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.