A non zero polynomial f(x) with real coefficients satisfy f(x) = f′(x)f′′(x) and f(0) = 96. If 'a' is the coefficient of highest degree term of f(x), then
A
a=118
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B
f1(−6)=6
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C
f11(3)=5
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D
f(1)=10
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Solution
The correct options are Aa=118 Cf11(3)=5 Let degree of f(x) be n. ∴n=(n−1)+(n−2)⇒n=3 Hence f(x) =ax3+bx2+cx+96[∵f(0)=96] Apply f(x) = f′(x).f′′(x)⇒a=118,b=2,c=24