Question

If the solutions for $\theta$ of $cosp\theta +cosq\theta =0$,
$p>q>0$ are in AP then numerically smallest common difference of A.P is

• A. $\frac{\pi }{\sqrt{p+q}}$
• B. $\frac{2\pi }{\sqrt{p+q}}$
• C. $\frac{2\pi }{(p+q)}$
• D. $\frac{1}{\sqrt{p+q}}$

Answer 1

Answer: (C)

$\frac{2\pi }{(p+q)}$

$cos\left(p\theta \right)=-cos\left(q\theta \right)$
Or
$cos\left(p\theta \right)=cos(\pi \pm q\theta )$
Or
$p\theta =\pi \pm q\theta$
Or
$p\theta =(2n-1)\pi +q\theta$ and $p\theta =(2n-1)\pi -q\theta$.
Hence
$\theta =\frac{(2n-1)\pi }{p-q}$ ...(i) and $\theta =\frac{(2n-1)\pi }{p+q}$ where $nϵN$
Hence
${\theta }_2^{\prime }\left(s\right)=\frac{\pi }{p-q},\frac{3\pi }{p-q},\frac{5\pi }{p-q}...$
And
${\theta }_1^{\prime }\left(s\right)=\frac{\pi }{p+q},\frac{3\pi }{p+q},\frac{5\pi }{p+q}...$
Now
$p>q>0$
Hence
$p-q<p+q$.
Hence ${\theta }_1^{\prime }s$ have a lesser magnitude.
Thus minimum difference of the A.P is
$=\frac{\left(2\right(n+1)-1)\pi }{p+q}-\frac{(2n-1)\pi }{p+q}$
$=\frac{\pi }{p+q}[2n+1-2n+1]$
$=\frac{2\pi }{p+q}$.