Limits, Continuity & Differentiability
Limit at Root of Quadratic — Trigonometric Form
nta_pyq_2024_apr
Grade 12
Question:
Let $a>0$ be a root of the equation $2x^2+x-2=0$. If $\displaystyle\lim_{x\to\frac{1}{a}}\frac{16(1-\cos(2+x-2x^2))}{(1-ax)^2}=\alpha+\beta\sqrt{17}$, where $\alpha,\beta\in\mathbb{Z}$, then $\alpha+\beta$ is equal to
Step-by-Step Solution
Key Concept: Roots of $2x^2+x-2=0$: $x=\frac{-1\pm\sqrt{17}}{4}$. Since $a>0$, $a=\frac{-1+\sqrt{17}}{4}$, so $1/a=\frac{4}{-1+\sqrt{17}}$. Use $1-\cos\theta\approx\theta^2/2$ near the root where $2+x-2x^2=0$.
$\alpha=153$, $\beta=17$. $\alpha+\beta=170$.
Correct Answer: 170