Question
Question: If for \(x\ne 0\), \(af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5\), where \(a\ne...
If for x=0, af(x)+bf(x1)=x1−5, where a=b, then 1∫2xf(x)dx is equal to
A. 9(a2−b2)b−9a
B. b(a2−b2)b−9a
C. 6(a2−b2)b−9a
D. None of these
Solution
Hint : We first express the main function f(x). We replace the value of x with x1 and place in the equation. From two equations we find the general form of the f(x). We then complete the integration part to find the solution after multiplying with x.
Complete step-by-step answer :
We first use the given function of af(x)+bf(x1)=x1−5 to find the main function of f(x).
We replace the value of x with x1 and place in the equation of af(x)+bf(x1)=x1−5.
So, af(x1)+bf(1/x1)=1/x1−5⇒af(x1)+bf(x)=x−5.
We find the value of f(x1) from the equation of af(x1)+bf(x)=x−5.
af(x1)+bf(x)=x−5⇒f(x1)=ax−5−bf(x)
We replace the value in the equation of af(x)+bf(x1)=x1−5.
⇒a2f(x)−b2f(x)=a(x1−5)−b(x−5)
Now taking f(x) common we have,
⇒f(x)[a2−b2]=a(x1−5)−b(x−5)
⇒f(x)[a2−b2]=a(x1−5)−b(x−5)
Now dividing by a2−b2 on both sides
⇒f(x)=a2−b2a(x1−5)−a2−b2b(x−5)
We found the main function of f(x).
We multiply with x to find the integral form for 1∫2xf(x)dx.