Let f be a polynomial function such that f (3x)=f '(x) ⋅ f ''(x), for all x belongs to R. Then the correct option is?

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1 Answer
Feb 26, 2018

f(x) = 3/2x^3

and the correct option is (2)

Explanation:

If f(x) is of first degree its second derivative is identically null, so also f(x) would have to be identically null. to satisfy the equation f(3x) = f'(x)f''(x)

Let then f(x) be a generic polynomial of degree n >= 2. Then f'(x) will have degree (n-1) and f''(x) degree (n-2)

Now, the product f'(x)*f''(x) is a polynomial of degree (n-1)+(n-2) and as two polynomials can be equal for every x only if they have the same degree:

(2) " " f(3x) = f'(x)*f''(x)

implies:

n = (n-1)+(n-2)

n = 2n -3

n=3

that is f(x) must be of third degree:

f(x) =ax^3+bx^2+cx+d

f'(x) = 3ax^2+2bx+c

f''(x) = 6ax+2b

then (2) becomes:

27ax^3+9bx^2+3cx+d = (3ax^2+2bx+c)(6ax+2b)

27ax^3+9bx^2+3cx+d = 18a^2x^3+12abx^2+6acx+ 6abx^2+4b^2x+2bc

27ax^3+9bx^2+3cx+d = 18a^2x^3+18abx^2+(6ac+4b^2)x+ 2bc

Equating the coefficients of the same degree we get:

27a = 18a^2

and so as a!=0

a=3/2

Then:

9b = 18ab = 27b

b=0

at the first degree:

3c = 6ac+4b^2

3c = 9c

c = 0

and finally:

d= 2bc = 0

The polynomial which satisfies the equation is then:

f(x) = 3/2x^3

so that:

f'(x) = 9/2x^2

f''(x) = 9x

f(2) = 12

f'(2) = 18

f''(2) = 18

and only the second statement is correct.