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

• A. $-4$
• B. $-3$
• C. $-2$
• D. $-1$

Answer 1

The correct option is B $-3$

${\sin }^{2}x+a\cos x+a^2>1+\cos x$
$\Rightarrow {\cos }^{2}x+(1-a)\cos x-a^2<0$
Now let $\cos x=t$
$\therefore t\in [-1,1],\forall x\in R$
So, $t^2+(1-a)t-a^2<0\cdots \left(i\right)$
For the given inequality to be true, $[-1,1]$ should lie in between the roots of equation $\left(i\right)$

Hence required condition are
$\left(1\right)f(-1)<0\Rightarrow a-a^2<0\Rightarrow a(a-1)>0\Rightarrow a<0\text{or}a>1\left(2\right)f\left(1\right)<0\Rightarrow 2-a-a^2<0\Rightarrow a^2+a-2>0\Rightarrow (a+2)(a-1)>0\Rightarrow a<-2\text{or}a>1$

So from the above condtitions, we get
$a\in (-\infty ,-2)\cup (1,\infty )$

Hence, the largest negative integral value of $a$ is $-3$