Question

Let \(f(x) = 1-\frac{1}{x},g_2(x) = f(f(x)),g_3(x) = f(f(f(x)))\) and so on. If \(\int x\cdot g_{2026}(x)\,dx = \int g_{2025}(x)\,dx+h(x)+c\), then \(h(x) =\)...

• A. \(x\)
• B. \(-x\)
• C. \(logx\)
• D. \(-logx\)

Hint:

Compute g1, g2, g3 and notice that g3(x) = x, so the sequence repeats every 3 steps.

Answer 1

The Correct Option is B

Step 1: Pattern
Compute by substitution: $f(x) = \frac{x-1}{x}$, $f(f(x)) = \frac{(x-1)/x - 1}{(x-1)/x} = \frac{-1}{x-1}$, and the third iterate returns $x$. So $g_n = g_{n \bmod 3}$ with $g_0 = x$.

Step 2: Indices
$2026 \equiv 1$ and $2025 \equiv 0 \pmod 3$, so the integrand on the left is $x \cdot \frac{x-1}{x} = x - 1$, and on the right it is $x$.

Step 3: Difference
The left integral minus $\int g_{2025}\,dx$ is $\int (x - 1 - x)\,dx = \int -1\,dx = -x + C$.

Step 4: Result
So $h(x) = -x$, option (B).

Final Answer:
h(x) equals -x. This is option (B). \[ \boxed{\text{(B) }-x} \]