Question

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

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

Answer 1

Answer: (B)

$-3$

${\sin }^{2}x=1-{\cos }^{2}x$
So, $1-{\cos }^{2}x+a\cos x+a^2>1+\cos x$
i.e. ${\cos }^{2}x+\cos x(1-a)-a^2<0$
$\Rightarrow 1+1-a-a^2<0$ or $a^2+a-2>0$ or $(a+2)(a-1)>0$
So, the largest possible negative integral value of a comes out to be $-3$.