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I took upon following question-

Find the values of $m$ for which given equation has real roots

$\sin^2 x-(m-3)\sin x+m=0 $

So I started by first satisfying $ D>=0$ and found that $m$ should range from $(-\infty,1] \cup [9,\infty)$

However after this, I found roots by quadratic formula as

$\sin x = (m-3)/2 \pm \sqrt{(m-1)(m-9)/4}$

However this value should be within range of sin function and I'm unable to solve the inequality so formed for values of $m$.

How to solve the inequality( actually the $\pm$ term is confusing me)?

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  • $\begingroup$ What do you denote $D$? $\endgroup$
    – Bernard
    Commented May 22, 2021 at 9:23
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    $\begingroup$ @Bernard Presumably the discriminant $\endgroup$ Commented May 22, 2021 at 9:28
  • $\begingroup$ I think it means $b^2-4ac$ from the quadratic rules $\endgroup$
    – 00GB
    Commented May 22, 2021 at 9:29
  • $\begingroup$ $sin$ is supposed to be within the range $[-1, 1]$, so solve the inequality for that too. $\endgroup$ Commented May 22, 2021 at 9:30
  • $\begingroup$ @ShubhamJohri: I guessed so, but I pointed that when one uses a non-standard notation, it has to be explained. $\endgroup$
    – Bernard
    Commented May 22, 2021 at 9:32

1 Answer 1

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When $m=1$, both roots of the quadratic are $-1$ so a solution for $x$ exists. When $m\in(-\infty,1)$, the smaller root $r_1=(m-3)/2-\frac12\sqrt{(m-1)(m-9)}<-1$ as $(m-3)/2<-1$. So we should enforce that the larger root $r_2=(m-3)/2+\frac12\sqrt{(m-1)(m-9)}\in[-1,1]$, i.e.$$\begin{align*}&-1\le(m-3)/2+\frac12\sqrt{(1-m)(9-m)}\le1\\& \iff1-m\le\sqrt{(1-m)(9-m)}\le5-m\end{align*}$$Note that the first inequality is always true since $9-m\ge1-m\ge0$ and hence $(1-m)(9-m)\ge(1-m)^2$. The second inequality on squaring gives$$m^2-10m+9\le m^2+25-10m$$which is true. So all values of $m\le1$ will work.


When $m\ge9$, the larger root $r_2\ge3$. So we must enforce $-1\le r_1\le 1$, i.e. $$\begin{align*}&-1\le(m-3)/2-\frac12\sqrt{(m-1)(m-9)}\le1\\& \iff1-m\le-\sqrt{(m-1)(m-9)}\le5-m\\& \iff m-5\le\sqrt{(m-1)(m-9)}\le m-1 \end{align*}$$ Correspondingly note that the second inequality is always true this time around. Squaring the first inequality would yield $25\le9$, which is never true. So no $m\ge9$ works.

Our final answer is $m\le1.~\blacksquare$

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  • $\begingroup$ I understood the correct way to solve is split the domain and then check $\endgroup$ Commented May 22, 2021 at 15:15
  • $\begingroup$ @lalittolani That's good. If my answer helped you, consider accepting it by clicking the tick-mark button next to it. $\endgroup$ Commented May 22, 2021 at 15:26
  • $\begingroup$ +1 for the same also $\endgroup$ Commented May 22, 2021 at 16:11

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