Let f and g be twice differentiable functions on such that f ' ' ( x ) = g ' ' ( x ) + 6 x , f ' ( 1 ) = 4 g ' ( 1 ) - 3 = 9 , f ( 2 ) = 3 g ( 2 ) = 12 . Then which of the following is NOT true? [2023]

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Solution

Q.
Let $f$ and $g$ be twice differentiable functions on $ℝ$ such that

$f'' (x) = g'' (x) + 6 x,$

$f' (1) = 4 g' (1) - 3 = 9,$

$f (2) = 3 g (2) = 12 .$

Then which of the following is NOT true [2023]

1 $| f' (x) - g' (x) |$ $< 6 \Rightarrow - 1 < x < 1$

2 If $- 1 < x < 2$ , then $| f (x) - g (x) |$ $< 8$

3 There exists $x_{0} \in (1, 3 / 2)$ such that $f (x_{0}) = g (x_{0})$

4 $g (- 2) - f (- 2) = 20$

Ans.
(2)

Suppose $f$ and $g$ be twice differentiable on $ℝ$ such that

$f'' (x) = g'' (x) + 6 x$ ...(1)

$f' (1) = 4 g' (1) - 3 = 9$ ...(2)

$f (2) = 3 g (2) = 12$ ...(3)

Firstly, we integrate equation (1).

So, $f' (x) = g' (x) + 6 \cdot \frac{x^{2}}{2} + C$ At $x$ = 1, we have

$f' (1) = g' (1) + 3 + C,$ where C is the constant of integration.

$\Rightarrow 9 = 3 + 3 + C \Rightarrow C = 3 (Using equation (2))$

$∴$ $f' (x) = g' (x) + 3 x^{2} + 3$

Again, by integrating, $f (x) = g (x) + \frac{3 x^{3}}{3} + 3 x + D,$ D is another constant of integration.

At $x$ = 2, we have $f (2) = g (2) + 8 + 6 + D$

$\Rightarrow 12 = 4 + 8 + 6 + D \Rightarrow D = - 6 (from (3))$

So, $f (x) = g (x) + x^{3} + 3 x - 6$

At $x = - 2$ , $\Rightarrow g (- 2) - f (- 2) = 20$

Hence, option (4) is true.

Now, for $- 1 < x < 2$

Let $b (x) = f (x) - g (x) = x^{3} + 3 x - 6 \Rightarrow b' (x) = 3 x^{2} + 3$

So, $b (- 1) < b (x) < b (2) \Rightarrow - 10 < b (x) < 8 \Rightarrow | b (x) | < 10$

So, option (2) is not true.

Now, $b' (x) = f' (x) - g' (x) = 3 x^{2} + 3$

If $| b' (x) | < 6 \Rightarrow$ $| 3 x^{2} + 3 |$ $< 6$

$\Rightarrow 3 x^{2} + 3 < 6 \Rightarrow x^{2} < 1 \Rightarrow - 1 < x < 1$

So, option (1) is also true.

For $x \in (- 1, 1)$ , we have $| f' (x) - g' (x) |$ $< 6$

We have to solve the equation $f (x) - g (x) = 0$

$\Rightarrow x^{3} + 3 x - 6 = 0 \Rightarrow b (x) = x^{3} + 3 x - 6 = 0$

$b (x) = x^{3} + 3 x - 6$

Here we have $b (1) = - ve$ and $b (\frac{3}{2}) = + ve$

$So, there exists x_{0} \in (1, \frac{3}{2}) such that f (x_{0}) = g (x_{0})$

Hence, option (3) is also true.

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