1

I did see the explanation [here] (Whats going on in this quadratic?). Unfortunately, I am unable to grasp the explanation for why 1 was chosen but also have some additional questions.

  1. The substitution $t = sin(x)$ completely changes the graph of the original equation when plotted in desmos. The graphs for $f(x) = sin^2(x) -ksin(x) -3$ and $f(t) = t^2-kt-3$ don't align for $0 < x < \pi $.

  2. The answer is given as $k < -2$. But this condition is true for $t^2-kt-3$ and not $sin^2(x) -ksin(x) -3$. Since we are asked to find the possible values of $k$ for the original equation: $sin^2(x) -ksin(x) -3$. When a value of $k = -2.1$ is substituted, f(t) > 0 but f(x) < 0 (This can be seen in the graph as well). So how can $k < -2$ be the correct answer ?

limbo1927
  • 23
  • 4

1 Answers1

0

Let

$$f_k(x) = \sin^2 x - k \sin x -3$$ the map defined on $[0, \pi]$. As $f_k(\pi - x) = f_k(x)$, $f_k$ has exactly two roots on $[0,\pi]$ if and only if $f_k$ has exactly one root on $I = [0, \pi/2)$. On $I$, $\sin x$ is stricly increasing from $0$ to $1$. As $x \mapsto \sin x$ is continuous, $f_k$ will have exactly one root on $[0, \pi/2)$ is and only if $g_k(t) = t^2 - kt -3$ has exactly one root on $[0,1)$.

As $g_k(0) = -3 \lt 0$, we must have $g_k(1) \gt 0$, i.e. $k \lt -2$.

Conversely, suppose that $k \lt -2$. We have $g^\prime_k(x) = 2t-k \gt 0$. Which implies that $g_k$ is strictly increasing from $-3$ to $-k-2 \gt 0$ and has only one root on $[0,1)$.

Conclusion: $k \lt -2$ is the right answer.

mathcounterexamples.net
  • 70,018
  • So, I should be looking at $x = \pi /2$ when $t = 1$ because of the mapping? Also, I had messed up the sign in the question, I have edited it since, so it should be $k<-2$. Thank you for the explanation. – limbo1927 Feb 21 '22 at 09:25
  • I updated the question and the answer to your question is yes. – mathcounterexamples.net Feb 21 '22 at 09:45