JEE MAIN - Mathematics (2023 - 29th January Evening Shift - No. 15)

Let $$f$$ and $$g$$ be the twice differentiable functions on $$\mathbb{R}$$ such that

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

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

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

Then which of the following is NOT true?

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

There exists $$x_0\in(1,3/2)$$ such that $$f(x_0)=g(x_0)$$

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

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

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

$$ \Rightarrow f'(x) = g'(x) + 3{x^2} + C$$

$$f'(1) = g'(1) + 3 + C$$

$$ \Rightarrow g = 3 + 3 + C \Rightarrow C = 3$$

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

$$ \Rightarrow f(x) = g(x) + {x^2} + 3x + C'$$

$$x = 2$$

$$f(2) = g(2) + 14 + C'$$

$$12 = 4 + 14 + C'$$

$$ \Rightarrow C' = - 6$$

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

$$f( - 2) = g( - 2) - 8 - 6 - 6$$

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

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

$$x \in ( - 1,1)$$

$$3{x^2} + 3 \in (0,6)$$

$$ \Rightarrow f'(x) - g'(x) \in (0,6)$$

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

At $$x = - 1$$

$$|f( - 1) - g( - 1)| = 10$$

$$\therefore$$ Option (4) is false.