单选题 （共 6 题 ），每题只有一个选项正确
设 $M=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin x}{1+x^{2}} \cos ^{4} x \mathrm{~d} x, N=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin ^{3} x+\cos ^{4} x\right) \mathrm{d} x, P=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^{2} \sin ^{3} x-\cos ^{4} x\right) \mathrm{d} x$, 则有
$\text{A.}$ $N < P < M$.
$\text{B.}$ $M < P < N$.
$\text{C.}$ $N < M < P$.
$\text{D.}$ $P < M < N$.
设在 $[0,1]$ 上 $f^{\prime \prime}(x)>0$, 则 $f^{\prime}(0), f^{\prime}(1), f(1)-f(0)$ 或 $f(0)-f(1)$ 的大小顺序是
$\text{A.}$ $f^{\prime}(1)>f^{\prime}(0)>f(1)-f(0)$.
$\text{B.}$ $f^{\prime}(1)>f(1)-f(0)>f^{\prime}(0)$.
$\text{C.}$ $f(1)-f(0)>f^{\prime}(1)>f^{\prime}(0)$.
$\text{D.}$ $f^{\prime}(1)>f(0)-f(1)>f^{\prime}(0)$.
设 $f(x)$ 连续， $F(x)=\int_0^{x^2} f\left(t^2\right) \mathrm{d} t$ ，则 $F^{\prime}(x)$ 等于
$\text{A.}$ $f\left(x^4\right)$
$\text{B.}$ $x^2 f\left(x^4\right)$
$\text{C.}$ $2 x f\left(x^4\right)$
$\text{D.}$ $2 x f\left(x^2\right)$
若 $f(x)$ 的导数是 $\sin x$ ，则 $f(x)$ 有一个原函数为
$\text{A.}$ $1+\sin x$
$\text{B.}$ $1-\sin x$
$\text{C.}$ $1+\cos x$
$\text{D.}$ $1-\cos x$
已知 $f(x)=\left\{\begin{array}{ll}x^2 & 0 \leq x < 1 \\ 1 & 1 \leq x \leq 2\end{array}\right.$ ，设
$$
F(x)=\int_1^x f(t) \mathrm{d} t(0 \leq x \leq 2) ，
$$
则 $f(x)$ 为
$\text{A.}$ $\left\{\begin{array}{l}\frac{1}{3} x^3, 0 \leq x < 1 \\ x, 1 \leq x \leq 2\end{array}\right.$
$\text{B.}$ $\left\{\begin{array}{l}\frac{1}{3} x^3-\frac{1}{3}, 0 \leq x < 1 \\ x, 1 \leq x \leq 2\end{array}\right.$
$\text{C.}$ $\left\{\begin{array}{l}\frac{1}{3} x^3, 0 \leq x < 1 \\ x-1,1 \leq x \leq 2\end{array}\right.$
$\text{D.}$ $\begin{cases}\frac{1}{3} x^3-\frac{1}{3} & 0 \leq x < 1 \\ x-1 & 1 \leq x \leq 2\end{cases}$
设 $f(x)$ 为连续函数，且 $F(x)=\int_{\frac{1}{x}}^{\ln x} f(t) \mathrm{d} t$ ，则 $F^{\prime}(x)$ 等于
$\text{A.}$ $\frac{1}{x} f(\ln x)+\frac{1}{x^2} f\left(\frac{1}{x}\right)$
$\text{B.}$ $f(\ln x)+f\left(\frac{1}{x}\right)$
$\text{C.}$ $\frac{1}{x} f(\ln x)-\frac{1}{x^2} f\left(\frac{1}{x}\right)$
$\text{D.}$ $f(\ln x)-f\left(\frac{1}{x}\right)$