CAT 2020 Set-2 | Question: 72

Question

Let $f(x) = x^{2} + ax + b $ and $g(x) = f(x+1) – f(x-1).$ If $ f(x) \geq 0 $ for all real $x,$ and $ g(20) = 72,$ then the smallest possible value of $b$ is 

• A. $1$
• B. $16$
• C. $0$
• D. $4$

Answer 1

Given that, $ f(x) = x^{2} + ax + b \quad \longrightarrow (1) $

$ g(x) = f(x+1) – f(x-1) \quad \longrightarrow (2) $

From equation $(1),$ we can write,

• $ f(x+1) = (x+1)^{2} + a(x+1) +b $
• $ f(x-1) = (x-1)^{2} + a(x-1) + b $

Now, from equation $(2),$

$ g(x) = (x+1)^{2} + a(x+1) + b – [ (x-1)^{2} + a(x-1) + b] $

$ \Rightarrow g(x) = x^{2} + 2x + 1 + ax + a + b – [ x^{2} – 2x + 1 + ax – a + b] $

$ \Rightarrow g(x) = x^{2} + 2x + 1 + ax + a + b – x^{2} + 2x - 1 - ax + a - b $

$ \Rightarrow g(x) = 4x + 2a \quad \longrightarrow (3) $

we have, $ g(20) = 72 $

So, $ g(20) = 4 (20) + 2a $

$ \Rightarrow 80 + 2a = 72 $

$ \Rightarrow 2a = -8 $

$ \Rightarrow \boxed{a = – 4} $

We have $ f(x) \geq 0 $

$ x^{2} – 4x + b \geq 0 $

For this one graph will be 

In this case graph can touch $x – \text{axis},$  so it can have at most one root.

Thus, discriminant $ \boxed{ \text{D} \leq 0} $

$ \Rightarrow \boxed{ b^{2} – 4ac \leq 0} $

$ \Rightarrow (-4)^{2} – 4 (1)(b) \leq 0 $

$ \Rightarrow 16 – 4b \leq 0 $

$ \Rightarrow 16 \leq 4b $

$ \Rightarrow 4b \geq 16 $

$ \Rightarrow \boxed{b \geq 4} $

So, least value of $b=4.$

$\therefore$ The smallest possible value of $b$ is $4.$

Correct Answer$: \text{D}$