Question

If $f\left(x\right),g\left(x\right)$ are two differentiable 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. $|f\left(x\right)-g\left(x\right)|<2$ if $-2<x<0$
• C. $f\left(x\right)-g\left(x\right)=2x$ has real roots
• D. $f\left(2\right)=g\left(2\right)$ if $x=-1$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $f\left(4\right)-g\left(4\right)=10$
B $|f\left(x\right)-g\left(x\right)|<2$ if $-2<x<0$
D $f\left(2\right)=g\left(2\right)$ if $x=-1$

We have $f^{\prime \prime }\left(x\right)=g^{\prime \prime }\left(x\right)$.On integration, we get
$f^{\prime }\left(x\right)=g^{\prime }\left(x\right)+C$
Putting $x=1$, we get
$f^{\prime }\left(1\right)=g^{\prime }\left(1\right)+C$
$4=2+C$
$C=2$
$f^{\prime }\left(x\right)=g^{\prime }\left(x\right)+\mathrm{2......}\left(1\right)$
Integrating w.r.t. $x$ we get
$f\left(x\right)=g\left(x\right)+2x+c$
Putting $x=2$
$f\left(2\right)=g\left(2\right)+4+c$
$9=3+4+c$
$c=2$
$f\left(x\right)=g\left(x\right)+2x+\mathrm{2.......}\left(2\right)$

Now put $x=4$ in eqn $\left(2\right)$
$f\left(4\right)-g\left(4\right)=10$

also $|f\left(x\right)-g\left(x\right)|<2$ or $|2x+2|<2$ or $|x+1|<1$ or $-2<x<0$

Also, if $f\left(2\right)=g\left(2\right)$ then $x=-1$
$f\left(x\right)-g\left(x\right)=2x$ has no real roots