Answer

Verified

85.8k+ views

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

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

Recently Updated Pages

Name the scale on which the destructive energy of an class 11 physics JEE_Main

Write an article on the need and importance of sports class 10 english JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main

What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main

A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main

Other Pages

A scooterist sees a bus 1km ahead of him moving with class 11 physics JEE_Main

Derive an expression for maximum speed of a car on class 11 physics JEE_Main

A cylinder of 10 Lcapacity at 300 Kcontaining the Hegas class 11 chemistry JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

The process requiring the absorption of energy is A class 11 chemistry JEE_Main