Question

Let $f,g:[-2,3]\to R$ be continuous functions which are twice differentiable on the interval $(-2,3)$. Let the values of $f$ and $g$ at the points $-2,0,1$ and $3$ be as given in table

	$x=-2$	$x=0$	$x=1$	$x=3$
$f\left(x\right)$	11	-4	6	16
$g\left(x\right)$	1	-2	0	2

In each of the intervals $(-2,0),(0,1)$ and
$(1,3)$ the function $(f-5g{)}^{\prime \prime }$ never vanishes. Then the correct statement(s) is/are

• A. $f^{\prime }\left(x\right)-5g^{\prime }\left(x\right)=0$ has exactly three solutions in$(-2,0)\cup (0,1)\cup (1,3)$
• B. $f^{\prime }\left(x\right)-5g^{\prime }\left(x\right)=0$ has exactly one solution in $(0,1)$
• C. $f^{\prime }\left(x\right)-5g^{\prime }\left(x\right)=0$ has atleast one solution in $(1,3)$
• D. $f^{\prime }\left(x\right)-5g^{\prime }\left(x\right)=0$ has atleast two solutions in $(-2,0)$

Answer 1

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

The correct options are
A $f^{\prime }\left(x\right)-5g^{\prime }\left(x\right)=0$ has exactly three solutions in$(-2,0)\cup (0,1)\cup (1,3)$
B $f^{\prime }\left(x\right)-5g^{\prime }\left(x\right)=0$ has exactly one solution in $(0,1)$
C $f^{\prime }\left(x\right)-5g^{\prime }\left(x\right)=0$ has atleast one solution in $(1,3)$

Let $h\left(x\right)=f\left(x\right)-5g\left(x\right)$
$h\left(x\right)$ is continuous, differentiable and $f^{\prime }\left(x\right)$ is also continuous, differentiable
$h(-2)=h\left(0\right)=h\left(3\right)=h\left(1\right)=6$
and $(f-5g{)}^{\prime \prime }$ never vanishes
$\Rightarrow h^{\prime \prime }\left(x\right)\ne 0$
$\Rightarrow h^{\prime }\left(x\right)$ never changes its nature in $(-2,0),(0,1),(1,3)$.
So using rolle's theorem there exists exactly one solution for $h^{\prime }\left(x\right)=0$ in each of the three intervals $(-2,0),(0,1)$ and $(1,3)$.