Question

The least value of a is

• A. 4
• B. 6
• C. 7
• D. 5

Answer 1

Answer: (D)

5

Answer 2

Given the quadratic equation $ax^2 - bx + c = 0$ with $a, b, c \in N$ and two distinct real roots in the interval (1, 2), we need to find the least value of $a$. The conditions are:

1. $f(1) = a - b + c > 0$
2. $f(2) = 4a - 2b + c > 0$
3. $1 < \frac{b}{2a} < 2 \implies 2a < b < 4a$
4. $c < \frac{b^2}{4a}$
5. $2 < \frac{b}{a} < 4$ and $1 < \frac{c}{a} < 4 \implies c > a$

By trying different values of $a$, we find that the smallest possible value for $a$ is 5. In this case, $b = 15$ and $c = 11$. $f(x) = 5x^2 - 15x + 11$ $f(1) = 5 - 15 + 11 = 1 > 0$ $f(2) = 20 - 30 + 11 = 1 > 0$ The vertex is at $x = \frac{15}{10} = 1.5$, which is in the interval (1, 2). $f(1.5) = 5(1.5)^2 - 15(1.5) + 11 = 5(2.25) - 22.5 + 11 = 11.25 - 22.5 + 11 = -0.25 < 0$ The discriminant is $15^2 - 4(5)(11) = 225 - 220 = 5 > 0$, so the roots are distinct. Therefore, the least value of $a$ is 5.