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Let $f, g:[-1,2] \rightarrow R$ 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:
| $x=-1$ | $x=0$ | $x=2$ | |
| $f(x)$ | $3$ | $6$ | $0$ |
| $g(x)$ | $0$ | $1$ | $-1$ |
In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)
$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$
$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$
$(A,B)$
$(B,D)$
$(A,D)$
$(B,C)$
Solution
Let $H(x)=f(x)-3 g(x)$
$H(-1)=H(0)=H(2)=3 \text {. }$
Applying Rolle's Theorem in the interval $[-1,0]$ $H^{\prime}(x)=f^{\prime}(x)-3 g^{\prime}(x)=0$ for atleast one $c \in(-1,0)$.
As $H ^{\prime \prime}( x )$ never vanishes in the interval
$\Rightarrow$ Exactly one $c \in(-1,0)$ for which $H^{\prime}(x)=0$
Similarly, apply Rolle's Theorem in the interval $[0,2]$.
$\Rightarrow H ^{\prime}( x )=0$ has exactly one solution in $(0,2)$
