Applications of Derivatives
Comparison of differentiable functions using integration
nta_pyq_2023_jan
Grade 12
Question:
Let $f$ and $g$ be 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?
(1) $g(-2) - f(-2) = 20$
(2) If $-1 < x < 2$, then $|f(x) - g(x)| < 8$
(3) $|f'(x) - g'(x)| < 6 \Rightarrow -1 < x < 1$
(4) There exists $x_0 \in \left(1, \frac{3}{2}\right)$ such that $f(x_0) = g(x_0)$
g(-2) - f(-2) = 20
If -1 < x < 2, then |f(x) - g(x)| < 8
|f'(x) - g'(x)| < 6 \Rightarrow -1 < x < 1
There exists x_0 \in (1, 3/2) such that f(x_0) = g(x_0)
Step-by-Step Solution
Key Concept: Integrate $f''(x) - g''(x) = 6x$ twice using the given conditions to find $f(x) - g(x) = x^3 + 3x - 6$, then check each option.
Integrating $f''(x) = g''(x) + 6x$: $f'(x) = g'(x) + 3x^2 + C_1$. At $x=1$: $f'(1) = 9$, $g'(1) = 3$, so $9 = 3 + 3 + C_1 \Rightarrow C_1 = 3$. Integrating: $f(x) = g(x) + x^3 + 3x + D$. At $x=2$: $f(2)=12$, $g(2)=4$, so $12 = 4 + 8 + 6 + D \Rightarrow D = -6$. Thus $f(x) - g(x) = x^3 + 3x - 6$. At $x=-2$: $g(-2)-f(-2) = 20$ (true). For $-1<x<2$: range is $(-10,8)$, so $|f-g|<10$, not $<8$. Option (2) is NOT true. Answer: (2).
Correct Answer: 2