Solving $y = \frac{{a{x^2} - 3x + 5}}{{5{x^2} - 3x + a}}$ If $y = \frac{{a{x^2} - 3x + 5}}{{5{x^2} - 3x + a}}$, then number of possible integral values of 'a' for which y may be capable of all values $\forall x \in R$ is
(1) 4
(2) 5
(3) 6
(4) none of these
My approach is as follow $5{x^2} - 3x + a > 0 \Rightarrow 9 - 20a < 0 \Rightarrow a > \frac{9}{{20}}$
For this case we get the following range that $x\in \mathbb R$, but not able to approach
 A: Cross-multiply and simplify to get: $$(5y-a)x^2+(3-3y)x+(ay-5)=0\tag{1}$$
The statement “y may be capable of all values…” suggests that $(1)$ has at least one real solution of $x$ for all $y$. Thus, the discriminant of $(1)$ must always be non-negative $\forall y, a\in\mathbb R$: $$(3-3y)^2-4(ay-5)(5y-a)\geq 0$$ which simplifies to $$(9y^2+9-18y)-4(5ay^2-(25+a^2)y+5a)\geq 0$$ or $$(9-20a)y^2+(82+4a^2)y+(9-20a)\geq0$$ $\forall y\in \mathbb R$. But a quadratic polynomial in y always being greater than $0$ when the domain of $y$ is all reals, means that the discriminant of THAT polynomial must be negative. Thus, $$(82+4a^2)^2-4(9-20a)^2\lt 0$$ or $$(82+4a^2)^2-(18-40a)^2\lt 0$$$$\implies (4a^2+40a+64)(4a^2-40a+100)\lt 0$$$$\implies (a^2+10a+16)(a^2-10a+25)\lt 0$$$$\implies (a^2+10a+16)(a-5)^2\lt 0$$ Now, the second factor is always non-negative by virtue of being a perfect square. This means that $a$ must satisfy $(a^2+10a+16)\lt0\implies a\in (-8,-2)$ which gives us $a=-3,-4,-5,-6,-7$. Now to just cap it off, we check the  boundary values whether $a=5,-2,-8$ satisfy the conditions and then we are done.
