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Question: If\[f(x) = {x^3} + \alpha {x^2} + \beta x + \gamma \], where \[\alpha \],\[\beta \],\[\gamma \] are ...

Iff(x)=x3+αx2+βx+γf(x) = {x^3} + \alpha {x^2} + \beta x + \gamma , where α\alpha ,β\beta ,γ\gamma are rational numbers and two roots of f(x)=0f(x) = 0 and eccentricities of a parabola and a rectangular hyperbola, then α+β+γ\alpha + \beta + \gamma is equal to:
A.-1
B.0
C.1
D.2

Explanation

Solution

Hint : By the given equation and condition we find the roots of the given polynomial. After finding the roots of the polynomial we solve this and comparing with the given polynomial, we obtain the α\alpha ,β\beta ,γ\gamma values.

Complete step-by-step answer :
We know that the eccentricity of a parabola is 11.
Also, the eccentricity of a rectangular hyperbola can be determined and it’s shown below:
Rectangular hyperbola is a hyperbola for which the asymptotes are perpendicular, also called an equilateral hyperbola or right hyperbola. This occurs when the si-major and semi- major are equal.
So a=ba = b
We know that the eccentricity of hyperbola
e=(a2+b2)12ae = \dfrac{{{{({a^2} + {b^2})}^{\dfrac{1}{2}}}}}{a}
In rectangular hyperbola we have a=ba = b
⇒e=(2a2)12a\Rightarrow e = \dfrac{{{{(2{a^2})}^{\dfrac{1}{2}}}}}{a}
⇒e=2aa\Rightarrow e = \dfrac{{\sqrt 2 a}}{a}
Hence, e=2e = \sqrt 2
Thus the root of polynomial f(x)f(x) is 11, 2\sqrt 2 and −2- \sqrt 2
Now f(x)=(x−1)(x−2)(x+2)f(x) = (x - 1)(x - \sqrt 2 )(x + \sqrt 2 )
Expanding the first and second brackets,
⇒f(x)=(x2−x2−x+2)(x+2)\Rightarrow f(x) = ({x^2} - x\sqrt 2 - x + \sqrt 2 )(x + \sqrt 2 )
Again expanding barkers, we get a polynomial,
⇒f(x)=x3−x22−x2+2x+x22−2x−2x+2\Rightarrow f(x) = {x^3} - {x^2}\sqrt 2 - {x^2} + \sqrt 2 x + {x^2}\sqrt 2 - 2x - \sqrt 2 x + 2
Cancelling terms and rearranging,
⇒f(x)=x3−x2−2x+2\Rightarrow f(x) = {x^3} - {x^2} - 2x + 2
Comparing with the coefficients given polynomial,
f(x)=x3+αx2+βx+γf(x) = {x^3} + \alpha {x^2} + \beta x + \gamma
We get that α=−1\alpha = - 1, β=−2\beta = - 2 and γ=2\gamma = 2
Now we need the value of α+β+γ\alpha + \beta + \gamma , and we know the individual values,
Substituting these we get,
⇒α+β+γ=−1−2+1\Rightarrow \alpha + \beta + \gamma = - 1 - 2 + 1
⇒α+β+γ=−1\Rightarrow \alpha + \beta + \gamma = - 1
So, the correct answer is “Option A”.

Note : In conic section, there is a locus of a point in which the distance to the point and the line are in the constant ratio. That ratio is known as eccentricity. It is denoted byee. We choose another root as −2- \sqrt 2 because if one root is irrational it occurs in a conjugate pair.