Mathematics

Mathematics

Problem

If f(x)=∫x2 d(x2−1x2)f(x)=\int \frac{x}{2}\,d\left(\frac{x^2-1}{x^2}\right) and f(2)=12f(2)=\frac{1}{2} and g1(x)=f(f(f(…f(x)… )))g_1(x)=f\big(f(f(\dots f(x)\dots))\big) i.e., g1(x)=f(x)g_1(x)=f(x), g2(x)=f(f(x))g_2(x)=f(f(x)) and so on then identify the correct statement(s). (a) ddx(g3n−2(x))=1\frac{d}{dx}\big(g_{3n-2}(x)\big)=1 whenever exists, n∈Nn\in\mathbb{N} (b) ddx(g3n(x))=1\frac{d}{dx}\big(g_{3n}(x)\big)=1 whenever exists, n∈Nn\in\mathbb{N} (c) lim⁡x→1∑r=1100(g3r(x))r+x−101x−1=5050\lim_{x\to 1}\frac{\sum_{r=1}^{100}(g_{3r}(x))^r+x-101}{x-1}=5050 (d) Slope of the tangent to the graph of the function y=g80(x)y=g_{80}(x) at x=12x=\frac{1}{2} is 44.
Intuitive explanation

This problem chains together integration by parts, function composition, and iteration. First, you'll use integration by parts on the given integral—recognize that d(x2−1x2)d\left(\frac{x^2-1}{x^2}\right) is your differential, so integrate x2\frac{x}{2} against it. Once you find f(x)f(x) using the boundary condition f(2)=12f(2)=\frac{1}{2}, the real insight is spotting a pattern: composing ff with itself repeatedly (gng_n means applying ff exactly nn times) often reveals a fixed point, a cycle, or a closed form that lets you predict gn(x)g_n(x) without computing every iteration.

Intuitive explanation

This problem chains together integration by parts, function composition, and iteration. First, you'll use integration by parts on the given integral—recognize that d(x2−1x2)d\left(\frac{x^2-1}{x^2}\right) is your differential, so integrate x2\frac{x}{2} against it. Once you find f(x)f(x) using the boundary condition f(2)=12f(2)=\frac{1}{2}, the real insight is spotting a pattern: composing ff with itself repeatedly (gng_n means applying ff exactly nn times) often reveals a fixed point, a cycle, or a closed form that lets you predict gn(x)g_n(x) without computing every iteration.

Browse the problem library