📖 generic · 12th TN - English Medium · MATHEMATICS-VOLUME 1 · Page 133question

3.9.2 Attainment of bounds · Part 3

Chapter 5: Chapter 3 · MATHEMATICS-VOLUME 1

sign changes for P x ( ) and P are zero and hence it has no positive roots and no negative roots. Clearly zero is not a root. Thus the polynomial has no real roots and hence all roots of the polynomial are imaginary roots. (ii) The number of sign changes for P x ( ) and P are and respectively.

Hence it has at most two positive roots and at most one negative root.Since the difference between number of sign changes in coefficients of P and the number of negative roots is even, we cannot have zero negative roots. So the number of negative roots is . Since the difference between number of sign changes in coefficient of P x ( ) and the number of positive roots must be even, we must have either zero or two positive roots. But as the sum of the coefficients is zero, is a root.

Thus we must have two and only two positive roots. Obviously the other two roots are imaginary numbers.

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