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Question

# If f:R→R and f(x) is a polynomial function of degree eleven and f(x)=0 has all real and distinct roots. Then the equation (f′(x))2−f(x)f′′(x)=0 has

A
no real roots
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B
10 real roots
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C
11 real roots
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D
21 real roots
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Solution

## The correct option is A no real rootsf(x)=a(x−x1)(x−x2)...(x−x11) lnf(x)=lna+ln(x−x1)+ln(x−x2)...+ln(x−x11) f′(x)f(x)=1x−x1+1x−x2 ... +1x−x11 f′′(x) f(x)−(f′(x))2(f(x))2 =−[1(x−x1)2+1(x−x2)2 .......+1(x−x11)2] Therefore f′′(x) f(x)−(f′(x))2 =−(f(x))2.[1(x−x1)2+ .......+1(x−x11)2] ⇒(f′(x))2−f(x) f′′(x)=0 has no real roots.

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