Binomial Theorem 1 Question 22
24. The expression $\left[x+\left(x^{3}-1\right)^{1 / 2}\right]^{5}+\left[x-\left(x^{3}-1\right)^{1 / 2}\right]^{5}$ is a polynomial of degree
(1992, 2M)
(a) 5
(b) 6
(c) 7
(d) 8
Show Answer
Solution:
- We know that,
$$ \begin{gathered} (a+b)^{5}+(a-b)^{5}={ }^{5} C _0 a^{5}+{ }^{5} C _1 a^{4} b+{ }^{5} C _2 a^{3} b^{2} \\ +{ }^{5} C _3 a^{2} b^{3}+{ }^{5} C _4 a b^{4}+{ }^{5} C _5 b^{5}+{ }^{5} C _0 a-{ }^{5} C _1 a^{4} b \\ +{ }^{5} C _2 a^{3} b^{2}-{ }^{5} C _3 a^{2} b^{3}+{ }^{5} C _4 a b^{4}-{ }^{5} C _5 b^{5} \\ =2\left[a^{5}+10 a^{3} b^{2}+5 a b^{4}\right] \\ \therefore \quad\left[x+\left(x^{3}-1\right)^{1 / 2}\right]^{5}+\left[x-\left(x^{3}-1\right)^{1 / 2}\right]^{5} \\ =2\left[x^{5}+10 x^{3}\left(x^{3}-1\right)+5 x\left(x^{3}-1\right)^{2}\right] \end{gathered} $$
Therefore, the given expression is a polynomial of degree 7.
