JEE Advance - Mathematics (2015 - Paper 2 Offline - No. 13)
Let $$f, g :$$ $$\left[ { - 1,2} \right] \to R$$ 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 $$(f-3g)''$$ never vanishes. Then the correct statement(s) is (are)
$$f'\left( x \right) - 3g'\left( x \right) = 0$$ has exactly three solutions in $$\left( { - 1,0} \right) \cup \left( {0,2} \right)$$
$$f'\left( x \right) - 3g'\left( x \right) = 0$$ has exactly one solution in $$(-1, 0)$$
$$f'\left( x \right) - 3g'\left( x \right) = 0$$ has exactly one solution in $$(0, 2)$$
$$f'\left( x \right) - 3g'\left( x \right) = 0$$ has exactly two solutions in $$(-1, 0)$$ and exactly two solutions in $$(0, 2)$$
Explanation
Let $$F(x) = f(x) - 3g(x)$$
$$\therefore$$ $$F( - 1) = 3$$, $$F(0) = 3$$ and $$F(2) = 3$$
So, $$F'(x)$$ will vanish at least twice in
$$( - 1,0) \cup (0,2)$$.
$$\because$$ $$F''(x) > 0$$ or $$ < 0$$, $$\forall x \in ( - 1,0) \cup (0,2)$$
Hence, $$f'(x) - 3g'(x) = 0$$ has exactly one solution in $$( - 1,0)$$ and one solution in $$(0,2)$$.
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