Answer
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Hint: In this question, we need to determine the value of the function at the value of x as 3 such that \[f\left( {f\left( x \right)} \right) \cdot \left( {1 + f\left( x \right)} \right) = - f\left( x \right)\] should be satisfied. For this, we will use the simple arithmetic operations and the rearrangements of the given expression to get the result.
Complete step-by-step answer:
The given expression is \[f\left( {f\left( x \right)} \right) \cdot \left( {1 + f\left( x \right)} \right) = - f\left( x \right)\]
Taking the term \[\left( {1 + f\left( x \right)} \right)\] to the right-hand side, we get
\[
\Rightarrow f\left( {f\left( x \right)} \right) \cdot \left( {1 + f\left( x \right)} \right) = - f\left( x \right) \\
\Rightarrow f\left( {f\left( x \right)} \right) = \dfrac{{ - f\left( x \right)}}{{1 + f\left( x \right)}} - - - - (i) \\
\]
According to the question, the function f(x) has been defined on a real number, so let the function f(x) be ‘x’ such that x belongs to the set of the real numbers.
Substituting the value of the function f(x) as x in the equation (i), we get
\[
\Rightarrow f\left( {f\left( x \right)} \right) = \dfrac{{ - f\left( x \right)}}{{1 + f\left( x \right)}} \\
\Rightarrow f(x) = \dfrac{{ - x}}{{1 + x}} - - - - (ii) \\
\]
Now, we have been asked to determine the value of the function at x=3. So, substitute x=3 in the equation (ii) to evaluate the value of the function.
\[
\Rightarrow f(x) = \dfrac{{ - x}}{{1 + x}} \\
\Rightarrow f(3) = \dfrac{{ - 3}}{{1 + 3}} \\
\]
Simplifying the above equation, we get
\[
\Rightarrow f(3) = \dfrac{{ - 3}}{{1 + 3}} \\
= \dfrac{{ - 3}}{4} \\
\]
Hence, the value of the function f(x) at x=3 is given as \[\dfrac{{ - 3}}{4}\] such that the function f(x) has been defined on real numbers and \[f\left( {f\left( x \right)} \right) \cdot \left( {1 + f\left( x \right)} \right) = - f\left( x \right)\].
Option A is correct.
Note: It is interesting to note here that the range and the domain of the function include 3 and so as our results should come within the given range only. Real numbers consist of zero (0), the positive and negative integers (-3, -1, 2, 4), and all the fractional and decimal values in between (0.4, 3.1415927, 1/2). Real numbers are divided into rational and irrational numbers.
Complete step-by-step answer:
The given expression is \[f\left( {f\left( x \right)} \right) \cdot \left( {1 + f\left( x \right)} \right) = - f\left( x \right)\]
Taking the term \[\left( {1 + f\left( x \right)} \right)\] to the right-hand side, we get
\[
\Rightarrow f\left( {f\left( x \right)} \right) \cdot \left( {1 + f\left( x \right)} \right) = - f\left( x \right) \\
\Rightarrow f\left( {f\left( x \right)} \right) = \dfrac{{ - f\left( x \right)}}{{1 + f\left( x \right)}} - - - - (i) \\
\]
According to the question, the function f(x) has been defined on a real number, so let the function f(x) be ‘x’ such that x belongs to the set of the real numbers.
Substituting the value of the function f(x) as x in the equation (i), we get
\[
\Rightarrow f\left( {f\left( x \right)} \right) = \dfrac{{ - f\left( x \right)}}{{1 + f\left( x \right)}} \\
\Rightarrow f(x) = \dfrac{{ - x}}{{1 + x}} - - - - (ii) \\
\]
Now, we have been asked to determine the value of the function at x=3. So, substitute x=3 in the equation (ii) to evaluate the value of the function.
\[
\Rightarrow f(x) = \dfrac{{ - x}}{{1 + x}} \\
\Rightarrow f(3) = \dfrac{{ - 3}}{{1 + 3}} \\
\]
Simplifying the above equation, we get
\[
\Rightarrow f(3) = \dfrac{{ - 3}}{{1 + 3}} \\
= \dfrac{{ - 3}}{4} \\
\]
Hence, the value of the function f(x) at x=3 is given as \[\dfrac{{ - 3}}{4}\] such that the function f(x) has been defined on real numbers and \[f\left( {f\left( x \right)} \right) \cdot \left( {1 + f\left( x \right)} \right) = - f\left( x \right)\].
Option A is correct.
Note: It is interesting to note here that the range and the domain of the function include 3 and so as our results should come within the given range only. Real numbers consist of zero (0), the positive and negative integers (-3, -1, 2, 4), and all the fractional and decimal values in between (0.4, 3.1415927, 1/2). Real numbers are divided into rational and irrational numbers.
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