The conjugate of the multiplicative inverse of the complex number $\boldsymbol{z}=\frac{\mathbf{1}+\mathbf{7} \boldsymbol{i}}{\mathbf{3}+\boldsymbol{i}}$ is:
$\frac{2}{5}+\frac{1}{5} i$
$\frac{1}{5}-\frac{2}{5} i$
$\frac{1}{5}+\frac{2}{5} i$
$1+2 i$
The absolute maximum and minimum values of the function $f(x)=\sin x+\sqrt{3} \cos x$ in $[0, \pi]$ are
Minimum value $=-\frac{1}{\sqrt{3}}$, maximum value $=2$
Minimum value $=\frac{1}{\sqrt{3}}$, maximum value $=2$
Minimum value $=\sqrt{3}$, maximum value $=2$
Minimum value $=-\sqrt{3}$, , maximum value $=2$
Every term of a geometric progression is positive, and every term is the sum of the two preceding terms. Then the common ratio of the geometric progression is:
$\frac{1+\sqrt{5}}{2}$
$\frac{\sqrt{5}-1}{2}$
1
$\frac{1-\sqrt{5}}{2}$
$$ \text { If } y=\tan ^{-1}\left(\frac{\sqrt{1+x^3}+\sqrt{1-x^3}}{\sqrt{1+x^3}-\sqrt{1-x^3}}\right) \text { then } \frac{\boldsymbol{d} \boldsymbol{y}}{\boldsymbol{d x}}= $$
$-\frac{3 x^2}{2 \sqrt{1-x^6}}$
$-\frac{6 x^2}{\sqrt{1-x^6}}$
$\frac{6 x^2}{\sqrt{1-x^6}}$
$\frac{3 x^2}{\sqrt{1-x^6}}$
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