JEE Main 2020 · 6 September · evening · Single correct · +4 / −1
If the tangent to the curve, y = f (x) = xlogₑ x,
(x > 0) at a point (c, f(c)) is parallel to the line-segment
joining the points (1, 0) and (e, e), then c is equal to :
- (a) (e - 1)/e
- (b) e^(( 1/(1 - e) ))
- (c) e^(( 1/(e - 1) ))
- (d) 1/(e - 1)
Answer and solution
Answer (c) e^(( 1/(e - 1) ))
Solution
Maths sir · 4 steps
Match the tangent's slope to the chord's slope, then undo the log to get c.
- Slope of the chord. The segment joins (1, 0) and (e, e), so its slope is (e - 0)/(e - 1) = e/(e - 1).
- Slope of the tangent. By the product rule, f'(x) = ln x + x · 1/x = ln x + 1. At x = c the tangent has slope ln c + 1.
ln c + 1 = e/(e - 1) ⇒ ln c = e/(e - 1) - 1 = 1/(e - 1)
- Undo the log. Exponentiate both sides: c = e^(1/(e-1)). As a check, 1/(e-1) ≈ 0.58, so c ≈ 1.79. This lies between 1 and e, as the Mean Value Theorem says it must, since the curve passes through both (1, 0) and (e, e).
- So. c = e^((1/(e-1))), which is option (c).
The idea to keep. Parallel lines have equal slopes. When the chord joins two points of the same curve, this is Lagrange's Mean Value Theorem, and c must lie between the two x-values.
Answer from the NTA final answer key. Solution by PrepWiser.