Question

Let $f$ and $g$ be twice differentiable functions on $R$ such that
$f^{\prime \prime }(x)=g^{\prime \prime }(x)+6x$
$f^{\prime }(1)=4g^{\prime }(1)-3=9$
$f(2)=3g(2)=12$
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)

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