Question

In the expansion of \[{{\left( x+{{x}^{{{\log }_{10}}x}} \right)}^{5}}\], the third term is \[{{10}^{6}}\], then x =
(a) 1
(b) 2
(c) 10
(d) 100

Answer 1

Hint: First of all consider the given expression. Now, find its third term by using \[{{T}_{r+1}}={{\text{ }}^{n}}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}\] which is the general term of the expansion of \[{{\left( x+y \right)}^{n}}\]. Now, equate this third term to \[{{10}^{6}}\]. Now take \[{{\log }_{10}}\] both sides and substitute \[{{\log }_{10}}x=t\] and solve the quadratic equation. From this find the value of x by again substituting \[{{\log }_{10}}x\] in place of t and mark the correct option.

Complete step-by-step answer:

We are given that in the expansion of \[{{\left( x+{{x}^{{{\log }_{10}}x}} \right)}^{5}}\], the third term is \[{{10}^{6}}\], we have to find the value of x. Let us consider the expression given in the question.
\[E={{\left( x+{{x}^{{{\log }_{10}}x}} \right)}^{5}}\]
We know that the general term or the (r + 1)th term in the expansion of \[{{\left( x+y \right)}^{n}}\] is given by \[{{T}_{r+1}}={{\text{ }}^{n}}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}\]. By using this, we get the general term in the expansion of the above expression as
\[{{T}_{r+1}}={{\text{ }}^{5}}{{C}_{r}}{{\left( x \right)}^{5-r}}{{\left( {{x}^{{{\log }_{10}}x}} \right)}^{r}}\]
By substituting r = 2, we get, the (r + 1)the term that is the third term of the expansion as,
\[{{T}_{2+1}}={{T}_{3}}={{\text{ }}^{5}}{{C}_{2}}{{\left( x \right)}^{5-2}}{{\left( {{x}^{{{\log }_{10}}x}} \right)}^{2}}\]
We know that \[{{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}\]. By using this, we get,
\[{{T}_{3}}={{\text{ }}^{5}}{{C}_{2}}{{x}^{3}}.{{x}^{2{{\log }_{10}}x}}\]
We know that \[{{a}^{x}}.{{a}^{y}}={{a}^{x+y}}\]. By using this, we get,
\[{{T}_{3}}={{\text{ }}^{5}}{{C}_{2}}{{x}^{3+2{{\log }_{10}}x}}\]
We are given that the third term of the expansion is \[{{10}^{6}}\]. So, by equating the above expression by \[{{10}^{6}}\], we get,
\[^{5}{{C}_{2}}{{x}^{3+2{{\log }_{10}}x}}={{10}^{6}}\]
We know that, \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]. By using this, we get,
\[\dfrac{5!}{2!3!}{{x}^{3+2{{\log }_{10}}x}}={{10}^{6}}\]
\[\dfrac{5\times 4\times 3!}{2\times 3!}{{x}^{3+2{{\log }_{10}}x}}={{10}^{6}}\]
\[10{{x}^{3+2{{\log }_{10}}x}}={{10}^{6}}\]
\[{{x}^{3+2{{\log }_{10}}x}}=\dfrac{{{10}^{6}}}{10}\]
\[{{x}^{3+2{{\log }_{10}}x}}={{10}^{5}}\]
By taking \[{{\log }_{10}}\] on both the sides of the above equation, we get,
\[{{\log }_{10}}{{x}^{3+2{{\log }_{10}}x}}={{\log }_{10}}{{10}^{5}}\]
We know that, \[\log {{a}^{b}}=b\log a\].By using this, we get,
\[\left( 3+2{{\log }_{10}}x \right)\left( {{\log }_{10}}x \right)=5{{\log }_{10}}10\]
We know that, \[{{\log }_{a}}a=1\]. By using this, we get,
\[\left( 3+2{{\log }_{10}}x \right)\left( {{\log }_{10}}x \right)=5\]
By taking \[{{\log }_{10}}x=t\], we get,
\[\left( 3+2t \right)t=5\]
\[3t+2{{t}^{2}}=5\]
\[2{{t}^{2}}+3t-5=0\]
We can also write the above equation as,
\[2{{t}^{2}}+5t-2t-5=0\]
\[t\left( 2t+5 \right)-t\left( 2t+5 \right)=0\]
By taking out (2t + 5) common, we get,
\[\left( 2t+5 \right)\left( t-1 \right)=0\]
So, we get,
\[t=\dfrac{-5}{2},t=1\]
By substituting \[t={{\log }_{10}}x\], we get,
\[{{\log }_{10}}x=\dfrac{-5}{2},{{\log }_{10}}x=1\]
We know that when \[{{\log }_{a}}b=c\], then \[b={{\left( a \right)}^{c}}\], by using this we get,
\[x={{\left( 10 \right)}^{\dfrac{-5}{2}}};x=10\]
Hence, the option (c) is the right answer.

Note: In this question, students must note that if \[{{\left( 10 \right)}^{\dfrac{-5}{2}}}\] would have been in the options, then that would also be the correct answer. Also, some students make the mistake of taking the third term as \[^{n}{{C}_{3}}{{\left( x \right)}^{n-3}}{{\left( y \right)}^{3}}\] which is wrong because we have than an expression for (r + 1)th term. So, to get the third term, we must substitute r = 2 in it and not r = 3. So, the correct third term would be \[^{n}{{C}_{2}}{{\left( x \right)}^{n-2}}{{\left( y \right)}^{2}}\].