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Relations and Functions

Multiple Choice Questions

261.

For the circuit show below, the Boolean polynomial is

$(~ p \lor q) \lor (p \lor ~ q)$
$(~ p \land q) \lor (p \land q)$
$(~ p \land ~ q) \lor (q \land p)$
$(~ p \land q) \lor (p \land ~ q)$

262.

In a Boolean Algebra B, for all x, y in B, $x \land (x \lor y)$ is equal to

263.

The value of $(1 + ∆) (1 - \nabla)$ is

0
- 1
1
None of the above

264.

f : R $\to$ R, then f(x) = $x |x|$ will be

many-one-onto
one-one-onto
many-one-into
one-one-into

265.

The inverse ofthe function y = $\frac{2^{x}}{1 + 2^{x}}$ is

$x = \log_{2} (\frac{1}{1 - 2^{y}})$
$x = \log_{2} (1 - \frac{1}{y})$
$x = \log_{2} (\frac{1}{1 - y})$
$x = \log_{2} (\frac{y}{1 - y})$

266.
The domain of the definition of the function
$y = \frac{1}{\log_{10} (1 - x)} + \sqrt{(x + 2)}$ is

$x \geq - 2$

$- 3 < x \leq - 2$

$- 2 \leq x < 0$

$- 2 \leq x < 1$

$- 2 \leq x < 0$

$Given, y = \frac{1}{\log_{10} (1 - x)} + \sqrt{(x + 2)} y is defined for \log_{10} (1 - x) \neq 0 and x + 2 \geq 0 \Rightarrow 1 - x \neq 0 and 1 - x \neq 0 x \neq 0 and 1 - x > 0 and 1 - x \neq 0 x < 1 and x \neq 1 ∴ y is defined for values x for which both the terms of the functon is defined ∴ y is defined, when - 2 \leq x < 0 and 0 < x < 1$