Question

Let $f\left(x\right)=x^2+x{g}^{\prime }\left(1\right)+g^{\prime \prime }\left(2\right)$ and $g\left(x\right)=x^2+x{f}^{\prime }\left(2\right)+f^{\prime \prime }\left(3\right)$ then which of the following option/s is/are correct?

• A. $f^{\prime }\left(1\right)=4+f^{\prime }\left(2\right)$
• B. $g^{\prime }\left(2\right)=8+g^{\prime }\left(1\right)$
• C. $g^{\prime \prime }\left(2\right)+f^{\prime \prime }\left(3\right)=4$
• D. $g^{\prime }\left(2\right)=6+g^{\prime }\left(1\right)$

Answer 1

The correct option is A $f^{\prime }\left(1\right)=4+f^{\prime }\left(2\right)$

$\begin{array}{cc}\text{given}:& f\left(x\right)=x^2+x{g}^{\prime }\left(1\right)+g^{\prime \prime }\left(2\right)\\ g\left(x\right)& =x^2+x{f}^{\prime }\left(2\right)+f^{\prime \prime }\left(3\right)\end{array}$

$\begin{array}{c}{f}^{\prime }\left(x\right)=2x+g^{\prime }\left(1\right)\\ f^{\prime \prime }\left(x\right)=2\\ f^{\prime }\left(1\right)=2+g^{\prime }\left(1\right)\\ f^{\prime }\left(2\right)=4+g^{\prime }\left(1\right)\\ \Rightarrow g^{\prime }\left(1\right)=f^{\prime }\left(2\right)-4\\ \therefore f^{\prime }\left(1\right)=-2+f^{\prime }\left(2\right)\end{array}$

$\begin{array}{c}{g}^{\prime }\left(x\right)=2x+f^{\prime }\left(2\right)\\ g^{\prime \prime }\left(x\right)=2\\ g^{\prime }\left(1\right)=2+f^{\prime }\left(2\right)\\ g^{\prime }\left(2\right)=4+f^{\prime }\left(2\right)\\ \Rightarrow f^{\prime }\left(2\right)=4+g^{\prime }\left(1\right)\\ \Rightarrow g^{\prime }\left(2\right)=8+g^{\prime }\left(1\right)\end{array}$

Hence, B is correct