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Question
When the tangent to 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
e1/1−e
e(e−1)(2e−1)
\[e^\frac{2e - 1}{e - 1}\]
\[\frac{e - 1}{e}\]
Solution
e1/1−e
Given: \[y = f\left( x \right) = x\log x\]
Differentiating the given function with respect to x, we get
Slope of the chord joining the points
\[\Rightarrow \frac{e}{e - 1} - 1 = \log x\]
\[ \Rightarrow \frac{e - e + 1}{e - 1} = \log x\]
\[ \Rightarrow \frac{1}{e - 1} = \log x\]
\[ \Rightarrow x = e^\frac{1}{e - 1}\]
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