Question

If \[\alpha ,\beta ,\gamma \] are the roots of cubic equation \[{x^3} - 3{x^2} + 2x + 4 = 0\] and \[y = 1 + \dfrac{\alpha }{{x - \alpha }} + \dfrac{{\beta x}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}} + \dfrac{{\gamma {x^2}}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)}}\] then the value of \[y\] at \[x = 2\], is
A.0
B.1
C.2
D.3

Answer 1

Hint: Here we will firstly solve the equation of \[y\] and simplify it by taking the common denominator from all the terms in the equation. Then we will use the concept of roots and replace the equation of the roots with the equation \[{x^3} - 3{x^2} + 2x + 4 = 0\]. We will then put the value of \[x = 2\] in the equation of \[y\] to get its value.

Complete step-by-step answer:
The given equation of \[y\] is \[y = 1 + \dfrac{\alpha }{{x - \alpha }} + \dfrac{{\beta x}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}} + \dfrac{{\gamma {x^2}}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)}}\].
First, we will simplify this equation of \[y\] by simply taking the common denominator and solve it. Therefore taking the common denominator in the first two terms of the equation, we get
\[ \Rightarrow y = \dfrac{{\left( {x - \alpha } \right) + \alpha }}{{x - \alpha }} + \dfrac{{\beta x}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}} + \dfrac{{\gamma {x^2}}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)}}\]
\[ \Rightarrow y = \dfrac{x}{{x - \alpha }} + \dfrac{{\beta x}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}} + \dfrac{{\gamma {x^2}}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)}}\]
Again taking the common denominator for the first two terms of the equation, we get
\[ \Rightarrow y = \dfrac{{x\left( {x - \beta } \right) + \beta x}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}} + \dfrac{{\gamma {x^2}}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)}} = \dfrac{{{x^2} - \beta x + \beta x}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}} + \dfrac{{\gamma {x^2}}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)}}\]
Simplifying the expression, we get
\[ \Rightarrow y = \dfrac{{{x^2}}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)}} + \dfrac{{\gamma {x^2}}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)}}\]
Now again taking the common denominator in the equation, we get

\[ \Rightarrow y = \dfrac{{{x^2}\left( {x - \gamma } \right) + \gamma {x^2}}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)}} = \dfrac{{{x^3} - \gamma {x^2} + \gamma {x^2}}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)}}\]
On further simplification, we get
\[ \Rightarrow y = \dfrac{{{x^3}}}{{\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)}}\]…………………….\[\left( 1 \right)\]
It is given that \[\alpha ,\beta ,\gamma \] are the roots of the equation \[{x^3} - 3{x^2} + 2x + 4 = 0\] which means that
\[{x^3} - 3{x^2} + 2x + 4 = \left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)\].
So substituting \[{x^3} - 3{x^2} + 2x + 4 = \left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)\] in the equation \[\left( 1 \right)\], we get
\[ \Rightarrow y = \dfrac{{{x^3}}}{{{x^3} - 3{x^2} + 2x + 4}}\]
Now we will find the value of \[y\] at \[x = 2\]. So we will put the value of \[x = 2\] in the equation of \[y\]. Therefore, we get
\[\begin{array}{l} \Rightarrow y = \dfrac{{{2^3}}}{{{2^3} - 3 \times {2^2} + 2 \times 2 + 4}}\\ \Rightarrow y = \dfrac{8}{{8 - 3 \times 4 + 4 + 4}}\end{array}\]
Multiplying the terms, we get
\[ \Rightarrow y = \dfrac{8}{{8 - 12 + 4 + 4}}\]
Adding the terms, we get
\[ \Rightarrow y = \dfrac{8}{{16 - 12}}\]
On further simplification, we get
\[ \Rightarrow y = \dfrac{8}{4} = 2\]
Hence, the value of \[y\] at \[x = 2\] is 2.
So, the option C is the correct option.

Note:Here we should note that the cubic equations are the equation in which the highest exponent of the variable is 3. Also in an equation the number of its roots is equal to the value of the highest exponent of the variable of that equation which means for a cubic equation there are three roots of the equation. Also we should know that the quadratic equation is the equation in which the highest exponent of the variable is 2. So there are only two roots of the quadratic equation.
Quadratic equation: \[a{x^2} + bx + c = 0\]
Cubic equation: \[a{x^3} + b{x^2} + cx + d = 0\]