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?
Best Answer
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 $a\in[3,4)$