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Which of the following is correct for the nature of the roots x^5 – a0x^4 + 3ax^3 + bx^2 + cx + d = 0 if it is given that 2a0^2 < 15 and a0, a, b, c, d are real?(a) Can’t be real(b) Equal(c) Real(d) Depends on the value of xThe question was posed to me by my school teacher while I was bunking the class.The origin of the question is Second Order Derivative topic in chapter Limits and Derivatives of Mathematics – Class 11

Answer» RIGHT option is (a) Can’t be real

Easiest explanation: LET f(x) = x^5 – a0x^4 + 3ax^3 + bx^2 + cx + d so that,

Now, f’(x) = 5x^5 – 4a0x^3 + 9ax^2 + 2bx + c

And, f”(x) = 20x^3 – 12a0x^2 + 18ax + 2b

And, f”(x) = 60x^2 – 24a0x + 18a

= 6(10x^2 – 4a0x + 3a)

Now, DISCRIMINANT of 10x^2 – 4a0x + 3a = 16a0^2 – 4

And it is given 8(2a0^2 – 15a) < 0

Hence, root of f”’(x) can’t be real

=>all the roots can’t be real.


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