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) $8$
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Answer:
Correct Answer: 24. (c)
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.
