When the tangent the curve y=x log (x) is parallel to the chord joining the points (1,0) and (e,e) the value of x , is

Options

A. `1//1-e`
B. `e^((e-1)(2e-1))`
C. `e^((2e-1)/(e-1))`
D. `(e-1)/(e)`

Answer

To solve the problem, we need to find the value of \( x \) for which the tangent to the curve \( y = x \log(x) \) is parallel to the chord joining the points \( (1, 0) \) and \( (e, e) \).

### Step-by-Step Solution:

1. **Find the slope of the chord joining the points (1, 0) and (e, e)**:
The slope \( m \) of a line joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For points \( (1, 0) \) and \( (e, e) \):
\[ m = \frac{e - 0}{e - 1} = \frac{e}{e - 1} \]

**Hint**: Use the slope formula for two points to find the slope of the chord.

2. **Differentiate the curve \( y = x \log(x) \)**:
To find the slope of the tangent to the curve, we need to differentiate \( y \) with respect to \( x \):
\[ y' = \frac{d}{dx}(x \log(x)) = \log(x) + 1 \]
(Using the product rule: \( u = x \) and \( v = \log(x) \))

**Hint**: Remember to apply the product rule when differentiating products of functions.

3. **Set the slopes equal**:
Since the tangent is parallel to the chord, we set the slope of the tangent equal to the slope of the chord:
\[ \log(x) + 1 = \frac{e}{e - 1} \]

**Hint**: When two lines are parallel, their slopes are equal.

4. **Solve for \( \log(x) \)**:
Rearranging the equation gives:
\[ \log(x) = \frac{e}{e - 1} - 1 \]
Simplifying the right-hand side:
\[ \log(x) = \frac{e - (e - 1)}{e - 1} = \frac{1}{e - 1} \]

**Hint**: To isolate \( \log(x) \), move the constant to the other side of the equation.

5. **Exponentiate to find \( x \)**:
To solve for \( x \), we exponentiate both sides:
\[ x = e^{\frac{1}{e - 1}} \]

**Hint**: Remember that if \( \log(x) = a \), then \( x = e^a \).

### Final Answer:
The value of \( x \) is:
\[ x = e^{\frac{1}{e - 1}} \]