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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)

best_answer

$f(x)$ 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

Posted by

vishal kumar

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