Question

If $f\left(x\right),g\left(x\right)$ be twice differential functions on $[0,2]$ satisfying $f^{\prime \prime }\left(x\right)=g^{\prime \prime }\left(x\right)$, $f^{\prime }\left(1\right)=2g^{\prime }\left(1\right)=4$ and $f\left(2\right)=3g\left(2\right)=9$, then

• A. $f\left(4\right)-g\left(4\right)=10$
  
• B. $|f\left(x\right)-g\left(x\right)|<2\Rightarrow -2<x<0$
  
• C. $f\left(2\right)=g\left(2\right)\Rightarrow x=-1$
  
• D. $f\left(x\right)-g\left(x\right)=2x$ has real root

Answer 1

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

The correct options are
A $f\left(4\right)-g\left(4\right)=10$

B $|f\left(x\right)-g\left(x\right)|<2\Rightarrow -2<x<0$

C $f\left(2\right)=g\left(2\right)\Rightarrow x=-1$


We have $f^{\prime \prime }\left(x\right)=g^{\prime \prime }\left(x\right)$. On integration, we get
$f^{\prime }\left(x\right)=g^{\prime }\left(x\right)+C...\left(1\right)$
Putting $x=1$, we get
$f^{\prime }\left(1\right)=g^{\prime }\left(1\right)+C$ or $4=2+C$ or $C=2$
$\therefore f^{\prime }\left(x\right)=g^{\prime }\left(x\right)+2$

Integrating w.r.t. $x$, we get $f\left(x\right)=g\left(x\right)+2x+c_1$ $\left(2\right)$
Putting $x=2$, we get
$f\left(2\right)=g\left(2\right)+4+c_1$ or $9=3+4+c_1$ or $c_1=2$
$\therefore f\left(x\right)=g\left(x\right)+2x+2$
Putting $x=4$, we get $f\left(4\right)-g\left(4\right)=10$
$|f\left(x\right)-g\left(x\right)|<2$ or $|2x+2|<2$ or $|x+1|<1$ or $-2<x<0$
Also, $f\left(2\right)=g\left(2\right)$ or $x=1$
$f\left(x\right)-g\left(x\right)=2x$ has no solution.