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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

1. $1$
2. $16$
3. $0$
4. $4$

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

DIAGRAM

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$: \text{D}$

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