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KEAM 2016 Math Paper

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Question : 13 of 120

Marks: +1, -0

If the roots of the quadratic equation

mx^3-nx+k=0

are tan 330 and tan 120 then the value of

\frac{m+n+k}{m}

is equal to

0
1
2
3
4

Solution:

Given, quadratic equation is

m x^{2}-n x+k=0

Roots of the equation are

\tan 33^{\circ}

and

\tan 12^{\circ}

\therefore \;\; \tan 33^{\circ}+\tan 12^{\circ}=\frac{n}{m}

.....(i)

and

\;\; \tan 33^{\circ} \times \tan 12^{\circ}=\frac{k}{m}

.....(ii)

Value of

\;\frac{2 m+n+k}{m}

is

\;\frac{2 m+n+k}{m}=\;\frac{2 m}{m}+\;\frac{n}{m}+\;\frac{k}{m}

=2+(\tan 33^{\circ}+\tan 12^{\circ})

+(\tan 33^{2} \times \tan 12^{\circ})

.....(iii)

Let

\;\; (\tan 45^{\circ})=\tan (33^{\circ}+12^{\circ})

1=\;\frac{\tan 33^{\circ}+\tan 12^{\circ}}{1-\tan 33^{\circ} \tan 12^{\circ}}

\Rightarrow 1-\tan 33^{\circ} \tan 12^{\circ}=\tan 33^{\circ}+\tan 12^{\circ}

\Rightarrow \;\; \tan 33^{\circ}+\tan 12^{\circ}+\tan 33^{\circ} \times \tan 12^{\circ}=1

.....(iv)

By putting the value from Eq. (iv) into Eq. (in)

\;\frac{2 m+n+k}{m}=2+1=3

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