Question

Let $f$ and $g$ be twice differentiable functions on $R$ 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 ?

• A. $g (-2)- f (-2)=20$
• B. If $-1< x< 2$, then $|f(x)-g(x)|< 8$
• C. $\left|f^{\prime}(x)-g^{\prime}(x)\right|< 6 \Rightarrow-1< x< 1 \mid$
• D. There exists $x _0 \in\left(1, \frac{3}{2}\right)$ such that $f \left( x _0\right)= g \left( x _0\right)$

Answer 1

Answer: (B)

$ f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x $....(1)
$ f^{\prime}(1)=4 g^{\prime}(1)-3=9$....(2)
$ f(2)=3 g(2)=12$ ....(3)
By integrating (1)
$f^{\prime}(x)=g^{\prime}(x)+6 \frac{x^2}{2}+C$
At $x =1$,
$ f ^{\prime}(1)= g ^{\prime}(1)+3+ C$
$\Rightarrow 9=4+3+ C \Rightarrow C =3$
$\therefore f ^{\prime}( x )= g ^{\prime}( x )+3 x ^2+3$
Again by integrating,
$f(x)=g(x)+\frac{3 x^3}{3}+3 x+D $
$ \text { At } x=2$
$ f(2)=g(2)+8+3(2)+D $
$ \Rightarrow 12=4+8+6+D \Rightarrow D=-6$
So, $f(x)=g(x)+x^3+3 x-6$
$\Rightarrow f(x)-g(x)=x^3+3 x-6$
At $x=-2$,
$\Rightarrow g (-2)- f (-2)=20 $ (Option (1) is true)
Now, for $-1< x , 2$
$ h(x)=f(x)-g(x)=x^3+3 x-6 $
$ \Rightarrow h^{\prime}(x)=3 x^2+3$
$ \Rightarrow h(x) \uparrow$
So, $h (-1)< h ( x )< h (2)$
$ \Rightarrow-10< h(x)< 8 $
$ \Rightarrow|h(x)|<10 $ (option (2) is NOT true)
Now, $h^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)=3 x^2+3$
If $\left|h^{\prime}(x)\right|<6 \Rightarrow\left|3 x^2+3\right|<6$
$ \Rightarrow 3 x ^2+3<6$
$ \Rightarrow x ^2< 1$
$ \Rightarrow-1< x< 1 $ (option (3) is True)
If $x \in(-1,1)\left|f^{\prime}(x)-g^{\prime}(x)\right|<6$
option (3) is true and now to solve
$ f(x)-g(x)=0 $
$ \Rightarrow x^3+3 x-6=0 $
$ h(x)=x^3+3 x-6$
here, $h(1)=-$ ve and $h\left(\frac{3}{2}\right)=+$ ve
So there exists $x _0 \in\left(1, \frac{3}{2}\right)$ such that $f \left( x _0\right)= g \left( x _0\right)$
(option (4) is true)