If x ∈ R, then roots of equation x 4 + 4x 3 – 8x 2 + k = 0 when
(i) k ∈ [0, 3] are-
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Ans.
(i)
Sol. x 4 + 4x 3 – 8x 2 + k = 0
x 4 + 4x 3 – 8x 2 = – k
Let f(x) = x 4 + 4x 3 –8x 2 & g(x) = –k
= x 2 (x 2 + 4x –8)
When 0 ≤ k ≤ 3 then – k ∈ [–3, 0]
∴ y = –k will cut graph at four real points.
(ii)
Sol . x 4 + 4x 3 – 8x 2 + k = 0
x 4 + 4x 3 – 8x 2 = – k
Let f(x) = x 4 + 4x 3 –8x 2 & g(x) = –k
= x 2 (x 2 + 4x –8)
k ∈ (3, 4]
then –k ∈ [–4, –3)
In this case y = –k cuts graph at two real points.
(iii)
Sol. x 4 + 4x 3 – 8x 2 + k = 0
x 4 + 4x 3 – 8x 2 = – k
Let f(x) = x 4 + 4x 3 –8x 2 & g(x) = –k
= x 2 (x 2 + 4x –8)
When k ∈ (4, ∞ )
⇒ –k ∈ (– ∞ , –4)
∴ No real root.
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