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Signum Function is one of the important functions of Relations and Functions. It can be understood using a formula where f(x) signifies signum function. Or this function can also be understood through a graph. The signum function returns values of -1, 0, or 1 when x is negative, zero, or positive respectively. In this article, we will learn about signum function, its graphical representation and the equation associated with it. Along with this, the article also includes a set of sample questions for better understanding of the topic.
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Keyterms: Relation and Functions, Identity Function, Constant Function, Polynomial Function, Rational Function, Modulus Function
Read Also: Complex Numbers and Quadratic Equations
What is a Signum Function?
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Signum function along with the other kinds of special functions, such as, the identity function, constant function, polynomial function, rational function and the modulus function is an important part of Mathematics. The Signum function is denoted through f(x).
Although, mathematically the signum function is defined as:
F(x) = x/|x| ---- (Equation of Signum Function)
In the above equation,
F(x) represents Signum Function which means that,
If x<0, then F(x) = -1
If x=0, then F(x) = 0
If x>0, then F(x) = 1
In other words, this means that, if the value of x is negative, the signum function will also be negative (-1), while if the value of x is positive, then, f (x) shall also be positive (1) and if the value of x is zero, the signum function shall be zero. Therefore, the value of x and f (x) are directly proportional to each other.
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Graphical Representation of Signum Function
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The graph drawn above shows a break in the curve where the value of x is zero. Whereas, apart from that break, the f(x) is seen in continuation for every value of x other than zero. Therefore, for a signum function f(x):
Domain: x ∈ R and
Range: {-1, 0, 1}
Signum Function for Real Number
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The signum function is also called as the derivative of the absolute value function. Therefore, as a result, any real number is eligible to be represented as the product of its absolute value. For instance,
x = sgn(x).|x|
Hence, if the value of x is not equal to zero, then,
sgn(x) = x/|x|
Signum Function for Complex Arguments
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The signum function for any complex number shall be defined as the following.
Let the complex number be ‘a’.
Therefore,
If, a = 0, then,
sgn(a) = 0
If, a is not equal to 0, then,
sgn(a) = a/|a|
Hence, when a ≠ 0, sgn(a) would be projection of a onto a unit circle as:
a ∈ C| |a| = 1
Therefore, for real arguments, the complex signum function tends to reduce itself to real signum function in such a way that you will have:
a sgna- = |a|
where a- refers to the complex conjugate of a.
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Things to Remember
- The signum functions consisting of both the real number and the complex numbers have the tendency to get reduced to the real signum function.
- Another name for the signum function is the sign function. The main usage of the signum function derivation is to extract the sign of the real number.
- The signum function has the property of getting differentiated with the derivative 0 except at 0.
- Some important properties of the signum function are represented as
If x<0, then F(x) = -1
If x=0, then F(x) = 0
If x>0, then F(x) = 1
- The signum function also has a property to be written while using the floor function and the functions of the absolute value. For instance,
Sgn x = {x/|x| + 1 } – { -x /|-x| + 1}
Previous Year Questions
- (10101101)2=(................)10
- A binary sequence is an array of 0′s0′s and 1′s1′s. The number of nn -digit binary sequences which contain even number of 0′s0′s is
- The number of values of rr satisfying the equation 39C3r−1−39Cr2=39Cr2−1−39C3r39C3r−1−39Cr2=39Cr2−1−39C3r is
- On Q−{1}Q−{1} such that a∗b=a+b−aba∗b=a+b−ab. Find the identity element.
- Suppose that the number of elements in set AA is pp, the number of elements in set BB is qq and the number of elements in A×BA×B is 77. Then p2+q2=
- The inverse of the function y=10x−10−x10x+10−xy=10x−10−x10x+10−x is equal to :
- The number of binary operations that can be defined on a set of 2 elements is
- The number of bijective function from a set A to itself when A contains 106 elements is
- If R is a relation on a set R of all real numbers defined by aRb, if |a-b| ≤ 1. Then R is? [BITSAT 2013]
- Use binary multiplication to solve the function. [ KCET 2006]
- Prove the associative and commutative property for the binary operations. [KCET 2006]
- If A={a,b,c} then number of binary operations on A is? [KCET 2020]
- In the group G={1,3,7,9} under multiplication, the inverse of 3 is? [KCET 2005]
- In the group G={1,5,7,11}G={1,5,7,11} under multiplication modulo 1212, the solution of 7−1⊗12(x⊗1211)=57−1⊗12(x⊗1211)=5 is xx =...[KCET 2006]
- In the group (Q+,⋆)(Q+,⋆) of positive rational numbers w.r.t. the binary operation ⋆⋆ defined by a⋆b=ab3,∀a,b∈Q+a⋆b=ab3,∀a,b∈Q+ the solution of the equation 5⋆x=4−15⋆x=4−1 in Q+Q+ is….[KCET 2005]
- The image of the interval [-1, 3] under the mapping f:R→Rf:R→R given by f(x)=4x3−12xf(x)=4x3−12x is
Sample Questions
Ques.From the function f: A →B, which is defined as f(x) = (x - 2)/(x -3). Find out if f is one-one and onto? Here, Let A = R {3} and B = R – {1}. Also, justify your answer. (5 marks)
Ans. The Function is as follows:
f(x) = (x-2)/(x-3)
Now, checking for one-one function.
