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If the inequality $${\sin}^2 x+a\cos x+{a}^2>1+\cos x$$ holds for all $x \in \Bbb R$ then what's the smallest positive integral value of $a$?

Here's my approach to the problem $$\cos^2 x+(1-a)\cos x-a^2<0$$ Let us consider this as a quadratic form respect to $a$.

Applying the quadratic formula $a=\frac{-\cos x\pm\sqrt{5\cos^2 x+4\cos x}}2 $ and substituting $\cos x$ with $1$ and $-1$ we get 3 values of where the graph should touch the x axis $-2,0,1$ How should I proceed now?

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    1. Make a substitution $t = cos(x), t \in[-1, 1]$ 2. Find union of intervals $[\frac{t-\sqrt{5t^2+4t}}{2},\frac{t+\sqrt{5t^2+4t}}{2}]$ where $0 \le 5t^2 + 4t$ – 2017-02-24
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    If you could elaborate further please? – 2017-02-24

4 Answers 4

2

We need that the inequality $\cos^2x+(1-a)\cos{x}-a^2<0$ will be true for all real $x$.

Let $\cos{x}=t$.

Thus, we need to find a smallest natural $a$ for which the inequality $$t^2+(1-a)t-a^2<0$$ is true for all $t\in[-1,1],$ for which we need $$(-1)^2+(1-a)(-1)-a^2<0$$ and $$1^2+(1-a)\cdot1-a^2<0,$$ which is $$a^2+a-2>0$$ and $$a^2-a>0,$$ which gives $$a\in(-\infty,-2)\cup(1,+\infty)$$ and we got the answer: $2$.

3

The smallest positive integer is $1$. That doesn't satisfy the condition, because $$ \sin^2 x + \cos x + 1 > 1 + \cos x $$ has equality when $\sin^2x = 0$, which happens, among other places, at $x=0$.

The next positive integer is $2$. Does that work? $$ \sin^2 x + 2\cos x + 4 > 1 + \cos x $$ holds if and only if $$ \sin^2 x + \cos x + 3 > 0 $$ which is easily true -- since $\sin^2 x$ is never less than $0$ and $\cos x$ is never less than $-1$, the left-hand side is always $\ge 2$.

So the answer is $$ \Huge 2 $$

2

Write sin$^2$x=1-cos$^2$x, and factorize the resultant inequation.

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    Factorize this expression. I think you should try to figure that by yourself. If you stay stuck for too long, I'll post the solution. Until then, keep at it. – 2017-02-24
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$$ \cos^2{x}+(1-a)\cos{x}-a^2\,\lt0 \quad\&\quad \left|\,\cos{x}\,\right|\,\le1 \\[6mm] -1\le\,\cos{x}=\frac12\left(\,-(1-a)\pm\sqrt{(1-a)^2+4a^2}\,\right)\,\le+1 \\ -1-a\,\le\,\pm\sqrt{5a^2-2a+1}\,\le\,+3-a \\[6mm] \begin{align} &\text{For}\,\colon\,\,\,\pm\sqrt{5a^2-2a+1}\,\ge\,-1-a \implies 5a^2-2a+1=a^2+2a+1 \\ &\implies 4a^2-4a=a^2-a=0\implies a=0,\,a=1\quad\color{red}{a\in[0,1]} \\[6mm] &\text{For}\,\colon\,\,\,\pm\sqrt{5a^2-2a+1}\,\le\,+3-a \implies 5a^2-2a+1=a^2-6a+9 \\ &\implies 4a^2+4a-8=a^2+a-2=0\implies a=-2,\,a=1\quad\color{blue}{a\in[-2,1]} \\[6mm] \end{align} $$ Hence, for the inequality $\,\{\cos^2{x}+(1-a)\cos{x}-a^2\,\lt0\}\,$ to hold for all $\,x\in\mathbb{R}\,$ and the operator without equal (less than only), Then: $$ \boxed{ \quad \color{blue}{a \in (-\infty,\,-2)\,\wedge\,(1,\infty) } \quad } $$ And for $\,a\in\mathbb{N}^{+}\implies\color{red}{a=2}\,$


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