3

Let $f(x)=x^2 + ax + b$ such that $f(2)\times f(3)= \frac{1}{2}$ and $1< f(2) + f(3)< 2$, then the equation $f(x)=1$ has $(a,b \in \mathbb{R})$

  1. Both roots real and distinct

  2. Both roots real and equal

  3. Non real roots

  4. Roots whose nature depends on value of a & b

Only one of the options is correct.

My attempt

I tried doing various manipulations to $f(x)$, I tried putting the values of $f(2)$ and $f(3)$ in the equations and inequality qiven above but that derived nothing. I tried thinking that either $f(2)$ or $f(3)$ must be less than 1 but came to a point where I disagreed myself because $f(2)$ and $f(3)$ can both have values less than 1. I doubt if this question is really tough or I can't approach the question correctly.

Community
  • 1
Prajwal Turkar
  • 57

2 Answers2

3

We have $f(2)=2a+b+4$ and $f(3)=3a+b+9$. So, $a=f(3)-f(2)-5$ and $b=3f(2)-2f(3)+6$.

The discriminant of the equation $f(x)=1$ is

\begin{align*} a^2-4(b-1)&=(f(3)-f(2)-5)^2-4(3f(2)-2f(3)+5)\\ &=(f(3)-f(2))^2-10(f(3)-f(2))+25-12f(2)+8f(3)-20\\ &=(f(3)+f(2))^2-4f(3)f(2)-2f(3)-2f(2)+5\\ &=(f(3)+f(2))^2-2f(3)-2f(2)+5-2\\ &=(f(3)+f(2)-1)^2+2\\ &>0 \end{align*}

The equation has two distinct real roots.

CY Aries
  • 23,393
3

Answer is 1.

If $f(x)=1$ does not have two distinct real roots, it means $\forall x,f(x)\ge 1$. Then, $f(2)f(3)\neq \frac{1}{2}$ and $1<f(2)+f(3)<2$.

Therefore, we just need to use 'either' of the conditions, not 'both', to say that the answer is 1.

Besides, I expect that these auxiliary condition can give us some more information about further properties of roots, but I will post that if asked.

Moonshine
  • 2,088