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AP EAPCET 23-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 11 of 160

Marks: +1, -0

If

\alpha, \beta, \gamma

are roots of the equation

x^{3}+4 x-19=0

. Then the value of

\frac{a^{3}}{19-4\alpha}+\frac{\beta^{3}}{19-4\beta}+\frac{\gamma^{3}}{19-4\gamma}=

0
3
-3
2

Solution:

\alpha, \beta

and

\gamma

are roots of

x^{3}+4 x-19=0

Then,

\alpha+\beta+\gamma=0......(i)

\alpha\beta+\beta\gamma+g\alpha=4.....(ii)

\alpha\beta\gamma=19.....(iii)

From Eq. (ii),

\alpha(\beta+\gamma)+\beta\gamma=4

=\alpha(-\alpha)+\beta\gamma=4

[using Eq. (i)]

\Rightarrow -\alpha^{2}+\frac{19}{\alpha}=4

[using Eq. (iii)]

\Rightarrow -\alpha^{3}+19=4\alpha

or

\alpha^{3}=19-4\alpha

This gives

\frac{\alpha^{3}}{19-4\alpha}=1

...(iv)

Similarly, from Eq. (ii),

\beta(\alpha+\gamma)+\gamma\alpha=4

\Rightarrow \beta(-\beta)+\gamma\alpha=4

\text{[using Eq.(i)]}

\Rightarrow -\beta^{2}+\frac{19}{\beta}=4

\text{[using Eq. (iii)]}

\Rightarrow \beta^{3}=19-4\beta

\Rightarrow \frac{\beta^{3}}{19-4\beta}=1.....(v)

And, from Eq. (ii)

\gamma(\alpha+\beta)+\alpha\beta=4

\gamma(-\gamma)+\frac{19}{\gamma}=4

\Rightarrow \frac{\gamma^{3}}{19-4\gamma}=1......(vi)

Adding Eqs.(iv), (v) and (vi), we get

\frac{\alpha^{3}}{19-4\alpha}+\frac{\beta^{3}}{19-4\beta}+\frac{\gamma^{3}}{19-4\gamma}=3

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