Question

If x is very large, positive and $\sqrt[3]{{{x}^{3}}+1}=x+\dfrac{a}{{{x}^{2}}}+\dfrac{b}{{{x}^{5}}}$ , then
A. $a+b=0$
B. $3a+b=0$
C. $a+3b=0$
D. $a-3b=0$

Answer 1

Hint: At first, we take ${{x}^{3}}$ common from the entire expression. The expression becomes $x\sqrt[3]{1+\dfrac{1}{{{x}^{3}}}}$ . After that, we expand the expression within the bracket according to Taylor series expansion. After a little arithmetic simplification, the final expression becomes $x+\dfrac{1}{3{{x}^{2}}}-\dfrac{1}{9{{x}^{5}}}$ . Comparing it with $x+\dfrac{a}{{{x}^{2}}}+\dfrac{b}{{{x}^{5}}}$ , we get, $a=\dfrac{1}{3},b=-\dfrac{1}{9}$ and eventually $a+3b=0$ .

Complete step-by-step answer:
The expression that we are given in this problem is $\sqrt[3]{{{x}^{3}}+1}$ . This expression can be expressed as a series summation only if we implement the Taylor series expansion. In mathematics, the Taylor series of a function is an infinite sum of terms that are expressed in terms of the function’s derivatives at a single point. For most common functions, the function and the sum of its Taylor’s series are equal near this point. The Taylor series expansion of a function at $x=0$ is,
$f\left( x \right)=f\left( 0 \right)+\dfrac{f'\left( 0 \right)}{1!}+\dfrac{f''\left( 0 \right)}{2!}+\dfrac{f'''\left( 0 \right)}{3!}+...$
We now manipulate the expression $\sqrt[3]{{{x}^{3}}+1}$ to prepare it form Taylor series expansion. We take ${{x}^{3}}$ common from the entire expression and get,
$\Rightarrow x\sqrt[3]{1+\dfrac{1}{{{x}^{3}}}}$
This can be written as \[\Rightarrow x{{\left( 1+\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{1}{3}}}\] . The expression inside the bracket, that is, \[1+\dfrac{1}{{{x}^{3}}}\] can be expanded in accordance with Taylor series expansion as,
$\Rightarrow x\left( 1+\dfrac{1}{3{{x}^{3}}}-\dfrac{2}{9.2!{{x}^{6}}}+... \right)$
This can be simplified as $\Rightarrow x+\dfrac{1}{3{{x}^{2}}}-\dfrac{1}{9{{x}^{5}}}+...$
Since it is said that the value of x is very large, we can ignore the higher order terms as those tend to smaller and smaller values. So, the expansion becomes $x+\dfrac{1}{3{{x}^{2}}}-\dfrac{1}{9{{x}^{5}}}$ . Comparing it with the given equation $x+\dfrac{a}{{{x}^{2}}}+\dfrac{b}{{{x}^{5}}}$ , we get, $a=\dfrac{1}{3},b=-\dfrac{1}{9}$ .
Thus, we can conclude that $a+3b=0$ which is option c.

So, the correct answer is “Option C”.

Note: In series expansions, we must be careful of the signs of the terms and should avoid messing with them. Also, since the expression involves fractional terms, we must be extra cautious. We can also expand binomial expansion instead of Taylor series expansion.