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Post a LessonAnswered on 13 Apr Learn Unit I: Relations and Functions
Nazia Khanum
A relation R:A→AR:A→A is said to be reflexive if, for every element aa in the set AA (where AA is a non-empty set), the ordered pair (a,a)(a,a) belongs to the relation RR. In simpler terms, reflexive relations include every element paired with itself in the set.
Answered on 13 Apr Learn Unit I: Relations and Functions
Nazia Khanum
A relation R:A→AR:A→A is said to be symmetric if for every pair of elements a,ba,b in set AA, whenever (a,b)(a,b) is in RR, then (b,a)(b,a) must also be in RR. In other words, if aa is related to bb, then bb must be related to aa as well, for all a,ba,b in AA.
read lessAnswered on 13 Apr Learn Unit I: Relations and Functions
Nazia Khanum
A universal relation in the context of relational databases refers to a relation (or table) that contains all possible combinations of tuples from the sets involved. In simpler terms, it includes every possible pair of elements from its constituent sets.
For example, let's consider a universal relation that represents the Cartesian product of the sets A = {1, 2} and B = {x, y}. The universal relation would contain all possible combinations of elements from A and B:
Universal Relation: (1, x) (1, y) (2, x) (2, y)
In this example, the universal relation contains all possible combinations of elements from set A and set B.
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Answered on 13 Apr Learn Unit I: Relations and Functions
Nazia Khanum
To prove that the function f:R→Rf:R→R given by f(x)=2xf(x)=2x is one-to-one (injective), we need to show that if f(x1)=f(x2)f(x1)=f(x2), then x1=x2x1=x2 for all x1,x2x1,x2 in the domain.
Let's assume f(x1)=f(x2)f(x1)=f(x2): 2x1=2x22x1=2x2
Now, we'll solve for x1x1 and x2x2: x1=x2x1=x2
Since x1=x2x1=x2, it means that for any two inputs x1x1 and x2x2 that produce the same output under the function f(x)=2xf(x)=2x, those inputs must be the same. This proves that the function f(x)=2xf(x)=2x is one-to-one.
Answered on 13 Apr Learn Unit I: Relations and Functions
Nazia Khanum
To prove that (f+g)∘h=f∘h+g∘h(f+g)∘h=f∘h+g∘h, let's start by understanding what (f+g)∘h(f+g)∘h means:
(f+g)∘h(x)=(f+g)(h(x))=f(h(x))+g(h(x))(f+g)∘h(x)=(f+g)(h(x))=f(h(x))+g(h(x))
Now, let's find (f∘h+g∘h)(x)(f∘h+g∘h)(x):
f∘h(x)+g∘h(x)=f(h(x))+g(h(x))f∘h(x)+g∘h(x)=f(h(x))+g(h(x))
This expression is identical to what we found for (f+g)∘h(x)(f+g)∘h(x). Hence, we can conclude that (f+g)∘h=f∘h+g∘h(f+g)∘h=f∘h+g∘h.
Answered on 13 Apr Learn Unit I: Relations and Functions
Nazia Khanum
To find (g∘f)(x)(g∘f)(x), which is the composition of g(x)g(x) with f(x)f(x), we substitute f(x)f(x) into g(x)g(x) wherever we see xx.
Given:
f(x)=∣x∣f(x)=∣x∣ g(x)=∣5x+1∣g(x)=∣5x+1∣
We first find f(x)f(x):
f(x)=∣x∣f(x)=∣x∣
And then substitute it into g(x)g(x):
g(f(x))=∣5(∣x∣)+1∣g(f(x))=∣5(∣x∣)+1∣
Now, ∣x∣∣x∣ can be either xx if x≥0x≥0 or −x−x if x<0x<0.
So, ∣5(∣x∣)+1∣∣5(∣x∣)+1∣ will be:
If x≥0x≥0: g(f(x))=∣5x+1∣g(f(x))=∣5x+1∣
If x<0x<0: g(f(x))=∣−5x+1∣g(f(x))=∣−5x+1∣
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Answered on 06 Apr Learn Unit III: Calculus
Sadika
To find the principal value of tan−1(1)tan−1(1), we need to determine the angle whose tangent is equal to 1.
