Consider f(x)=(a-1) x2-8 x+1, a ∈ R, a ≠ 1, x ∈ R and g(x)=x2-8 x+13, x ∈ R The least integral value of ' a ' for which minimum finite value of f(x) minimum value of g(x) is

Question

Consider f(x)=(a-1) x2-8 x+1, a ∈ R, a ≠ 1, x ∈ R and g(x)=x2-8 x+13, x ∈ R The least integral value of ' a ' for which minimum finite value of f(x)> minimum value of g(x) is (A) 0 (B) 1 (C) 5 (D) 6

Tardigrade Admin · Accepted Answer

f(x)=(a-1) x2-8 x+1, g(x)=x2-8 x+13 -(64-4(a-1)/4(a-1)) >-3 ⇒ (a-1-16/a-1) >-3 ⇒ (a-17+3 a-3/a-1) >0 ⇒ (a-5/a-1)>0 ⇒ a ∈(-∞, 1) ∪(5, ∞) For the existance of minimum finite value of f ( x ) a-1 >0 ⇒ a >1 ∴ least possible integral value of ' a ' is 6 .

Q. Consider f(x)=(a−1)x2−8x+1,a∈R,a=1,x∈R and g(x)=x2−8x+13,x∈R The least integral value of ' a ' for which minimum finite value of f(x)> minimum value of g(x) is

0

1

5

6

f(x)=(a−1)x2−8x+1,g(x)=x2−8x+13 −4(a−1)64−4(a−1)​>−3⇒a−1a−1−16​>−3⇒a−1a−17+3a−3​>0 ⇒a−1a−5​>0⇒a∈(−∞,1)∪(5,∞) For the existance of minimum finite value of f(x) a−1>0⇒a>1 ∴ least possible integral value of ' a ' is 6 .

Q. Consider $f (x) = (a - 1) x^{2} - 8 x + 1, a \in R, a \neq = 1, x \in R$ and $g (x) = x^{2} - 8 x + 13, x \in R$
The least integral value of ' $a$ ' for which minimum finite value of $f (x) >$ minimum value of $g (x)$ is