrt6$. Thus the line $y=3$ meets the curve at $$x=\pm \sqrt6,\quad x=\pm 2\sqrt3.$$ 4. **Identify the bounded regions** Because of symmetry about the $y$-axis, the total bounded area is twice the area on the right side. On the right side: - from $x=\sqrt6$ to $x=3$, the curve is $y=9-x^2$, which lies above $y=3$, - from $x=3$ to $x=2\sqrt3$, the curve is $y=x^2-9$, which lies below $y=3$. Hence right-side area is $$\int_{\sqrt6}^{3} \big[(9-x^2)-3\big]dx + \int_{3}^{2\sqrt3} \big[3-(x^2-9)\big]dx.$$ So total area: $$A=2\left[\int_{\sqrt6}^{3} (6-x^2)\,dx + \int_{3}^{2\sqrt3} (12-x^2)\,dx\right].$$ 5. **Evaluate the first integral** $$\int (6-x^2)dx=6x-\frac{x^3}{3}.$$ Therefore, $$\int_{\sqrt6}^{3}(6-x^2)dx= \left[6x-\frac{x^3}{3}\right]_{\sqrt6}^{3}.$$ At $x=3$: $$6(3)-\frac{27}{3}=18-9=9.$$ At $x=\sqrt6$: $$6\sqrt6-\frac{(\sqrt6)^3}{3}=6\sqrt6-\frac{6\sqrt6}{3}=6\sqrt6-2\sqrt6=4\sqrt6.$$ So, $$\int_{\sqrt6}^{3}(6-x^2)dx=9-4\sqrt6.$$ 6. **Evaluate the second integral** $$\int (12-x^2)dx=12x-\frac{x^3}{3}.$$ Thus, $$\int_{3}^{2\sqrt3}(12-x^2)dx= \left[12x-\frac{x^3}{3}\right]_{3}^{2\sqrt3}.$$ At $x=2\sqrt3$: $$(2\sqrt3)^3=24\sqrt3,$$ so $$12(2\sqrt3)-\frac{24\sqrt3}{3}=24\sqrt3-8\sqrt3=16\sqrt3.$$ At $x=3$: $$12(3)-\frac{27}{3}=36-9=27.$$ Hence, $$\int_{3}^{2\sqrt3}(12-x^2)dx=16\sqrt3-27.$$ 7. **Total area** So, $$A=2\Big[(9-4\sqrt6)+(16\sqrt3-27)\Big].$$ $$A=2(16\sqrt3-4\sqrt6-18).$$ $$A=32\sqrt3-8\sqrt6-36.$$ Factorizing, $$A=4(8\sqrt3-2\sqrt6-9).$$ 8. **Compare with options** The options are: - A: $4(2\sqrt3+\sqrt6-4)$ - B: $4(4\sqrt3+\sqrt6-4)$ - C: $8(4\sqrt3+3\sqrt6-9)$ - D: $8(4\sqrt3+2\sqrt6-9)$ Our result $$A=32\sqrt3-8\sqrt6-36=8(4\sqrt3-\sqrt6-\tfrac{9}{2}),$$ which matches **none** of the given options. 9. **Check for algebra/sign mistakes** A quick numerical check: $$\sqrt3\approx 1.732,\quad \sqrt6\approx 2.449.$$ Then $$A\approx 32(1.732)-8(2.449)-36\approx 55.424-19.592-36\approx -0.168?$$ This indicates an arithmetic slip in simplification, so let us recompute carefully. From step 6: $$16\sqrt3-27\approx 27.712-27=0.712,$$ and from step 5: $$9-4\sqrt6\approx 9-9.798=-0.798.$$ That is impossible for area, so the first integrand was chosen on the wrong interval sign-wise. 10. **Correct region setup** For $|x|<3$, curve is $y=9-x^2$. This is above $y=3$ when $$9-x^2>3 \Rightarrow x^2<6 \Rightarrow |x|<\sqrt6.$$ So the bounded central region is from $x=-\sqrt6$ to $x=\sqrt6$. For $|x|>3$, curve is $y=x^2-9$. This is below $y=3$ when $$x^2-9<3 \Rightarrow x^2<12 \Rightarrow |x|<2\sqrt3.$$ Together with $|x|>3$, this gives side bounded regions on $$3\le |x|\le 2\sqrt3.$$ Thus total area is sum of: - central region between $x=-\sqrt6$ and $x=\sqrt6$, - two side regions between $x=3$ and $x=2\sqrt3$ and symmetrically on the left. 11. **Compute correctly** Central area: $$A_1=\int_{-\sqrt6}^{\sqrt6}\big[(9-x^2)-3\big]dx =\int_{-\sqrt6}^{\sqrt6}(6-x^2)dx.$$ Using symmetry, $$A_1=2\int_0^{\sqrt6}(6-x^2)dx =2\left[6x-\frac{x^3}{3}\right]_0^{\sqrt6}.

