Question : What is the value of
$
\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}^8+\frac{1}{\mathrm{k}^8}\right)\left(\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}\right)\left(\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}\right)?
$

Option 1: $\frac{\mathrm{k}^{64}-\frac{1}{\mathrm{k}^{64}}}{\mathrm{k}+\frac{1}{\mathrm{k}}}$

Option 2: $\frac{\mathrm{k}^{32}-\frac{1}{\mathrm{k}^{32}}}{\mathrm{k}-\frac{1}{\mathrm{k}}}\\$

Option 3: $\frac{\mathrm{k}^{32}-\frac{1}{\mathrm{k}^{32}}}{\mathrm{k}+\frac{1}{\mathrm{k}}}\\$

Option 4: $\frac{\mathrm{k}^{32}+\frac{1}{\mathrm{k}^{32}}}{\mathrm{k}+\frac{1}{\mathrm{k}}}$

Question : Which of the following statements is correct?
I. The value of $100^2-99^2+98^2-97^2+96^2-95^2+$ $94^2-93^2+\ldots \ldots+22^2-21^2$ is 4840.
II. The value of $\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right)$ is $\mathrm{k}^{16} - \frac{1}{\mathrm{k}^{16}}$.

Option 1: Neither I nor II

Option 2: Only II

Option 3: Only I

Option 4: Both I and II

Question : Which of the following statements is correct?
I. The value of $100^2-99^2+98^2-97^2+96^2-95^2+94^2$ $- 93^2+\ldots \ldots+22^2-21^2$ is 4840.
II. The value of
$\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4-\frac{1}{\mathrm{k}^4}\right) \text { is } \mathrm{k}^{16}-\frac{1}{\mathrm{k}^{16}}$.

Option 1: Neither I nor II

Option 2: Both I and II

Option 3: Only II

Option 4: Only I

Question : The value of $\left(1 \frac{1}{3} \div 2 \frac{6}{7}\right.$ of $\left.5 \frac{3}{5}\right) \times\left(6 \frac{2}{5} \div 4 \frac{1}{2}\right.$ of $\left.5 \frac{1}{3}\right) \div\left(\frac{3}{4} \times 2 \frac{2}{3} \div \frac{5}{9}\right.$ of $\left.1 \frac{1}{5}\right)=k$, where $\mathrm{k}$ lies between:

Option 1: 0.07 and 0.08

Option 2: 0.007 and 0.008

Option 3: 0.0007 and 0.0008

Option 4: 0.7 and 0.8

Question : If $\frac{1}{x^2+a^2}=x^2-a^2$, then the value of $x$ is:

Option 1: $\left(1-a^4\right)^\frac{1}{4}$

Option 2: $a$

Option 3: $\left(a^4-1\right)^\frac{1}{4}$

Option 4: $\left(a^4+1\right)^\frac{1}{4}$