Question

If a real value of fraction f of a real variable x is such that \[\dfrac{1}{{[(1{\text{ }} + {\text{ }}{x^2}){\text{ }}(1{\text{ }} + {\text{ }}x)]}} = \dfrac{A}{{[1 + x]}} + \dfrac{{f(x)}}{{(1{\text{ }} + {\text{ }}{x^2})}}\]
1) \[\dfrac{{[1 - x]}}{2}\]
2) \[\dfrac{{[{x^2} + 1]}}{2}\]
3) \[1 - x\]
4) None of these

Answer 1

Hint: We are given a real variable x and we need to find the value of f(x). First, we will solve the given equation by taking LCM and then simplify it. We will assume any value of x, let say x = -1 and then substitute it in the simplify equation to get the value of A. After getting this value of A, we will again substitute this value in the simplify equation to the final output.

Complete step-by-step answer:
Given that,
\[\dfrac{1}{{[(1{\text{ }} + {\text{ }}{x^2}){\text{ }}(1{\text{ }} + {\text{ }}x)]}} = \dfrac{A}{{[1 + x]}} + \dfrac{{f(x)}}{{(1{\text{ }} + {\text{ }}{x^2})}}\]
Taking LCM of the RHS, we get,
\[ \Rightarrow \dfrac{1}{{[(1{\text{ }} + {\text{ }}{x^2}){\text{ }}(1{\text{ }} + {\text{ }}x)]}} = \dfrac{{A(1{\text{ }} + {\text{ }}{x^2}) + f(x)(1 + x)}}{{(1{\text{ }} + {\text{ }}x)(1{\text{ }} + {\text{ }}{x^2})}}\]
\[ \Rightarrow \dfrac{{(1{\text{ }} + {\text{ }}x)(1{\text{ }} + {\text{ }}{x^2})}}{{[(1{\text{ }} + {\text{ }}{x^2}){\text{ }}(1{\text{ }} + {\text{ }}x)]}} = A(1{\text{ }} + {\text{ }}{x^2}) + f(x)(1 + x)\]
\[ \Rightarrow 1 = A(1{\text{ }} + {\text{ }}{x^2}) + f(x)(1 + x)\] ---- (1)
Since, x is a real variable.
Let us take x = -1
Substituting this value of x in the equation (1), we will get,
\[ \Rightarrow 1 = A(1 + {( - 1)^2}) + f(x)(1 + ( - 1))\]
Removing the brackets, we will get,
\[ \Rightarrow 1 = A(1 + 1) + f(x)(1 - 1)\]
On evaluating this above equation, we will get,
\[ \Rightarrow 1 = A(2) + f(x)(0)\]
\[ \Rightarrow 1 = 2A + 0\]
\[ \Rightarrow 1 = 2A\]
\[ \Rightarrow A = \dfrac{1}{2}\]
Substituting the value of A in the equation (1), we will get,
\[ \Rightarrow 1 = \dfrac{1}{2}(1{\text{ }} + {\text{ }}{x^2}) + f(x)(1 + x)\]
On evaluating this above equation, we will get,
\[ \Rightarrow 2 = (1{\text{ }} + {\text{ }}{x^2}) + f(x)(1 + x)\]
\[ \Rightarrow 2 - 1{\text{ - }}{x^2} = f(x)(1 + x)\]
\[ \Rightarrow 1 - {x^2} = f(x)(1 + x)\]
\[ \Rightarrow \dfrac{{1 - {x^2}}}{{1 + x}} = f(x)\]
Rearrange this above equation, we will get,
\[ \Rightarrow f(x) = \dfrac{{1 - {x^2}}}{{1 + x}}\]
We know that, \[{a^2} - {b^2} = (a - b)(a + b)\].
Applying this rule, we will get,
\[ \Rightarrow f(x) = \dfrac{{(1 - x)(1 + x)}}{{1 + x}}\]
Simplify this, we will get,
\[ \Rightarrow f(x) = 1 - x\]
Hence, for a real value of fraction f of a real variable x, then the value of \[f(x) = 1 - x\] .
So, the correct answer is “Option 3”.

Note: If the highest power of the variable of an equation in one variable is 2, then that equation is called a quadratic equation. The quadratic equation will always have two roots. The nature of roots may be either real or imaginary. A quadratic polynomial, when equated to zero, becomes a quadratic equation. The values of x satisfying the equation are called the roots of the quadratic equation.