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प्रश्न
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 ______.
पर्याय
`e^(1/(1−e))`
`e^((e−1)(2e−1))`
\[e^\frac{2e - 1}{e - 1}\]
\[\frac{e - 1}{e}\]
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उत्तर
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 `bb(e^(1/(1−e)))`.
Explanation:
Given: \[y = f\left( x \right) = x\log x\]
Differentiating the given function with respect to x, we get
\[f'\left( x \right) = 1 + \log x\]
\[\Rightarrow\] Slope of the tangent to the curve = \[1 + \log x\]
Also,
Slope of the chord joining the points
\[\left( 1, 0 \right) \text { and } \left( e, e \right)\] (m) = \[\frac{e}{e - 1}\]
The tangent to the curve is parallel to the chord joining the points \[\left( 1, 0 \right) \text { and } \left( e, e \right)\].
\[\therefore m = 1 + \log x\]
\[\Rightarrow \frac{e}{e - 1} = 1 + \log x\]
\[\Rightarrow \frac{e}{e - 1} - 1 = \log x\]
\[ \Rightarrow \frac{e - e + 1}{e - 1} = \log x\]
\[ \Rightarrow \frac{1}{e - 1} = \log x\]
\[ \Rightarrow x = e^\frac{1}{e - 1}\]
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