JEE Quadratic Equations 1

Conditions on problem translate with the following inequalities

$\textrm{GM}(a,c) < \frac{b}{2} < \textrm{AM}(c,a)$

with $c < a$

Set $S = \left. \left\{ \dfrac{n}{2} \, \right\vert \ \ n \in \mathbb N \right\} \subset \mathbb Q$ .

Anytime there is an element of $S$ between GM and AM we have a parabola fitting the criteria.

Since $f(x) = \sqrt{x}$ is obviously monotone increasing, and by the simple fact that for positive reals $x,y,z,t$ we have $x \leq y, z \leq t \Rightarrow xz \leq yt$ , our problem to minimize $abc$ is then split in two easier problem.

FIRST: Find $a$ and $c$ so that their product $ac$ is the minimum value such that

$S_{a,c} := S \cap \left( \textrm{GM}(a,c) , \textrm{AM}(c,a) \right) \not= \emptyset$

SECOND: Once such minimum $ac$ is found simply set $b= 2\cdot \min S_{a,c}$ .

To solve the first problem we can start thinking that way. We have to minimize $ac$ so we keep $ac$ as low as possible in a way that $a+c$ is as large as possible. This happens when $c$ and $a$ have max distance.

So examine initially the case when $c=1$ , and the least $a$ we find so that $S_{a,1} \not = \emptyset$ is $a= 5$ . In this case we have

$\textrm{GM}(1,5) = \sqrt 5 \approx 2.236 \quad \textrm{AM}(1,5) = 3 \quad S_{5,1} = \left\{ \dfrac{5}{2}\right\}$

and $ac = 5$ .

In the cases where $c > 1$ , since it must always be $c < a$ , the lowest choice possible is $c = 2, \quad a = 3$ that gives a product $ac = 6$ that is greater than $5$ we obtained previously. This proves that we found the minimum value for $ac$ .

By the second problem, since the minimum of $S_{5,1}$ is $2.5$ , we take $b = 5$ and the product $abc = 5\cdot 5 \cdot 1 = 25$ .

And the answer is $4$ .

abc must belong to N,now if abc=1,by checking D<0,now if is 5 then b must be 5 to make D>0,but after checking we get roots greater than 1,so for 25 a=5,b=5,c=1 the conditions satisfy(D is the discriminant)