Question

If the inequality ${\sin }^{2}x+a\cos x+a^2>1+\cos x$ holds for any $x\in R$, then the largest negative integral value of $a$ is

Answer 1

${\sin }^{2}x+a\cos x+a^2>1+\cos x\forall x\in R$
So, putting $x=0$, we get
$a+a^2>2$
$\Rightarrow (a+2)(a-1)>0$
$\Rightarrow a\in (-\infty ,-2)\cup (1,\infty )$
Therefore, the largest negative integral value of $a$ is $-3$

Alternate solution:
${\sin }^{2}x+a\cos x+a^2>1+\cos x$
$\Rightarrow a^2+a\cos x-{\cos }^{2}x-\cos x>0$
When inequality hold
$\Rightarrow a^2+a\cos x-{\cos }^{2}x-\cos x=0$
$\Rightarrow a=\frac{-\cos x\pm \sqrt{5{\cos }^{2}x+4\cos x}}{2}$
Now, $\sqrt{5{\cos }^{2}x+4\cos x}$ is maximum when $\cos x=1$
$\Rightarrow a=1,-2$ at $\cos x=1$

$\therefore (a+2)(a-1)>0$
$\Rightarrow a\in (-\infty ,-2)\cup (1,\infty )$
Therefore, the largest negative integral value of $a$ is $-3$