If $\underset{x→∞}{\lim}(\sqrt{x^2-x+1}-ax-b)=0$, then for k ≥ 2, $\underset{x→∞}{\lim}\sec^{2n}(k!πb)=$

$a$

$\underset{x→∞}{\lim}(\sqrt{x^2-x+1}-ax-b)=0$  [Here a > 0 for if a ≤ 0, then limit = ∞]

$⇒\underset{x→∞}{\lim}\frac{x^2-x+1-(ax-b)^2}{\sqrt{x^2-x+1}+ax+b}=0$

$⇒\underset{x→∞}{\lim}\frac{(1-a^2)x^2-(1+2ab)x+1-b^2}{\sqrt{x^2-x+1}+ax+b}=0$

This is possible only when $1-a^2=0$ and 1 + 2ab = 0

$∴ a = 1 (∵ a > 0)$ and $b = −1/2$

Now $k!\, πb = k!\, π (−1/2)$ = an integral multiple of $π$ as k ≥ 2

$∴ sec^2k!\, πb = 1$

$∴\underset{x→∞}{\lim}(\sec^{2n}k!πb)=1=a$