The expression $${\left( {x + {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5} + {\left( {x - {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5}$$ is a polynomial of degree
A.
5
B.
6
C.
7
D.
8
Answer :
7
Solution :
The given expression is
$${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5}$$
We know by binomial theorem, that
$${\left( {x + a} \right)^n} + {\left( {x - a} \right)^n} - 2\left[ {^n{C_0}{x^n} + {\,^n}{C_2}{x^{n - 2}}{a^2} + {\,^n}{C_4}{x^{n - 4}}{a^4} + ......} \right]$$
∴ The given expression is equal to
$$2\left[ {^5{C_0}{x^5} + {\,^5}{C_2}{x^3}\left( {{x^3} - 1} \right) + {\,^5}{C_4}x{{\left( {{x^3} - 1} \right)}^2}} \right]$$
Max. power of $$x$$ involved here is 7, also only +ve integral powers of $$x$$ are involved, therefore given expression is a polynomial of degree 7.
Releted MCQ Question on Algebra >> Binomial Theorem
Releted Question 1
Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then
If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is