JEE Main 2023 — Application of Derivatives Question with Solution

Given:

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

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

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

Now,

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

$\Rightarrow g\left(x\right)=f\left(1\right)x^2+x\left[2x+g^{\text{'}}\left(1\right)\right]+2$

$\Rightarrow g\left(x\right)=\left[f\left(1\right)+2\right]x^2+g^{\text{'}}\left(1\right)x+2$

$\Rightarrow g^{\text{'}}\left(x\right)=2f\left(1\right)x+4x+g^{\text{'}}\left(1\right) ....\left(1\right)$

$\Rightarrow g^{"}\left(x\right)=2f\left(1\right)+4 ...\left(2\right)$

Put $x=1$ in $\left(1\right)$, then we get

$2f\left(1\right)+4=0$

$\Rightarrow f\left(1\right)=-2$

$\therefore g^{"}\left(2\right)=2f\left(1\right)+4=0$

Now,

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

Put $x=1$, then we get

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

$\Rightarrow g^{\text{'}}\left(1\right)=-3$

So,

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

$f\left(x\right)=x^2-3x+c$

Since, $f\left(1\right)=-2\Rightarrow c=0$

So,

$f\left(x\right)=x^2-3x$

And, 

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

$\Rightarrow g\left(x\right)=-2x^2+x\left(2x-3\right)+2$

$\Rightarrow g\left(x\right)=-3x+2$

Hence, $f\left(4\right)-g\left(4\right)=4+10=14$