Question 4:
If \( \mathrm{f}^{\prime \prime}(\mathrm{x})=8 \sin (4 \mathrm{x}) ; \mathrm{f}^{\prime}(2 \pi)=1 \), and \( \mathrm{f}^{(0)}=2 \), what function satisfy the given conditions?
A) \( f(x)=\sin (4 x)+3 x \)
B) \( f(x)=-\sin (4 x)+3 x \)
C) \( f(x)=-\sin (4 x)-3 x \)
D) None of them

Answer

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Answer

f(x)= -2\int\cos(4x) dx + \int C dx = -\frac{1}{2}\sin(4x) + Cx + D

Steps

Step 1 ：\int 8\sin(4x) dx = 8 \int \sin(4x) dx

Step 2 ：f'(x)= 8 \int \sin(4x) dx = -2\cos(4x) + C

Step 3 ：\int -2\cos(4x) + C dx = -2\int\cos(4x) dx + \int C dx

Step 4 ：f(x)= -2\int\cos(4x) dx + \int C dx = -\frac{1}{2}\sin(4x) + Cx + D