Question

For positive $a$ and $b$ define $\theta ={\cos }^{-1}\left(\frac{a}{2}\right)+{\cos }^{-1}\left(\frac{b}{3}\right).\underset{\theta \to 0^{+}}{lim}\frac{9a^2-12ab\cos \theta +4b^2}{{\theta }^2}$ is equal to

• A. $9$
• B. $4$
• C. $18$
• D. $36$

Answer 1

The correct option is A $9$

$\theta =co{s}^{-1}\left(\alpha /2\right)+cosb/3{lim}_{\theta \to 0}\frac{9a^2-12abcos\theta +4b^2}{{\theta }^2}$

% form

$cos\theta =1$

$\theta =0$

$=9a^2-12abcos\theta +4b^2$

$=(3a-4b{)}^2$

$=9$