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 = 1 |
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 - 3g)'' never vanishes. Then the correct statements) is (are):
Answer & explanation
Correct answer: option 4
Let $F(x)=f(x)-3 g(x)$ for all $x \in[-1,2]$
We find that
$F(-1)=f(-1)-3 g(-1)=3-3 \times 0=3$
$F(0)=f(0)-3 g(0)=6-3 \times 1=3$
$F(2)=f(2)-3 g(2)=0-3 \times(-1)=3$
By Rolle's theorem, F'(x) vanishes at least once in each of the intervals [-1, 0] and [0, 2]
Now, $F(x)=f(x)-3 g(x)$
$\Rightarrow F''(x)=f''(x)-3 g''(x)=\left(f''-3 g''\right)(x)=(f-3 g)''(x)$
It is given that $(f-3 g)''(x)$ i.e. $F''(x)$ never vanishes. Therefore, $F''(x)>0$ or, $F''(x)<0$ for all $x \in(-1,0) \cup(0,2)$.
$\Rightarrow F'(x)$ is either increasing or decreasing in $(-1,0)$ and $(0,2)$.
$\Rightarrow F'(x)=0$ has exactly one solution in $(-1,0)$ and one solutions in $(0,2)$.
$\Rightarrow f'(x)-3 g'(x)=0$ has exactly one solution in $(-1,0)$ and exactly one solution in $(0,2)$.