JEE Main 2023 · 25 January · morning · Single correct · +4 / −1
Let f:(0,1)→ℝ be a function defined f(x) = 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) Both (I) and (II) are true
- (b) Neither (I) nor (II) is true
- (c) Only (II) is true
- (d) Only (I) is true
Answer and solution
Answer (a) Both (I) and (II) are true
Solution
Maths sir · 4 steps
Put both terms over the common denominator eˣ - 1 and g becomes a constant plus one simple fraction.
- Rewrite both terms. Multiply top and bottom by eˣ: f(x) = 1/(1-e⁻ˣ) = eˣ/(eˣ-1). Replacing x by -x gives f(-x) = 1/(1-eˣ) = -1/(eˣ-1).
g(x) = -1/(eˣ-1) - eˣ/(eˣ-1) = -(eˣ+1)/(eˣ-1) = -1 - 2/(eˣ-1)
- Check statement (I). Differentiate: g'(x) = 2eˣ/(eˣ-1)². On (0,1) both eˣ and (eˣ-1)² are positive, so g'(x) > 0. That means g is strictly increasing, so (I) is true.
- Check statement (II). A strictly increasing function never takes the same value twice, because x₁ < x₂ forces g(x₁) < g(x₂). So g is one-one and (II) is also true.
- So. Both statements hold, and the answer is (a).
The idea to keep. To test whether a function is one-one on an interval, check the sign of its derivative. If g' keeps one strict sign throughout, g is strictly monotonic and therefore one-one.
Answer from the NTA final answer key. Solution by PrepWiser.