If the minimum value of y=(x-2)(x-4)(x-6)(x-8)+16, x ∈ R is m then (m+3) equals

Question

If the minimum value of y=(x-2)(x-4)(x-6)(x-8)+16, x ∈ R is m then (m+3) equals (A) 0 (B) 3 (C) 6 (D) 15

Tardigrade Admin · Accepted Answer

y=(x-2)(x-6)(x-8)(x-4)+16 =(x2-10 x+16)(x2-10 x+24)+16 =(t+16)(t+24)+16, text where t ∈ x2-10 x = t 2+40 t +400=( x -5)2-25 =( t +20)2 t ≥-25 .y| min =0= m ∴ m +3=3

Q. If the minimum value of $y = (x - 2) (x - 4) (x - 6) (x - 8) + 16, x \in R$ is $m$ then $(m + 3)$ equals

0

3

6

15

$y = (x - 2) (x - 6) (x - 8) (x - 4) + 16$
$= (x^{2} - 10 x + 16) (x^{2} - 10 x + 24) + 16$
$= (t + 16) (t + 24) + 16, where t \in x^{2} - 10 x$
$= t^{2} + 40 t + 400 = (x - 5)^{2} - 25$
$= (t + 20)^{2} t \geq - 25$
$y ∣_{m i n} = 0 = m$
$∴ m + 3 = 3$

Q. If the minimum value of y=(x−2)(x−4)(x−6)(x−8)+16,x∈R is m then (m+3) equals

0

3

6

15

y=(x−2)(x−6)(x−8)(x−4)+16 =(x2−10x+16)(x2−10x+24)+16 =(t+16)(t+24)+16, where t∈x2−10x =t2+40t+400=(x−5)2−25 =(t+20)2t≥−25 y∣min​=0=m ∴m+3=3

Q. If the minimum value of $y = (x - 2) (x - 4) (x - 6) (x - 8) + 16, x \in R$ is $m$ then $(m + 3)$ equals

$y = (x - 2) (x - 6) (x - 8) (x - 4) + 16$
$= (x^{2} - 10 x + 16) (x^{2} - 10 x + 24) + 16$
$= (t + 16) (t + 24) + 16, where t \in x^{2} - 10 x$
$= t^{2} + 40 t + 400 = (x - 5)^{2} - 25$
$= (t + 20)^{2} t \geq - 25$
$y ∣_{m i n} = 0 = m$
$∴ m + 3 = 3$