211/18.......
it'll b a cubic poly.......
I guess this hint shud help!!
A non zero polynomial with real coefficients has the property that
f(x)=f'(x).f"(x). The leading coefficient of f(x) is?
A.) 1/6
B.)1/9
C.)1/12
D.)1/18
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4 Answers
21
341First try to match degrees of the polynomials on both sides. If the degree of f(x) is n, f'(x) f"(x) will have degree 2n-3.
So n = 3 and hence its a cubic polynomial as tapan has pointed out.
But, to solve the problem you need one more information that can be extracted from the equation.
Can someone point out what that is?
21no its simple......
let f(x) = ax^3 + bx^2......
f'(x) = 3ax2 + 2bx +.....
f"(x) = 6ax + 2b
COMPARE CO-EFF OF X3
f'(x).F"(X) = F(X)
3ax2*6ax = ax3
matlab 18a2 = a
matlab a = 1/18
341ok. what i meant was f"(x) and f(x) have a common root. this means, if f"(α) = 0, then α is a root that is repeated at least thrice in f(x). But, f(x) is of degree three. Hence f(x) = a(x-α)3
So that f'(x) f"(x) = 18a2(x-α)3 = a(x-α)3 = f(x)
Hence a = 1/18