Since tangent is defined as the ratio of the opposite side to the adjacent side in a right triangle, we can consider a right triangle where the angle whose tangent is 1 is one of its acute angles.
In a right triangle, if the ratio of the opposite side to the adjacent side is 1, then the opposite side and the adjacent side are equal in length. Therefore, we have a triangle with legs of equal length.
The angle whose tangent is 1 corresponds to a 45-degree angle (or π44π radians) in standard position.
So, the principal value of tan−1(1)tan−1(1) is π44π radians.
In LaTeX code: tan−1(1)=π4tan−1(1)=4π
Answered on 06 Apr Learn Unit III: Calculus
Sadika
To solve the equation tan−1(2x)+tan−1(3x)=π4tan−1(2x)+tan−1(3x)=4π, we'll use the tangent addition formula:
tan(α+β)=tanα+tanβ1−tanα⋅tanβtan(α+β)=1−tanα⋅tanβtanα+tanβ
Let α=tan−1(2x)α=tan−1(2x) and β=tan−1(3x)β=tan−1(3x). Then, we have:
tan(α+β)=2x+3x1−2x⋅3xtan(α+β)=1−2x⋅3x2x+3x
tan(α+β)=5x1−6x2tan(α+β)=1−6x25x
Given that tan(α+β)=π4tan(α+β)=4π, we have:
5x1−6x2=π41−6x25x=4π
Cross-multiply:
5x⋅4=π(1−6x2)5x⋅4=π(1−6x2)
20x=π−6πx220x=π−6πx2
6πx2+20x−π=06πx2+20x−π=0
Now, solve this quadratic equation for xx. We can use the quadratic formula:
x=−b±b2−4ac2ax=2a−b±b2−4ac
where a=6πa=6π, b=20b=20, and c=−πc=−π:
x=−20±(20)2−4⋅6π⋅(−π)2⋅6πx=2⋅6π−20±(20)2−4⋅6π⋅(−π)
x=−20±400+24π212πx=12π−20±400+24π2
x=−20±400+24π212πx=12π−20±400+24π2
So, the solutions for xx are:
x=−20+400+24π212πx=12π−20+400+24π2
x=−20−400+24π212πx=12π−20−400+24π2
These are the solutions for the given equation.
In LaTeX code:
x=−20+400+24π212πx=12π−20+400+24π2
x=−20−400+24π212πx=12π−20−400+24π2
Answered on 06 Apr Learn Matrices
Sadika
In a skew-symmetric matrix, also known as an antisymmetric matrix, the diagonal elements are all equal to zero.
Formally, a matrix AA is skew-symmetric if it satisfies the condition AT=−AAT=−A, where ATAT denotes the transpose of matrix AA.
In a skew-symmetric matrix, for any diagonal element aiiaii, it must satisfy aii=−aiiaii=−aii. The only number that satisfies this condition is zero.
Therefore, every diagonal element of a skew-symmetric matrix is zero.
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Answered on 06 Apr Learn Matrices
Sadika
If a matrix AA is both symmetric and skew-symmetric, then AA will be the zero matrix.
Let's denote AA as the matrix:
A=[aij]A=[aij]
Symmetric Matrix: A matrix is symmetric if it is equal to its transpose. Mathematically, AT=AAT=A. In other words, for every ii and jj, aij=ajiaij=aji.
Skew-Symmetric Matrix: A matrix is skew-symmetric if its transpose is equal to the negative of itself. Mathematically, AT=−AAT=−A. In other words, for every ii and jj, aij=−ajiaij=−aji.
Combining these two conditions, we have aij=ajiaij=aji and aij=−ajiaij=−aji.
The only number that satisfies both conditions simultaneously is aij=0aij=0, because it's the only number that is equal to its negative.
Therefore, in this case, every element of matrix AA must be zero, making AA the zero matrix.
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