Question

Values of x for which the sixth term of the expansion of $E = {\left( {{3^{{{\log }_3}\sqrt {9|x - 2|} }} + {7^{\left( {\dfrac{1}{5}} \right){{\log }_7}\left[ {(4){{.3}^{|x - 2|}} - 9} \right]}}} \right)^7}$ is 567, are
(). 1
(). 2
(). 3
(). None of these

Answer 1

Hint: The given binomial expansion is of the form ${(a + b)^7}$, hence the r+1, term of the binomial expansion is given by- \[{T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}\]. Put r = 5, in this equation to find the 6th term.

Complete step-by-step answer:
We have been given in the question the binomial expansion, $E = {\left( {{3^{{{\log }_3}\sqrt {9|x - 2|} }} + {7^{\left( {\dfrac{1}{5}} \right){{\log }_7}\left[ {(4){{.3}^{|x - 2|}} - 9} \right]}}} \right)^7}$.
Also, the 6th term of the expansion is 567.
Therefore, using the hint, the value of the 6th term of the expansion can be found out by using the formula,
\[{T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}\]
Now we have to find 6th term, so keep r = 5, n = 7, $a = {3^{{{\log }_3}\sqrt {9|x - 2|} }}$, $b = {7^{\left( {\dfrac{1}{5}} \right){{\log }_7}\left[ {(4){{.3}^{|x - 2|}} - 9} \right]}}$ in the given equation.
We get-
\[
  {T_{5 + 1}} = {}^7{C_5}{\left[ {{3^{{{\log }_3}\sqrt {9|x - 2|} }}} \right]^{7 - 5}}{\left[ {{7^{\left( {\dfrac{1}{5}} \right){{\log }_7}\left[ {(4){{.3}^{|x - 2|}} - 9} \right]}}} \right]^5} \\
  {T_6} = {T_{5 + 1}} = {}^7{C_5}{\left[ {{3^{{{\log }_3}\sqrt {9|x - 2|} }}} \right]^2}{\left[ {{7^{\left( {\dfrac{1}{5}} \right){{\log }_7}\left[ {(4){{.3}^{|x - 2|}} - 9} \right]}}} \right]^5} \\
   \Rightarrow {T_6} = \dfrac{{7!}}{{5!.2!}}{\left[ {\sqrt {{9^{|x - 2|}}} } \right]^2}{\left[ {{7^{{{\log }_7}\left[ {(4){{.3}^{|x - 2|}} - 9} \right]}}^{\dfrac{1}{5}}} \right]^5} \\
   = 21{\left[ {\sqrt {{9^{|x - 2|}}} } \right]^2}\left[ {\left[ {(4){{.3}^{|x - 2|}} - 9} \right]} \right] \\
   = {21.9^{|x - 2|}}.\left\{ {{{4.3}^{|x - 2|}} - 9} \right\} \\
   = {21.3^{2|x - 2|}}.\left\{ {{{4.3}^{|x - 2|}} - 9} \right\} \\
 \]
Now, we have been given that the 6th term is 567. Therefore keeping, ${T_6} = 567$, we get-
\[
  {T_6} = {21.3^{2|x - 2|}}\left\{ {{{4.3}^{|x - 2|}} - 9} \right\} = 567 \\
   \Rightarrow {21.3^{2|x - 2|}}\left\{ {{{4.3}^{|x - 2|}} - 9} \right\} = 21 \times 27 \\
   \Rightarrow {3^{2|x - 2|}}\left\{ {{{4.3}^{|x - 2|}} - 9} \right\} = 27 \\
   \Rightarrow {4.3^{3|x - 2|}} - (9){3^{2|x - 2|}} = 27 \\
 \]
Put $u = {3^{|x - 2|}}$, we get-
$4{u^3} - 9{u^2} - 27 = 0$
Now, we can see u = 3 satisfies the equation, so we can write-
$
  {3^{|x - 2|}} = 3 \\
   \Rightarrow |x - 2| = 1 \\
   \Rightarrow x - 2 = \pm 1 \\
   \Rightarrow x = 2 \pm 1 \\
   \Rightarrow x = 3,1 \\
 $
Therefore, we have two values of x for which the 6th term of the expansion is 567.
Hence, the correct options are [A] and [C].

Note: Whenever solving such types of questions, always write down the information provided in the question, and then use the standard formula of binomial expansion, as mentioned in the solution, i.e., \[{T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}\], to find the 6th term of the given expansion, and then equate it to 567 to find the values of x.