Question

Let $\cos \theta ,\cos \frac{\theta }{2}$ and $\cos \frac{\theta }{4}$ are the roots of the equation $x^3+a{x}^2+bx+c=0$. If $\underset{\theta \to 0}{lim}\left[\frac{1+a+b+c}{{\theta }^6}\right]=\frac{1}{k},$ then the value of $k$ is

Answer 1

$x^3+a{x}^2+bx+c=0$
$\text{Let}f\left(x\right)=x^3+a{x}^2+bx+c\cdots \left(1\right)$
Given that $\cos \theta ,\cos \frac{\theta }{2}$ and $\cos \frac{\theta }{4}$ are the zeros of $f\left(x\right)$
$f\left(x\right)=(x-\cos \theta )(x-\cos \frac{\theta }{2})(x-\cos \frac{\theta }{4})$
$f\left(1\right)=(1-\cos \theta )(1-\cos \frac{\theta }{2})(1-\cos \frac{\theta }{4})\text{and}f\left(1\right)=1+a+b+c\left[\text{from eqn}\right(1\left)\right]$

$\Rightarrow 1+a+b+c=(1-\cos \theta )(1-\cos \frac{\theta }{2})(1-\cos \frac{\theta }{4})\Rightarrow 1+a+b+c=\left(2{\sin }^2\frac{\theta }{2}\right)\left(2{\sin }^2\frac{\theta }{4}\right)\left(2{\sin }^2\frac{\theta }{8}\right)$

$\text{Now,}\underset{\theta \to 0}{lim}\left[\frac{1+a+b+c}{{\theta }^6}\right]=\underset{\theta \to 0}{lim}[2^3{\left(\frac{\sin \theta /2}{\theta /2}\right)}^2{\left(\frac{\sin \theta /4}{\theta /4}\right)}^2{\left(\frac{\sin \theta /8}{\theta /8}\right)}^2\times \frac{1}{2^{12}}]$

$=2^{-9}\times 1^2\times 1^2\times 1^2=\frac{1}{512}$
$\therefore k=512$