Question

If $\frac{d}{d x}(f(x))=g(x)$ and $\frac{d}{d x}(g(x))=f\left(x^2\right)$ then $\frac{d^2}{d x^2}\left(f\left(x^3\right)\right)$ can be expressed in the form $k$ $x ^{ a } f \left( x ^{ b }\right)+ px g\left( x ^{ c }\right)$ where $a , b , c , k , p \in N$, then the value of $( k + p + a + b + c )$ equals

• A. 26
• B. 27
• C. 28
• D. 30

Answer 1

Answer: (C)

Given $ f ^{\prime}( x )= g ( x ) $
$g ^{\prime}( x )= f \left( x ^2\right)= f ^{\prime \prime}( x ) $
$\text { let } y=f\left( x ^3\right)$
$\frac{ dy }{ dx }= f ^{\prime}\left( x ^3\right) \cdot 3 x ^2$
$\frac{d^2 y}{d x^2}=3\left[3 x^2 x^2 \cdot f^{\prime \prime}\left(x^3\right)+2 x \cdot f^{\prime}\left(x^3\right)\right]=9 x^4 f^{\prime \prime}\left(x^3\right)+6 x f^{\prime}\left(x^3\right)=9 x^4 \cdot f\left(x^6\right)+6 x g\left(x^3\right) $
$k x^a f\left(x^b\right)+p x g\left(x^c\right) $
$\therefore (k+p+a+b+c)=9+4+6+6+3=28$