Question

If $498[16\cos x+12\sin x]=2k+60$, then the maximum value of $k$ is

Answer 1

$498[16\cos x+12\sin x]=2k+60$
$\Rightarrow 1992(4\cos x+3\sin x)=2k+60$
Consider, $4\cos x+3\sin x$
Put $4=r\cos \alpha ,3=r\sin \alpha$
$\Rightarrow r=5,\tan \alpha =\frac{3}{4}$
So, $4\cos x+3\sin x=5\cos (x-\alpha )$
Since, $|\cos (x-\alpha )|\le 1$
$\Rightarrow 5\cos (x-\alpha )\le 5$
$\Rightarrow 4\cos x+3\sin x\le 5$
$\Rightarrow 1992(4\cos x+3\sin x)\le 9960$
$\Rightarrow 2k+60\le 9960$
$\Rightarrow k\le 4950$
Hence, maximum value of $k$ is $4950$.