Modulus Function
y = |x|
It is defined as
y = x; if x >0
y = âˆ’x; if x <0
y = 0; if x = 0
Despite that in the above equations, we are finding a negative value of x if x < 0, its absolute value can never be negative. This can be seen through the following example:
We are finding out the value of y = |âˆ’5|.
Assuming that â€“ 5 = x, so y = |x|
Now, since x <0, so y = âˆ’x = âˆ’(âˆ’5) = 5
Graphical representation of modulus function
This is the graph of y = |x|
It is observed that for every value of x, be it +ve or â€“ve, the value of y cannot be negative.
Example-1
What is the value of x if |2x + 3| = 9?
Solution
In the questions involving modulus, first the value of expression under the modulus is taken as a positive and then as a negative.
Case 1 When (2x + 3) > 0, or, x > , then |2x + 3|
= 2x + 3
So, 2x + 3 = 9 or, 2x = 6.
So, x = 3
Case 2 When (2x + 3) < 0, or, x < , then |2x + 3|
= âˆ’(2x + 3)
So, âˆ’(2x + 3) = 9, or, âˆ’2x = 12
So, x = âˆ’6
Example-2
What is the value of â€˜xâ€™ if x^{2} + 5|x| + 6 = 0?
Solution
Taking x > 0, x^{2} + 5|x| + 6 = x^{2} + 5x + 6 = (x + 2) (x + 3) = 0
Or, x = âˆ’2 and x = âˆ’3
But as we have assumed that x>0, so x = âˆ’2 and â€“3 are not admissible.
Taking x < 0, x^{2} + 5|x| + 6 = x^{2} âˆ’ 5x + 6 = (x âˆ’ 2) (x âˆ’ 3) = 0
Or, x = 2 and x = 3
But as we have assumed that x<0, so x = 2 and 3 are not admissible.
So, there is no real value of x which can satisfy this equation.
Alternatively, it can be seen that x^{2} and 5|x| and 6, are all positive values. So the sum of these three can never be equal to 0, so no real value of x is possible.
Example-3
If |x^{2} â€“ 5x + 6| > x^{2} â€“ 5x + 6, then find the values of x?
Solution
If |N| > N, then N < 0. (It can be understood by assuming the values)
So, x^{2} â€“ 5x + 6 < 0
Or, (x â€“ 2) (x â€“ 3) < 0
So, 2 < x < 3
Greatest Integer Value Function
y = [x]
It is defined as the largest integral value of x which is less than or equal to x.
It is given y = [3.23] and we have to find the greatest integer value of y.
Taking the second part of the definition, i.e., the integer less than or equal to 3.23, we get a set of integers less than or equal to 3.23 â‡’ 3, 2, 1, 0, âˆ’1,â€¦and so on.
The largest integer among all these integers = 3. So the greatest integer value of [3.23] = 3
Similarly, if we find the greatest integer value of y = [âˆ’2.76], then all the integers less than this value (â€“2.76) = {âˆ’3, âˆ’4, âˆ’5, âˆ’6,â€¦}.
Now the greatest integer among all these integers given in the above set = âˆ’3
It can also be seen through the tabular presentation:
x |
y |
0â€“1 (excluding at x = 1) |
0 |
1â€“2 (excluding at x = 2) |
1 |
2â€“3 (excluding at x = 3) |
2 |
â€‹
Example-4
What is the value of x in the following expression?
[x]^{2} â‰¤ 16?
Solution
4 â‰¤ [x] â‰¤ 4
âˆ’4 â‰¤ [x], or, âˆ’4 â‰¤ x
And [x] â‰¤ 4, or, x < 5.
So, the value of x is: âˆ’4 â‰¤ x < 5.
Logarithmic Function
y = log_{a}xIt is known that the value of x has to be positive here. However, y can have negative values.
Since with an increase in the value of x, y always increases; so y = log_{a}x is an increasing function.
We will discuss more about this function in the logarithm chapter.
Exponential Function
y = e^{x}
^{}
^{}
Exponential function is the inverse of a logarithmic function.
Again it can be observed that the value of y cannot be negative, whatever be the value of x in y = e^{x}
Constant Function
â€‹f(x) = k, (where k is any constant) is known as a constant function.