Question

Consider $f\left(x\right)=1-2p\cos x+p^2=g\left(p\right);$ $h\left(x\right)=g(-p).g\left(p\right);$ $k\left(p\right)=\underset{0}{\overset{\pi }{\int }}\ln g\left(p\right)dx$ for $p,x\in R.$ Then which of the following is/are correct ?

• A. $k\left(x\right)=\frac{1}{4}k\left(x^2\right)$
• B. $k\left(p\right)$ is an even function
• C. $g\left(p^2\right)=h\left(\frac{x}{2}\right)$
• D. $k\left(x\right)=\frac{1}{2}k\left(x^2\right)$

Answer 1

Answer: (B), (C), (D)

$k\left(x\right)=\frac{1}{2}k\left(x^2\right)$

$g\left(p\right)=1-2p\cos x+p^2$
$g(-p)=1+2p\cos x+p^2$
$h\left(x\right)=g\left(p\right)g(-p)$
$=(1+p^2{)}^2-4p^2{\cos }^{2}x$
$=1-2p^2(2{\cos }^{2}x-1)+p^4$
$h\left(x\right)=1-2p^2\cos 2x+p^4...\left(1\right)$

Also, $g\left(p^2\right)=1-2p^2\cos x+p^4...\left(2\right)$
From eqn $\left(1\right),$
$h\left(\frac{x}{2}\right)=1-2p^2\cos x+p^4$
$\therefore g\left(p^2\right)=h\left(\frac{x}{2}\right)$

Now, $k\left(p\right)=\underset{0}{\overset{\pi }{\int }}\ln g\left(p\right)dx$
$\Rightarrow k\left(p\right)=\underset{0}{\overset{\pi }{\int }}\ln (1-2p\cos x+p^2)dx...\left(3\right)$
Applying $\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx$
$\Rightarrow k\left(p\right)=\underset{0}{\overset{\pi }{\int }}\ln (1+2p\cos x+p^2)dx...\left(4\right)$
Adding eqn $\left(3\right)$ and $\left(4\right),$ we get
$2k\left(p\right)=\underset{0}{\overset{\pi }{\int }}\ln ((1+p^2{)}^2-4p^2{\cos }^{2}x)dx$
$\Rightarrow 2k\left(p\right)=\underset{0}{\overset{\pi }{\int }}\ln (1-2p^2\cos 2x+p^4)dx$
Put $2x=t\Rightarrow 2dx=dt$
$\Rightarrow k\left(p\right)=\frac{1}{4}\underset{0}{\overset{2\pi }{\int }}\ln (1-2p^2\cos t+p^4)dt$
$\Rightarrow k(-p)=\frac{1}{4}\underset{0}{\overset{2\pi }{\int }}\ln (1-2p^2\cos t+p^4)dt$
$\therefore k\left(p\right)=k(-p)$

Hence, $k\left(p\right)$ is even.

$k\left(p\right)+k(-p)=\frac{1}{2}\underset{0}{\overset{2\pi }{\int }}\ln (1-2p^2\cos t+p^4)dt$
$\Rightarrow 2k\left(p\right)=\underset{0}{\overset{\pi }{\int }}\ln (1-2p^2\cos t+p^4)dt$
$\Rightarrow 2k\left(p\right)=\underset{0}{\overset{\pi }{\int }}\ln g\left(p^2\right)dt=k\left(p^2\right)$
$\therefore k\left(p\right)=\frac{1}{2}k\left(p^2\right)$
$\Rightarrow k\left(x\right)=\frac{1}{2}k\left(x^2\right)$