Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given functions are:

$f^{"}\left(x\right)=g^{"}\left(x\right)+6x \dots \left(1\right)$

$f^{\text{'}}\left(1\right)=4g^{\text{'}}\left(1\right)-3=9 \dots \left(2\right)$

$f\left(2\right)=3 g\left(2\right)=12 \dots \left(3\right)$

By integrating $\left(1\right)$, we get
$f^{\text{'}}\left(x\right)=g^{\text{'}}\left(x\right)+3x^2+C$

At $x=1$

$f^{\text{'}}\left(1\right)=g^{\text{'}}\left(1\right)+3+C$

$\Rightarrow 9=3+3+C\Rightarrow C=3$

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

Again by integrating, we get

$f\left(x\right)=g\left(x\right)+x^3+3x+D$

At $x=2$, we get

$f\left(2\right)=g\left(2\right)+8+3\left(2\right)+D$

$\Rightarrow 12=4+8+6+D\Rightarrow D=-6$

So,

$f\left(x\right)=g\left(x\right)+x^3+3x-6$

Option $\left(A\right)$:

At $x=-2$,

$\Rightarrow g\left(-2\right)-f\left(-2\right)=20$

So, this option is true.

Option $\left(B\right)$

If $-1<x<2$

Let $h\left(x\right)=f\left(x\right)-g\left(x\right)=x^3+3x-6$

$\Rightarrow h^{\text{'}}\left(x\right)=3x^2+3$

$\Rightarrow h\text{'}\left(x\right)>0$ for all values of $x$.

So, $h\left(-1\right)<h\left(x\right)<h\left(2\right)$

$\Rightarrow -10<h\left(x\right)<8$

$\Rightarrow \left|h\left(x\right)\right|<10$     

So, this option is NOT true.

Option $\left(C\right)$

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

If $\left|h^{\text{'}}\left(x\right)\right|<6\Rightarrow \left|3x^2+3\right|<6$

$\Rightarrow 3x^2+3<6$ and $-6<3x^2+3$

$\Rightarrow x^2<1$ and $x^2>-3\left(\mathrm{always} \mathrm{true}\right)$

$\Rightarrow -1<x<1$        

So, If $x\in \left(-1,1\right)$ then $\left|f^{\text{'}}\left(x\right)-g^{\text{'}}\left(x\right)\right|<6$

So, this option is true.

Option $\left(D\right)$

$f\left(x\right)-g\left(x\right)=0$

$\Rightarrow x^3+3x-6=0$

$h\left(x\right)=x^3+3x-6$

Here, $h\left(1\right)=-\mathrm{ve}$ and $h\left(\frac{3}{2}\right)=+\mathrm{ve}$

So, there exists $x_0\in \left(1,\frac{3}{2}\right)$ such that $f\left(x_0\right)=g\left(x_0\right)$ 

So, this option is true.