A polynomial is given as F x - a 0- a 1 x - a 2 x 2- - a n -1 x n -1 with integer coefficients- If F 0 and F 1 are odd integers then the rootsareA- Even integersB- Odd integersC- Non integersD- Both odd and even integers

The correct option is C Non integersF(0)→ odd integer ⇒ a_0 → odd integer. F(1)→ odd integer ⇒ a_0+a_1+...+a_n 1→ odd nbsp; nbsp; ∴ a_1+a_2+...+a_n 1→ even ...

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Standard XII

Mathematics

Composite Function

A polynomial ...

Question

A polynomial is given as $F (x) = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{n - 1} x^{n - 1}$ with integer coefficients. If $F (0)$ and $F (1)$ are odd integers then the roots are

Even integers

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Odd integers

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Non integers

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Both odd and even integers

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Solution

The correct option is C Non integers
$F (0) \to$ odd integer $\Rightarrow a_{0} \to$ odd integer.
$F (1) \to$ odd integer
$\Rightarrow a_{0} + a_{1} + . . . + a_{n - 1} \to$ odd
$∴ a_{1} + a_{2} + . . . + a_{n - 1} \to$ even.

Now, suppose roots are even integers $x = 2 m$
$∴ F (2 m) = a_{0} + a_{1} (2 m) + a_{2} (2 m)^{2} + . . . . + a_{n - 1} (2 m)^{n - 1}$
$Odd Even Even ... Even$
So, $F (2 m) =$ odd $\neq 0$
$\Rightarrow$ Even integers roots is not possible.

Now, suppose roots are odd integers $x = 2 m + 1$
$∴ F (2 m + 1) = a_{0} + a_{1} (2 m + 1) + a_{2} (2 m + 1)^{2} + . . . + a_{n - 1} (2 m + 1)^{n - 1}$
$= a_{0} + (e v e n + 1) a_{1} + (e v e n + 1)^{2} a_{2} + . . . . . + (e v e n + 1)^{n} a_{n} + (2 m + 1)^{n}$
$= e v e n (a_{1} + a_{2} + a_{3} + . . . . + a_{n - 1}) ($ even $) + a_{0} + a_{1} + a_{2} + a_{3} + . . . . + a_{n - 1} ($ even $) +$ odd
$f (2 m + 1) = odd \neq 0$
$\Rightarrow$ Roots cannot be odd integers.
Hence, roots are non-integers.

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