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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)?

Bernard
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Lalit Tolani
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  • What do you denote $D$? – Bernard May 22 '21 at 09:23
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    @Bernard Presumably the discriminant – Shubham Johri May 22 '21 at 09:28
  • I think it means $b^2-4ac$ from the quadratic rules – 00GB May 22 '21 at 09:29
  • $sin$ is supposed to be within the range $[-1, 1]$, so solve the inequality for that too. – Jan Safronov May 22 '21 at 09:30
  • @ShubhamJohri: I guessed so, but I pointed that when one uses a non-standard notation, it has to be explained. – Bernard May 22 '21 at 09:32
  • I didn't think much about but I think you need to do for two, I mean take the first inequality with positive between -1 and 1 and solve it and take negative and do the same . and both cases your m also should give real root – 00GB May 22 '21 at 09:47
  • Does this help you? https://www.doubtnut.com/question/plastration-2107-find-all-the-values-of-m-for-which-theequation-sin-m-3-sinx-m-0-has-real-roots-18632792 –  May 22 '21 at 09:49
  • @JitendraSingh At around 3:08 in the video, the guy is squaring the whole expression but $a\le b\ge0$ does not imply $a^2\le b^2$ unless $a\ge0$, which in this case happens when $1-m\ge0$, which should have been an assumption before squaring, not a deduction after it – Shubham Johri May 22 '21 at 10:09

1 Answers1

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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 $m\le1.~\blacksquare$

Shubham Johri
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