Let the parabola y=a x2+b x+c has vertex at M(4,2) and a ∈[1,3]. If the difference between the extreme values of abc is equal to N, then find the digit at unit place of N.

Question

Let the parabola y=a x2+b x+c has vertex at M(4,2) and a ∈[1,3]. If the difference between the extreme values of abc is equal to N, then find the digit at unit place of N. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

2=16 a +4 b + c text and (-b/2 a)=4 ⇒ b=-8 a text , hence, c=16 a+2 ∵ 1 ≤ a ≤ 3 ∴ 18 ≤ c ≤ 50 text when a =1, b =-8 text , when a =3, b =-24 ∴ N max =18 × 1 ×-8 N min =50 × 3 ×-24 ∴ N text diff =3456

Q. Let the parabola $y = a x^{2} + b x + c$ has vertex at $M (4, 2)$ and $a \in [1, 3]$ . If the difference between the extreme values of abc is equal to $N$ , then find the digit at unit place of $N$ .

$2 = 16 a + 4 b + c$
$and \frac{- b}{2 a} = 4$
$\Rightarrow b = - 8 a, hence, c = 16 a + 2$
$∵ 1 \leq a \leq 3$
$∴ 18 \leq c \leq 50$
$when a = 1, b = - 8, when a = 3, b = - 24$
$∴ N_{m a x} = 18 \times 1 \times - 8$
$N_{m i n} = 50 \times 3 \times - 24$
$∴ N_{diff} = 3456$

Q. Let the parabola y=ax2+bx+c has vertex at M(4,2) and a∈[1,3]. If the difference between the extreme values of abc is equal to N, then find the digit at unit place of N.

2=16a+4b+c and 2a−b​=4 ⇒b=−8a, hence, c=16a+2 ∵1≤a≤3 ∴18≤c≤50 when a=1,b=−8, when a=3,b=−24 ∴Nmax​=18×1×−8 Nmin​=50×3×−24 ∴Ndiff ​=3456

Q. Let the parabola $y = a x^{2} + b x + c$ has vertex at $M (4, 2)$ and $a \in [1, 3]$ . If the difference between the extreme values of abc is equal to $N$ , then find the digit at unit place of $N$ .

$2 = 16 a + 4 b + c$
$and \frac{- b}{2 a} = 4$
$\Rightarrow b = - 8 a, hence, c = 16 a + 2$
$∵ 1 \leq a \leq 3$
$∴ 18 \leq c \leq 50$
$when a = 1, b = - 8, when a = 3, b = - 24$
$∴ N_{m a x} = 18 \times 1 \times - 8$
$N_{m i n} = 50 \times 3 \times - 24$
$∴ N_{diff} = 3456$