Determine all polynomial $P (x)$ with rational coefficient so that for all $x$ with $| x | \leq 1$ ; $P (x) = P (\frac{- x + \sqrt{3 - 3 x^{2}}}{2})$ .

P (x) = a_{0} + a_{1} (3 x - 4 x^{3}) + a_{2} (3 x - 4 x^{3})^{2} + . . . . + (3 x - 4 x^{3})^{k} . k (x)

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P (x) = a_{0} + a_{1} (3 x - 4 x^{2}) + a_{2} (3 x - 4 x^{2})^{2} + . . . . . + (3 x - 4 x^{2})^{k} . k (x)

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P (x) = a_{0} + a_{1} (3 x - 4 x^{4}) + a_{2} (3 x - 4 x^{4}) 2 + . . . . + (3 x - 4 x^{4})^{k} . k (x)

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Solution

The correct option is A $P (x) = a_{0} + a_{1} (3 x - 4 x^{3}) + a_{2} (3 x - 4 x^{3})^{2} + . . . . + (3 x - 4 x^{3})^{k} . k (x)$
Here, $P (x) = P (\frac{- x + \sqrt{3 - 3 x^{2}}}{2})$ for $| x | \leq 1$ (i)

Let $x = 0$ ,

$⟹ P (0) = P (\frac{\sqrt{3}}{2})$

Which shows, $P (x) - P (0)$ is divisible by $x (x - \frac{\sqrt{3}}{2})$ .

Since, $P (x) - P (0)$ has rational coefficients and $\frac{\sqrt{3}}{2}$ is one of the roots,

$∴ - \frac{\sqrt{3}}{2}$ is also a root of $P (x) - P (0)$ .

Thus,

$\displaystyle

x\left(x-\dfrac{\sqrt{3}}{2}\right)\left(x+\dfrac{\sqrt{3}}{2}\right)=x^3-\dfrac{3}{4}x=\dfrac{4x^3-3x}{4}$

is factor of $P (x) - P (0)$ .

$∴ P (x) = P (0) + (3 x - 4 x^{3}) P_{1} (x)$ for $| x | \leq 1$ (ii)

as $P (x) = P (\frac{- x + \sqrt{3 - 3 x^{2}}}{2})$

$⟹ 3 (\frac{- x + \sqrt{3 - 3 x^{2}}}{2}) - 4 (\frac{- x + \sqrt{3 - 3 x^{2}}}{2}) = 3 x - 4 x^{3}$

$∴ P_{1} (x) = P_{1} (\frac{- x + \sqrt{3 - 3 x^{2}}}{2})$

$∴ P_{1} (x) = P_{1} (0) + (3 x - 4 x^{3}) P_{2} (x)$ [using equation (ii)]
$⟹ P (x) = P (0) + (3 x - 4 x^{3}) (P_{1} (0)) + (3 x - 4 x^{3}) P_{2} (x)$
$= P (0) + (3 x - 4 x^{3}) P_{1} (0) + {(3 x - 4 x^{3})}^{2} P_{2} (x)$

Thus, in general,
$P (x) = a_{0} + a_{1} (3 x - 4 x^{3}) + a_{2} {(3 x - 4 x^{3})}^{2} + \dots + {(3 x - 4 x^{3})}^{k} . k (x)$
where $k (x)$ is a polynomial with rational coefficient.

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