The polynomial equation x3 – 3ax2 + (27a2 + 9)x + 2016 = 0 has -(A)�

1.	The polynomial equation x3 – 3ax2 + (27a2 + 9)x + 2016 = 0 has -(A) exactly one real root for any real a(B) three real roots for any real a(C) three real roots for any a 0, and exactly one real root for any a < 0(D) three real roots for any a ≤ 0, and exactly one real root for any a > 0
Answer» Correct option (A) exactly one real root for any real a Explanation: f'(x) = 3x₂ – 6ax + 27a² + 9 = 3[x² – 2ax + 9a² + 3] = 3((x – a)² + 8a² + 3) f'(x) is + ve for x ∈ R so f(x) is monotonic↑ for x ∈ R.

The polynomial equation x3 – 3ax2 + (27a2 + 9)x + 2016 = 0 has -(A) exactly one real root for any real a(B) three real roots for any real a(C) three real roots for any a 0, and exactly one real root for any a < 0(D) three real roots for any a ≤ 0, and exactly one real root for any a > 0

Answer»

Correct option (A) exactly one real root for any real a

Explanation:

f'(x) = 3x₂ – 6ax + 27a² + 9

= 3[x² – 2ax + 9a² + 3] = 3((x – a)² + 8a² + 3)

f'(x) is + ve for x ∈ R so f(x) is monotonic↑

for x ∈ R.

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The polynomial equation x3 – 3ax2 + (27a2 + 9)x + 2016 = 0 has -(A) exactly one real root for any real a(B) three real roots for any real a(C) three real roots for any a 0, and exactly one real root for any a &lt; 0(D) three real roots for any a ≤ 0, and exactly one real root for any a &gt; 0

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The polynomial equation x3 – 3ax2 + (27a2 + 9)x + 2016 = 0 has -(A) exactly one real root for any real a(B) three real roots for any real a(C) three real roots for any a 0, and exactly one real root for any a < 0(D) three real roots for any a ≤ 0, and exactly one real root for any a > 0