Now $6\sqrt6-\frac{6\sqrt6}{3}=6\sqrt6-2\sqrt6=4\sqrt6,$ so $A_1=2(4\sqrt6)=8\sqrt6.$

Side areas:

=2\int_3^{2\sqrt3}(12-x^2)dx.$$ From above, $$\int_3^{2\sqrt3}(12-x^2)dx=16\sqrt3-27,$$ therefore $$A_2=2(16\sqrt3-27)=32\sqrt3-54.$$ Hence total area: $$A=A_1+A_2=8\sqrt6+32\sqrt3-54.$$ Factor out $2$: $$A=2(4\sqrt6+16\sqrt3-27).$$ Or more neatly, $$A=2\big(16\sqrt3+4\sqrt6-27\big).$$ 12. **Compare with options again** Option D is $$8(4\sqrt3+2\sqrt6-9)=32\sqrt3+16\sqrt6-72,$$ which is **not equal** to our result. Our computed area is $$\boxed{32\sqrt3+8\sqrt6-54}.$$ This matches **none** of the listed options, so the stored answer appears incorrect.

More from Area Under the Curves

The area of the region bounded by y2 = 8x and y2 = 16(3 $-$ x) is equal to:2022 · MCQ
The area of the smaller region enclosed by the curves $y^{2} = 8 x + 4$ and $x^{2} + y^{2} + 43 x - 4 = 0$ is equal to2022 · MCQ
The area of the region enclosed by $y \leq 4 x^{2}, x^{2} \leq 9 y$ and $y \leq 4$ , is equal to :2022 · MCQ
Consider a curve $y = y (x)$ in the first quadrant as shown in the figure. Let the area $A_{1}$ is twice the area $A_{2}$ . Then the normal to the curve perpendicular to the line $2 x - 12 y = 15$ does NOT pass through the point. Includes diagram2022 · MCQ
Let $A_{1} = {(x, y) : ∣ x ∣ \leq y^{2}, ∣ x ∣ + 2 y \leq 8}$ and $A_{2} = {(x, y) : ∣ x ∣ + ∣ y ∣ \leq k}$ . If 27 (Area A1) = 5 (Area A2), then k is equal to :2022 · Numerical
If the area of the region ${(x, y) : x^{\frac{2}{3}} + y^{\frac{2}{3}} \leq 1, x + y \geq 0, y \geq 0}$ is A, then $\frac{256 A}{π}$ is equal to .2022 · Numerical
The area enclosed by the curves $y = lo g_{e} (x + e^{2}), x = lo g_{e} (\frac{2}{y})$ and $x = lo g_{e} 2$ , above the line $y = 1$ is:2022 · MCQ
The area of the region S = {(x, y) : y2 $\leq$ 8x, y $\geq 2$ x, x $\geq$ 1} is2022 · MCQ

