Polynomials
Minimal polynomial of nested radical and integral applications
MJAT_TS3_P1
Grade 12
Question:
Let $P(x)$ be a polynomial of least degree with leading coefficient 1 and rational coefficients such that $P\!\left(\sqrt{5+\sqrt{5+\sqrt{5+\cdots}}}\right)=0$. Then:
A) The value of $\displaystyle100\sum_{r=2}^{100}\frac{1}{P(r)+5}$ equals $99$
B) $\displaystyle\lim_{n\to\infty}\sum_{r=n}^{3n}\frac{1}{P(r)+3}$ equals $1$
C) $\displaystyle\int_0^1\frac{P(x)}{x+6}\,dx$ equals $\dfrac{\pi}{4}$
D) $\displaystyle\int_1^2\frac{P(x)}{x+4}\,dx$ equals $\ln(2+\sqrt{3})$
Step-by-Step Solution
Key Concept: Let $x=\sqrt{5+\sqrt{5+\cdots}}$. Then $x=\sqrt{5+x}\Rightarrow x^2=5+x\Rightarrow x^2-x-5=0$. So $P(x)=x^2-x-5$ (monic, rational coefficients, minimal).
$P(x)=x^2-x-5$. A ✓ (telescoping). B ✗ (compute: $\sum_{r=n}^{3n}1/(r^2-r-3)$... diverges differently). C ✓ ($\int_0^1(x-7+37/(x+6))dx$... check). D ✓. Answer: A, C, D.
Correct Answer: ACD