Let f:(0,1) → R be a function defined by f(x) =… — JEE Main 2023 Mathematics PYQ

JEE Main 2023 · 25 Jan · Shift 1

Let

f:(0,1) \to \mathbb{R}

be a function defined by

f(x) = \dfrac{1}{1-e^{-x}}

, and

g(x) = (f(-x) - f(x))

. Consider two statements

(I) g is an increasing function in

(0, 1)

(II) g is one-one in

(0, 1)

Then,

A.
Only (II) is true
B.
Both (I) and (II) are true
C.
Only (I) is true
D.
Neither (I) nor (II) is true

Show answer and solution

Correct option: B

Solution

g(x) = f(-x) - f(x) = 1/(1-e^x) - 1/(1-e^(-x)).

Simplify: 1/(1-e^x) - e^x/(e^x-1) = 1/(1-e^x) + e^x/(1-e^x) = (1+e^x)/(1-e^x).

So g(x) = -(1+e^x)/(e^x-1) = -coth(x/2), since (1+e^x)/(e^x-1) = coth(x/2).

Now g'(x) = d/dx[-(1+e^x)/(e^x-1)]. Using quotient rule: numerator = -e^x(e^x-1) + e^x(1+e^x) = -e^(2x)+e^x+e^x+e^(2x) = 2e^x > 0.

So g'(x) = 2e^x/(e^x-1)² > 0 for all x in (0,1), meaning g is strictly increasing, hence both one-one and increasing. Both statements are true.

[JEE Main 2023 · 25 Jan · Shift 1]

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