
Calculate the value of $\left\{ \left( 1+x+{{x}^{2}} \right)+\dfrac{{{x}^{3}}}{1-x} \right\}$ at $x=0.50$ \[\]
Answer
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Hint: The question is asking for the fractional part of the given function. Put the value $x=0.5$ in the function inside the bracket and then use the standard definition of the fractional part function to arrive at the result where you shall require the definition of the greatest integer function.
Complete step-by-step solution:
The given function is
$f\left( x \right)=\left\{ \left( 1+x+{{x}^{2}} \right)+\dfrac{{{x}^{3}}}{1-x} \right\}....(1)$
We shall denote the function inside the curly bracket as $g\left( x \right)=1+x+{{x}^{2}}+\dfrac{{{x}^{3}}}{1-x}$
Putting the value $x=0.5$ in $g\left( x \right)$ we get,
\[g\left( 0.5 \right)=1+0.5+{{0.5}^{2}}+\dfrac{{{0.5}^{3}}}{1-0.5}=1.75+.25=2\]
We get,
$g\left( 0.5 \right)=\left\{ 2 \right\}...(2)$
The definition of the fractional part function which returns the fractional part of any decimal number is
\[\left\{ x \right\}=x-\left[ x \right]\]
Where $ \left[ x \right]$ is the greatest integer function otherwise also known as step function.
The definition of the greatest integer function is given by $ \left[ x \right]=a $, if $x$ lies in the interval $a\le x\le b$. It means that the greatest integer function returns the greatest integer smaller than the number.
The value obtained to calculate fractional part is 2. So let us calculate the fractional part of 2 from the equation (2)
$f\left( 0.5 \right)=\left\{ 2 \right\}=2-\left[ 2 \right]$
Now we shall determine the greatest integer value for 2. Here the interval is \[\left[ 2,\infty \right)\]. So the greatest integer value is $\left[ 2 \right]=2$
$f\left( 0.5 \right)=\left\{ 2 \right\}=2-\left[ 2 \right]=2-2=0$
So the fractional part of the given expression is 0.
Note: You need to be careful while substituting the value in the given expression and also you need to take care of the confusion between the greatest function which returns a greatest integer less than the number and the smallest integer function which returns a greatest integer less than the number.
Complete step-by-step solution:
The given function is
$f\left( x \right)=\left\{ \left( 1+x+{{x}^{2}} \right)+\dfrac{{{x}^{3}}}{1-x} \right\}....(1)$
We shall denote the function inside the curly bracket as $g\left( x \right)=1+x+{{x}^{2}}+\dfrac{{{x}^{3}}}{1-x}$
Putting the value $x=0.5$ in $g\left( x \right)$ we get,
\[g\left( 0.5 \right)=1+0.5+{{0.5}^{2}}+\dfrac{{{0.5}^{3}}}{1-0.5}=1.75+.25=2\]
We get,
$g\left( 0.5 \right)=\left\{ 2 \right\}...(2)$
The definition of the fractional part function which returns the fractional part of any decimal number is
\[\left\{ x \right\}=x-\left[ x \right]\]
Where $ \left[ x \right]$ is the greatest integer function otherwise also known as step function.
The definition of the greatest integer function is given by $ \left[ x \right]=a $, if $x$ lies in the interval $a\le x\le b$. It means that the greatest integer function returns the greatest integer smaller than the number.
The value obtained to calculate fractional part is 2. So let us calculate the fractional part of 2 from the equation (2)
$f\left( 0.5 \right)=\left\{ 2 \right\}=2-\left[ 2 \right]$
Now we shall determine the greatest integer value for 2. Here the interval is \[\left[ 2,\infty \right)\]. So the greatest integer value is $\left[ 2 \right]=2$
$f\left( 0.5 \right)=\left\{ 2 \right\}=2-\left[ 2 \right]=2-2=0$
So the fractional part of the given expression is 0.
Note: You need to be careful while substituting the value in the given expression and also you need to take care of the confusion between the greatest function which returns a greatest integer less than the number and the smallest integer function which returns a greatest integer less than the number.
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