Question

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

Answer 1

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

${\left( {a + b} \right)^n} = {}^n{C_0}{\left( a \right)^n} + {}^n{C_1}{\left( a \right)^{n - 1}}\left( b \right) + {}^n{C_2}{\left( a \right)^{n - 2}}{\left( b \right)^2} + ....... + {}^n{C_{n - 1}}\left( a \right){\left( b \right)^{n - 1}} + {}^n{C_n}{\left( b \right)^n}$
and ${\left( {a - b} \right)^n} = {}^n{C_0}{\left( a \right)^n} - {}^n{C_1}{\left( a \right)^{n - 1}}\left( b \right) + {}^n{C_2}{\left( a \right)^{n - 2}}{\left( b \right)^2} - ....... + {\left( { - 1} \right)^{n - 1}}{}^n{C_{n - 1}}\left( a \right){\left( b \right)^{n - 1}} + {\left( { - 1} \right)^n}{}^n{C_n}{\left( b \right)^n}$

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

Now consider ${\left[ {x + {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5}$ and here replace $x$ by $a$ and ${\left( {{x^3} - 1} \right)^{\dfrac{1}{2}}}$ by $b$, we have

  $
  {\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 ${\left[ {x - {{\left( {{x^3} - 1} \right)}^{\dfrac{1}{2}}}} \right]^5}$ and here replace $x$ by $a$ and ${\left( {{x^3} - 1} \right)^{\dfrac{1}{2}}}$ by $b$, we have
  $
  {\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
$
  {\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, $
  {}^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 $a$ by $x$ and $b$ by ${\left( {{x^3} - 1} \right)^{\dfrac{1}{2}}}$, equation (3) becomes
$
   \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 $x$) of the polynomial obtained.