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Area Under the Curves question

2023 · 11 Apr · Shift 2 · Q40
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Area Under the Curves question

2023 · 11 Apr · Shift 2 · Q40

JEE MainMathematicsArea Under the CurvesNumerical+4 / −1
If A is the area in the first quadrant enclosed by the curve C:2x2−y+1=0\mathrm{C: 2 x^{2}-y+1=0}C:2x2−y+1=0, the tangent to C\mathrm{C}C at the point (1,3)(1,3)(1,3) and the line x+y=1\mathrm{x}+\mathrm{y}=1x+y=1, then the value of 60 A60 \mathrm{~A}60 A is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 16

  1. Write the given curve in explicit form

The curve is 2x2−y+1=0  ⟹  y=2x2+1.2x^2-y+1=0 \implies y=2x^2+1.2x2−y+1=0⟹y=2x2+1.

So CCC is the parabola y=2x2+1.y=2x^2+1.y=2x2+1.

  1. Find the tangent to CCC at (1,3)(1,3)(1,3)

Since y=2x2+1,y=2x^2+1,y=2x2+1, we have dydx=4x.\frac{dy}{dx}=4x.dxdy​=4x.

At x=1x=1x=1, slope of tangent is m=4.m=4.m=4.

Using point-slope form at (1,3)(1,3)(1,3): y−3=4(x−1).y-3=4(x-1).y−3=4(x−1). So, y=4x−1.y=4x-1.y=4x−1.

Thus, the three boundary curves are:

  • parabola: y=2x2+1y=2x^2+1y=2x2+1
  • tangent: y=4x−1y=4x-1y=4x−1
  • line: x+y=1  ⟹  y=1−xx+y=1 \implies y=1-xx+y=1⟹y=1−x
  1. Find the intersection points

(i) Parabola and line y=1−xy=1-xy=1−x

2x2+1=1−x2x^2+1=1-x2x2+1=1−x 2x2+x=02x^2+x=02x2+x=0 x(2x+1)=0.x(2x+1)=0.x(2x+1)=0.

So x=0x=0x=0 or x=−12x=-\frac12x=−21​.

In the first quadrant, the relevant point is (0,1).(0,1).(0,1).

(ii) Tangent and line

4x−1=1−x4x-1=1-x4x−1=1−x 5x=25x=25x=2

\quad y=1-\frac25=\frac35.$$ So intersection point is $$\left(\frac25,\frac35\right).$$ ### (iii) Parabola and tangent $$2x^2+1=4x-1$$ $$2x^2-4x+2=0$$ $$x^2-2x+1=0$$ $$(x-1)^2=0.$$ Hence they touch at $$(1,3).$$ --- 4. **Understand the enclosed region in the first quadrant** The closed region is bounded by: - from $(0,1)$ to $\left(\frac25,\frac35\right)$ along the line $y=1-x$, - from $\left(\frac25,\frac35\right)$ to $(1,3)$ along the tangent $y=4x-1$, - from $(1,3)$ back to $(0,1)$ along the parabola $y=2x^2+1$. For vertical strips: - on $0 \le x \le \frac25$, top curve is parabola and bottom curve is $y=1-x$, - on $\frac25 \le x \le 1$, top curve is parabola and bottom curve is tangent $y=4x-1$. So, $$A=\int_0^{2/5}\Big[(2x^2+1)-(1-x)\Big]dx+\int_{2/5}^{1}\Big[(2x^2+1)-(4x-1)\Big]dx.$$ Simplify: $$A=\int_0^{2/5}(2x^2+x)dx+\int_{2/5}^{1}(2x^2-4x+2)dx.$$ --- 5. **Evaluate the integrals** ### First integral $$\int (2x^2+x)dx=\frac{2x^3}{3}+\frac{x^2}{2}.$$ Thus, $$\int_0^{2/5}(2x^2+x)dx =\left[\frac{2x^3}{3}+\frac{x^2}{2}\right]_0^{2/5}.$$ At $x=\frac25$: $$\frac{2}{3}\left(\frac{8}{125}\right)+\frac12\left(\frac{4}{25}\right) =\frac{16}{375}+\frac{2}{25} =\frac{16}{375}+\frac{30}{375} =\frac{46}{375}.$$ So, $$I_1=\frac{46}{375}.$$ ### Second integral $$\int (2x^2-4x+2)dx=\frac{2x^3}{3}-2x^2+2x.$$ Thus, $$\int_{2/5}^{1}(2x^2-4x+2)dx =\left[\frac{2x^3}{3}-2x^2+2x\right]_{2/5}^{1}.$$ At $x=1$: $$\frac23-2+2=\frac23.$$ At $x=\frac25$: $$\frac{2}{3}\cdot \frac{8}{125}-2\cdot \frac{4}{25}+2\cdot \frac25 =\frac{16}{375}-\frac{8}{25}+\frac45.$$ Convert to denominator $375$: $$\frac{16}{375}-\frac{120}{375}+\frac{300}{375}= rac{196}{375}.$$ Hence, $$I_2=\frac23-\frac{196}{375}.

Now, 23=250375,\frac23=\frac{250}{375},32​=375250​, so I2=250−196375=54375=18125.I_2=\frac{250-196}{375}=\frac{54}{375}=\frac{18}{125}.I2​=375250−196​=37554​=12518​.

  1. Total area

A=I1+I2=46375+54375=100375=415.A=I_1+I_2=\frac{46}{375}+\frac{54}{375}=\frac{100}{375}=\frac{4}{15}.A=I1​+I2​=37546​+37554​=375100​=154​.

Therefore, 60A=60⋅415=16.60A=60\cdot \frac{4}{15}=16.60A=60⋅154​=16.

  1. Comparison with stored answer

Derived answer is 161616, which matches the stored correct answer.

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