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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

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### 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?

#### APPEARS IN

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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