Let f ( x ) = x 3 + x 2 f ' ( 1 ) + x f " ( 2 ) + f ''' ( 3 ) , x ∈ R . Then f ' ( 10 ) is equal to

Given: $f\left(x\right)=x^3+x^2{f}^{\text{'}}\left(1\right)+x{f}^{"}\left(2\right)+f^{\text{'}\text{'}\text{'}}\left(3\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=3x^2+2x{f}^{\text{'}}\left(1\right)+f^{"}\left(2\right) ...\left(i\right)$

$\Rightarrow f^{"}\left(x\right)=6x+2f^{\text{'}}\left(1\right) ...\left(ii\right)$

$\Rightarrow f^{\text{'}\text{'}\text{'}}\left(x\right)=6$

$\Rightarrow f^{\text{'}\text{'}\text{'}}\left(3\right)=6 ...\left(iii\right)$

Using $\left(ii\right)$,

$\Rightarrow f^{"}\left(2\right)=12+2f^{\text{'}}\left(1\right)$

Putting this value in equation $\left(i\right)$,

$\Rightarrow f^{\text{'}}\left(1\right)=3+2f^{\text{'}}\left(1\right)+12+2f^{\text{'}}\left(1\right)$

$\Rightarrow 0=15+3f^{\text{'}}\left(1\right)$

$\Rightarrow f^{\text{'}}\left(1\right)=-5 ...\left(iv\right)$

$\Rightarrow f^{"}\left(2\right)=12+2\left(-5\right)$

$\Rightarrow f^{"}\left(2\right)=2 ...\left(v\right)$

$\Rightarrow f\left(x\right)=x^3+x^2·(-5)+x·\left(2\right)+6$

$\Rightarrow f^{\text{'}}\left(x\right)=3x^2-10x+2$

$\Rightarrow f^{\text{'}}\left(10\right)=300-100+2$

$\Rightarrow f^{\text{'}}\left(10\right)=202$