MCQ
The value of $(1+\text{i})(1+\text{i}^2)(1+\text{i}^3)(1+\text{i}^4)$ is
- A2
- B0
- C1
- Di
Solution:
$(1+\text{i})(1+\text{i}^2)(1+\text{i}^3)(1+\text{i}^4)$
$=(1+\text{i})(1-1)(1-\text{i})(1+1) \ \big(\because\text{i}^2=-1, \beta=-\text{i and} \ \text{i}^4=1\big)$
$=(1+\text{i})(0)(1-\text{i})(2)$
$=0$
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$\lim\limits_{\text{x} \rightarrow0}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}$ is:
Choose the correct answer.
If the middle term of $\Big(\frac{1}{\text{x}}+\text{x}\sin\text{x}\Big)^{10}$ is equal to $7\frac{7}{8},$ then value of x is:
$2\text{n}\pi+\frac{\pi}{6}.$
$\text{n}\pi+\frac{\pi}{6}.$
$\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}.$
$\text{n}\pi+(-1)^\text{n}\frac{\pi}{3}.$
Hint: $\text{T}_6=\ ^{10}\text{C}_5\frac{1}{\text{x}^5}.\text{x}^5\ \sin^5\text{x}=\frac{63}{8}\Rightarrow\sin^5\text{x}=\frac{1}{2^5}\sin\frac{1}{2}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}$
The value of $\cos12^\circ+\cos84^\circ+\cos156^\circ+\cos132^\circ$ is:
$\frac{1}{2}$
$1$
$-\frac{1}{2}$
$\frac{1}{8}$