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Question: What is the highest degree in the expansion of \({\left[ {x + {{\left( {{x^3} - 1} \right)}^{\dfrac{...

What is the highest degree in the expansion of [x+(x3−1)12]5+[x−(x3−1)12]5{\left[ {x + {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} + {\left[ {x - {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} -
(A)8
(B)7
(C)6
(D)5

Explanation

Solution

Hint- Here in the expansion of the given function some terms will cancel out with each other.
As we know that according to binomial theorem of expansion, we have

(a+b)n=nC0(a)n+nC1(a)n−1(b)+nC2(a)n−2(b)2+.......+nCn−1(a)(b)n−1+nCn(b)n{\left( {a + b} \right)^n} = {}^n{C_0}{\left( a \right)^n} + {}^n{C_1}{\left( a \right)^{n - 1}}\left( b \right) + {}^n{C_2}{\left( a \right)^{n - 2}}{\left( b \right)^2} + ....... + {}^n{C_{n - 1}}\left( a \right){\left( b \right)^{n - 1}} + {}^n{C_n}{\left( b \right)^n}
and (a−b)n=nC0(a)n−nC1(a)n−1(b)+nC2(a)n−2(b)2−.......+(−1)n−1nCn−1(a)(b)n−1+(−1)nnCn(b)n{\left( {a - b} \right)^n} = {}^n{C_0}{\left( a \right)^n} - {}^n{C_1}{\left( a \right)^{n - 1}}\left( b \right) + {}^n{C_2}{\left( a \right)^{n - 2}}{\left( b \right)^2} - ....... + {\left( { - 1} \right)^{n - 1}}{}^n{C_{n - 1}}\left( a \right){\left( b \right)^{n - 1}} + {\left( { - 1} \right)^n}{}^n{C_n}{\left( b \right)^n}

where nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}

Now consider [x+(x3−1)12]5{\left[ {x + {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} and here replace xx by aa and (x3−1)12{\left( {{x^3} - 1} \right)^{\dfrac{1}{2}}} by bb, we have

