f(x) and g(x) are two differentiable function | vedclass

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(D) Given $f''(x) - g''(x) = 0$.Integrating with respect to $x$,we get $f'(x) - g'(x) = c$,where $c$ is a constant.At $x = 1$,$f'(1) - g'(1) = c \impl

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$-5$

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(D) Given $f''(x) - g''(x) = 0$.Integrating with respect to $x$,we get $f'(x) - g'(x) = c$,where $c$ is a constant.At $x = 1$,$f'(1) - g'(1) = c \implies 2 - 4 = c \implies c = -2$.Thus,$f'(x) - g'(x) = -2$.Integrating again with respect to $x$,we get $f(x) - g(x) = -2x + c_1$,where $c_1$ is a constant.At $x = 2$,$f(2) - g(2) = -2(2) + c_1 \implies 3 - 9 = -4 + c_1 \implies -6 = -4 + c_1 \implies c_1 = -2$.Therefore,$f(x) - g(x) = -2x - 2$.At $x = 3/2$,$f(3/2) - g(3/2) = -2(3/2) - 2 = -3 - 2 = -5$.

$f(x)$ and $g(x)$ are two differentiable functions on $[0, 2]$ such that $f''(x) - g''(x) = 0$,$f'(1) = 2$,$g'(1) = 4$,$f(2) = 3$,and $g(2) = 9$. Then $f(x) - g(x)$ at $x = 3/2$ is:

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