Algebraic Methods Of Solving A Pair Of Linear Equations

Step 1: Find the limit of as approaches 0 from the left: Using the series expansion of and considering the behavior around , this limit evaluates to 1.
Step 2: Find the limit of as approaches 0 from the right: Since near 0 approximates as , the limit becomes . Set for continuity.
Step 3: For continuity at , must equal the limits from both sides, which have been found to be 0. Thus, ensures is continuous at .

Step 1: Set up the binomial expansions.
- We are given , and we are asked to find when the coefficients of and are equal.
Step 2: Use the binomial theorem to find the general term.
The general term in the expansion of is: Thus, the coefficient of is: and the coefficient of is: Step 3: Equate the coefficients of and .
Equating the two coefficients: Simplify: Solve for : Step 4: Find the coefficient of in .
Now, we need to find the coefficient of in the expansion of . The general term in this expansion is: For , the term is: Substitute into this expression: Thus, the coefficient of is .

We are given the function: and we are asked to find , where is the period of the function.
Step 1: To find the period of the function, we need to find the periods of the individual trigonometric functions in the numerator and denominator:
- The period of is ,
- The period of is ,
- The period of is .
The period of the function is the least common multiple (LCM) of these three periods. To find the LCM, we consider the denominators of the periods: - .
The LCM of 5, 3, and 3 is 15. Therefore, the period of the function is:
Step 2: Now that we know , we can find . We substitute into the function: Substitute into the equation: After evaluating the trigonometric functions, we find that:

We are given that is the cube root of unity. The cube roots of unity satisfy the following relations: This implies that:
Step 1: We need to evaluate the expression: Using the fact that , we can substitute this in the numerator and denominator: Thus, the expression becomes:
Step 2: Now, we evaluate the second part of the equation: Similarly, applying , we get: Thus, this expression simplifies to:
Step 3: Since both expressions are now in the same form, it follows that: This implies that the value of the expression is .

Given that , , and are the roots of the equation . If , , and are the roots of the equation , then find the value of .

Let the given cubic equation be

$ \alpha \beta \gamma \alpha(\beta+\gamma) \beta(\gamma+\alpha) \gamma(\alpha+\beta) x^3 + px^2 + qx + r = 0 q \boldsymbol{28}$.

Step 1: Analyze the function under the square roots.
- For , must be non-negative.
- For , must be non-negative.
Step 2: Solve the inequalities.
- implies .
- implies .
Step 3: Determine the intersection of these intervals to find the domain.
- The intersection is .

We are given that is a root of the quadratic equation: Since the coefficients of the quadratic equation are real numbers, the complex roots of the equation occur in conjugate pairs. Thus, the other root of the equation must be .
Step 1: The sum and product of the roots of a quadratic equation are related to the coefficients as follows: - The sum of the roots is , - The product of the roots is . Let the roots of the equation be and . We can now calculate the sum and product of the roots:
Step 2: From the sum of the roots, we know that: Thus, the value of is .

We are given the function: To determine the domain of this function, we need to ensure that all the logarithmic expressions are valid, i.e., the argument of each logarithm must be positive.
Step 1:
The argument of the innermost logarithm, , must be positive: This is a quadratic expression. The discriminant of the quadratic is: Since the discriminant is negative, the quadratic does not have real roots and is always positive for all . Therefore, the argument of is always positive.
Step 2:
Next, the argument of must also be positive. For this to hold, we need: This implies: Solving this inequality: The discriminant of is: Since the discriminant is negative, for all . Thus, the argument of is also always positive.
Step 3:
Finally, the argument of must also be positive, which holds if: This implies: which simplifies to: Solving this inequality: The discriminant of is: The roots of the equation are: Thus, the solution to the inequality is or .
Conclusion:
The domain of the function is .

We are given the function: To find the range of this function, we need to analyze the expression and determine the possible values that can take.
Step 1: Let , then: Multiply both sides by the denominator: This expands to: Rearrange the terms to bring everything to one side: This is a quadratic equation in . For real values of , the discriminant must be greater than or equal to zero.
Step 2: The discriminant of a quadratic equation is given by: In our case, , , and . Therefore: Simplify the discriminant expression: Now expand and simplify further: After simplifying this discriminant, we find that the discriminant must be non-negative for real values of . The discriminant analysis yields the range values for , specifically and . Thus, the range of the function is .

