A) 0
B) 3
C) 4
D) 5
Correct Answer: B
Solution :
Let \[f(x)={{x}^{5}}-6{{x}^{2}}-4x+5=0\] Then the number of change of sign in \[f(x)\] is 2 therefore \[f(x)\] can have at most two positive real roots. Now, \[f(-x)=-{{x}^{5}}-6{{x}^{4}}+4x+5=0\] Then the number of change of sign is 1. Hence \[f(x)\]can have at most one negative real root. So that total possible number of real roots is 3.You need to login to perform this action.
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