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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 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:
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.
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 \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), \(\tan^{-1}(x)\), \(\cot^{-1}(x)\), \(\sec^{-1}(x)\), 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.
read lessAnswered on 06 Apr Learn Unit III: Calculus
Sadika
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Answered on 06 Apr Learn Unit III: Calculus
Sadika
To determine the principal value of
Answered on 06 Apr Learn Unit III: Calculus
Sadika
Let's denote \( \tan^{-1} \left( \frac{x - 1}{x - 2} \right) \) as \( \alpha \) and \( \tan^{-1} \left( \frac{x + 1}{x + 2} \right) \) as \( eta \).
Given that \( \tan^{-1} \left( \frac{x - 1}{x - 2} \right) + \tan^{-1} \left( \frac{x + 1}{x + 2} \right) = \frac{\theta}{4} \), we can use the tangent addition formula:
\[ \tan(\alpha + eta) = \frac{\tan \alpha + \tan eta}{1 - \tan \alpha \cdot \tan eta} \]
Substitute \( \tan \alpha = \frac{x - 1}{x - 2} \) and \( \tan eta = \frac{x + 1}{x + 2} \):
\[ \tan(\alpha + eta) = \frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \frac{x - 1}{x - 2} \cdot \frac{x + 1}{x + 2}} \]
\[ \tan(\alpha + eta) = \frac{\frac{(x - 1)(x + 2) + (x + 1)(x - 2)}{(x - 2)(x + 2)}}{1 - \frac{(x - 1)(x + 1)}{(x - 2)(x + 2)}} \]
\[ \tan(\alpha + eta) = \frac{x^2 + x - 2 + x^2 - x - 2}{(x - 2)(x + 2) - (x^2 - 1)} \]
\[ \tan(\alpha + eta) = \frac{2x^2 - 4}{x^2 + 4 - x^2 + 1} \]
\[ \tan(\alpha + eta) = \frac{2x^2 - 4}{5} \]
Given that \( \tan(\alpha + eta) = \frac{\theta}{4} \), we have:
\[ \frac{2x^2 - 4}{5} = \frac{\theta}{4} \]
\[ 8x^2 - 16 = 5\theta \]
\[ 8x^2 = 5\theta + 16 \]
\[ x^2 = \frac{5\theta + 16}{8} \]
\[ x = \pm \sqrt{\frac{5\theta + 16}{8}} \]
So, the value of \( x \) depends on the value of \( \theta \).
In LaTeX code:
\[ x = \pm \sqrt{\frac{5\theta + 16}{8}} \]
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 together.
Since cos−1cos−1 gives an angle whose cosine is equal to the given value, we need to find an angle θθ such that cosθ=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 cosθ=3−2cosθ=3−2. Hence, cos−1(3−2)cos−1(3−2
) is undefined.
Similarly, −32−23 is outside the range of the cosine function, and there is no real angle ϕϕ for which cosϕ=−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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Answered on 06 Apr Learn Matrices
Sadika
The identity matrix of order n is denoted by I∩
It is a square matrix with dimensions n×nn×n where all the elements on the main diagonal (from the top left to the bottom right) are 1, and all other elements are 0.
Answered on 06 Apr Learn Matrices
Sadika
A square matrix is a matrix that has the same number of rows and columns. In other words, it is a matrix where the number of rows is equal to the number of columns.
For example, a 3×33×3 matrix and a 4×44×4 matrix are both square matrices because they have 3 rows and 3 columns, and 4 rows and 4 columns, respectively.
Square matrices are commonly encountered in various mathematical contexts, such as linear algebra, where they are used to represent linear transformations, systems of linear equations, and many other mathematical structures.
read lessAnswered on 06 Apr Learn Matrices
Sadika
To find the number of all possible matrices of order 3×33×3 with each entry being either 0 or 1, we can consider that each entry in the matrix has 2 choices (0 or 1), and there are a total of 3×3=93×3=9 entries in the matrix.
Therefore, the total number of possible matrices is 2929, because for each entry, there are 2 choices, and we multiply these choices together for all 9 entries.
So, the number of all possible matrices of order 3×33×3 with each entry being either 0 or 1 is 29=51229=512.
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Answered on 06 Apr Learn Matrices
Sadika
Two matrices A=[aij]A=[aij] and B=[bij]B=[bij] are said to be equal if they have the same dimensions (i.e., the same number of rows and columns) and if each corresponding entry of matrix AA is equal to the corresponding entry of matrix BB.
Formally, two matrices AA and BB are equal if and only if:
In other words, matrices AA and BB are equal if they have the same size and if the corresponding entries in each position are equal.
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