Let f:R → R and g:R → R be two functions defined by f(x) = log ₑ(x² + 1) - e⁻ˣ + 1 and g(x) = (1 - 2e²ˣ)/eˣ. Then, for which of the following range of α, the inequality f( g( (α - 1)²/3 ) ) > f( g( α -5/3 ) ) holds ?

Solution

Maths sir · 3 steps

Do not compose anything. Check which way f and g each move, then compare the two inputs directly.

1. f is increasing. f'(x) = 2x/(x²+1) + e⁻ˣ. For x ≥ 0 both terms are non-negative and e⁻ˣ > 0. For x < 0, 2x/(x²+1) ≥ -1 while e⁻ˣ > 1, so f'(x) > 0 for every real x.
2. g is decreasing. g(x) = e⁻ˣ - 2eˣ, so g'(x) = -e⁻ˣ - 2eˣ < 0 for every x. An increasing f applied to a decreasing g makes f ° g decreasing, so f(g(a)) > f(g(b)) exactly when a < b.

   (α-1)²/3 < α - 5/3 ⇒ α² - 5α + 6 < 0 ⇒ (α-2)(α-3) < 0

3. So. The quadratic is negative only between its roots, so 2 < α < 3. The answer is (a).

The idea to keep. For f(g(a)) > f(g(b)), find the sign of f' and g' first. If one function decreases, the composition decreases and the inequality on the inputs flips.

Answer from the NTA final answer key. Solution by PrepWiser.