Step 1: Analyze the original determinant, known to be a Vandermonde determinant.
- The determinant simplifies to , which is the product of the differences between the variables.
Step 2: Compare each option's determinant.
- Option (A), (B), and (C) give non-zero determinants as they involve transformations that preserve the structure of the Vandermonde determinant, and none of them result in linear dependency. - Option (D) is notably different because it results in a matrix with linearly dependent rows, as each element in the third row is the constant .
- This makes the determinant zero, which does not match the original determinant unless .

To find the cubic equation whose roots are the squares of the roots of the given equation , we proceed as follows: Step 1: Let the roots of the original equation Let the roots of the original equation be . Then: Step 2: Find the roots of the new equation The roots of the new equation are the squares of the roots of the original equation, i.e., . We need to find the sums and products of these new roots. #Sum of the new roots: #Sum of the products of the new roots: #Product of the new roots: Step 3: Construct the new equation The new cubic equation with roots is: Substituting the values from Step 2: To eliminate fractions, multiply through by : Step 4: Match with the options The equation matches option (3). Final Answer:

Step 1: We are given the equation . First, let's express the equation in terms of tangents using the given condition. We will use trigonometric identities to simplify.
Step 2: Using the tangent addition formula, we know: Next, substitute these expressions into the original equation.
Step 3: Assume symmetry or specific angle relationships that simplify the expressions. By solving the system of trigonometric identities and evaluating the expressions, we find that the sum of the two tangents simplifies to zero: Step 4: Thus, the correct answer is:

Step 1: Start with the given expression: We need to determine the value of such that: This implies the expression is equal to some constant value. We begin by simplifying the given quadratic expression.
Step 2: Simplify the expression. The expression is a ratio of two quadratic polynomials, so we need to check for when the numerator and denominator satisfy the condition that results in the value . For , substitute into the expression: Simplifying the numerator: The denominator simplifies to: Thus, the expression simplifies to: Since the condition for is satisfied when , the correct answer is .

Step 1: Given the argument condition, we rewrite the expression: This equation suggests that the phase difference between and is , implying a rotation in the complex plane.
Step 2: Express the equation in terms of : Multiplying both sides by yields: Expanding and simplifying: Rearrange to isolate : Step 3: Solve for explicitly if necessary or evaluate directly from the established relationship, considering the geometric interpretation of the movements in the complex plane: This calculation is confirmed by plugging in the coordinates derived for and calculating the Euclidean distance to .

Step 1: Differentiating each expression
1. For : Since , we get: Thus, (1) is true. 2. For : Since , we get: Thus, (2) is true. 3. For : Since , we get: Thus, (3) is true. 4. For : Since simplifies to for values where it is defined, its derivative is: Since , this statement is false.
Step 2: Conclusion The false statement is:

Step 1: Determine the total number of four-digit numbers containing the digits 3 and 7 exactly once.
- We need to fill four positions with the digits 3 and 7 appearing exactly once in any two of those positions.
Step 2: Choose the positions for the digits 3 and 7.
- There are ways to select two positions for digits 3 and 7 in the four-digit number.
Step 3: For the remaining two positions, choose digits from the remaining digits (0-9, excluding 3 and 7).
- The first remaining position can be filled with any digit from 0-9, excluding 3 and 7, so there are 8 possible choices for the first remaining digit. - The second remaining position can then be filled with any remaining digit, excluding the previously chosen ones, so there are 7 possible choices for the second remaining digit.
Step 4: Calculate the total number of four-digit numbers.
- The total number of possibilities is . This gives us the total number of ways to assign digits to the four positions.
Step 5: Adjust for the constraint on the first position (thousands place).
- We must ensure that the thousands place (the first digit) cannot be 0, as that would make the number a three-digit number. So, if 0 is chosen for one of the remaining two positions, we need to calculate the possible configurations where 0 is not in the first position. - If 0 is selected for the second position (thousands place), there are 7 choices left for the third and fourth digits. The total number of possibilities for these configurations is .
Step 6: After carefully adjusting for constraints and recalculating, the correct total number is .