Here,
f(x1) = (x1 – 2)/(x1 – 3)
f(x2) = (x2 - 2)/(x2 - 3)
Therefore, on putting f(x1) = f(x2)
(x1 - 2)/(x1 - 3) = (x2 - 2)/(x2 - 3)
(x1 - 2)(x2 – 3) = (x1 – 3)(x2 - 2)
x1(x2 – 3) - 2(x2 - 3) = x1(x2 – 2) – 3(x2 – 2)
x1x2 - 3x1 - 2x2 + 6 = x1x2 – 2x1 - 3x2 + 6
- 3x1 – 2x2 = – 2x1 – 3x2
3x2 - 2x2 = – 2x1 + 3x1
x1 = x2
Hence,
If, f (x1) = f (x2),
Then, x1 = x2
Thus, the function f is one - one function.
Now, Checking for onto function:
f (x) = (x-2)/(x-3)
Let f(x) = y such that y B or y ∈ R – {1}
So, y = (x -2) / (x- 3)
y(x -3) = x - 2
xy - 3y = x-2
xy-x = 3y-2
x(y – 1) = 3y- 2
x = (3y – 2)/(y-1)
For y=1, x isn’t defined
But it is given that. y ∈ R – {1}
Therefore,
x = (3y - 2)/(y- 1) ∈ R - {3}
Hence, f is onto.
Ques. Let f : N → Y be a function which is defined as f (x) = 4x + 3, now, here, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. (5 marks)
Ans. Firstly, checking for Inverse:
f(x) = 4x + 3
Let f(x) = y
y = 4x + 3
y – 3 = 4x
4x = y – 3
x = (y − 3)/4
Let g (y) = (y − 3) / 4
where g: Y → N
Now for gof,
Gof = g(f(x))
= g(4x + 3) = [(4x + 3) − 3]/4
= [4x + 3 − 3]/4
=4x/4
=> x = IN
Now for fog,
Fog = f(g(y))
= f[(y − 3)/4]
=4[(y − 3)/4] +3
= y – 3 + 3
= y + 0
= y = Iy
Thus, gof = IN and fog = Iy,
Hence, it shows that f is invertible
Ques. Prove the following. Signum Function f: R→R, given by F(x) = 1, 0, -1 (4 Marks)
Ans. Firstly, check for one - to - one function:
F(0) = 0
F(-1) = -1
F(1) = 1
F(2) = 1
F(3) = 1
Since, F(1), F(2) and F(3), shows the same image,
Therefore, the function is not a one-to-one function.
Check for Onto Function:
For the function, f: R →R
If, x > 0
F(x) = 1
If, x = 0
F(x) = 0
If, x < 0
F(x) = -1
Now, here, the value of f(x) is defined only if x = 1, 0, -1
Therefore, considering any other real numbers such as, y = 2, y = 100, there is no corresponding element x.
Thus, the function “f” is not an onto function.
Hence, it is proven that the given function “f” is neither one - one function nor onto function.
Ques. Given is R {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (2, 3)} and R is the relation in the set (1, 2, 3, 4). Prove that R is reflexive and transitive but not symmetric. (3 Marks)
Ans. R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (2, 3)}
Since, (a, a) ∈ R,
Therefore, for every a ∈ (1, 2, 3, 4)
Hence, this shows that R is reflexive.
Now, (1, 2) ∈ R but (2, 1) is not ∈ R
Hence, R is not symmetric.
And, (a, b), (b, c) ∈ R = (a, c) ∈ R for all (a, b, c) ∈(1, 2, 3, 4)
Therefore, R is transitive.
Hence Proved.
Ques. Illustrate the signum function for complex arguments. (2 Marks)
Ans. The signum function for any complex number shall be defined as the following.
Let the complex number be ‘a’.
Therefore,
If, a = 0, then,
sgn (a) = 0
If, a is not equal to 0, then,
sgn (a) = a /|a|
Ques. Illustrate the signum function for real argument. (2 Marks)
Ans. The signum function is also called as the derivative of the absolute value function. Therefore, as a result any real number is eligible to be represented as the product of its absolute value. For instance,
X = sgn (x).|x|
Hence, if the value of x is not equal to zero, then,
sgn (x) = x/|x|
Ques. Given is R {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (2, 3)} and R is the relation in the set (1, 2, 3, 4). Find out the correct option from the following. (1 mark)
(A) R is reflexive but not transitive
(B) R is an equivalence relation
(C) R is reflexive but not symmetric
(D) R is symmetric but not reflexive
Ans. C) R is reflexive but not symmetric is the correct answer.
Ques. If f: R → R is defined by f(x) = x2 − 3x + 2, find F(f(x)). (3 Marks)
Ans.Given:
f(x) = x2 − 3x + 2
Therefore, to find F(f(x))
F(f(x)) = f(x)2 − 3f(x) + 2.
= (x2–3x+2)2 – 3(x2–3x+2) + 2
Now, by applying the formula
(a – b + c)2 = a2 + b2 + c2 - 2ab + 2ac - 2ab,
We get,
= (x2)2 + (3x)2 + 22– 2x2 (3x) + 2x2(2) – 2x2(3x) – 3(x2 – 3x + 2) + 2
Now, on substituting the values
= x4 + 9x2 + 4 – 6x3 – 12x + 4x2 – 3x2 + 9x – 6 + 2
= x4 – 6x3 + 9x2 + 4x2 – 3x2 – 12x + 9x – 6 + 2 + 4
therefore, we get,
F(f(x)) = x4 – 6x3 + 10x2 – 3x
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