According to Newton-Raphson method, the value of $\sqrt{12}$. upto three places of decimal will be

• 3.463

• 3.462

• 3.467

• None of these

If for all x, y $\in$ N, there exists a function f(x) satisfying f(x + y) = f(x) · f(y) such that f(1) = 3 and $\sum _{\mathrm{x}=1}^{\mathrm{n}}\mathrm{f}\left(\mathrm{x}\right) = 120$, then value of n will be

• 4

• 5

• 6

• None of these

If f(x) = ${\log }_{\mathrm{e}}\left(6 - \left|{\mathrm{x}}^2 + \mathrm{x} - 6\right|\right)$, then domain of f(x) has how many integral values of x ?

• 5

• 4

• infinite

• None of these

$\mathrm{If} \frac{\mathrm{x} - 4}{{\mathrm{x}}^2 - 5\mathrm{x} - 2\mathrm{k}} = \frac{2}{\mathrm{x} - 2} - \frac{1}{\mathrm{x} + \mathrm{k}\text{'}},\mathrm{then} \mathrm{k} \mathrm{is} \mathrm{equal} \mathrm{to}$

• - 3

• - 2

• 2

• 3

55.

${20}^{2 - 3{\mathrm{x}}^2} = {\left(40\sqrt{5}\right)}^{3{\mathrm{x}}^2 - 2}$, then x is equal to

• $\pm \sqrt{\frac{3}{2}}$

• $\pm \sqrt{\frac{2}{3}}$

• $\pm \sqrt{\frac{4}{3}}$

• $\pm \sqrt{\frac{5}{4}}$

Let A = $\left\{\mathrm{x} \in \mathrm{R}, \mathrm{x} \ne 0, - 4 \le \mathrm{x} \le 4\right\}$ and $\mathrm{f} : \mathrm{A} \to \mathrm{R}$ defined by f(x) = $\frac{\left|\mathrm{x}\right|}{\mathrm{x}}$ for x $\in$ A. Then, the range of f is

• {1, - 1}

• $\left\{\mathrm{x} : 0 \le \mathrm{x} \le 1\right\}$

• (1)

• $\left\{\mathrm{x} : - 4\le \mathrm{x} \le 0\right\}$

If $\log \left(2\right) = \mathrm{a}, \log \left(3\right) = \mathrm{b}, \log \left(7\right) = \mathrm{c}$ and 6^x = 7^{x + 4} then x is equal to

• $\frac{4\mathrm{b}}{\mathrm{c} + \mathrm{a} - \mathrm{b}}$

• $\frac{4\mathrm{c}}{\mathrm{a} + \mathrm{b} - \mathrm{c}}$

• $\frac{4\mathrm{c}}{\mathrm{c} - \mathrm{a} - \mathrm{b}}$

• $\frac{4\mathrm{a}}{\mathrm{a} + \mathrm{b} - \mathrm{c}}$

In the sequence, (1, 2, 3), (4, 5, 6), (7, 8, 9, 10)... of sets, the sum of elements in the 50th set is

• 62525

• 65225

• 56255

• 55765

If f : $\mathrm{R} \to \mathrm{R} \mathrm{and} \mathrm{g} : \mathrm{R} \to \mathrm{R}$ are defined by f(x) = 2x + 3 and g(x) = x² + 7, then the values of x such that g(x) = x² + 7, then values ofx such that g(f(x)) = 8 are

• 1, 2

• - 1, 2

• - 1, - 2

• 1, - 2

Suppose f : [- 2, 2] $\to$ R is defined by

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{- 1, \mathrm{for} - 2 \le \mathrm{x} \le 0 \\ \mathrm{x} - 1, \mathrm{for} 0 \le \mathrm{x} \le 2 \\ \left\{\mathrm{x} \in \left[- 2, 2\right] : \mathrm{x} \le 0\right\} \mathrm{and} \mathrm{f}\left(\left|\mathrm{x}\right|\right) = \mathrm{x}\right\} \end{gathered}$$

is equal to

• {- 1}

• 0

• $\left\{- \frac{1}{2}\right\}$

• $\mathrm{\varphi }$