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Question: If \[f\left( x \right) = \dfrac{{ax + b}}{{cx + d}}\] then \[fof\left( x \right) = x\] provided that...

If f(x)=ax+bcx+df\left( x \right) = \dfrac{{ax + b}}{{cx + d}} then fof(x)=xfof\left( x \right) = x provided that
A. d=-a
B. a=b=c=d=1
C. d=a
D. a=b=1

Explanation

Solution

We are given a function in the form of fraction. We are given a condition such that if the same function is placed in another function such that it is equal to x, we have to find the condition that satisfies the equation above. For that we will place the value of the function given in the place of x in the function itself. On solving it we will get the condition necessary for the satisfaction.

Complete step-by-step solution:
Given that,
f(x)=ax+bcx+df\left( x \right) = \dfrac{{ax + b}}{{cx + d}}
Now we have to find fof(x)fof\left( x \right) that is nothing but f(f(x))f\left( {f\left( x \right)} \right) that is we will replace the value of x by the function given.
f(f(x))=a(ax+bcx+d)+bc(ax+bcx+d)+df\left( {f\left( x \right)} \right) = \dfrac{{a\left( {\dfrac{{ax + b}}{{cx + d}}} \right) + b}}{{c\left( {\dfrac{{ax + b}}{{cx + d}}} \right) + d}}
Now we will solve these ratios or fractions.
First we will take the LCM as well as we will multiply the brackets,
f(f(x))=a(ax+b)+b(cx+d)cx+dc(ax+b)+d(cx+d)cx+df\left( {f\left( x \right)} \right) = \dfrac{{\dfrac{{a\left( {ax + b} \right) + b\left( {cx + d} \right)}}{{cx + d}}}}{{\dfrac{{c\left( {ax + b} \right) + d\left( {cx + d} \right)}}{{cx + d}}}}
f(f(x))=a2x+ab+bcx+bdcx+dacx+bc+cdx+d2cx+df\left( {f\left( x \right)} \right) = \dfrac{{\dfrac{{{a^2}x + ab + bcx + bd}}{{cx + d}}}}{{\dfrac{{acx + bc + cdx + {d^2}}}{{cx + d}}}}
Now cancelling the denominators of both the numerator and denominator,
f(f(x))=a2x+bcx+ab+bdacx+cdx+bc+d2f\left( {f\left( x \right)} \right) = \dfrac{{{a^2}x + bcx + ab + bd}}{{acx + cdx + bc + {d^2}}}
Taking x common from the first two terms of numerator and first two terms of denominator.
f(f(x))=x(a2+bc)+b(a+d)x(ac+cd)+bc+d2f\left( {f\left( x \right)} \right) = \dfrac{{x\left( {{a^2} + bc} \right) + b\left( {a + d} \right)}}{{x\left( {ac + cd} \right) + bc + {d^2}}}
Now this is a simplified form. We will take the conditions from the options one by one that should satisfy the given condition.
fof(x)=xfof\left( x \right) = x
So we will take first condition d=-a
Putting this in the equation above,
f(f(x))=x(a2+bc)+b(a−a)x(ac−ac)+bc+(−a)2f\left( {f\left( x \right)} \right) = \dfrac{{x\left( {{a^2} + bc} \right) + b\left( {a - a} \right)}}{{x\left( {ac - ac} \right) + bc + {{\left( { - a} \right)}^2}}}
On solving this we get,
f(f(x))=x(a2+bc)bc+a2f\left( {f\left( x \right)} \right) = \dfrac{{x\left( {{a^2} + bc} \right)}}{{bc + {a^2}}}
On cancelling the common bracket we get,
f(f(x))=xf\left( {f\left( x \right)} \right) = x
This is the condition so given. Thus the condition given in option A is correct.

Note: Note that, here we just have to put the given function in place of x because there is no other function like g(x) that is to be placed as in general f(g(x)).
Note that, we chose the first condition at random like we used trial and error method. It is not like every time the first attempt will be the correct one. The only thing is we should satisfy the equation at its maximum possible terms and check for the options.
Also note that sometimes we miss or shuffle plus and product signs because of which we can miss the calculation. Note that anything multiplied by zero is zero but anything added to zero is that number itself.