The coordinates of the point on the curve `(x^(2)+ 1) `(y-3)=x where a tangent to the curve has the greatest slope are given by

Options

A. `(sqrt(3),3 +sqrt(3)//4)`
B. `(-sqrt(3),3-sqrt(3)//4)`
C. (0,3)
D. none of these

Answer

To find the coordinates of the point on the curve \((x^2 + 1)(y - 3) = x\) where the tangent to the curve has the greatest slope, we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation of the curve:
\[ (x^2 + 1)(y - 3) = x \]
We can rearrange this to express \(y\) in terms of \(x\):
\[ y - 3 = \frac{x}{x^2 + 1} \]
\[ y = \frac{x}{x^2 + 1} + 3 \]

### Step 2: Differentiate to find the slope
Next, we differentiate \(y\) with respect to \(x\) to find the slope of the tangent:
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x}{x^2 + 1}\right) \]
Using the quotient rule, where \(f(x) = x\) and \(g(x) = x^2 + 1\):
\[ \frac{dy}{dx} = \frac{f'(x)g(x) - g'(x)f(x)}{(g(x))^2} \]
Calculating \(f'(x)\) and \(g'(x)\):
\[ f'(x) = 1, \quad g'(x) = 2x \]
Now substituting into the quotient rule:
\[ \frac{dy}{dx} = \frac{(1)(x^2 + 1) - (2x)(x)}{(x^2 + 1)^2} \]
Simplifying this:
\[ \frac{dy}{dx} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} \]

### Step 3: Find critical points
To find where the slope is maximized, we set the derivative equal to zero:
\[ 1 - x^2 = 0 \]
Solving for \(x\):
\[ x^2 = 1 \implies x = \pm 1 \]
We also check the slope at \(x = 0\):
\[ \frac{dy}{dx} \text{ at } x = 0 \implies \frac{1 - 0^2}{(0^2 + 1)^2} = 1 \]

### Step 4: Determine maximum slope
To determine which of \(x = 1\) or \(x = -1\) gives the maximum slope, we can evaluate the second derivative or check the values of the first derivative around these points.

Calculating the slope at \(x = 1\):
\[ \frac{dy}{dx} \text{ at } x = 1 \implies \frac{1 - 1^2}{(1^2 + 1)^2} = 0 \]
Calculating the slope at \(x = -1\):
\[ \frac{dy}{dx} \text{ at } x = -1 \implies \frac{1 - (-1)^2}{((-1)^2 + 1)^2} = 0 \]

### Step 5: Evaluate \(y\) at critical points
Now we find \(y\) at these critical points:
1. For \(x = 1\):
\[ y = \frac{1}{1^2 + 1} + 3 = \frac{1}{2} + 3 = \frac{7}{2} \]
Thus, the point is \((1, \frac{7}{2})\).

2. For \(x = -1\):
\[ y = \frac{-1}{(-1)^2 + 1} + 3 = \frac{-1}{2} + 3 = \frac{5}{2} \]
Thus, the point is \((-1, \frac{5}{2})\).

### Step 6: Conclusion
The maximum slope occurs at \(x = 0\) where the slope is 1. Evaluating \(y\) at \(x = 0\):
\[ y = \frac{0}{0^2 + 1} + 3 = 3 \]
Thus, the coordinates of the point on the curve where the tangent has the greatest slope are:
\[ \boxed{(0, 3)} \]