Let P(x) be a polynomial with real coefficients such that ∫01 xmP(1−x)dx=0∀m∈N∪{0}, then

A
$P (x) = x^{n} (1 - x)^{n}$ for some $n \in N$
B
$P (x) = (1 - x)^{2 n}$ for some $n \in N$
C
$P (x) = 1 - x^{m} (1 - x)^{n}$ for some $m, n \in N$
D
P(x) = 0

Solution:

$0 = \int_{0}^{1} x^{m} P (1 - x) d x = \int_{0}^{1} (1 - x)^{m} P (x)$

Let $P (x) = \sum_{k = 0}^{n} a_{k} x^{k}$ where $a_{k} \in R$

$\begin{array}{l} = \sum_{k = 0}^{n} a_{k} [1 - (1 - x)]^{k} \\ = \sum_{k = 0}^{n} a_{k} (\sum_{j = 0}^{k} (- 1)^{k} (^{k} C_{j}) (1 - x)^{j}) \\ = \sum_{k = 0}^{n} b_{k} (1 - x)^{k} \end{array}$

where $b_{k} = \sum_{j = k}^{n} (- 1)^{k} (^{j} C_{k}) a_{j}$

Now $\int_{0}^{1} (P (x))^{2} d x = \int_{0}^{1} P (x) (\sum_{k = 0}^{n} b_{k} (1 - x)^{k}) d x$

$= \sum_{k = 0}^{n} b_{k} \int_{0}^{1} (1 - x)^{k} P (x) d x = 0$

As P(x) is a polynomial,

$\begin{array}{l} P (x) \equiv 0 \forall x \in [0, 1] \\ \Rightarrow P (x) \equiv 0 \end{array}$

Let P(x) be a polynomial with real coefficients such that $\int_{0}^{1} x^{m} P (1 - x) d x = 0 \forall m \in N \cup {0}$ , then

Solution:

Related content