Question

If $ y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{bx}}{{\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{c}{{\left( {x - c} \right)}} + 1 $ then $ \dfrac{{y'}}{y} = \dfrac{k}{x}\left( {\dfrac{a}{{a - x}} + \dfrac{b}{{b - x}} + \dfrac{c}{{c - x}}} \right) $ where $ k = $ ( $ a,b,c $ are non-real numbers and $ x \ne a,b,c $ ).
A. $ - 1 $
B. $ 1 $
C. $ 2 $
D. $ - 3 $

Answer 1

Hint: In order to find the value of $ k $ , initiate by simplifying the equation $ y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{bx}}{{\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{c}{{\left( {x - c} \right)}} + 1 $ into its simplest term possible, then differentiate it with respect to $ x $ .Then compare it with the second equation given and get the result.
Formula used:
 $ \log \dfrac{a}{b} = \log a - \log b $
 $ \log a.b = \log a + \log b $
 $ \log {a^n} = n\log a $
\[\dfrac{{d\log y}}{{dx}} = \dfrac{1}{y}\dfrac{{dy}}{{dx}}\]
 $ \dfrac{{d\log x}}{{dx}} = \dfrac{1}{x} $

Complete step by step solution:
We are given with an equation $ y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{bx}}{{\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{c}{{\left( {x - c} \right)}} + 1 $ .
Starting with simplifying the equation.
Solving the last two operands of the right-hand side, multiplying and dividing $ 1 $ by $ \left( {x - c} \right) $ in order to get the same denominator. And by this we get:
 $ y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{bx}}{{\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{c}{{\left( {x - c} \right)}} + \dfrac{{1\left( {x - c} \right)}}{{\left( {x - c} \right)}} $
Since, the last two have same denominator, so adding them, we get:
 $
   \Rightarrow y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{bx}}{{\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{c + x - c}}{{\left( {x - c} \right)}} \\
   \Rightarrow y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{bx}}{{\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{x}{{\left( {x - c} \right)}} \\
  $
Similarly, multiplying and dividing $ \dfrac{x}{{\left( {x - c} \right)}} $ by $ \left( {x - b} \right) $ in order to get the same denominator as it’s previous operand, and we get:
 $ \Rightarrow y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{bx}}{{\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{x}{{\left( {x - c} \right)}} \times \dfrac{{\left( {x - b} \right)}}{{\left( {x - b} \right)}} $
Simplifying the last operand, we get:
 $ \Rightarrow y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{bx}}{{\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{x\left( {x - b} \right)}}{{\left( {x - c} \right)\left( {x - b} \right)}} $
Opening the parenthesis of the last operand:
 $ \Rightarrow y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{bx}}{{\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{{x^2} - bx}}{{\left( {x - c} \right)\left( {x - b} \right)}} $
Since, the 2nd and 3rd operand have the same denominator, so adding the numerator terms and we get:
 $
   \Rightarrow y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{bx + {x^2} - bx}}{{\left( {x - b} \right)\left( {x - c} \right)}} \\
   \Rightarrow y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{{x^2}}}{{\left( {x - b} \right)\left( {x - c} \right)}} \;
  $
Similarly, multiplying and dividing $ \dfrac{{{x^2}}}{{\left( {x - b} \right)\left( {x - c} \right)}} $ by $ \left( {x - a} \right) $ , in order to have the same denominator as the previous operand:
 $ \Rightarrow y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{{x^2}}}{{\left( {x - b} \right)\left( {x - c} \right)}} \times \dfrac{{\left( {x - a} \right)}}{{\left( {x - a} \right)}} $
Simplifying the second operand:
 $ \Rightarrow y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{{x^2}\left( {x - a} \right)}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} $
Opening the parenthesis of the last operand:
 $ \Rightarrow y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{{x^3} - a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} $
Since, the 1st and 2nd operand have the same denominator, so adding the numerator terms and we get:
 $
   \Rightarrow y = \dfrac{{a{x^2} + {x^3} - a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} \\
   \Rightarrow y = \dfrac{{{x^3}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} \;
  $
Therefore, the simplified term of $ y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{bx}}{{\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{c}{{\left( {x - c} \right)}} + 1 $ is $ y = \dfrac{{{x^3}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} $ .
Taking $ \log $ both the sides of the obtained equation:
 $ \log y = \log \left( {\dfrac{{{x^3}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}}} \right) $
From the logarithmic rules, we know that $ \log \dfrac{a}{b} = \log a - \log b $ , $ \log a.b = \log a + \log b $ , and $ \log {a^n} = n\log a $ , so writing the above equation accordingly, we get:
 $ \log y = \log {x^3} - \log \left( {\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)} \right) $
