Let a- b- c be rational numbers and f- be a function given by f x - ax2 - bx - c- Then a - b is

We have, nbsp;f (x) = ax^2 + bx + c, ∀ x∈ℤ f(x) is an integer, ∀ nbsp;x∈ℤ ⇒ f(0) and f(1) are integers ⇒[f(1) f(0)] is an integer Here, f(1)=a(1)^2+b()1 ...

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Let a, b, c b...

Question

Let $a$ , $b$ , $c$ be rational numbers and $f : Z \to Z$ be a function given by $f (x) = a x^{2} + b x + c$ . Then $a + b$ is

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Solution

We have, $f (x) = a x^{2} + b x + c, \forall$ $x \in Z$

$f (x)$ is an integer, $\forall$ $x \in Z$

$\Rightarrow f (0)$ and $f (1)$ are integers

$\Rightarrow [f (1) - f (0)]$ is an integer

Here, $f (1) = a (1)^{2} + b () 1 + c = a + b + c$

And, $f (0) = a (0)^{2} + b (0) + c = c$

$\Rightarrow [(a + b + c) - (c)]$ is an integer

$\Rightarrow (a + b)$ is an integer

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Let a, b, c be rational numbers and f:Z→Z be a function given by f(x)=ax2+bx+c. Then a+b is

We have, f(x)=ax2+bx+c,∀ x∈Z f(x) is an integer, ∀ x∈Z ⇒f(0) and f(1) are integers ⇒[f(1)−f(0)] is an integer Here, f(1)=a(1)2+b()1+c=a+b+c And, f(0)=a(0)2+b(0)+c=c ⇒[(a+b+c)−(c)] is an integer ⇒(a+b) is an integer

Let $a$ , $b$ , $c$ be rational numbers and $f : Z \to Z$ be a function given by $f (x) = a x^{2} + b x + c$ . Then $a + b$ is