Let $f, g : [- 2, 3] \to R$ be continuous function 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 the following table:
In each of the intervals $(- 2, 0), (0, 1) a n d (1, 3)$ the function $(f - 5 g) "$ never vanishes. Then the correct statement(s) is (are)

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$f' (x) - 5 g' (x) = 0$ has atleast one solution in $(1, 3)$

$f' (x) - 5 g' (x) = 0$ has exactly one solution in $(0, 1)$

$f' (x) - 5 g' (x) = 0$ has exactly three solutions in $(- 2, 0) \cup (0, 1) \cup (1, 3)$

$f' (x) - 5 g' (x) = 0$ has atleast two solutions in $(- 2, 0)$

answer is A, B, C.

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Detailed Solution

Let $h (x) = f (x) - 5 g (x)$
$h (x)$ is continuous, differentiable & $f' (x)$ is
Also, continuous, differentiable
$h (2) = 6 = h (0) = h (1) = h (3) & (f - 5 g) "$ never vanishes
$\Rightarrow$ Exactly one root in $(- 2, 0), (0, 1) & (1, 3)$

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