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# 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

Last updated date: 13th Jun 2024
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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.

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 $^{n}{{C}_{2}}{{\left( x \right)}^{n-2}}{{\left( y \right)}^{2}}$.