Let $f(x)=3 x^2+4 x g^{\prime}(1)+g^{\prime \prime}(2)$ and, $g(x)=2 x^2+3 x f^{\prime}(2)+f^{\prime \prime}(3)$ for all $x \in R$. Then,
Answer & explanation
Correct answer: option 4
We have,
$f(x)=3 x^2+4 x g^{\prime}(1)+g^{\prime \prime}(2)$ ......(i)
$g(x)=2 x^2+3 x f^{\prime}(2)+f^{\prime \prime}(3)$ ......(ii)
∴ $f^{\prime}(x)=6 x+4 g^{\prime}(1)$ ......(iii)
and, $g^{\prime}(x)=4 x+3 f^{\prime}(2)$ ......(iv)
$\Rightarrow f^{\prime \prime}(x)=6$ ......(v)
and, $g^{\prime \prime}(x)=4$ ......(vi)
Putting x = 1 in (iii) and (iv), we get
$f^{\prime}(1) =6+4 g^{\prime}(1)$ and $g^{\prime}(1)=4+3 f^{\prime}(2)$
$\Rightarrow f^{\prime}(1) =6+4\left\{4+3 f^{\prime}(2)\right\}$
$\Rightarrow f^{\prime}(1) =22+12 f^{\prime}(2)$
Putting x = 2 in (iv), we get
$g^{\prime}(2)=8+3 f^{\prime}(2)$
$\Rightarrow g^{\prime}(2)=8+3\left\{12+4 g^{\prime}(1)\right\}$ $\left[\begin{array}{l}\text { Putting } x=2 \text { in (iii) we get } \\ f^{\prime}(2)=12+4 g^{\prime}(1)\end{array}\right]$
$\Rightarrow g^{\prime}(2)=44+12 g^{\prime}(1)$
Putting x = 3 in (v) and x = 2 in (vi), we get
$f^{\prime \prime}(3)=6$ and $g^{\prime \prime}(2)=4$
$\Rightarrow f^{\prime \prime}(3)+g^{\prime \prime}(2)=6+4=10$