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]

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]

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

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

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)

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

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

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 7and 7 R 3
But 3 is not related to 3 as | 3 -3 | = 0.