Let f : (–1, ∞) → R be defined by f(0) = 1 and
f(x) = 1/xlog ₑ( 1 + x ), x ≠ 0. Then the function f :

Solution

Maths sir · 5 steps

Differentiate, then treat the messy numerator as its own function: its largest value is 0, at x = 0.

1. Differentiate. For x ≠ 0, use the quotient rule on f(x) = ln(1+x)/x. The denominator x² is always positive, so the sign of f'(x) is the sign of the numerator.

   f'(x) = g(x)/x², g(x) = x/(1+x) - ln(1+x)

2. Study the numerator. g(0) = 0 and g'(x) = 1/(1+x)² - 1/(1+x) = -x/(1+x)². So g rises on (-1, 0) and falls on (0, ∞), which makes x = 0 its maximum.
3. Sign of the derivative. The largest value of g is g(0) = 0, so g(x) < 0 for every x ≠ 0. That gives f'(x) < 0 on both (-1, 0) and (0, ∞).
4. Join at zero. lim(x → 0) ln(1+x)/x = 1 = f(0), so f is continuous at 0. Near 0, f(x) ≈ 1 - x/2, so f'(0) = -1/2, which is also negative. The two decreasing pieces therefore join into one decreasing function.
5. So. f decreases on the whole interval (-1, ∞), which is option (a).

The idea to keep. When the numerator of f' is hard to sign, call it g(x) and differentiate again. Find where g is largest or smallest, then use that value to fix the sign of g everywhere.

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