Question

Let $f\left(x\right)=x^2+x{g}^{{}^{\prime }}\left(1\right)+g^{\prime \prime }\left(2\right)$ and $g\left(x\right)=f\left(1\right).x^2+x{f}^{{}^{\prime }}\left(x\right)+f^{\prime \prime }\left(x\right)$ then

• A. $f^{{}^{\prime }}\left(1\right)+f^{{}^{\prime }}\left(2\right)=0$
• B. $g^{{}^{\prime }}\left(2\right)=g^{{}^{\prime }}\left(1\right)$
• C. $g^{{}^{\prime \prime }}\left(2\right)+f^{{}^{\prime \prime }}\left(3\right)=6$
• D. none of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f^{{}^{\prime }}\left(1\right)+f^{{}^{\prime }}\left(2\right)=0$
B $g^{{}^{\prime }}\left(2\right)=g^{{}^{\prime }}\left(1\right)$
C $g^{{}^{\prime \prime }}\left(2\right)+f^{{}^{\prime \prime }}\left(3\right)=6$

Let $g^{\prime }\left(1\right)=a$ and $g^{\prime }\left(2\right)=b$ --------(1)
Then
$f\left(x\right)=x^2+ax+b,f\left(1\right)=1+a+b{f}^{\prime }\left(x\right)=2x+a,f^{\prime \prime }\left(x\right)=2\therefore g\left(x\right)=(1+a+b)x^2+(2x+a)x+2=x^2(3+a+b)+ax+2\Rightarrow g^{\prime }\left(x\right)=2x(3+a+b)+a$

Hence
$g^{\prime }\left(1\right)=2(3+a+b)+a$ --------(2)

$g^{\prime }\left(2\right)=4(3+a+b)+a$ --------(3)

From (1),(2) and (3)
$a=2(3+a+b)+a$ and $b=2(3+a+b)$

$\Rightarrow 3+a+b=0$ and $b+2a+6=0$

$\therefore f\left(x\right)=x^2-3x$ and $g\left(x\right)=-3x+2$

$\therefore f^{\prime }\left(1\right)+f^{\prime }\left(2\right)=0g^{\prime \prime }\left(2\right)+f^{\prime \prime }\left(3\right)=6$