Cho F(x) là một nguyên hàm của hàm số f(x)=1/cos^2x. Biết F(π/4+kπ)=k, ∀k∈Z. Tính F(0)+F(π)+F(2π)+…+F(10π)

Hướng dẫn giải:

Đáp án B.

Ta có:  \( \int{f(x)dx}=\int{\frac{1}{{{\cos }^{2}}x}dx}=\tan x+C  \)

Suy ra: \(F(x)=\left\{ \begin{align}  & \tan x+{{C}_{0}},x\in \left( -\frac{\pi }{2};\frac{\pi }{2} \right) \\  & \tan x+{{C}_{1}},x\in \left( \frac{\pi }{2};\frac{3\pi }{2} \right) \\  & \tan x+{{C}_{2}},x\in \left( \frac{3\pi }{2};\frac{5\pi }{2} \right) \\  & … \\  & \tan x+{{C}_{9}},x\in \left( \frac{17\pi }{2};\frac{19\pi }{2} \right) \\  & \tan x+{{C}_{10}},x\in \left( \frac{19\pi }{2};\frac{21\pi }{2} \right) \\ \end{align} \right.\)\(\Rightarrow \left\{ \begin{align}  & F\left( \frac{\pi }{4}+0\pi  \right)=1+{{C}_{0}}=0\Rightarrow {{C}_{0}}=-1 \\  & F\left( \frac{\pi }{4}+\pi  \right)=1+{{C}_{1}}=1\Rightarrow {{C}_{0}}=0 \\  & F\left( \frac{\pi }{4}+2\pi  \right)=1+{{C}_{2}}=2\Rightarrow {{C}_{0}}=1 \\  & … \\  & F\left( \frac{\pi }{4}+9\pi  \right)=1+{{C}_{9}}=9\Rightarrow {{C}_{9}}=8 \\  & F\left( \frac{\pi }{4}+10\pi  \right)=1+{{C}_{10}}=10\Rightarrow {{C}_{10}}=9 \\ \end{align} \right.\)

Vậy  \( F(0)+F(\pi )+F(2\pi )+…+F(10\pi ) \) \( =\tan 0-1+\tan \pi +\tan 2\pi +1+…+\tan 10\pi +9=44 \)