Let g(x) be a linear function and f(x)={g(x), , x ≤ 0; ((1+x)/(2+x))^(1/x), , x>0, is continuous at x=0. If f′(1)=f(-1), then the value g(3) is

Solution

Maths sir · 5 steps

The base tends to 1/2, not 1, so the limit is 0. That makes g(x) = ax, and log differentiation finds a.

1. Continuity at zero. As x → 0⁺, the base (1+x)/(2+x) → 1/2 and the exponent 1/x → +∞, so this is not a 1^∞ form. The right-hand limit is (1/2)^∞ = 0. Continuity gives g(0) = 0, so the linear function is g(x) = ax.
2. Take logs. For x > 0, write ln f = 1/x[ln(1+x) - ln(2+x)] and differentiate both sides.

   f'(x)/f(x) = -1/x²ln(1+x)/(2+x) + 1/x(1/(1+x) - 1/(2+x))

3. Value at one. f(1) = 2/3, and the right side becomes ln 3/2 + (1/2 - 1/3) = ln 3/2 + 1/6. So f'(1) = 2/3 ln 3/2 + 1/9.
4. Find the slope. f(-1) = g(-1) = -a. Setting -a = f'(1) gives a = -2/3 ln 3/2 - 1/9 = 1/3 ln 4/9 - 1/9.
5. So. g(3) = 3a = ln 4/9 - 1/3 = ln 4/9e^(1/3), which is option (c).

The idea to keep. Before using the 1^∞ formula, check what the base actually tends to. When x sits in both the base and the exponent, take logs first and then differentiate.

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