Definite Integration
Arithmetic mean of g at dyadic points
MJAT_TS6_P2
Grade 12
Question:
**Paragraph II (continued):** $g(x)$ continuous with $\int_0^1 g(x)(4x^2-g(x))dx=\frac{4}{5}$. If AM of $g(1/2),g(1/4),\ldots,g(1/2^{10})$ is $\frac{1}{m}\!\left(1-\frac{1}{2^n}\right)$, $m,n\in\mathbb{N}$, then $m+n=$
Step-by-Step Solution
Key Concept: From $\int_0^1[g(x)-2x^2]^2dx=\int_0^1 g(4x^2-g)dx-\int_0^14x^4dx+\int_0^14x^4dx=4/5-4/5+0=0\Rightarrow g(x)=2x^2$.
$m+n=\mathbf{35}$.
Correct Answer: 35