[x+(x3−1)12]5=(a+b)5=5C0(a)5+5C1(a)5−1(b)+5C2(a)5−2(b)2+5C3(a)5−3(b)3+5C4(a)5−4(b)4+5C5(b)5 ⇒[x+(x3−1)12]5=(a+b)5=5C0(a)5+5C1(a)4(b)+5C2(a)3(b)2+5C3(a)2(b)3+5C4(a)(b)4+5C5(b)5 →(1)  {\left[ {x + {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} = {\left( {a + b} \right)^5} = {}^5{C_0}{\left( a \right)^5} + {}^5{C_1}{\left( a \right)^{5 - 1}}\left( b \right) + {}^5{C_2}{\left( a \right)^{5 - 2}}{\left( b \right)^2} + {}^5{C_3}{\left( a \right)^{5 - 3}}{\left( b \right)^3} + {}^5{C_4}{\left( a \right)^{5 - 4}}{\left( b \right)^4} + {}^5{C_5}{\left( b \right)^5} \\\ \Rightarrow {\left[ {x + {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} = {\left( {a + b} \right)^5} = {}^5{C_0}{\left( a \right)^5} + {}^5{C_1}{\left( a \right)^4}\left( b \right) + {}^5{C_2}{\left( a \right)^3}{\left( b \right)^2} + {}^5{C_3}{\left( a \right)^2}{\left( b \right)^3} + {}^5{C_4}\left( a \right){\left( b \right)^4} + {}^5{C_5}{\left( b \right)^5}{\text{ }} \to {\text{(1)}} \\\
Now consider [x−(x3−1)12]5{\left[ {x - {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} and here replace xx by aa and (x3−1)12{\left( {{x^3} - 1} \right)^{\dfrac{1}{2}}} by bb, we have
[x−(x3−1)12]5=(a−b)5=5C0(a)5−5C1(a)5−1(b)+5C2(a)5−2(b)2−5C3(a)5−3(b)3+5C4(a)5−4(b)4−5C5(b)5 ⇒[x−(x3−1)12]5=(a−b)5=5C0(a)5−5C1(a)4(b)+5C2(a)3(b)2−5C3(a)2(b)3+5C4(a)(b)4−5C5(b)5 →(2)  {\left[ {x - {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} = {\left( {a - b} \right)^5} = {}^5{C_0}{\left( a \right)^5} - {}^5{C_1}{\left( a \right)^{5 - 1}}\left( b \right) + {}^5{C_2}{\left( a \right)^{5 - 2}}{\left( b \right)^2} - {}^5{C_3}{\left( a \right)^{5 - 3}}{\left( b \right)^3} + {}^5{C_4}{\left( a \right)^{5 - 4}}{\left( b \right)^4} - {}^5{C_5}{\left( b \right)^5} \\\ \Rightarrow {\left[ {x - {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} = {\left( {a - b} \right)^5} = {}^5{C_0}{\left( a \right)^5} - {}^5{C_1}{\left( a \right)^4}\left( b \right) + {}^5{C_2}{\left( a \right)^3}{\left( b \right)^2} - {}^5{C_3}{\left( a \right)^2}{\left( b \right)^3} + {}^5{C_4}\left( a \right){\left( b \right)^4} - {}^5{C_5}{\left( b \right)^5}{\text{ }} \to {\text{(2)}} \\\
Now using equations (1) and (2), we have
[x+(x3−1)12]5+[x−(x3−1)12]5=5C0(a)5+5C1(a)4(b)+5C2(a)3(b)2+5C3(a)2(b)3 \+5C4(a)(b)4+5C5(b)5+5C0(a)5−5C1(a)4(b)+5C2(a)3(b)2−5C3(a)2(b)3+5C4(a)(b)4−5C5(b)5 ⇒[x+(x3−1)12]5+[x−(x3−1)12]5=2[5C0(a)5]+2[5C2(a)3(b)2]+2[5C4(a)(b)4] →(3)  {\left[ {x + {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} + {\left[ {x - {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} = {}^5{C_0}{\left( a \right)^5} + {}^5{C_1}{\left( a \right)^4}\left( b \right) + {}^5{C_2}{\left( a \right)^3}{\left( b \right)^2} + {}^5{C_3}{\left( a \right)^2}{\left( b \right)^3} \\\ \+ {}^5{C_4}\left( a \right){\left( b \right)^4} + {}^5{C_5}{\left( b \right)^5} + {}^5{C_0}{\left( a \right)^5} - {}^5{C_1}{\left( a \right)^4}\left( b \right) + {}^5{C_2}{\left( a \right)^3}{\left( b \right)^2} - {}^5{C_3}{\left( a \right)^2}{\left( b \right)^3} + {}^5{C_4}\left( a \right){\left( b \right)^4} - {}^5{C_5}{\left( b \right)^5} \\\ \Rightarrow {\left[ {x + {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} + {\left[ {x - {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} = 2\left[ {{}^5{C_0}{{\left( a \right)}^5}} \right] + 2\left[ {{}^5{C_2}{{\left( a \right)}^3}{{\left( b \right)}^2}} \right] + 2\left[ {{}^5{C_4}\left( a \right){{\left( b \right)}^4}} \right]{\text{ }} \to {\text{(3)}} \\\
Also, 5C0=5!0!(5−0)!=5!0!5!=1  [∵0!=1],5C2=5!2!(5−2)!=5.4.3!2.1!3!=5×42=10 5C4=5!4!(5−4)!=5.4!4!1!=51=5  {}^5{C_0} = \dfrac{{5!}}{{0!\left( {5 - 0} \right)!}} = \dfrac{{5!}}{{0!5!}} = 1\;\left[ {\because 0! = 1} \right],{}^5{C_2} = \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}} = \dfrac{{5.4.3!}}{{2.1!3!}} = \dfrac{{5 \times 4}}{2} = 10 \\\ {}^5{C_4} = \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}} = \dfrac{{5.4!}}{{4!1!}} = \dfrac{5}{1} = 5 \\\
Substituting above values and replacing aa by xx and bb by (x3−1)12{\left( {{x^3} - 1} \right)^{\dfrac{1}{2}}}, equation (3) becomes
⇒[x+(x3−1)12]5+[x−(x3−1)12]5=2[1×x5]+2[10×x3×((x3−1)12)2]+2[5x((x3−1)12)4] ⇒[x+(x3−1)12]5+[x−(x3−1)12]5=2x5+20x3(x3−1)+10x(x3−1)2=2x5+20x6−20x3+10x(x6+1−2x3) ⇒[x+(x3−1)12]5+[x−(x3−1)12]=2x5+20x6−20x3+10x7+10x−20x4  \Rightarrow {\left[ {x + {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} + {\left[ {x - {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} = 2\left[ {1 \times {x^5}} \right] + 2\left[ {10 \times {x^3} \times {{\left( {{{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right)}^2}} \right] + 2\left[ {5x{{\left( {{{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right)}^4}} \right] \\\ \Rightarrow {\left[ {x + {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} + {\left[ {x - {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} = 2{x^5} + 20{x^3}\left( {{x^3} - 1} \right) + 10x{\left( {{x^3} - 1} \right)^2} = 2{x^5} + 20{x^6} - 20{x^3} + 10x\left( {{x^6} + 1 - 2{x^3}} \right) \\\ \Rightarrow {\left[ {x + {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5} + \left[ {x - {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right] = 2{x^5} + 20{x^6} - 20{x^3} + 10{x^7} + 10x - 20{x^4} \\\
Clearly from the above equation, we can see that the highest degree in the expansion of the given function is 7.
Hence, option B is correct.

Note- These types of problems are solved by using binomial theorem of expansion and simplifying the given function and then finally checking the highest degree (highest power of variable xx) of the polynomial obtained.