MCQ Relations and Functions -

The domain of definition of the function $f(x)=\sqrt{\log_{10}(\frac{5x-x^2}{4})}$ is
(A) [1,4]
(B) [1,0]
(C) [0,5]
(D) [5,0]

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Solution: (A)
Given that $f(x)=\sqrt{\log_{10}(\frac{5x-x^2}{4})}$
For domain of $f(x)$ , $\log_{10}(\frac{5x-x^2}{4})\geq 0$
$\Rightarrow \frac{5x-x^2}{4} \geq 1$
$\Rightarrow x^2-5x+4 \leq 0$
$\Rightarrow (x-1)(x-4) \leq 0$
$\Rightarrow x \in [1,4]$

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Let $R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}$ be a relation on the set $A = \{1, 2, 3, 4\}.$ The relation $R$ is
(A) a function
(B) transitive
(C) not symmetric
(D) reflexive

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Solution: (C)
Since, $(2, 4)\in R$ and $(2, 3)\in R$ So, R is not a function.
Since, $(1, 3)\in R$ and $(3, 1)\in R$ but $(1, 1)\notin R$ So, R is not transitive
Since, $(2, 3)\in R$ but $(3, 2)\notin R$ so, R is not symmetric
Since, $(1, 1), (2,2), (3,3), (4,4) \notin R$ so, R is not reflexive

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The domain of the function $f(x)=\frac{1}{\sqrt{|x|-x}}$
(A) $(0, \infty)$
(B) $(-\infty ,0)$
(B) $(-\infty ,\infty)-\{0\}$
(B) $(-\infty ,\infty)$

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Answer: (B)
Explanation:
We must have $|x|-x >0$
$\Rightarrow |x| >x$
This is only possible when $x<0$

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Let R be a relation defined as $a R b$ if $| a -b | > 0$ . Then, the relation $R$ is
(A) Reflexive
(B) Symmetric
(C) Transitive
(D) None of these

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Answer: (B)
$R$ is not reflexive since $| a-a | = 0$
$R$ is symmetric since if $| a - b | > 0$ , then
$| b -a | = | a -b | > 0.$
Thus $a R b \Rightarrow b R a$
R is not transitive. For example,
consider the numbers 3, 7, 3.
since $| 3 - 7 | = 4 > 0$ and $| 7 -3 | = 4 > 0$ , so we have $3 R 7$ and $7 R 3$
But 3 is not related to 3 as $| 3 -3 | = 0$ .

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