Question

Let $f,g:[-1,2]\to R$ be continuous functions which are twice differential 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:
$f\left(x\right)=\{\begin{array}{c}3;x=-1\\ 6;x=0\\ 0;x=2\end{array}$ and $g\left(x\right)=\{\begin{array}{c}0;x=-1\\ 1;x=0\\ 0;x=-1\end{array}$
In each of the intervals $(-1,0)$ 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 solutions in $(-1,0)$
• C. $f^{\prime }\left(x\right)-3g^{\prime }\left(x\right)=0$ has exactly one solutions 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 solutions in $(-1,0)$
$f^{\prime }\left(x\right)-3g^{\prime }\left(x\right)=0$ has exactly one solutions in $(0,2)$

$h\left(x\right)=f\left(x\right)-3g\left(x\right)$
$h(-1)=f(-1)-3g(-1)=3-0=3$
$h\left(0\right)=f\left(0\right)-3g\left(0\right)=6-3=3$
$h\left(2\right)=f\left(2\right)-3g\left(2\right)=3$
$\therefore$ According to rolle's theorem $f^{\prime }\left(x\right)-3g^{\prime }\left(x\right)=0$ in $(-1,0)$ and $(0,2)$
Hence, options B and C.