Let f(x) = 1 - frac{1}{x}, g_2(x) = f(f(x)), | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Step $1$: Find the pattern of $g_n(x)$.$f(x) = 1 - \frac{1}{x} = \frac{x-1}{x}$.$g_2(x) = f(f(x)) = 1 - \frac{1}{1 - 1/x} = 1 - \frac{x}{x-1} = \f

Vedclass · Accepted Answer

$-x$

Vedclass · Answer

(B) Step $1$: Find the pattern of $g_n(x)$.$f(x) = 1 - \frac{1}{x} = \frac{x-1}{x}$.$g_2(x) = f(f(x)) = 1 - \frac{1}{1 - 1/x} = 1 - \frac{x}{x-1} = \frac{x-1-x}{x-1} = \frac{-1}{x-1} = \frac{1}{1-x}$.$g_3(x) = f(g_2(x)) = 1 - \frac{1}{1/(1-x)} = 1 - (1-x) = x$.Step $2$: Determine $g_{2025}(x)$ and $g_{2026}(x)$.Since $g_3(x) = x$, the functions cycle with period $3$.$2025$ is divisible by $3$, so $g_{2025}(x) = g_3(x) = x$.$g_{2026}(x) = f(g_{2025}(x)) = f(x) = 1 - \frac{1}{x}$.Step $3$: Solve the integral equation.$\int x \cdot (1 - \frac{1}{x}) \, dx = \int (x - 1) \, dx = \frac{x^2}{2} - x + c$.$\int g_{2025}(x) \, dx = \int x \, dx = \frac{x^2}{2} + c$.Substituting into the given equation: $\frac{x^2}{2} - x + c = \frac{x^2}{2} + c + h(x)$.Thus, $h(x) = -x$.

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) = $

Similar Questions

$\int \frac{dx}{2+\cos x} = $ (Where $C$ is a constant of integration.)

$\int \sqrt{x^2 - 8x + 7} \, dx = $

The value of $\int \frac{d x}{(x+1)^{3 / 4}(x-2)^{5 / 4}}$ is equal to

If $\int(\sqrt{\operatorname{cosec} x+1}) d x=k \tan ^{-1}(f(x))+c$, then $\frac{1}{k} f\left(\frac{\pi}{6}\right)=$

If $f(x) = \int \csc^5 x \ dx$,then $f(\frac{\pi}{4}) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module