Let f(x)=x13+x11+x9+x7+x5+x3+x+19. Then f(x)=0 has
A
13 real roots
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B
only one positive and only two negative real roots
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C
not more than one real root
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D
has two positive and one negative real root
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Solution
The correct option is C not more than one real root f′(x)=12x12+11x10+9x8+7x6+5x4+3x2+1>0∀x∈R So f(x) is an increasing function and power is odd there will be exactly one real root