Let fx-ax2-bx-c- where a-b-c are rational- and f- Z- Z- where Z is the set of integers- Then a-b is

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Rational Function

Let fx=ax2+...

Question

Let $f (x) = a x^{2} + b x + c,$ where $a, b, c$ are rational, and $f : Z \to Z,$ where $Z$ is the set of integers. Then $a + b$ is

a negative integer

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an integer

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nonintegral rational number

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none of these

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Solution

The correct option is B an integer
$f : Z \to Z$ is defined as $f (x) = a x^{2} + b x + c$
which implies for integer inputs, the function gives integer outputs.
$\Rightarrow f (0) = c = Z_{1}$ ...(1) (where $Z_{1}$ is some integer)
Similarly, $f (1) = a + b + c = Z_{2}$ ...(2) (where $Z_{2}$ is some integer)
(2) - (1) gives $a + b = Z_{2} - Z_{1}$ , which is also an integer.

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Let f(x)=ax2+bx+c, where a,b,c are rational, and f:Z→Z, where Z is the set of integers. Then a+b is

The correct option is B an integerf:Z→Z is defined as f(x)=ax2+bx+cwhich implies for integer inputs, the function gives integer outputs.⇒f(0)=c=Z1 ...(1) (where Z1 is some integer)Similarly, f(1)=a+b+c=Z2 ...(2) (where Z2 is some integer)(2) - (1) gives a+b=Z2−Z1, which is also an integer.

Let $f (x) = a x^{2} + b x + c,$ where $a, b, c$ are rational, and $f : Z \to Z,$ where $Z$ is the set of integers. Then $a + b$ is