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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

$x \geq - 2$
$- 3 < x \leq - 2$
$- 2 \leq x < 0$
$- 2 \leq x < 1$

267.

Function f : N $\to$ N, f(x) = 2x + 3 is

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

268.
If domain of the function f(x) = x² - 6x + 7 is ( $- \infty, \infty$ ), then its range is

[ - 2, 3]

$(- \infty, 2)$

$(- \infty, \infty)$

$[- 2, \infty)$

$[- 2, \infty)$

$Let y = x^{2} - 6 x + 7 \Rightarrow x^{2} - 6 x + 7 - y = 0 On comparing with {ax}^{2} + bx + c = 0, we get a = 1, b = - 6, and c = (7 - y) Now, using Sridharacharya formula, x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{a} = \frac{6 \pm \sqrt{36 - 4 (7 - y)}}{2} = \frac{6 \pm \sqrt{36 - 28 + 4 y}}{2} = \frac{6 \pm \sqrt{4 y + 8}}{2} = \frac{6 \pm 2 \sqrt{y + 2}}{2} = 3 \pm \sqrt{y + 2} f (x) is defined only when y + 2 \geq 0 \Rightarrow y \geq - 2 ∴ Range of f (x) = [- 2, \infty)$