Let f(x)=x2+a x+b. If ∀ x ∈ R, there exist a real value of y such that f(y)=f(x)+y, then find the maximum value of 100 a.

Question

Let f(x)=x2+a x+b. If ∀ x ∈ R, there exist a real value of y such that f(y)=f(x)+y, then find the maximum value of 100 a. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given, f ( y )= f ( x )+ y ⇒ y2+a y+b=x2+a x+b+y ⇒ y2+y(a-1)-x2-a x=0 As, y ∈ R, so D ≥ 0 ∀ x ∈ R ⇒ (a-1)2+4(x2+a x) ≥ 0 ∀ x ∈ R ⇒ 4 x2+4 a x+a2-2 a+1 ≥ 0 ∀ x ∈ R Now, D ≤ 0 ⇒ 16 a2-16(a2-2 a+1) ≤ 0 ⇒ 2 a-1 ≤ 0 ⇒ a ≤ (1/2) So, a max .=(1/2) Hence, maximum value of 100 a =50

Q. Let f(x)=x2+ax+b. If ∀x∈R, there exist a real value of y such that f(y)=f(x)+y, then find the maximum value of 100a.

Given, f(y)=f(x)+y ⇒y2+ay+b=x2+ax+b+y⇒y2+y(a−1)−x2−ax=0 As, y∈R, so D≥0∀x∈R ⇒(a−1)2+4(x2+ax)≥0∀x∈R ⇒4x2+4ax+a2−2a+1≥0∀x∈R Now, D≤0 ⇒16a2−16(a2−2a+1)≤0⇒2a−1≤0⇒a≤21​ So, amax.​=21​ Hence, maximum value of 100a=50

Q. Let $f (x) = x^{2} + a x + b$ . If $\forall x \in R$ , there exist a real value of $y$ such that $f (y) = f (x) + y$ , then find the maximum value of $100 a$ .