Let f be a polynomial function such that\dpi{100} \small f(3x)= {f}'\left ( x \right ) \cdot{f}''\left ( x \right ), for all \small x\epsilon R Then :
Option: 1 f\left ( 2 \right )+{f}' \left ( 2 \right ) =28

Option: 7 {f}''\left ( 2 \right )-{f}'\left ( 2 \right )=0

Option: 13 {f}''\left ( 2 \right )-f\left ( 2 \right )=4

Option: 19 f\left ( 2 \right )-{f}'\left ( 2 \right )+{f\left }''\left ( 2 \right )=10
 

Answers (1)

f(x) is given a polynomial function

Let f(x) having n^{th} degree

f'(x) becomes (n-1)^{th} degree

f''(x) becomes (n-2 )^{th} degree

Now it is given that

f(3x)=f'(x)\cdot f''(x)

If we consider only degree

n=(n-1)+(n-2)\\\Rightarrow n=3

Assume

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

we get

f'(x)=3ax^2+2bx+c and f''(x)=6ax+2b

f(3x)=f'(x)\cdot f''(x)

a(3x)^3+b(3x)^2+c(3x)+d=(3ax^2+2bx+c)\cdot(6ax+2b)

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

Now Comparing coefficient of x^3

27a=18a^2

a=\frac32

Now Comparing coefficient of x^2

9b=18ab

Here a=0 or b=0

a cannot be zero so b=0

Now Comparing coefficient of x

c=0

Similarly d=0

Hence 

f(x)=\frac32x^3

f'(x)=\frac323x^2=\frac92x^2 and 

f''(x)=\frac326x=9x

f''(2)-f'(2)=\frac92\times4-9\times2=0

Option 2 is correct

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