0
$\begingroup$

The inequality is :$$\sin^2x+a\cos x+a^2>1+\cos x$$

I have simplified it to:$$\cos^2x+(1-a)\cos x-a^2<0$$

My approach was that since this is in the form of a parabola the minimum value for x must be $\frac{-b}{2a}$ form so after applying conditions I get : $\frac{a-1}{2}$ which should be equal to -1 since that is the minimum value of cos.

But I get answer -1 which is wrong, where have I flawed?

$\endgroup$
  • $\begingroup$ Can you elaborate on "which should be equal to -1 since that is the minimum value of cos."? Why is that something that one should do? What is it that you're trying to show? $\endgroup$ – Calvin Lin Nov 28 '20 at 4:03
  • $\begingroup$ @CalvinLin I have taken cosx as some variable and found the minimum of the parabola which is equal to the variable, but the variable itself has a domain restriction whose minimum is -1 so I equated it. $\endgroup$ – JustJohan Nov 28 '20 at 4:06
  • 1
    $\begingroup$ Alright, let me rephrase what you said. Suppose we want to find the minimum of an upward facing parabola on a restricted range, then the minimum has to occur at the lower endpoint of the restricted range. Do you agree that's a rephrasing of your statement? Does this statement make sense to you? $\endgroup$ – Calvin Lin Nov 28 '20 at 4:10
2
$\begingroup$

(Fill in the gaps. If you're stuck, show your work and explain why you're stuck.)

Hint:

  1. Your quadratic is in $ \cos x$, which is restricted to $ -1 \leq \cos x \leq 1$.

  2. If you want to prove that a upward opening quadratic $f(x) $ on a restricted domain of $ [p, q]$ is $< 0$, then we just need to show that $f(p) < 0$ and $ f(q) < 0$.
    In particular, why do we not need to check the turning point?

  3. Hence, conclude that the largest negative integral values of $a$ is -3.

$\endgroup$
  • $\begingroup$ sir, I don't understand the second hint, Is it something fundamental? Why do we have to show that? $\endgroup$ – JustJohan Nov 28 '20 at 4:11
  • $\begingroup$ What do you not understand about it? If you don't already know it, you can prove it. So, what have you tried? Where are you stuck at? E.g. If you consider $f(x) = x^2$ on various restricted domains, does it apply? $\endgroup$ – Calvin Lin Nov 28 '20 at 4:12
  • $\begingroup$ sir this is my understanding after reading your answer, The minimum value of the parabola is of no use to us it may or may not lie in the domain restriction and provides no information, we form an inequality in "a" at the min and max values of the domain, where we can get the largest negative value of "a" as -3. Is there any flaw here sir? $\endgroup$ – JustJohan Nov 28 '20 at 4:57
  • $\begingroup$ @JustJohan A) Not quite. To show that the upward facing parabola is < 0 on the restricted domain, we just need to check the endpoints. The minimum turning point, even if it's in the domain, will be smaller than either of the endpoints. B) Yes, checking $f(1) < 0 $ will give us $ a < -2$, hence $ a = -3$. $\endgroup$ – Calvin Lin Nov 28 '20 at 5:29
  • $\begingroup$ Thank you very much, sir. I have understood . $\endgroup$ – JustJohan Nov 28 '20 at 5:46
0
$\begingroup$

Hint $$\cos x = 1 \implies 1+(1-a)-a^2<0 \implies 0 < a^2+a-2=(a-1)(a+2) \implies a<-2 \text{ or } a> 1.$$ Therefore $a$ can't be $-1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.