View written solutionFree

Correct answer: NONE OF THE GIVEN OPTIONS, $32\SQRT{3}+8\SQRT{6}-54$

Given curves

We need the area bounded by $y=|x^2-9|$ and the line $y=3.$

Write the modulus function piecewise

Since

\begin{cases} 9-x^2, & |x|\le 3,\\ x^2-9, & |x|\ge 3. \end{cases}$$ So the curve is: - downward parabola $y=9-x^2$ between $x=-3$ and $x=3$, - upward branches $y=x^2-9$ for $|x|\ge 3$. 3. **Find points of intersection with $y=3$** We solve $$|x^2-9|=3.$$ This gives two cases: - $x^2-9=3 \Rightarrow x^2=12 \Rightarrow x=\pm 2\sqrt3$, - $x^2-9=-3 \Rightarrow x^2=6 \Rightarrow x=\pm \sqrt6$. Thus the line $y=3$ meets the curve at $$x=\pm \sqrt6,\quad x=\pm 2\sqrt3.$$ 4. **Identify the bounded regions** Because of symmetry about the $y$-axis, the total bounded area is twice the area on the right side. On the right side: - from $x=\sqrt6$ to $x=3$, the curve is $y=9-x^2$, which lies above $y=3$, - from $x=3$ to $x=2\sqrt3$, the curve is $y=x^2-9$, which lies below $y=3$. Hence right-side area is $$\int_{\sqrt6}^{3} \big[(9-x^2)-3\big]dx + \int_{3}^{2\sqrt3} \big[3-(x^2-9)\big]dx.$$ So total area: $$A=2\left[\int_{\sqrt6}^{3} (6-x^2)\,dx + \int_{3}^{2\sqrt3} (12-x^2)\,dx\right].$$ 5. **Evaluate the first integral** $$\int (6-x^2)dx=6x-\frac{x^3}{3}.$$ Therefore, $$\int_{\sqrt6}^{3}(6-x^2)dx= \left[6x-\frac{x^3}{3}\right]_{\sqrt6}^{3}.$$ At $x=3$: $$6(3)-\frac{27}{3}=18-9=9.$$ At $x=\sqrt6$: $$6\sqrt6-\frac{(\sqrt6)^3}{3}=6\sqrt6-\frac{6\sqrt6}{3}=6\sqrt6-2\sqrt6=4\sqrt6.$$ So, $$\int_{\sqrt6}^{3}(6-x^2)dx=9-4\sqrt6.$$ 6. **Evaluate the second integral** $$\int (12-x^2)dx=12x-\frac{x^3}{3}.$$ Thus, $$\int_{3}^{2\sqrt3}(12-x^2)dx= \left[12x-\frac{x^3}{3}\right]_{3}^{2\sqrt3}.$$ At $x=2\sqrt3$: $$(2\sqrt3)^3=24\sqrt3,$$ so $$12(2\sqrt3)-\frac{24\sqrt3}{3}=24\sqrt3-8\sqrt3=16\sqrt3.$$ At $x=3$: $$12(3)-\frac{27}{3}=36-9=27.$$ Hence, $$\int_{3}^{2\sqrt3}(12-x^2)dx=16\sqrt3-27.$$ 7. **Total area** So, $$A=2\Big[(9-4\sqrt6)+(16\sqrt3-27)\Big].$$ $$A=2(16\sqrt3-4\sqrt6-18).$$ $$A=32\sqrt3-8\sqrt6-36.$$ Factorizing, $$A=4(8\sqrt3-2\sqrt6-9).$$ 8. **Compare with options** The options are: - A: $4(2\sqrt3+\sqrt6-4)$ - B: $4(4\sqrt3+\sqrt6-4)$ - C: $8(4\sqrt3+3\sqrt6-9)$ - D: $8(4\sqrt3+2\sqrt6-9)$ Our result $$A=32\sqrt3-8\sqrt6-36=8(4\sqrt3-\sqrt6-\tfrac{9}{2}),$$ which matches **none** of the given options. 9. **Check for algebra/sign mistakes** A quick numerical check: $$\sqrt3\approx 1.732,\quad \sqrt6\approx 2.449.$$ Then $$A\approx 32(1.732)-8(2.449)-36\approx 55.424-19.592-36\approx -0.168?$$ This indicates an arithmetic slip in simplification, so let us recompute carefully. From step 6: $$16\sqrt3-27\approx 27.712-27=0.712,$$ and from step 5: $$9-4\sqrt6\approx 9-9.798=-0.798.$$ That is impossible for area, so the first integrand was chosen on the wrong interval sign-wise. 10. **Correct region setup** For $|x|<3$, curve is $y=9-x^2$. This is above $y=3$ when $$9-x^2>3 \Rightarrow x^2<6 \Rightarrow |x|<\sqrt6.$$ So the bounded central region is from $x=-\sqrt6$ to $x=\sqrt6$. For $|x|>3$, curve is $y=x^2-9$. This is below $y=3$ when $$x^2-9<3 \Rightarrow x^2<12 \Rightarrow |x|<2\sqrt3.$$ Together with $|x|>3$, this gives side bounded regions on $$3\le |x|\le 2\sqrt3.$$ Thus total area is sum of: - central region between $x=-\sqrt6$ and $x=\sqrt6$, - two side regions between $x=3$ and $x=2\sqrt3$ and symmetrically on the left. 11. **Compute correctly** Central area: $$A_1=\int_{-\sqrt6}^{\sqrt6}\big[(9-x^2)-3\big]dx =\int_{-\sqrt6}^{\sqrt6}(6-x^2)dx.$$ Using symmetry, $$A_1=2\int_0^{\sqrt6}(6-x^2)dx =2\left[6x-\frac{x^3}{3}\right]_0^{\sqrt6}.

Now $6\sqrt6-\frac{6\sqrt6}{3}=6\sqrt6-2\sqrt6=4\sqrt6,$ so $A_1=2(4\sqrt6)=8\sqrt6.$

Side areas:

=2\int_3^{2\sqrt3}(12-x^2)dx.$$ From above, $$\int_3^{2\sqrt3}(12-x^2)dx=16\sqrt3-27,$$ therefore $$A_2=2(16\sqrt3-27)=32\sqrt3-54.$$ Hence total area: $$A=A_1+A_2=8\sqrt6+32\sqrt3-54.$$ Factor out $2$: $$A=2(4\sqrt6+16\sqrt3-27).$$ Or more neatly, $$A=2\big(16\sqrt3+4\sqrt6-27\big).$$ 12. **Compare with options again** Option D is $$8(4\sqrt3+2\sqrt6-9)=32\sqrt3+16\sqrt6-72,$$ which is **not equal** to our result. Our computed area is $$\boxed{32\sqrt3+8\sqrt6-54}.$$ This matches **none** of the listed options, so the stored answer appears incorrect.

Area Under the Curves question

Area Under the Curves question

More from Area Under the Curves

Area Under the Curves question

Area Under the Curves question

More from Area Under the Curves