Question

If \[ y=\frac{ax+b}{cx+d} \] and \[ \frac{dx}{dy}=\frac{ad-bc}{Py^2+Qy+R} \] then \(P+Q+R=\)

• A. \((a+c)^2\)
• B. \((a-c)^2\)
• C. \(\dfrac{ad-bc}{a^2+c^2-2ac}\)
• D. \(\dfrac{1}{(a-c)^2}\)

Hint:

For fractional linear functions, first express \(x\) in terms of \(y\), then differentiate carefully using the quotient rule.

Answer 1

The Correct Option is B

Step 1: Express \(x\) in terms of \(y\).
Given, \[ y=\frac{ax+b}{cx+d} \] Cross multiplying, \[ y(cx+d)=ax+b \] \[ cxy+dy=ax+b \] Bring the terms containing \(x\) on one side, \[ cxy-ax=b-dy \] Taking \(x\) common, \[ x(cy-a)=b-dy \] Therefore, \[ x=\frac{b-dy}{cy-a} \] Multiplying numerator and denominator by \(-1\), \[ x=\frac{dy-b}{a-cy} \]

Step 2: Differentiate \(x\) with respect to \(y\).
Now, \[ x=\frac{dy-b}{a-cy} \] Using quotient rule, \[ \frac{dx}{dy} = \frac{(a-cy)\frac{d}{dy}(dy-b)-(dy-b)\frac{d}{dy}(a-cy)}{(a-cy)^2} \] Since, \[ \frac{d}{dy}(dy-b)=d \] and \[ \frac{d}{dy}(a-cy)=-c \] we get \[ \frac{dx}{dy} = \frac{d(a-cy)-(dy-b)(-c)}{(a-cy)^2} \] \[ \frac{dx}{dy} = \frac{ad-cdy+cdy-bc}{(a-cy)^2} \] \[ \frac{dx}{dy} = \frac{ad-bc}{(a-cy)^2} \]

Step 3: Compare with the given form.
Given, \[ \frac{dx}{dy}=\frac{ad-bc}{Py^2+Qy+R} \] From our result, \[ \frac{dx}{dy}=\frac{ad-bc}{(a-cy)^2} \] Therefore, \[ Py^2+Qy+R=(a-cy)^2 \] Expanding, \[ (a-cy)^2=a^2-2acy+c^2y^2 \] So, \[ Py^2+Qy+R=c^2y^2-2acy+a^2 \] Thus, \[ P=c^2,\quad Q=-2ac,\quad R=a^2 \]

Step 4: Find \(P+Q+R\).
\[ P+Q+R=c^2-2ac+a^2 \] \[ P+Q+R=a^2-2ac+c^2 \] \[ P+Q+R=(a-c)^2 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{(a-c)^2} \]