Again, applying the logarithmic rule in $ \log \left( {\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)} \right) $ and $ \log {x^3} $ , we get:
\[
  \log y = \log {x^3} - \log \left( {\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)} \right) \\
   \Rightarrow \log y = 3\log x - \left( {\log \left( {x - a} \right) + \log \left( {x - b} \right) + \log \left( {x - c} \right)} \right) \;
 \]
Opening the outer parenthesis:
 $ \log y = 3\log x - \log \left( {x - a} \right) - \log \left( {x - b} \right) - \log \left( {x - c} \right) $
Differentiating both sides of the above equation with respect to $ x $ , we get:
 $ \dfrac{{d\log y}}{{dx}} = 3\dfrac{{d\log x}}{{dx}} - \dfrac{{d\log \left( {x - a} \right)}}{{dx}} - \dfrac{{d\log \left( {x - b} \right)}}{{dx}} - \dfrac{{d\log \left( {x - c} \right)}}{{dx}} $
From differentiation formula, we know that: \[\dfrac{{d\log y}}{{dx}} = \dfrac{1}{y}\dfrac{{dy}}{{dx}}\] and $ \dfrac{{d\log x}}{{dx}} = \dfrac{1}{x} $ , since $ a,b,c $ are constants, so they will create any difference in differentiation, and we get:
 $
   \Rightarrow \dfrac{{d\log y}}{{dx}} = 3\dfrac{{d\log x}}{{dx}} - \dfrac{{d\log \left( {x - a} \right)}}{{dx}} - \dfrac{{d\log \left( {x - b} \right)}}{{dx}} - \dfrac{{d\log \left( {x - c} \right)}}{{dx}} \\
   \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{3}{x} - \dfrac{1}{{x - a}} - \dfrac{1}{{x - b}} - \dfrac{1}{{x - c}} \;
  $
We can expand the term $ \dfrac{3}{x} $ as $ \dfrac{3}{x} = \dfrac{1}{x} + \dfrac{1}{x} + \dfrac{1}{x} $ , so expanding the term and making a pair with other terms in a parenthesis, and we get:
 $ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{3}{x} - \dfrac{1}{{x - a}} - \dfrac{1}{{x - b}} - \dfrac{1}{{x - c}} $
 $ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {\dfrac{1}{x} - \dfrac{1}{{x - a}}} \right) + \left( {\dfrac{1}{x} - \dfrac{1}{{x - b}}} \right) + \left( {\dfrac{1}{x} - \dfrac{1}{{x - c}}} \right) $
Solving the parenthesis, for each operand, and we can do it so by making the denominator common by multiplying denominator to the alternate numerator and simplifying, like for $ \left( {\dfrac{1}{x} - \dfrac{1}{{x - a}}} \right) = \left( {\dfrac{{\left( {x - a} \right)}}{{x\left( {x - a} \right)}} - \dfrac{x}{{x\left( {x - a} \right)}}} \right) = \dfrac{{x - a - x}}{{x\left( {x - a} \right)}} = \dfrac{{ - a}}{{x - a}} $ :
Similarly, simplifying for all other operands, we get:
\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {\dfrac{{\left( {x - a} \right)}}{{x\left( {x - a} \right)}} - \dfrac{x}{{x\left( {x - a} \right)}}} \right) + \left( {\dfrac{{\left( {x - b} \right)}}{{x\left( {x - b} \right)}} - \dfrac{x}{{\left( {x - b} \right)}}} \right) + \left( {\dfrac{{\left( {x - c} \right)}}{{x\left( {x - c} \right)}} - \dfrac{x}{{\left( {x - c} \right)}}} \right)\]
Solving the values inside parentheses:
\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {\dfrac{{x - a - x}}{{x\left( {x - a} \right)}}} \right) + \left( {\dfrac{{x - b - x}}{{x\left( {x - b} \right)}}} \right) + \left( {\dfrac{{x - c - x}}{{x\left( {x - c} \right)}}} \right)\]
\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {\dfrac{{ - a}}{{x\left( {x - a} \right)}}} \right) + \left( {\dfrac{{ - b}}{{x\left( {x - b} \right)}}} \right) + \left( {\dfrac{{ - c}}{{x\left( {x - c} \right)}}} \right)\]
Taking \[ - 1\] common from \[a,\left( {x - a} \right)\] and others, we get:
\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {\dfrac{{ - a}}{{ - x\left( {a - x} \right)}}} \right) + \left( {\dfrac{{ - b}}{{ - x\left( {b - x} \right)}}} \right) + \left( {\dfrac{{ - c}}{{ - x\left( {c - x} \right)}}} \right)\]
\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {\dfrac{a}{{x\left( {a - x} \right)}}} \right) + \left( {\dfrac{b}{{x\left( {b - x} \right)}}} \right) + \left( {\dfrac{c}{{x\left( {c - x} \right)}}} \right)\]
Taking \[\dfrac{1}{x}\] common outside the parentheses:
\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{x}\left( {\dfrac{a}{{\left( {a - x} \right)}} + \dfrac{b}{{\left( {b - x} \right)}} + \dfrac{c}{{\left( {c - x} \right)}}} \right)\]
Since, we can write \[\dfrac{{dy}}{{dx}} = y'\] , so writing this, and we get:
\[ \Rightarrow \dfrac{{y'}}{y} = \dfrac{1}{x}\left( {\dfrac{a}{{a - x}} + \dfrac{b}{{b - x}} + \dfrac{c}{{c - x}}} \right)\] ………(1)
The other equation give to us is $ \dfrac{{y'}}{y} = \dfrac{k}{x}\left( {\dfrac{a}{{a - x}} + \dfrac{b}{{b - x}} + \dfrac{c}{{c - x}}} \right) $ ……(2)
Comparing equation (1) and (2), we get:
 $ k = 1 $
Which matches with the option 2.
Therefore, If $ y = \dfrac{{a{x^2}}}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{{bx}}{{\left( {x - b} \right)\left( {x - c} \right)}} + \dfrac{c}{{\left( {x - c} \right)}} + 1 $ then $ \dfrac{{y'}}{y} = \dfrac{k}{x}\left( {\dfrac{a}{{a - x}} + \dfrac{b}{{b - x}} + \dfrac{c}{{c - x}}} \right) $ where $ k = 1 $ .
Hence, Option B is Correct.
So, the correct answer is “Option B”.

Note: It’s always preferred to solve step by step for ease rather than solving at once, otherwise there is a huge chance of error.
It’s important to take logarithmic function’s both sides, we cannot directly differentiate the equation obtained.