A: \[\frac{1}{2}(A - E)\]
B: \[\frac{1}{2}(A + E)\]
C: \[\frac{1}{4}(A - E)\]
D: \[\frac{1}{4}(A + E)\]
举一反三
- 设`\n`阶方阵`\A`满足`\|A| = 2`，则`\|A^TA| = ,|A^{ - 1}| = ,| A^ ** | = ,| (A^ ** )^ ** | = ,|(A^ ** )^{ - 1} + A| = ,| A^{ - 1}(A^ ** + A^{ - 1})A| = `分别等于（ ） A: \[4,\frac{1}{2},{2^{n - 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^n},\frac{{{3^n}}}{2}\] B: \[2,\frac{1}{2},{2^{n - 1}},{2^{{{(n + 1)}^2}}},2{(\frac{3}{2})^n},\frac{{{3^n}}}{2}\] C: \[4,\frac{1}{2},{2^{n + 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^{n - 1}},\frac{{{3^n}}}{2}\] D: \[2,\frac{1}{2},{2^{n - 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^{n - 1}},\frac{{{3^n}}}{2}\]
- 积分$\int_0^1 x \arctan xdx=$（）。 A: $\frac{\pi}{4}+\frac{1}{2}$ B: $\frac{\pi}{4}$ C: $\frac{\pi}{4}-\frac{1}{2}$ D: $\frac{1}{2}$
- For the integral $\int_0^{+\infty}\frac{dx}{(x^2+p^2)(x^2+q^2)}$, which of the following statements are CORRECT? A: $\frac{1}{q^2-p^2}[\frac{1}{p}-\frac{1}{q}]\frac{\pi}{2},p>0 \ q>0;$ B: $\frac{1}{q^2-p^2}[\frac{1}{q}+\frac{1}{p}]\frac{\pi}{2}, -p>0 \ -q>0;$ C: $\frac{1}{q^2-p^2}[\frac{1}{p}-\frac{1}{q}]\frac{\pi}{2}, p>0 \ -q>0;$ D: $\frac{1}{p^2-q^2}[\frac{1}{q}+\frac{1}{p}]\frac{\pi}{2}, -p>0 \ q>0.$
- (10). 已知在5重贝努里试验中成功的次数 \( X \) 满足 \( P\{X=1\}=P\{X=2\} \)，则概率 \( P\{X=4\}= \)（ ）。 A: \(1- C_4^5 (\frac{1}{3})^4(\frac{2}{3}) \) B: \( C_5^4 (\frac{1}{3})^2(\frac{2}{3})^3 \) C: \( C_5^4 (\frac{1}{3})^4(\frac{2}{3})^4 \) D: \( C_5^4 (\frac{1}{3})^4(\frac{2}{3}) \)
- $\int_{0}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{4}}{[\cos (2t)\mathbf{i}+\sin (2t)\mathbf{j}+t\sin t\mathbf{k}]}\operatorname{dt}=$( ) A: $(\frac{1}{2},\frac{1}{2},\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ B: $(1,\frac{1}{2},\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ C: $(\frac{1}{2},1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ D: $(1,1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$
内容
- 0
Solve $\int_{-\frac{1}{2}}^1{1-x^2}dx=$? A: $\frac{\pi}{3}+\frac{\sqrt{3}}{8}$. B: $\frac{\pi}{2}$. C: $\frac{\pi}{6}+\frac{\sqrt{3}}{4}$. D: $\frac{\pi}{4}$.
- 1
将函数\(f(x)=\sin^4 x\)展开成Fourier级数为 ____ . A: \(f(x) = \frac{3}{8}-\frac{1}{2}\cos 2x +\frac{1}{8}cos 4x\) B: \(f(x) = \frac{1}{4}-\frac{1}{2}\cos x +\frac{3}{8}cos 4x\) C: \(f(x) = \frac{1}{4}-\frac{1}{2}\sin 2x -\frac{3}{8}cos 4x\) D: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)
- 2
已知$f(x),\ g(x)$互为反函数，且$f(1)=2,\ {g}'(2)=2,\ {g}''(2)=1$，则${f}''(1)=$( )。 A: $1$ B: $\frac{1}{2}$ C: $-\frac{1}{4}$ D: $-\frac{1}{8}$
- 3
方程${{x}^{2}}{{y}^{''}}-(x+2)(x{{y}^{'}}-y)={{x}^{4}}$的通解是( ) A: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$ B: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ C: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ D: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$
- 4
\(已知曲线弧L:y=\sqrt{1-x^2}(0\le x\le 1).则\int_{L}xyds=(\,)\) A: \[1\] B: \[\frac{1}{2}\] C: \[\frac{1}{3}\] D: \[\frac{1}{4}\]