JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 13)
Let the tangent at any point P on a curve passing through the points (1, 1) and $$\left(\frac{1}{10}, 100\right)$$, intersect positive $$x$$-axis and $$y$$-axis at the points A and B respectively. If $$\mathrm{PA}: \mathrm{PB}=1: k$$ and $$y=y(x)$$ is the solution of the differential equation $$e^{\frac{d y}{d x}}=k x+\frac{k}{2}, y(0)=k$$, then $$4 y(1)-6 \log _{\mathrm{e}} 3$$ is equal to ____________.
Answer
4
Explanation
Let the equation of tangent to the curve at $(x, y)$.
Whose slope is $\frac{d y}{d x}$ is
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$$ \begin{aligned} &\mathrm{Y}-y=\frac{d y}{d x}(\mathrm{X}-x)\\\\ &\begin{aligned} & \text { Putting } Y=0 \Rightarrow X=x-\frac{y}{\left(\frac{d y}{d x}\right)} \mathrm{t} \\\\ & \Rightarrow \alpha=x-\frac{y}{\left(\frac{d y}{d x}\right)} \end{aligned} \end{aligned} $$
and putting $\mathrm{X}=0 \Rightarrow \mathrm{Y}=y-x \frac{d y}{d x}$
$$ \Rightarrow \beta=y-x \frac{d y}{d x} $$
$\because \mathrm{P}$ divides $\mathrm{AB}$ in $1: k$
$$ x=\frac{k \alpha+0}{k+1} \text { and } y=\frac{k \times 0+\beta}{k+1} $$
$$ \begin{aligned} & \Rightarrow x(k+1)=k\left(x-\frac{y}{\frac{d y}{d x}}\right) \\\\ & \Rightarrow x k+x=x k-\frac{y k}{\frac{d y}{d x}} \\\\ &\Rightarrow x \frac{d y}{d x}=-y k \end{aligned} $$
$$ \begin{aligned} & \text { or } \int \frac{d y}{y}=-k \times \int \frac{1}{x} d x \\\\ & \Rightarrow \log y=-k \log x+\log \mathrm{C} \\\\ & \text { or } \log y \times x^k=\log \mathrm{C} \\\\ & \Rightarrow y x^k=\mathrm{C} \\\\ & \text { putting } x=1, y=1 \Rightarrow c=1 \\\\ & \text { so } y x^k=1 \end{aligned} $$
Putting $x=\frac{1}{10}, y=100 \Rightarrow 100 \times\left(\frac{1}{100}\right)^k=1 \Rightarrow k=2$
so $y x^2=1$ or $y=\frac{1}{x^2}$
Now $e^{\frac{d y}{d x}}=k x+\frac{k}{2}$
$$ \Rightarrow \frac{d y}{d x}=\log _e\left(k x+\frac{k}{2}\right)=\log _e(2 x+1) $$
On integrating
$$ \begin{aligned} & y=\int 1 \cdot \log _e(2 x+1) d x \\\\ & =x \log _e(2 x+1)-\int \frac{1 \times 2}{2 x+1} \times x d x \\\\ & =x \log _e(2 x+1)-\int 1-\frac{1}{2 x+1} d x \\\\ & y=x \log _e(2 x+1)-x+\frac{1}{2} \log _e(2 x+1)+c \end{aligned} $$
Put $x=0, y=k=2 \Rightarrow c=2$ putting $x=1$
$$ \begin{aligned} & y(1)=\log _e 3-1+\frac{1}{2} \log _e 3+2=\frac{3}{2} \log _e 3-1+2 \\\\ & \Rightarrow 4 y(1)=6 \log _e 3+4 \\\\ & \Rightarrow 4 y(1)-6 \log _e 3= 4 \end{aligned} $$
Whose slope is $\frac{d y}{d x}$ is
$$ \begin{aligned} &\mathrm{Y}-y=\frac{d y}{d x}(\mathrm{X}-x)\\\\ &\begin{aligned} & \text { Putting } Y=0 \Rightarrow X=x-\frac{y}{\left(\frac{d y}{d x}\right)} \mathrm{t} \\\\ & \Rightarrow \alpha=x-\frac{y}{\left(\frac{d y}{d x}\right)} \end{aligned} \end{aligned} $$
and putting $\mathrm{X}=0 \Rightarrow \mathrm{Y}=y-x \frac{d y}{d x}$
$$ \Rightarrow \beta=y-x \frac{d y}{d x} $$
$\because \mathrm{P}$ divides $\mathrm{AB}$ in $1: k$
$$ x=\frac{k \alpha+0}{k+1} \text { and } y=\frac{k \times 0+\beta}{k+1} $$
$$ \begin{aligned} & \Rightarrow x(k+1)=k\left(x-\frac{y}{\frac{d y}{d x}}\right) \\\\ & \Rightarrow x k+x=x k-\frac{y k}{\frac{d y}{d x}} \\\\ &\Rightarrow x \frac{d y}{d x}=-y k \end{aligned} $$
$$ \begin{aligned} & \text { or } \int \frac{d y}{y}=-k \times \int \frac{1}{x} d x \\\\ & \Rightarrow \log y=-k \log x+\log \mathrm{C} \\\\ & \text { or } \log y \times x^k=\log \mathrm{C} \\\\ & \Rightarrow y x^k=\mathrm{C} \\\\ & \text { putting } x=1, y=1 \Rightarrow c=1 \\\\ & \text { so } y x^k=1 \end{aligned} $$
Putting $x=\frac{1}{10}, y=100 \Rightarrow 100 \times\left(\frac{1}{100}\right)^k=1 \Rightarrow k=2$
so $y x^2=1$ or $y=\frac{1}{x^2}$
Now $e^{\frac{d y}{d x}}=k x+\frac{k}{2}$
$$ \Rightarrow \frac{d y}{d x}=\log _e\left(k x+\frac{k}{2}\right)=\log _e(2 x+1) $$
On integrating
$$ \begin{aligned} & y=\int 1 \cdot \log _e(2 x+1) d x \\\\ & =x \log _e(2 x+1)-\int \frac{1 \times 2}{2 x+1} \times x d x \\\\ & =x \log _e(2 x+1)-\int 1-\frac{1}{2 x+1} d x \\\\ & y=x \log _e(2 x+1)-x+\frac{1}{2} \log _e(2 x+1)+c \end{aligned} $$
Put $x=0, y=k=2 \Rightarrow c=2$ putting $x=1$
$$ \begin{aligned} & y(1)=\log _e 3-1+\frac{1}{2} \log _e 3+2=\frac{3}{2} \log _e 3-1+2 \\\\ & \Rightarrow 4 y(1)=6 \log _e 3+4 \\\\ & \Rightarrow 4 y(1)-6 \log _e 3= 4 \end{aligned} $$
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