Answer
397.2k+ views
Hint: Take the given expression, as the given function is with respect to “x” then create the expression in the form of $ \dfrac{1}{x} $ and then simplify both the equation finding the correlation between the two for the resultant answer.
Complete step-by-step answer:
Take the given expression –
$ af(x) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5 $ .... (A)
Replace $ x $ by $ \left( {\dfrac{1}{x}} \right) $ in the above equation –
$ \Rightarrow af\left( {\dfrac{1}{x}} \right) + bf\left( x \right) = x - 5 $ .... (B)
Multiply Equation (A) with “a” on both the sides of the equation –
$ a\left[ {af(x) + bf\left( {\dfrac{1}{x}} \right)} \right] = a\left[ {\dfrac{1}{x} - 5} \right] $
Simplify the above equation –
$ \Rightarrow \left[ {{a^2}f(x) + abf\left( {\dfrac{1}{x}} \right)} \right] = \left[ {\dfrac{a}{x} - 5a} \right] $ ..... (C)
Multiply Equation (B) with “b” on both the sides of the equation –
$ \Rightarrow b\left[ {af\left( {\dfrac{1}{x}} \right) + bf\left( x \right)} \right] = b\left[ {x - 5} \right] $
Simplify the above equation –
$ \Rightarrow \left[ {abf\left( {\dfrac{1}{x}} \right) + {b^2}f\left( x \right)} \right] = \left[ {bx - 5b} \right] $ ..... (D)
Subtract equation (D) from the equation (C)
$ \Rightarrow \left[ {{a^2}f(x) + abf\left( {\dfrac{1}{x}} \right)} \right] - \left[ {abf\left( {\dfrac{1}{x}} \right) + {b^2}f\left( x \right)} \right] = \left[ {\dfrac{a}{x} - 5a} \right] - \left[ {bx - 5b} \right] $
When you open the brackets and if there is a negative sign outside it, then the sign of all the terms inside the bracket changes. Positive term becomes negative and negative term becomes positive.
$ \Rightarrow {a^2}f(x) + abf\left( {\dfrac{1}{x}} \right) - abf\left( {\dfrac{1}{x}} \right) - {b^2}f\left( x \right) = \dfrac{a}{x} - 5a - bx + 5b $
Make pair of like terms in the above equation –
$ \Rightarrow \underline {{a^2}f(x) - {b^2}f\left( x \right)} + \underline {abf\left( {\dfrac{1}{x}} \right) - abf\left( {\dfrac{1}{x}} \right)} = \dfrac{a}{x} - 5a - bx + 5b $
Like terms with equal values and opposite sign cancel each other.
$ \Rightarrow \left( {{a^2} - {b^2}} \right)f\left( x \right) = \dfrac{a}{x} - 5a - bx + 5b $
The above equation can be written as –
$ \Rightarrow \left( {{a^2} - {b^2}} \right)f\left( x \right) = \left[ {\dfrac{a}{x} - 5a - bx + 5b} \right] $
When the term multiplicative on one side is moved to the opposite side then it goes to the denominator on the opposite side.
$ \Rightarrow f\left( x \right) = \dfrac{{\left[ {\dfrac{a}{x} - 5a - bx + 5b} \right]}}{{\left( {{a^2} - {b^2}} \right)}} $
This is the required solution.
So, the correct answer is “ $ \Rightarrow f\left( x \right) = \dfrac{{\left[ {\dfrac{a}{x} - 5a - bx + 5b} \right]}}{{\left( {{a^2} - {b^2}} \right)}} $ ”.
Note: Be very careful while simplifying the terms. When you open the brackets and there is a negative sign outside the bracket then the sign of the terms inside the bracket will be changed. Positive terms will become negative and negative terms become positive. Also, remember when there is a positive sign outside the bracket then the sign of the terms inside the bracket do not change.
Complete step-by-step answer:
Take the given expression –
$ af(x) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5 $ .... (A)
Replace $ x $ by $ \left( {\dfrac{1}{x}} \right) $ in the above equation –
$ \Rightarrow af\left( {\dfrac{1}{x}} \right) + bf\left( x \right) = x - 5 $ .... (B)
Multiply Equation (A) with “a” on both the sides of the equation –
$ a\left[ {af(x) + bf\left( {\dfrac{1}{x}} \right)} \right] = a\left[ {\dfrac{1}{x} - 5} \right] $
Simplify the above equation –
$ \Rightarrow \left[ {{a^2}f(x) + abf\left( {\dfrac{1}{x}} \right)} \right] = \left[ {\dfrac{a}{x} - 5a} \right] $ ..... (C)
Multiply Equation (B) with “b” on both the sides of the equation –
$ \Rightarrow b\left[ {af\left( {\dfrac{1}{x}} \right) + bf\left( x \right)} \right] = b\left[ {x - 5} \right] $
Simplify the above equation –
$ \Rightarrow \left[ {abf\left( {\dfrac{1}{x}} \right) + {b^2}f\left( x \right)} \right] = \left[ {bx - 5b} \right] $ ..... (D)
Subtract equation (D) from the equation (C)
$ \Rightarrow \left[ {{a^2}f(x) + abf\left( {\dfrac{1}{x}} \right)} \right] - \left[ {abf\left( {\dfrac{1}{x}} \right) + {b^2}f\left( x \right)} \right] = \left[ {\dfrac{a}{x} - 5a} \right] - \left[ {bx - 5b} \right] $
When you open the brackets and if there is a negative sign outside it, then the sign of all the terms inside the bracket changes. Positive term becomes negative and negative term becomes positive.
$ \Rightarrow {a^2}f(x) + abf\left( {\dfrac{1}{x}} \right) - abf\left( {\dfrac{1}{x}} \right) - {b^2}f\left( x \right) = \dfrac{a}{x} - 5a - bx + 5b $
Make pair of like terms in the above equation –
$ \Rightarrow \underline {{a^2}f(x) - {b^2}f\left( x \right)} + \underline {abf\left( {\dfrac{1}{x}} \right) - abf\left( {\dfrac{1}{x}} \right)} = \dfrac{a}{x} - 5a - bx + 5b $
Like terms with equal values and opposite sign cancel each other.
$ \Rightarrow \left( {{a^2} - {b^2}} \right)f\left( x \right) = \dfrac{a}{x} - 5a - bx + 5b $
The above equation can be written as –
$ \Rightarrow \left( {{a^2} - {b^2}} \right)f\left( x \right) = \left[ {\dfrac{a}{x} - 5a - bx + 5b} \right] $
When the term multiplicative on one side is moved to the opposite side then it goes to the denominator on the opposite side.
$ \Rightarrow f\left( x \right) = \dfrac{{\left[ {\dfrac{a}{x} - 5a - bx + 5b} \right]}}{{\left( {{a^2} - {b^2}} \right)}} $
This is the required solution.
So, the correct answer is “ $ \Rightarrow f\left( x \right) = \dfrac{{\left[ {\dfrac{a}{x} - 5a - bx + 5b} \right]}}{{\left( {{a^2} - {b^2}} \right)}} $ ”.
Note: Be very careful while simplifying the terms. When you open the brackets and there is a negative sign outside the bracket then the sign of the terms inside the bracket will be changed. Positive terms will become negative and negative terms become positive. Also, remember when there is a positive sign outside the bracket then the sign of the terms inside the bracket do not change.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)