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Q.

If α,β,γ are the roots of cubic equation x3-3x2+2x+4=0 and 


y=1+x-+βx(x-α)(x-β)+γx2(x-α)(x-β)(x-γ)  then the value of y at x=2 is


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a

0

b

1

c

2

d

3 

answer is B.

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Detailed Solution

The given equation of y is y=1+x-+βx(x-α)(x-β)+γx2(x-α)(x-β)(x-γ)
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
y=(x-)x-+βx(x-α)(x-β)+γx2(x-α)(x-β)(x-γ)
 y=xx-+βx(x-α)(x-β)+γx2(x-α)(x-β)(x-γ)
Again taking the common denominator for the first two terms of the equation, we get
y=xx-β+βx(x-α)(x-β)+γx2(x-α)(x-β)(x-γ)
= x2-βx+βx(x-α)(x-β)+γx2(x-α)(x-β)(x-γ)
Simplifying the expression, we get
y=x2(x-α)(x-β)+γx2(x-α)(x-β)(x-γ)
Now again taking the common denominator in the equation, we get
 y=x2x-γ+γx2(x-α)(x-β)(x-γ)=x3-γx2+γx2(x-α)(x-β)(x-γ)

On further simplification, we get
y=x3(x-α)(x-β)(x-γ)…………………….(1)
It is given that α,β,γ are the roots of the equation x3-3x2+2x+4=0
 which means that x3-3x2+2x+4=(x−α)(x−β)(x−γ).
So substituting x3-3x2+2x+4=0=(x−α)(x−β)(x−γ)
 in the equation (1), we get
y=x3x3-3x2+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
y=2323-3×22+2×2+4
y=88-3×4+4+4
Multiplying the terms, we get
y=88-12+4+4
Adding the terms, we get
y=816-12
On further simplification, we get
y=84=2
Hence, the value of y at x=2 is 2.
So, the option 3 is the correct option
 
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If α,β,γ are the roots of cubic equation x3-3x2+2x+4=0 and y=1+∝x-∝+βx(x-α)(x-β)+γx2(x-α)(x-β)(x-γ)  then the value of y at x=2 is