Question

If $x^4+3\cos (a{x}^2+bx+c)=2(x^2-2)$ has two solutions with $a,b,c\in (2,5),$ then which of the following can be TRUE ?

• A. $a+b+c=\pi$
• B. $a-b+c=\pi$
• C. $a+b+c=3\pi$
• D. $a-b+c=3\pi$

Answer 1

Answer: (B), (C)

$a+b+c=3\pi$

$x^4+3\cos (a{x}^2+bx+c)=2(x^2-2)$
$\Rightarrow x^4-2x^2+1=-3(1+\cos (a{x}^2+bx+c\left)\right)$
$\Rightarrow (x^2-1{)}^2=-3(1+\cos (a{x}^2+bx+c)]$
LHS $\ge 0$ but RHS $\le 0$
Hence, the equation is true only if LHS $=$ RHS $=0$
$\Rightarrow x\pm 1$ and $\cos (a{x}^2+bx+c)=-1$

When $x=1,$
$\cos (a+b+c)=-1$
$\Rightarrow a+b+c=(2n+1)\pi ,n\in Z$

Since $a,b,c\in (2,5),$
$a+b+c\in (6,15)$ and $a-b+c\in (-1,8)$
Hence, $a+b+c=3\pi$

When $x=-1,$
$\cos (a-b+c)=-1$
$\Rightarrow a-b+c=(2m+1)\pi ,m\in Z$
Hence, $a-b+c=\pi$