Question:
The expression $[x+(x^3-1)^{1/2}]^5 +[x-(x^3-1)^{1/2}]^5$ is a polynomial of degree
The expression $[x+(x^3-1)^{1/2}]^5 +[x-(x^3-1)^{1/2}]^5$ is a polynomial of degree
Updated On: Jun 14, 2022
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The Correct Option is C
Solution and Explanation
We know that,
$(a+b)^5+(a-b)^5= \, ^5C_0 a^5+ \, ^5C_1 a^4b+ \, ^5C_2 a^3b^2$
$ \, \, \, \, \, + ^5C_3a^2b^3+ \, ^5C_4ab^4+ \, ^5C_5b^5+^5C_0 a^5- \, ^5C_1 a^4b$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, + ^5 C_2 a^3b^2- \, ^5C_4 ab^4+ \, ^5C_3 a^2b^3+ \, ^5C_4 ab^4- \, ^5C_5 b^5$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =2[a^5+10a^3+b^2+5ab64]$
$\therefore \, [x+(x^3-1)^{1/5}]^5 + [ x-(x^3-1)^{1/2}]^5$
=2 $[x^5+10x^3(x^3-1)+5x(x^3-1)^2]$
Therefore, the given expression is a polynomial of degree 7.
$(a+b)^5+(a-b)^5= \, ^5C_0 a^5+ \, ^5C_1 a^4b+ \, ^5C_2 a^3b^2$
$ \, \, \, \, \, + ^5C_3a^2b^3+ \, ^5C_4ab^4+ \, ^5C_5b^5+^5C_0 a^5- \, ^5C_1 a^4b$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, + ^5 C_2 a^3b^2- \, ^5C_4 ab^4+ \, ^5C_3 a^2b^3+ \, ^5C_4 ab^4- \, ^5C_5 b^5$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =2[a^5+10a^3+b^2+5ab64]$
$\therefore \, [x+(x^3-1)^{1/5}]^5 + [ x-(x^3-1)^{1/2}]^5$
=2 $[x^5+10x^3(x^3-1)+5x(x^3-1)^2]$
Therefore, the given expression is a polynomial of degree 7.
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Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.