Question

If \[af\left( x \right)+bf\left( \dfrac{1}{x} \right)=x-1,\text{ }x\ne 0,\text{ }\] and \[\text{a}\ne \text{b, }\] then \[f\left( 2 \right)=\]
 \[(1)\text{ }\left[ \dfrac{\left( 2a+b \right)}{\left( 2\left( {{a}^{2}}+{{b}^{2}} \right) \right)} \right] \]
 \[(2)\text{ }\left[ \dfrac{\left( 2a+b \right)}{\left( 2\left( {{a}^{2}}-{{b}^{2}} \right) \right)} \right] \]
 \[(3)\text{ }\left[ \dfrac{\left( a+2b \right)}{\left( {{a}^{2}}+{{b}^{2}} \right)} \right] \]
 \[(4)\text{ }\left[ \dfrac{\left( a+2b \right)}{\left( {{a}^{2}}+{{b}^{2}} \right)} \right] \]

Answer 1

Hint: In this type of question we have to put the required value of \[x\] in the given function. After substituting the value of \[x\] we will get equations and when we get those equations we have to check which arithmetic operation can be performed on those equations to get our required result.

Complete step-by-step answer:
Function defines a relationship between one set of variables to another set of variables i.e., the independent variable and dependent variable. This relationship is commonly symbolized as \[y=f(x)\] . \[f(x)\] is used to represent functions of the independent variable \[x\] . One set is of input and another set is of output in which each input is exactly related to only one output i.e. , no two or more outputs can be related to one input but one output can be related to two or more inputs. Functions are of different types such as one-one, many one. Function is also called a “map”.
The concept here is we think of eliminating the \[\dfrac{1}{x}\] term and this becomes possible because function is having reciprocal value of any \[x\] , then our equation will become way more simple as it becomes the equation of single variable i.e. is \[f\left( x \right)\]
Given equation: \[af\left( x \right)+bf\left( \dfrac{1}{x} \right)=x-1\]
We substitute the value of \[x\] by the value of \[2\] in the given equation to find the value of \[f(2)\]
After substituting the value of \[x\] equation becomes,
 \[af\left( 2 \right)+bf\left( \dfrac{1}{2} \right)=2-1\]
After solving further, the equation becomes
 \[af\left( 2 \right)+bf\left( \dfrac{1}{2} \right)=1..........Equation(1)\]
Now, we will substitute the value of \[\dfrac{1}{x}\] by \[2\] in the given equation to eliminate \[f\left( \dfrac{1}{2} \right)\]
After substituting the value of \[x\] , our equation becomes:
 \[af\left( \dfrac{1}{2} \right)+bf\left( 2 \right)=\dfrac{1}{2}-1\]
Now after solving, we get the equation as:
 \[af\left( \dfrac{1}{2} \right)+bf\left( 2 \right)=-\dfrac{1}{2}........(Equation2)\]
Now, we have \[E{{q}^{n}}(1)\] and \[E{{q}^{n}}(2)\] :
To find the value of \[f(2)\] we need to eliminate the value of \[f\left( \dfrac{1}{2} \right)\] by solving both equation
Now, we multiply \[E{{q}^{n}}(1)\] by \[a\] and \[E{{q}^{n}}(2)\] by \[b\] :
 \[af\left( 2 \right)+bf\left( \dfrac{1}{2} \right)=1\text{ }\times a\]
 \[af\left( \dfrac{1}{2} \right)+bf\left( 2 \right)=-\dfrac{1}{2}\text{ }\times \text{b}\]
After multiplication of the equation with suitable values, the equation becomes as \[{{a}^{2}}f\left( 2 \right)+abf\left( \dfrac{1}{2} \right)=a......Equation(3)\]
 \[abf\left( \dfrac{1}{2} \right)+{{b}^{2}}f\left( 2 \right)=-\dfrac{1}{2}\text{b}.........\text{Equation(4)}\]
Now we subtract \[E{{q}^{n}}(4)\] from the \[E{{q}^{n}}(3)\] ,
 \[{{a}^{2}}f\left( 2 \right)+abf\left( \dfrac{1}{2} \right)-abf\left( \dfrac{1}{2} \right)-{{b}^{2}}f\left( 2 \right)=a+\dfrac{1}{2}\text{b}\]
 \[{{a}^{2}}f\left( 2 \right)-{{b}^{2}}f\left( 2 \right)=a+\dfrac{1}{2}\text{b}\]
 \[\left( {{a}^{2}}-{{b}^{2}} \right)f\left( 2 \right)=a+\dfrac{1}{2}\text{b}\]
 \[\left( {{a}^{2}}-{{b}^{2}} \right)f\left( 2 \right)=\dfrac{2a+b}{2}\]
 \[f\left( 2 \right)=\dfrac{2a+b}{\left( 2\left( {{a}^{2}}-{{b}^{2}} \right) \right)}\]
So, the final answer is option \[2\] which is \[\dfrac{2a+b}{\left( 2\left( {{a}^{2}}-{{b}^{2}} \right) \right)}\] .
So, the correct answer is “Option 2”.

Note: Functions can be of many types, one of them is one-one onto in which each distinct input in domain relates to a unique element in range , and the set of outputs produced by the domain is equal to the range set, i.e. Co-domain is equal to the Range. And this type of function is invertible.