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Question: Let a function is given as \(f\left( x \right) = \dfrac{{ax + b}}{{cx + d}},{\text{ then }}fof\left(...

Let a function is given as f(x)=ax+bcx+d, then fof(x)=x,f\left( x \right) = \dfrac{{ax + b}}{{cx + d}},{\text{ then }}fof\left( x \right) = x, provided that

A. d=−a B. d=a C. a=b=1 D. a=b=c=d=1  A.{\text{ }}d = - a \\\ B.{\text{ }}d = a \\\ C.{\text{ }}a = b = 1 \\\ D.{\text{ }}a = b = c = d = 1 \\\
Explanation

Solution

Hint: In order to solve such questions use the basic formula for function of function and then proceed to find an algebraic equation. Finally solve the algebraic equation found out from the function to get the answer.

Complete step-by-step answer:

Given that: f(x)=ax+bcx+d, and fof(x)=x,f\left( x \right) = \dfrac{{ax + b}}{{cx + d}},{\text{ and }}fof\left( x \right) = x,
As we know that fof\left( x \right) = f\left\\{ {f\left( x \right)} \right\\}
So, we will proceed by putting the value of f(x)f\left( x \right) in the above equation.
\Rightarrow fof\left( x \right) = f\left\\{ {f\left( x \right)} \right\\} \\\ \Rightarrow fof\left( x \right) = f\left\\{ {\dfrac{{ax + b}}{{cx + d}}} \right\\} \\\ \Rightarrow fof\left( x \right) = \dfrac{{a\left\\{ {\dfrac{{ax + b}}{{cx + d}}} \right\\} + b}}{{c\left\\{ {\dfrac{{ax + b}}{{cx + d}}} \right\\} + d}} \\\
Now let us solve the above equation by taking LCM and also we will replace  fof(x) by x,{\text{ }}fof\left( x \right){\text{ by }}x,
\Rightarrow fof\left( x \right) = \dfrac{{a\left\\{ {\dfrac{{ax + b}}{{cx + d}}} \right\\} + b}}{{c\left\\{ {\dfrac{{ax + b}}{{cx + d}}} \right\\} + d}} = x \\\ \Rightarrow \dfrac{{\dfrac{{a\left( {ax + b} \right) + b\left( {cx + d} \right)}}{{\left( {cx + d} \right)}}}}{{\dfrac{{c\left( {ax + b} \right) + d\left( {cx + d} \right)}}{{\left( {cx + d} \right)}}}} = x \\\ \Rightarrow \dfrac{{a\left( {ax + b} \right) + b\left( {cx + d} \right)}}{{c\left( {ax + b} \right) + d\left( {cx + d} \right)}} = x \\\
Further simplifying the above equation, we obtain:
⇒a(ax+b)+b(cx+d)=x[c(ax+b)+d(cx+d)] ⇒a2x+ab+bcx+bd=x[acx+bc+dcx+d2] ⇒a2x+bcx+ab+bd=acx2+bcx+dcx2+d2x ⇒(a2+bc)x+(ab+bd)=(ac+cd)x2+(bc+d2)x  \Rightarrow a\left( {ax + b} \right) + b\left( {cx + d} \right) = x\left[ {c\left( {ax + b} \right) + d\left( {cx + d} \right)} \right] \\\ \Rightarrow {a^2}x + ab + bcx + bd = x\left[ {acx + bc + dcx + {d^2}} \right] \\\ \Rightarrow {a^2}x + bcx + ab + bd = ac{x^2} + bcx + dc{x^2} + {d^2}x \\\ \Rightarrow \left( {{a^2} + bc} \right)x + \left( {ab + bd} \right) = \left( {ac + cd} \right){x^2} + \left( {bc + {d^2}} \right)x \\\
In order to obtain the relation between the terms, let us compare the coefficient of constant terms on both the sides
⇒ab+bd=0 ⇒bd=−ab ⇒d=−a  \Rightarrow ab + bd = 0 \\\ \Rightarrow bd = - ab \\\ \Rightarrow d = - a \\\
Hence, the relation d=−ad = - a holds true.
So option A is the correct option.

Note: In order to solve such questions related to function of function the formula for function of function is very important and must be remembered. With the help of problem statements try to obtain some algebraic equation, as it becomes easier to find the relation. Relation between the terms could also have been found out at the earlier steps by direct substitution of the provided options. But comparing coefficients is a better way.