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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 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 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), and then multiply them together.
Since sin−1sin−1 gives an angle whose sine is equal to the given value, we need to find an angle θθ such that sinθ=−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 sinθ=−3−2sinθ=−3−2. Hence, sin−1(−3−2)sin−1(−3−2
) is undefined.
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 sinϕ=−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 sinϕ=−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 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.
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
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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Answered on 06 Apr Learn Matrices
Sadika
Matrices AA and BB will be inverses of each other only if their product is the identity matrix.
The product of matrices AA and BB, denoted as ABAB, is defined as the matrix obtained by multiplying each row of matrix AA by each column of matrix BB.
If AA and BB are inverses of each other, then AB=IAB=I, where II is the identity matrix.
Similarly, BA=IBA=I must also hold true.
So, matrices AA and BB will be inverses of each other if and only if their product is the identity matrix:
AB=BA=IAB=BA=I
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