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Solution for 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 (A) E1/1−E (B) E(E−1)(2e−1) (C) E 2 E − 1 E − 1 (D) E − 1 E - CBSE (Commerce) Class 12 - Mathematics

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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 (ee), the value of x is

(a) e1/1−e

(b) e(e−1)(2e−1)

(c) \[e^\frac{2e - 1}{e - 1}\]

(d) \[\frac{e - 1}{e}\]

Solution

(a) e1/1−e
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 \[\left( 1, 0 \right) \text { and } \left( e, e \right)\].
\[\therefore m = 1 + \log x\]
\[\Rightarrow \frac{e}{e - 1} = 1 + \log x\]

\[\Rightarrow \frac{e}{e - 1} - 1 = \log x\]

\[ \Rightarrow \frac{e - e + 1}{e - 1} = \log x\]

\[ \Rightarrow \frac{1}{e - 1} = \log x\]

\[ \Rightarrow x = e^\frac{1}{e - 1}\]

  Is there an error in this question or solution?
Solution for 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 (A) E1/1−E (B) E(E−1)(2e−1) (C) E 2 E − 1 E − 1 (D) E − 1 E concept: Maximum and Minimum Values of a Function in a Closed Interval. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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