Answer
Verified
423.6k+ views
Hint: Expand the given entity. It gives the expansion form of ${{e}^{\log 2\ }}and\ {{e}^{2\log 2}}$. Substitute the exponent values and simplify the expression obtained until you get a whole number.
Complete step-by-step answer:
Given us $4+2\left( 1+2 \right)\log 2+\dfrac{2\left( 1+{{2}^{2}} \right){{\left( \log 2 \right)}^{2}}}{2!}+\dfrac{2\left( 1+{{2}^{3}} \right){{\left( \log 2 \right)}^{3}}}{3!}$
Take the first number 4 it can split as 2 + 2;
Take 2nd value, $2\left( 1+2 \right)\log 2$ it can be split us as;
$\left( 2+2\times 2 \right)\log 2=2\log 2+4\log 2$
Similarly,
$\dfrac{2\left( 1+{{2}^{2}} \right){{\left( \log 2 \right)}^{2}}}{2!}=\dfrac{\left( 2+{{2}^{3}} \right){{\left( \log 2 \right)}^{2}}}{2!}=\dfrac{2{{\left( \log 2 \right)}^{2}}}{2!}+\dfrac{{{2}^{3}}{{\left( \log 2 \right)}^{2}}}{2!}$
The other values can be similarly splitted,
$\begin{align}
& 4+2\left( 1+2 \right)\log 2+\dfrac{2\left( 1+{{2}^{2}} \right){{\left( \log 2 \right)}^{2}}}{2!}+........ \\
& =2+2+2\log 2+4\log 2+\dfrac{2{{(\log 2)}^{2}}}{2!}+\dfrac{{{2}^{3}}{{\left( \log 2 \right)}^{2}}}{2!}+..... \\
& =\left( 2+2\log 2+\dfrac{2{{\left( \log 2 \right)}^{2}}}{2!}+..... \right)+\left( 2+4\log 2+\dfrac{{{2}^{3}}{{\left( \log 2 \right)}^{2}}}{2!}+.... \right) \\
\end{align}$
Taking 2 common from both brackets
$=2\left( 1+\log 2+\dfrac{{{\left( \log 2 \right)}^{2}}}{2!}+..... \right)+2\left( 1+2\log 2+\dfrac{{{\left( 2\log 2 \right)}^{2}}}{2!}+.... \right)$
This is the fourier expansion of general expansion of $_{e}{{\log }^{2}}$.
$\therefore {{\ }_{e}}{{\log }^{2}}=1+\log 2+\dfrac{{{\left( \log 2 \right)}^{2}}}{2!}+.....$
Similarly $1+2\log 2\dfrac{{{\left( 2\log 2 \right)}^{2}}}{2!}+....$ is the expansion of $_{e}{{\log }^{2\log 2}}$
$\therefore $ It can be written as,
$2\left( {{e}^{\log 2}} \right)+2\left( {{e}^{2\log 2}} \right)$
We know that ${{e}^{\log x}}=x$
$\begin{align}
& \therefore \ {{e}^{\log 2}}=2 \\
& {{e}^{2\log 2}}={{e}^{\log {{2}^{2}}}}={{2}^{2}}=4 \\
& \Rightarrow 2\times 2+2\times 4=4+8=12 \\
\end{align}$
Therefore, the correct answer is option B.
Note: Splitting of the entity is important in the beginning. So as to form the expansion form of ${{e}^{\log x}}$.
Complete step-by-step answer:
Given us $4+2\left( 1+2 \right)\log 2+\dfrac{2\left( 1+{{2}^{2}} \right){{\left( \log 2 \right)}^{2}}}{2!}+\dfrac{2\left( 1+{{2}^{3}} \right){{\left( \log 2 \right)}^{3}}}{3!}$
Take the first number 4 it can split as 2 + 2;
Take 2nd value, $2\left( 1+2 \right)\log 2$ it can be split us as;
$\left( 2+2\times 2 \right)\log 2=2\log 2+4\log 2$
Similarly,
$\dfrac{2\left( 1+{{2}^{2}} \right){{\left( \log 2 \right)}^{2}}}{2!}=\dfrac{\left( 2+{{2}^{3}} \right){{\left( \log 2 \right)}^{2}}}{2!}=\dfrac{2{{\left( \log 2 \right)}^{2}}}{2!}+\dfrac{{{2}^{3}}{{\left( \log 2 \right)}^{2}}}{2!}$
The other values can be similarly splitted,
$\begin{align}
& 4+2\left( 1+2 \right)\log 2+\dfrac{2\left( 1+{{2}^{2}} \right){{\left( \log 2 \right)}^{2}}}{2!}+........ \\
& =2+2+2\log 2+4\log 2+\dfrac{2{{(\log 2)}^{2}}}{2!}+\dfrac{{{2}^{3}}{{\left( \log 2 \right)}^{2}}}{2!}+..... \\
& =\left( 2+2\log 2+\dfrac{2{{\left( \log 2 \right)}^{2}}}{2!}+..... \right)+\left( 2+4\log 2+\dfrac{{{2}^{3}}{{\left( \log 2 \right)}^{2}}}{2!}+.... \right) \\
\end{align}$
Taking 2 common from both brackets
$=2\left( 1+\log 2+\dfrac{{{\left( \log 2 \right)}^{2}}}{2!}+..... \right)+2\left( 1+2\log 2+\dfrac{{{\left( 2\log 2 \right)}^{2}}}{2!}+.... \right)$
This is the fourier expansion of general expansion of $_{e}{{\log }^{2}}$.
$\therefore {{\ }_{e}}{{\log }^{2}}=1+\log 2+\dfrac{{{\left( \log 2 \right)}^{2}}}{2!}+.....$
Similarly $1+2\log 2\dfrac{{{\left( 2\log 2 \right)}^{2}}}{2!}+....$ is the expansion of $_{e}{{\log }^{2\log 2}}$
$\therefore $ It can be written as,
$2\left( {{e}^{\log 2}} \right)+2\left( {{e}^{2\log 2}} \right)$
We know that ${{e}^{\log x}}=x$
$\begin{align}
& \therefore \ {{e}^{\log 2}}=2 \\
& {{e}^{2\log 2}}={{e}^{\log {{2}^{2}}}}={{2}^{2}}=4 \\
& \Rightarrow 2\times 2+2\times 4=4+8=12 \\
\end{align}$
Therefore, the correct answer is option B.
Note: Splitting of the entity is important in the beginning. So as to form the expansion form of ${{e}^{\log x}}$.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE