JEE Mathematics 2024 Question with Solution

Given: The tangent to the curve $Y = Y(X)$ at $(x,y)$ is $Y-y=Y'(x)(X-x)$, and the area enclosed by this tangent and the coordinate axes is $A=-\frac{y^2}{2Y'(x)}+1$.

Find: The value of $12Y(2)$ given $Y(1)=1$.

For the tangent line, the intercepts are obtained as follows.

Setting $X=0$ gives the $Y$-intercept:

$$Y_{\text{int}} = y - xY'(x)$$

Setting $Y=0$ gives the $X$-intercept:

$$X_{\text{int}} = x - \frac{y}{Y'(x)} = \frac{xY'(x)-y}{Y'(x)}$$

Hence the area of the triangle formed with the coordinate axes is

$$A = \frac{1}{2}\left|X_{\text{int}}\,Y_{\text{int}}\right| = \frac{1}{2}\left|\left(\frac{xY'(x)-y}{Y'(x)}\right)(y-xY'(x))\right|$$

So,

$$A = \frac{(y-xY'(x))^2}{2|Y'(x)|}$$

Using the given area relation and the first-quadrant condition, we take the form used in the solution:

$$\frac{(y-xY'(x))^2}{-2Y'(x)} = -\frac{y^2}{2Y'(x)} + 1$$

Multiplying by $-2Y'(x)$,

$$(y-xY'(x))^2 = y^2 - 2Y'(x)$$

Expanding,

$$y^2 - 2xyY'(x) + x^2(Y'(x))^2 = y^2 - 2Y'(x)$$

Therefore,

$$-2xyY'(x) + x^2(Y'(x))^2 = -2Y'(x)$$

Since $Y'(x) \ne 0$, divide by $Y'(x)$:

$$-2xy + x^2Y'(x) = -2$$

That is,

$$x^2Y'(x) - 2xy = -2$$

Rewriting,

$$Y' - \frac{2}{x}y = -\frac{2}{x^2}$$

This is a first-order linear differential equation. Its integrating factor is

$$\text{I.F.} = e^{\int -\frac{2}{x}\,dx} = x^{-2}$$

Hence,

$$\frac{y}{x^2} = \int \left(-\frac{2}{x^2}\right)\left(\frac{1}{x^2}\right)dx + C$$

So,

$$\frac{y}{x^2} = \int -2x^{-4}\,dx + C = \frac{2}{3x^3} + C$$

Therefore,

$$Y(x) = \frac{2}{3x} + Cx^2$$

Using $Y(1)=1$,

$$1 = \frac{2}{3} + C$$

Thus,

$$C = \frac{1}{3}$$

Hence the curve is

$$Y(x) = \frac{2}{3x} + \frac{1}{3}x^2$$

Now,

$$Y(2) = \frac{2}{3\cdot 2} + \frac{1}{3}(2)^2 = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$$

Therefore,

$$12Y(2) = 12 \times \frac{5}{3} = 20$$

So, the required value is $20$.

Given: The tangent at $(x,y)$ to $Y=Y(X)$ forms a triangle with the coordinate axes whose area is prescribed.

Find: Evaluate $12Y(2)$.

The tangent line is

$$Y-y=Y'(x)(X-x)$$

Its intercepts are:

• with the $Y$-axis, $X=0$
• with the $X$-axis, $Y=0$

So,

$$Y_{\text{int}} = y-xY'(x)$$

and

$$X_{\text{int}} = x-\frac{y}{Y'(x)}$$

Hence triangle area:

$$A=\frac{1}{2}\left|\left(x-\frac{y}{Y'(x)}\right)(y-xY'(x))\right|$$

Using the expression given in the solution,

$$\frac{(y-xY'(x))^2}{-2Y'(x)}=-\frac{y^2}{2Y'(x)}+1$$

After multiplying through by $-2Y'(x)$ and simplifying, we get

$$x^2Y'(x)-2xy=-2$$

or

$$Y'-\frac{2}{x}y=-\frac{2}{x^2}$$

Now solve by integrating factor:

$$\text{I.F.}=e^{\int -2/x\,dx}=\frac{1}{x^2}$$

Then,

$$\frac{d}{dx}\left(\frac{y}{x^2}\right)=-\frac{2}{x^4}$$

Integrating,

$$\frac{y}{x^2}=\frac{2}{3x^3}+C$$

Thus,

$$Y(x)=\frac{2}{3x}+Cx^2$$

Apply $Y(1)=1$:

$$1=\frac{2}{3}+C \Rightarrow C=\frac{1}{3}$$

So,

$$Y(x)=\frac{2}{3x}+\frac{1}{3}x^2$$

At $x=2$,

$$Y(2)=\frac{2}{6}+\frac{4}{3}=\frac{5}{3}$$

Therefore,

$$12Y(2)=12\cdot\frac{5}{3}=20$$

Thus the answer is $20$.