Question

Let $f\left(x\right)={\cos }^{-1}\left(\cos x\right)$ and $g\left(x\right)=x^2-2\pi x+{\pi }^2+a.$ Then the equation $f\left(x\right)=g\left(x\right)$ has

• A. exactly $2$ solutions if $a<\pi$
• B. No solution if $a=4\pi$
• C. exactly one solution if $a=\pi$
• D. more than two solutions for $a<0$

Answer 1

Answer: (A), (B), (C)

exactly one solution if $a=\pi$

Given : $f\left(x\right)={\cos }^{-1}\left(\cos x\right),$ and
$g\left(x\right)=x^2-2\pi x+{\pi }^2+a$
$=(x-\pi {)}^2+a=h(x)+a$


From above graph, $f\left(x\right)=g\left(x\right)$ will have
$\left(1\right)$ two solutions if $a<\pi$
$\left(2\right)$ one solution if $a=\pi$
$\left(3\right)$ zero solution if $a>\pi$