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a, b, c, d are roots of the equation n− n- n2−1 = 0. If f(n) = n−  n− n− n− n, then f(a) + f(b) + f(c) + f(d) = ?
1. 1
2. 0
3. -1
4. 2

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Correct Answer - Option 3 : -1

Calculation:

Let g(n) = n4 − n3 − n2 – 1 = 0 so we have a, b, c and d as the roots of g(n) so a4 − a3 − a2 – 1 = 0 etc.

f(n) = n2{n4 – n3 − n2 – 1 − (1/n)},

So, f(a) = a2(a4 − a3 − a2 − 1 – 1/a) = −a.

Hence, f(a) + f(b) + f(c) + f(d) = – (a + b + c + d) = –1.

If α , β, γ, δ are roots of a biquadratic equation  ax4+ bx3 + cx2 + dx + e = 0, then

⇒ α + β + γ + δ = -b/a

⇒ αβ + βγ + γδ +αγ + αδ + βδ  = c/a

⇒ αβγ + αγδ +αβδ + βγδ  = -d/a

⇒ αβγδ = e/a

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