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Differential Equations question

2019 · Shift 1 · Q23
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Differential Equations question

2019 · Shift 1 · Q23

JEE AdvancedMathematicsDifferential EquationsMultiple correct+4 / −1
Let Γ\GammaΓ denote a curve y = y(x) which is in the first quadrant and let the point (1, 0) lie on it. Let the tangent to I` at a point P intersect the y-axis at YP. If PYP has length 1 for each point P on I`, then which of the following options is/are correct?
  1. A
    xy′+1−x2=0xy' + \sqrt {1 - {x^2}} = 0xy′+1−x2​=0
  2. B
    xy′−1−x2=0xy' - \sqrt {1 - {x^2}} = 0xy′−1−x2​=0
  3. C
    y=log⁡e(1+1−x2x)−1−x2y = {\log _e}\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}}y=loge​(x1+1−x2​​)−1−x2​
  4. D
    y=−log⁡e(1+1−x2x)+1−x2y = - {\log _e}\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) + \sqrt {1 - {x^2}}y=−loge​(x1+1−x2​​)+1−x2​
View written solutionFree

Correct answer: A, C

  1. Set up the tangent condition

Let P(x,y)P(x,y)P(x,y) be any point on the curve in the first quadrant.

The tangent at PPP has slope y′y'y′, so its equation is Y−y=y′(X−x).Y-y=y'(X-x).Y−y=y′(X−x).

To find where this tangent meets the yyy-axis, put X=0X=0X=0: Y−y=y′(0−x)=−xy′.Y-y=y'(0-x)=-xy'.Y−y=y′(0−x)=−xy′. So the yyy-intercept is Y0=y−xy′.Y_0=y-xy'.Y0​=y−xy′. Hence the point of intersection with the yyy-axis is

\,y-xy').$$ 2. **Use the given length condition** The length $PY$ is given to be $1$ for every point $P$. Distance between $P(x,y)$ and $(0,y-xy')$ is $$PY=\sqrt{(x-0)^2+\left(y-(y-xy')\right)^2}.$$ This becomes $$PY=\sqrt{x^2+(xy')^2}.$$ Given $PY=1$, we get $$x^2+x^2(y')^2=1.$$ Thus $$x^2(1+(y')^2)=1,$$ so $$1+(y')^2=\frac1{x^2}.$$ Hence $$(y')^2=\frac{1-x^2}{x^2}.$$ Therefore $$xy'=\pm\sqrt{1-x^2}.$$ So the differential equation can be either $$xy'-\sqrt{1-x^2}=0$$ or $$xy'+\sqrt{1-x^2}=0.$$ Now we use the extra condition that the curve lies in the first quadrant and passes through $(1,0)$. 3. **Determine the correct sign** Since the curve is in the first quadrant, we need $y\ge 0$ for relevant $x\in(0,1]$. From $$y'=\pm \frac{\sqrt{1-x^2}}{x},$$ we integrate both cases. --- ### Case 1: $$xy'+\sqrt{1-x^2}=0$$ Then $$y'=-\frac{\sqrt{1-x^2}}{x}.$$ Integrate: $$y=-\int \frac{\sqrt{1-x^2}}{x}\,dx + C.$$ A standard antiderivative is $$\int \frac{\sqrt{1-x^2}}{x}\,dx=\ln\left(\frac{1-\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}+C.$$ Equivalently, $$-\int \frac{\sqrt{1-x^2}}{x}\,dx=\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}+C.$$ So $$y=\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}+C.$$ Using $(1,0)$: $$0=\ln\left(\frac{1+0}{1}\right)-0+C=0+C,$$ so $C=0$. Thus $$y=\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}.$$ This is exactly **Option C**. Also, for $0<x\le 1$, this gives $y\ge 0$, so it matches the first quadrant condition. Hence **Option A** and **Option C** are correct. --- ### Case 2: $$xy'-\sqrt{1-x^2}=0$$ Then $$y'=\frac{\sqrt{1-x^2}}{x}.$$ Integrating, $$y=-\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}+C.$$ Using $(1,0)$ again gives $C=0$, so $$y=-\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}.$$ This is **Option D**. But for $0<x<1$, this expression is negative (for example near $x=1/2$), so the curve does **not** lie in the first quadrant. Hence this branch is rejected. Therefore **Option B** and **Option D** are not valid for the given curve. 4. **Final conclusion** The correct options are: $$\boxed{A,\ C}$$
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