The maximum number of positive real roots which f(x) = 0, (where f(x) is a polynomial) can have is equal to the change of sign in the expression f(x). The maximum number of negative real roots which f(x) = 0 can have is equal to the change of sign in the expression
f(–x) for e.g. f(x) = x 3 – x + 1 have two change of sign this implies f(x) = 0 can have maximum of two positive real roots. Further f(–x) = – x 3 + x + 1 which has 1 sign change so f(x) = 0 has at most one negative root.
(i) If both the critical points of f(x) = ax 3 + bx 2 + cx + d are negative then
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Ans.
(i)
Sol. f ′ (x) = 3ax 2 + 2bx + c
f ′ (x) = 0 has both roots negative if a, b, c are of same sign
(ii)
Sol. If a, b, c are of same sign (say positive) then for maximum number of real roots 'd' must be negative. So the equation has atmost two positive roots and no negative root. Hence at least four imaginary roots.
(iii)
Sol. Clear from the adjacent graph that f(x) = 0 has at least two negative roots
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