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Find set of all real values of $a$ for which the equation $(a-4)\sec^4x+(a-3)\sec^2x+1=0, (a\ne4)$ has real solutions.

Let $\sec^2x=t$. So, the equation becomes $(a-4)t^2+(a-3)t+1=0$.

Since $\sec^2x\ge1\implies t\ge1\implies$ both the roots of the quadratic equation are $\ge1$.

Now, if we have $f(x)=ax^2+bx+c=0$ and its roots are $\ge k$ then (i) $af(k)\ge0$ and (ii) discriminant $\ge0$ and (iii) $\frac{-b}{2a}\gt k$.

(i) gives me $(a-4)(a-4+a-3+1)\ge 0\implies(a-3)(a-4)\ge0\implies a\in(-\infty,3]\cup[4,\infty)$.

(ii) gives me $(a-3)^2-4(a-4)\ge0\implies a^2+9-6a-4a+16\ge0\implies a^2-10a+25\ge0$. This is always true.

(iii) gives me $\dfrac{-a+3}{2(a-4)}\gt1\implies\dfrac{-a+3}{2(a-4)}-1\gt0\implies\dfrac{-a+3-2a+8}{2(a-4)}\gt0\implies\dfrac{3a-11}{2(a-4)}\lt0\implies a\in(\frac{11}3,4)$

On taking intersection, I do not get any common region but the answer given is $[3,4)$.

What is my mistake?

Sebastiano
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aarbee
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  • It’s not clear to me why (I) is true. And (iii) need not be true (equality can occur). – MPW May 10 '21 at 11:45
  • @MPW for $ax^2+bx+c$, if $a\gt0$, it's an upward parabola and if both roots are greater than $k$ then $f(k)$ would be +ve. Similarly, if $a\lt0$, it's a downward parabola and if both roots are greater than $k$ then $f(k)$ would be -ve. Combining these two scenarios, we get $af(k)\gt0$. Also, if one root can be $k$ as well then $f(k)=0$. So, final condition: $af(k)\ge0$. – aarbee May 10 '21 at 15:22
  • @MPW I agree that for (iii), equality can occur too. – aarbee May 10 '21 at 15:25

1 Answers1

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I think you overdid it. It is quite neat

$$\sec^2x=\frac{-(a-3)\pm\sqrt{(a-3)^2-4(a-4)}}{2(a-4)}=-1,\frac{1}{4-a}$$

So, obviously $\sec^2x\geq 1$

So $$\frac{1}{4-a}\geq 1$$

which gives $\frac{a-3}{a-4}\leq 0$

Thus $a\in[3,4)$

Maverick
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  • Thanks. +1. Any chance you could point out where my mistake is? – aarbee May 10 '21 at 15:26
  • You assumed that both roots are greater than or equal to $1$ but the given equation can have solution even if one root is greater than or equal to $1$. That case you did not take and the answer would be obtained if you solve for the fact that $1$ lies between the roots of the equation and also if $1$ is root of the equation. – Maverick May 10 '21 at 15:50
  • Sorry I don't understand how one root can be less than $1$. – aarbee May 10 '21 at 16:04
  • See you took $sec^2x$ to be equal to $t$ but does $t$ know that it is supposed to be $sec^2x$. $t$ is going to behave like a normal real number. It is for you to invoke conditions on it. – Maverick May 10 '21 at 16:07
  • Thanks for your comment but I am still struggling to grasp it. In the answer above, you took $\sec^x\ge1$ i.e you too took both the roots to be greater than or equal to $1$. How is that different from my attempt? – aarbee May 10 '21 at 16:49
  • No. I never did that. Infact i rejected $-1$ because it was smaller than 1. – Maverick May 10 '21 at 16:50
  • Besides i never knew that i would get $-1$ as value of $sec^2x$ – Maverick May 10 '21 at 16:51
  • You appear to be too much into procedure and not doing logical thinking – Maverick May 10 '21 at 16:53
  • You rejected $-1$ because root can't be $-1$. If you had got one root as $0.9$, you would still have rejected it because $\sec^x$ can't be less than $1$. That is to say, you too are implying that both roots should be greater than or equal to $1$, aren't you? – aarbee May 10 '21 at 17:03
  • Because $t=sec^2x$. It was only after solving the equation completely that i rejected $-1$. Your issue was that from beginning you assumed that both roots are greater than or equal to 1 and that is why you did not get the answer. Had you considered the situation that the quadratic in t would yield two roots of which one could be smaller than 1 the answer would have been obtained by you – Maverick May 10 '21 at 17:09
  • I see your point now, thanks. – aarbee May 10 '21 at 17:36