Let f and g be twice differentiable functions | vedclass

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(A) Given $f^{\prime \prime}(x) - g^{\prime \prime}(x) = 6x$. Integrating once,we get $f^{\prime}(x) - g^{\prime}(x) = 3x^2 + C_1$.At $x=1$,$f^{\prime

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

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(A) Given $f^{\prime \prime}(x) - g^{\prime \prime}(x) = 6x$. Integrating once,we get $f^{\prime}(x) - g^{\prime}(x) = 3x^2 + C_1$.At $x=1$,$f^{\prime}(1) - g^{\prime}(1) = 4 - 3 = 1$. Thus,$3(1)^2 + C_1 = 1 \Rightarrow C_1 = -2$.So,$f^{\prime}(x) - g^{\prime}(x) = 3x^2 - 2$.Integrating again,$f(x) - g(x) = x^3 - 2x + C_2$.At $x=2$,$f(2) - g(2) = 12 - 4 = 8$. Thus,$(2)^3 - 2(2) + C_2 = 8 \Rightarrow 8 - 4 + C_2 = 8 \Rightarrow C_2 = 4$.Therefore,$f(x) - g(x) = x^3 - 2x + 4$.Check option $A$: $g(-2) - f(-2) = -((-2)^3 - 2(-2) + 4) = -(-8 + 4 + 4) = 0$. The original statement $g(-2)-f(-2)=20$ is false.Check option $B$: $h(x) = x^3 - 2x + 4$. $h^{\prime}(x) = 3x^2 - 2$. $h^{\prime}(x) = 0$ at $x = \pm \sqrt{2/3}$. The function is not bounded on $(-1, 2)$,so $|f(x)-g(x)| Check option $C$: $|f^{\prime}(x) - g^{\prime}(x)| = |3x^2 - 2| Note: The question asks for the $NOT$ true statement. Given the derivation $f(x)-g(x) = x^3-2x+4$,option $A$ is clearly false.

Let $f$ and $g$ be twice differentiable functions on $R$ such that
$f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x$
$f^{\prime}(1)=4, g^{\prime}(1)=3$
$f(2)=12, g(2)=4$
Then which of the following is $NOT$ true?

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Let $f$ and $g$ be twice differentiable functions on $R$ such that$f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x$$f^{\prime}(1)=4, g^{\prime}(1)=3$$f(2)=12, g(2)=4$Then which of the following is $NOT$ true?