Question

Let $f,g:[-1,2]\to R$ be continuous functions which are twice differentiable on the interval $(-1,2).$ Let the values of $f$ and $g$ at the points $-1,0$ and $2$ be as given in the following table:

	$x=-1$	$x=0$	$x=2$
$f\left(x\right)$	$3$	$6$	$0$
$g\left(x\right)$	$0$	$1$	$-1$

In each of the interval $(-1,0)$ and $(0,2)$ the function $(f-3g{)}^{\prime \prime }$ never vanishes. Then the correct statement(s) is(are)

• A. $f^{\prime }\left(x\right)-3g^{\prime }\left(x\right)=0$ has exactly three solutions in$(-1,0)\cup (0,2)$
• B. $f^{\prime }\left(x\right)-3g^{\prime }\left(x\right)=0$ has exactly one solution in $(-1,0)$
• C. $f^{\prime }\left(x\right)-3g^{\prime }\left(x\right)=0$ has exactly one solution in $(0,2)$
• D. $f^{\prime }\left(x\right)-3g^{\prime }\left(x\right)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$

Answer 1

Answer: (B), (C)

$f^{\prime }\left(x\right)-3g^{\prime }\left(x\right)=0$ has exactly one solution in $(0,2)$

$f(-1)=3,f\left(0\right)=6,f\left(2\right)=0$
$g(-1)=0,g\left(0\right)=1,g\left(2\right)=-1$
$h\left(x\right)=f\left(x\right)-3g\left(x\right)$
$h(-1)=3-0=3,h\left(0\right)=6-3=3,h\left(2\right)=0-(-3)=3$
By Rolle's theorem atleast one solution for $h^{\prime }\left(x\right)=0\text{in}(-1,0)$ and atleast one solution for $h^{\prime }\left(x\right)=0\text{in}(0,2)$
Now given in $(-1,0)$ and $(0,2)$
$(f-3g{)}^{\prime \prime }=h^{\prime \prime }\ne 0$
therefore $h^{\prime }\left(x\right)$ is monotonic so exactly one solution in $(-1,0)$ and exactly one solution in $(0,2)$ exists.