If the parabola $y=ax^2+bx+c$ has vertex at (4, 2) and $a \in [1, 3]$, then the difference between the extreme the extreme values of abc is equal to

3456

The correct answer is option (1) : 3456

The vertex of the parabola $y =ax^2 +bx +c $ is at

$\left(-\frac{b}{2a}, -\frac{b^2-4ac}{4a}\right)$

$∴-\frac{b}{2a}= 4 $ and $ -b^2-\frac{4ac}{4a}+2$

$⇒b=-8a$ and $c-\frac{b^2}{4a}=2$

$⇒b=-8a $ and $c=16a+2$

Now,

$f(a) = abc = -8a^2 (16a+2) = -16 (8a^3 +a^2)$

$\frac{df(a)}{da}= -16(24a^2+2a)$

Clearly,

$\frac{df(a)}{da}< 0 \, ∀ \, a \in [1, 3]$

$⇒f(a)$ is decreasing on [1,3]

∴ Min. value of $f(a)=f(3) = -16(8×3^3+3^2)=-3600$

Max. value of $f(a) =f(1) -16 (8+1) = -144$

Hence, required difference $=-144 + 3600= 3456$