Step 1: Consider the equation of the circle . Chords that subtend a right angle at the origin imply that the product of their slopes is .
Step 2: The distance from the origin to any point on the circle is 5 (since the radius of the circle is ).
Step 3: A chord subtending a right angle at the origin means the triangle formed by the chord and lines joining its endpoints with the origin is a right-angled isosceles triangle. Thus, the perpendicular from the origin to the chord bisects the chord.
Step 4: For a chord subtending a right angle at the origin, the midpoint of the chord lies on the locus described by the equation , where is related to the radius of the circle. - Since the radius of the circle is 5, and the locus describes the relationship of the midpoints of chords subtending right angles at the origin, we know that is half the square of the radius of the circle. This gives: Thus,

Step 1: First ensure each guest receives 3 apples, allocating apples and leaving apples to be distributed.
Step 2: Utilize the "stars and bars" theorem to distribute the remaining 5 apples among 4 guests without any restrictions.
- The formula for distributing identical items among groups is given by: - For our case, apples and guests, the formula becomes: Step 3: Calculate which simplifies to:

Step 1: Apply the binomial theorem for fractional exponents to the expression .
- The general term in the expansion is given by: Step 2: Find the fourth term, , by setting (since the term count starts from ).
- Compute: - Simplify using properties of binomial coefficients and powers.
Step 3: Evaluate the binomial coefficient and the negative exponent on .
- involves factorials and gamma functions for fractional parts, leading to:

Lagrange's Mean Value Theorem (LMVT) states that for a function continuous on a closed interval and differentiable on the open interval , there exists at least one in such that: Here, the function is , and the interval is . We are to find where the Mean Value Theorem holds.
Step 1: Calculate and . - For : - For : Step 2: Find the derivative of . Using the product rule, the derivative of is: Using the product rule for three functions: Simplifying each term: Combine like terms: Step 3: Apply the Mean Value Theorem. The Mean Value Theorem states: Substituting and : Thus, . Step 4: Solve for by setting . From , set this equal to 3: Simplifying: Divide through by 3: Use the quadratic formula: Simplifying the discriminant: Thus: So, the value of is:

To find the direction cosines of the line of intersection of the given planes, we compute the cross product of their normal vectors.
Step 1: The normals to the planes and are and , respectively.
Step 2: Compute the cross product using the determinant of the matrix formed by the components of the normal vectors: Thus, .
Step 3: Normalize the vector to obtain the direction cosines: Thus, the direction cosines are:

Step 1: The equation of the hyperbola is given by . To rewrite it in standard form, divide both sides by 20: This represents a hyperbola with and .
Step 2: The asymptotes of the hyperbola are given by the lines: These are the asymptotes of the hyperbola.
Step 3: For a point on the hyperbola, the perpendicular distances and from the point to the asymptotes can be calculated using the formula for the perpendicular distance from a point to a line. For the asymptotes, we have the equations , so the formula for the perpendicular distance from to these asymptotes is: Similarly, the other perpendicular distance is: Step 4: Now, compute and : Now, multiply these two expressions: Using the difference of squares, we get: Step 5: Given that the equation of the hyperbola is , substitute the values for and that satisfy this equation. After solving, we find:

Step 1: The equation of the directrix is given as , which simplifies to: The point lies on the hyperbola.
Step 2: For the hyperbola, the relationship between the point and the directrix is given by: where is a point on the hyperbola and the directrix is . The distance from the point to the directrix is: Step 3: Using the distance formula for the point , the left-hand side of the equation becomes: Now, equating both sides: Step 4: Solving for , we get: Finally, squaring :

Step 1: Complete the square to find the center and radius of the circle. For the equation: complete the square: which simplifies to: Step 2: Analyze the position of point relative to the circle. Substitute into the circle's equation: So, the radius squared is , and we need to ensure is inside the circle.
Step 3: Ensure the circle does not touch the axes. This means both and intercepts must be positive, i.e., the radius must be less than the distance from the center to either axis.
Step 4: Compute the valid range for : and since the center is at , the smallest radius to not touch the axes is the distance to the x-axis (4 units), hence: The correct conditions are and .