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Answered 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. read more

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.

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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 more

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.

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Answered 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... read more

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:

yaml
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... read more

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.

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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... read more

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.

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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,... read more

To find (g∘f)(x)(gf)(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 I: Relations and Functions

Sadika

A bijective function, also known as a bijection, is a type of function between two sets, where each element in the domain (input) set is paired with exactly one element in the codomain (output) set, and vice versa. In other words, a bijective function establishes a one-to-one correspondence between... read more

A bijective function, also known as a bijection, is a type of function between two sets, where each element in the domain (input) set is paired with exactly one element in the codomain (output) set, and vice versa. In other words, a bijective function establishes a one-to-one correspondence between the elements of the domain and the elements of the codomain.

Formally, a function f:A→Bf:A→B is bijective if and only if:

  1. Every element in set BB is mapped to by exactly one element in set AA, i.e., for every yy in BB, there exists a unique xx in AA such that f(x)=yf(x)=y.
  2. Every element in set AA is mapped to by exactly one element in set BB, i.e., for every xx in AA, there exists a unique yy in BB such that f(x)=yf(x)=y.

In simpler terms, a bijective function is both injective (or one-to-one) and surjective (or onto). It means that each element in the codomain is paired with exactly one element in the domain, and there are no "extra" elements left in either set without a corresponding match in the other set.

Graphically, a bijective function can be represented by a plot where each input value has a unique corresponding output value, and there are no horizontal or vertical line tests that intersect the graph at more than one point.

 
 
 
 
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Answered on 06 Apr Learn Unit III: Calculus

Sadika

Inverse trigonometric functions are functions that "undo" the effects of trigonometric functions. They provide a way to find the angle (or value) associated with a given trigonometric ratio. Inverse trigonometric functions are denoted by \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), \(\tan^{-1}(x)\), \(\cot^{-1}(x)\),... read more

Inverse trigonometric functions are functions that "undo" the effects of trigonometric functions. They provide a way to find the angle (or value) associated with a given trigonometric ratio. Inverse trigonometric functions are denoted by \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), \(\tan^{-1}(x)\), \(\cot^{-1}(x)\), \(\sec^{-1}(x)\), and \(\csc^{-1}(x)\), representing arcsine, arccosine, arctangent, arccotangent, arcsecant, and arccosecant, respectively.

Here's a brief explanation of each inverse trigonometric function:

1. **arcsin (or \(\sin^{-1}(x)\))**: Gives the angle whose sine is \(x\), where \(x\) is between -1 and 1.

2. **arccos (or \(\cos^{-1}(x)\))**: Gives the angle whose cosine is \(x\), where \(x\) is between -1 and 1.

3. **arctan (or \(\tan^{-1}(x)\))**: Gives the angle whose tangent is \(x\).

4. **arccot (or \(\cot^{-1}(x)\))**: Gives the angle whose cotangent is \(x\).

5. **arcsec (or \(\sec^{-1}(x)\))**: Gives the angle whose secant is \(x\), where \(x \geq 1\) or \(x \leq -1\).

6. **arccsc (or \(\csc^{-1}(x)\))**: Gives the angle whose cosecant is \(x\), where \(x \geq 1\) or \(x \leq -1\).

It's important to note that the range of inverse trigonometric functions is restricted to ensure that they are single-valued and have unique inverses. The specific range depends on the convention used, but commonly accepted ranges are as follows:

- For arcsin and arccos: \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\) (or \(-90^\circ \leq \theta \leq 90^\circ\)).
- For arctan: \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\) (or \(-90^\circ < \theta < 90^\circ\)).
- For arccot: \(0 < \theta < \pi\) (or \(0^\circ < \theta < 180^\circ\)).
- For arcsec and arccsc: \(0 \leq \theta < \frac{\pi}{2}\) and \(\frac{\pi}{2} < \theta \leq \pi\) (or \(0^\circ \leq \theta < 90^\circ\) and \(90^\circ < \theta \leq 180^\circ\)).

These functions are essential in solving trigonometric equations, modeling periodic phenomena, and various applications in science, engineering, and mathematics.

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Answered on 06 Apr Learn Unit III: Calculus

Sadika

To find the principal value of sin⁡−1(−3−2)sin⁡−1(−32)sin−1(−3−2)sin−1(−23), we'll start by finding the individual principal values of sin⁡−1(−3−2)sin−1(−3−2 ) and sin⁡−1(−32)sin−1(−23),... read more

To find the principal value of sin⁡−1(−3−2)sin⁡−1(−32)sin−1(−3−2)sin−1(−23), we'll start by finding the individual principal values of sin⁡−1(−3−2)sin−1(−3−2

) and sin⁡−1(−32)sin−1(−23), and then multiply them together.

  1. sin⁡−1(−3−2)sin−1(−3−2
  1. ):

Since sin⁡−1sin−1 gives an angle whose sine is equal to the given value, we need to find an angle θθ such that sin⁡θ=−3−2sinθ=−3−2

.

However, the sine function only returns values between -1 and 1. Therefore, −3−2−3−2

is outside the range of the sine function, and there is no real angle θθ for which sin⁡θ=−3−2sinθ=−3−2. Hence, sin⁡−1(−3−2)sin−1(−3−2

) is undefined.

  1. sin⁡−1(−32)sin−1(−23):

Again, since sin⁡−1sin−1 gives an angle whose sine is equal to the given value, we need to find an angle ϕϕ such that sin⁡ϕ=−32sinϕ=−23.

Similar to the previous case, −32−23 is outside the range of the sine function, and there is no real angle ϕϕ for which sin⁡ϕ=−32sinϕ=−23. Hence, sin⁡−1(−32)sin−1(−23) is also undefined.

Therefore, the principal value of sin⁡−1(−3−2)sin⁡−1(−32)sin−1(−3−2

)sin−1(−23) is undefined.

 
 
 
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Answered on 06 Apr Learn Unit III: Calculus

Sadika

To find the principal value of cos⁡−1(3−2)cos⁡−1(−32)cos−1(3−2)cos−1(−23), we'll start by finding the individual principal values of cos⁡−1(3−2)cos−1(3−2 ) and cos⁡−1(−32)cos−1(−23), and then multiply them... read more

To find the principal value of cos⁡−1(3−2)cos⁡−1(−32)cos−1(3−2)cos−1(23), we'll start by finding the individual principal values of cos⁡−1(3−2)cos−1(3−2

) and cos⁡−1(−32)cos−1(23), and then multiply them together.

  1. cos⁡−1(3−2)cos−1(3−2
  1. ):

Since cos⁡−1cos−1 gives an angle whose cosine is equal to the given value, we need to find an angle θθ such that cos⁡θ=3−2cosθ=3−2

.

However, the cosine function only returns values between -1 and 1. Therefore, 3−23−2

is outside the range of the cosine function, and there is no real angle θθ for which cos⁡θ=3−2cosθ=3−2. Hence, cos⁡−1(3−2)cos−1(3−2

) is undefined.

  1. cos⁡−1(−32)cos−1(23):

Similarly, −32−23 is outside the range of the cosine function, and there is no real angle ϕϕ for which cos⁡ϕ=−32cosϕ=−23. Hence, cos⁡−1(−32)cos−1(23) is also undefined.

Therefore, the principal value of cos⁡−1(3−2)cos⁡−1(−32)cos−1(3−2

)cos−1(23) is undefined.

 
 
 
 
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