Question

If $ x $ is so small that $ {x^2} $ and higher powers of $ x $ are neglected then $ \dfrac{{\sqrt {1 + x} + \sqrt[3]{{1 + 4x}}}}{{\left( {1 + {x^2}} \right)\sqrt[3]{{{{\left( {1 - 3x} \right)}^2}}}}} = $
A. $ 1 + \dfrac{{11x}}{{12}} $
B. $ 2 + \dfrac{{35x}}{6} $
C. $ 1 - \dfrac{{5x}}{{12}} $
D. $ 1 + \dfrac{{5x}}{{12}} $

Answer 1

Hint: Here given that the higher power of $ x $ should be neglected including $ {x^2} $ . To solve this problem we are going to use the concept of fractional index pension which neglects the $ {x^2} $ and the higher terms. One of the example is shown below:
Here \[{\left( {1 + x} \right)^{\dfrac{1}{n}}} = 1 + \dfrac{x}{n}\]

Complete step-by-step answer:
The fractional index pension describes that given a fractional index, then the denominator of the fraction is the root of the number or letter, and the numerator of the fraction is the power to raise the answer to.
Consider the given fraction of functions $ \dfrac{{\sqrt {1 + x} + \sqrt[3]{{1 + 4x}}}}{{\left( {1 + {x^2}} \right)\sqrt[3]{{{{\left( {1 - 3x} \right)}^2}}}}} $ , as given below:
 $ \Rightarrow \dfrac{{\sqrt {1 + x} + \sqrt[3]{{1 + 4x}}}}{{\left( {1 + {x^2}} \right)\sqrt[3]{{{{\left( {1 - 3x} \right)}^2}}}}} = \dfrac{{{{\left( {1 + x} \right)}^{\dfrac{1}{2}}} + {{\left( {1 + 4x} \right)}^{\dfrac{3}{2}}}}}{{\left( {1 + {x^2}} \right){{\left( {1 - 3x} \right)}^{\dfrac{2}{3}}}}} $
Here applying the concept of fractional index pension which was described above, the formula applied to the above expression is given below:
 \[ \Rightarrow \dfrac{{{{\left( {1 + x} \right)}^{\dfrac{1}{2}}} + {{\left( {1 + 4x} \right)}^{\dfrac{3}{2}}}}}{{\left( {1 + {x^2}} \right){{\left( {1 - 3x} \right)}^{\dfrac{2}{3}}}}} = \dfrac{{\left( {1 + \dfrac{1}{2}x} \right) + \left( {1 + \dfrac{4}{3}x} \right)}}{{\left( {1 + {x^2}} \right)\left( {1 - 3\left( {\dfrac{2}{3}} \right)x} \right)}}\]
On further simplification and further solving of the expression, as shown below, the described formula is applied below, which is given by: \[{\left( {1 + x} \right)^{\dfrac{1}{n}}} = 1 + \dfrac{x}{n}\]
 \[ \Rightarrow \dfrac{{\left( {1 + \dfrac{1}{2}x} \right) + \left( {1 + \dfrac{4}{3}x} \right)}}{{\left( {1 + {x^2}} \right)\left( {1 - 3\left( {\dfrac{2}{3}} \right)x} \right)}} = \dfrac{{1 + \dfrac{1}{2}x + 1 + \dfrac{4}{3}x}}{{\left( {1 + {x^2}} \right)\left( {1 - 2x} \right)}}\]
Here we have to neglect the term $ {x^2} $ , which means that $ {x^2} \simeq 0 $
 \[ \Rightarrow \dfrac{{1 + \dfrac{1}{2}x + 1 + \dfrac{4}{3}x}}{{\left( {1 + {x^2}} \right)\left( {1 - 2x} \right)}} = \dfrac{{1 + \dfrac{1}{2}x + 1 + \dfrac{4}{3}x}}{{\left( {1 + 0} \right)\left( {1 - 2x} \right)}}\]
 \[ \Rightarrow \dfrac{{1 + \dfrac{1}{2}x + 1 + \dfrac{4}{3}x}}{{1 - 2x}}\]
 \[ \Rightarrow \dfrac{{2 + \dfrac{{11}}{6}x}}{{1 - 2x}}\]
Now multiplying and dividing the expression with $ 1 + 2x $ in the numerator and denominator as shown:
 \[ \Rightarrow \dfrac{{\left( {2 + \dfrac{{11}}{6}x} \right)\left( {1 + 2x} \right)}}{{1 - 2x\left( {1 + 2x} \right)}}\]
 \[ \Rightarrow \dfrac{{2 + 4x + \dfrac{{11}}{6}x + \dfrac{{22}}{6}{x^2}}}{{1 - 4{x^2}}} = \dfrac{{2 + \dfrac{{35}}{6}x + \dfrac{{11}}{3}{x^2}}}{{1 - 4{x^2}}}\]
Neglecting the $ {x^2} $ term both in the numerator and the denominator, as given below:
 \[ \Rightarrow \dfrac{{2 + \dfrac{{35}}{6}x + \dfrac{{11}}{3}{x^2}}}{{1 - 4{x^2}}} = \dfrac{{2 + \dfrac{{35}}{6}x + \dfrac{{11}}{3}\left( 0 \right)}}{{1 - 4\left( 0 \right)}}\]
 \[ \Rightarrow 2 + \dfrac{{35}}{6}x\]

Final answer: On neglecting the term $ {x^2} $ and higher terms of $ x $ , $ \dfrac{{\sqrt {1 + x} + \sqrt[3]{{1 + 4x}}}}{{\left( {1 + {x^2}} \right)\sqrt[3]{{{{\left( {1 - 3x} \right)}^2}}}}} = 2 + \dfrac{{35}}{6}x $

Note:
Please note that while solving the given problem, as given in the question itself that we have to neglect the higher power of $ x $ , here the power of the $ x $ should be an integer, whenever the power of $ x $ is fractional then, the fractional index pension concept is applied to the given expression, understand the difference here.