Let $f$ and $g$ be twice differentiable function on $\mathbb{R}$ such that $f^{\prime \prime}(x)=g^{

Question

Let $f$ and $g$ be twice differentiable function on $\mathbb{R}$ such that $f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x$ $f^{\prime}(1)=4 g^{\prime}(1)-3=9$ $f(2)=3 g(2)=12$. Then which of the following is NOT true?

Professor PC Thomas Classes · Answer

(b) : Suppose $f$ and $g$ be twice differentiable on $\mathbb{R}$ such that $$ \begin{aligned} & f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x \ & f^{\prime}(1)=4 g^{\prime}(1)-3=9 \ & f(2)=3 g(2)=12 \end{aligned} $$ Firstly we integrate equation (1)\ So, $f^{\prime}(x)=g^{\prime}(x)+6 \cdot \frac{x^{2}}{2}+C$ at $x=1$,\ we have $f^{\prime}(1)=g^{\prime}(1)+3+C$, where $C$ is constant of integration\ $\Rightarrow \quad 9=3+3+C \Rightarrow C=3$\ (Using equation (2))\ $	herefore \quad f^{\prime}(x)=g^{\prime}(x)+3 x^{2}+3$\ Again by integrating,\ $f(x)=g(x)+\frac{3 x^{3}}{3}+3 x+D$,\ $D$ is another constant of integration.\ At $x=2$, we have $f(2)=g(2)+8+3(2)+D$\ $\Rightarrow 12=4+8+6+D \Rightarrow D=-6\{$ from (3) $\}$\ So, $f(x)=g(x)+x^{3}+3 x-6$\ At $x=-2, \Rightarrow g(-2)-f(-2)=20$\ Hence $u_{1}{ }^{	ext {tion }}$ (d) is true.\ Now, for -1 、 $x<2$\ Let $b(x)=f(x)-g(x)=x^{3}+3 x-6$\ $\Rightarrow \quad b^{\prime}(x)=3 x^{2}+3$\ So, $b(-1)<b(x)<b(2)$\ $\Rightarrow \quad-10<b(x)<8$\ $\Rightarrow \quad|b(x)|<10$\ So, option (b) is not true.\ Now, $b^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)=3 x^{2}+3$\ If $\left|b^{\prime}(x)ight|<6 \Rightarrow\left|3 x^{2}+3ight|<6$\ $\Rightarrow \quad 3 x^{2}+3<6 \Rightarrow x^{2}<1$\ $\Rightarrow \quad-1<x<1$\ So, option (a) is also true.\ For $x \in(-1,1)$, we have $\left|f^{\prime}(x)-g^{\prime}(x)ight|<6$\ We have to solve the equation $f(x)-g(x)=0$\ $\Rightarrow \quad x^{3}+3 x-6=0$\ $\Rightarrow b(x)=x^{3}+3 x-6=0$\ $b(x)=x^{3}+3 x-6$\ Here we have $b(1)=-$ ve and $b\left(\frac{3}{2}ight)=+\mathrm{ve}$\ So, there exists $x_{0} \in\left(1, \frac{3}{2}ight)$ such that\ $f\left(x_{0}ight)=g\left(x_{0}ight)$\ So, option (c) is also true.\