General Quiz Structure Overview

<p>1 / √1-x²</p> Signup and view all the answers

<p>± 1</p> Signup and view all the answers

<p>linear function</p> Signup and view all the answers

<p>P(A ∩ B) = 1/2 P(B)</p> Signup and view all the answers

<p>2, 1</p> Signup and view all the answers

<p>i + j – 2k</p> Signup and view all the answers

<p>√a² + c²</p> Signup and view all the answers

<p>3</p> Signup and view all the answers

<p>AB = O</p> Signup and view all the answers

<p>No, f(x) is not differentiable at x = 0.</p> Signup and view all the answers

<p>We have that y = √tan√x. To differentiate this function, we will use the chain rule. Using the chain rule repeatedly, we can write:</p> <p>d/dx(√tan√x) = (1/2√tan√x) * d/dx(tan√x)</p> <p>= (1/2√tan√x) * (sec²√x) * d/dx(√x)</p> <p>= (1/2√tan√x) * (sec²√x) * (1/2√x)</p> <p>= (sec²√x)/(4√x√tan√x)</p> <p>= (sec²√x)/(4√x√tan√x) * (sec²√x)/(sec²√x)</p> <p>= (sec⁴√x)/(4√x√tan√x * sec²√x)</p> <p>= (1 + tan²√x)²/(4√x√tan√x * sec²√x)</p> <p>= (1 + y⁴)/(4y)</p> <p>Therefore the result is verified.</p> Signup and view all the answers

<p>To find the maximum and minimum values of the function f(x), we must get the critical points and analyze them. To get the critical points, we have to find the roots of f'(x) = 0. If we derivative f(x), we will get:</p> <p>f'(x) = 12x² - 36x + 27</p> <p>This is a quadratic function and we can search for its root through Vieta's formula: a + b = 36/12 and a * b = 27/12</p> <p>Solving this system of equations, we get:</p> <p>a = 3/2 and b = 3/2.</p> <p>Thus, f'(x) = 12(x - 3/2)² and we can see that f'(x) is always greater or equal to zero, and it is zero only at x = 3/2.</p> <p>Therefore, we have found the only critical point of the function and its derivative f'(x) is always positive. This ensures that f(x) is always increasing and has no maximum or minimum values.</p> <p>Therefore, it is proven that the function f(x) = 4x³ – 18x² + 27x – 7 has neither maximum nor minimum values.</p> Signup and view all the answers

<p>To solve this integral, we will use integration by parts. First, we set u = x and dv = √1 + 2x dx. Thus, du = dx and v = (1/3)(1 + 2x)³/².</p> <p>Applying integration by parts, we obtain:</p> <p>∫ x√1+2x dx = x * (1/3)(1 + 2x)³/² - ∫ (1/3)(1 + 2x)³/² dx</p> <p>To solve the new integral, we can use the substitution method, setting u = 1 + 2x. This yields du = 2dx, and dx = du/2.</p> <p>Substituting, we get:</p> <p>∫ x√1+2x dx = x * (1/3)(1 + 2x)³/² - ∫ (1/3)(1 + 2x)³/² dx = x * (1/3)(1 + 2x)³/² - ∫ (1/3)u³/² * (du/2)</p> <p>= x * (1/3)(1 + 2x)³/² - (1/6) ∫ u³/² du</p> <p>= x * (1/3)(1 + 2x)³/² - (1/6) * (2/5)u⁵/² + C</p> <p>= x * (1/3)(1 + 2x)³/² - (1/15)(1 + 2x)⁵/² + C</p> <p>Therefore, the solution to the integral is: ∫ x√1+2x dx = x * (1/3)(1 + 2x)³/² - (1/15)(1 + 2x)⁵/² + C</p> Signup and view all the answers

<p>To solve this integral, we will use the substitution method. We set u = √x. This yields du = (1/2√x) dx, and dx = 2√x du.</p> <p>Now, we need to change the limits of integration:</p> <p>When x = 0, u = √0 = 0</p> <p>When x = π/2, u = √π/2.</p> <p>Substituting, we get:</p> <p>∫[0, π/2] sin√x dx = ∫[0, √π/2] sin(u) * 2√x du = 2 ∫[0, √π/2] u sin(u) du</p> <p>To solve the new integral, we can use integration by parts, setting u = u and dv = sin(u) du. Thus, du = du and v = -cos(u).</p> <p>Applying integration by parts, we obtain</p> <p>2∫[0, √π/2] u sin(u) du = - 2u cos(u) |_[0, √π/2] + 2∫[0, √π/2] cos(u) du</p> <p>= [- 2√π/2 * cos(√π/2) + 0] + [2sin(u) |_[0, √π/2]</p> <p>= - √π cos(√π/2) + 2sin(√π/2) - 0 = 2 sin(√π/2) - √π cos(√π/2) .</p> <p>Therefore, the solution to the integral is: ∫[0, π/2] sin√x dx = 2 sin(√π/2) - √π cos(√π/2).</p> Signup and view all the answers

<p>Given that (a + b) . a = 0, we can expand the dot product to get: a . a + b . a = 0</p> <p>Since the dot product is commutative, we can rewrite this as:</p> <p>a . a + a . b = 0</p> <p>Similarly, we can expand the dot product (2a + b) . b = 0 to get</p> <p>2a . b + b . b = 0</p> <p>Let's denote the magnitude of a as |a| and the magnitude of b as |b|. Using the properties of the dot product, we can simplify the above equations:</p> <p>|a|² + a . b = 0</p> <p>2a . b + |b|² = 0</p> <p>Solving the first equation for a . b, we get:</p> <p>a . b = -|a|²</p> <p>Substituting this value of a . b into the second equation, we get:</p> <p>2(-|a|²) + |b|² = 0</p> <p>-2|a|² + |b|² = 0</p> <p>|b|² = 2|a|²</p> <p>Taking the square root of both sides, we obtain:</p> <p>|b| = √2 |a|</p> <p>Therefore, it is proven that |b| = √2 |a|.</p> Signup and view all the answers

<p>In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal AC is the same as the midpoint of diagonal BD.</p> <p>Let's find the midpoint of diagonal BD. Using the midpoint formula, we get</p> <p>Midpoint of BD = [(2 + 3)/2, (-4 - 6)/2, (5 + 2)/2] = (5/2, -5, 7/2)</p> <p>Since the midpoint of AC is the same as the midpoint of BD, we have that the midpoint of AC is also (5/2, -5, 7/2).</p> <p>Now, we can use the midpoint formula to determine AD. Let the coordinates of AD be (x, y, z).</p> <p>Midpoint of AC = (2 + x)/2, (-4 + y)/2, (5 + z)/2</p> <p>Setting this equal to (5/2, -5, 7/2), we get</p> <p>(2 + x)/2 = 5/2</p> <p>(-4 + y)/2 = -5</p> <p>(5 + z)/2 = 7/2</p> <p>Solving for x, y, and z, we obtain:</p> <p>x = 3</p> <p>y = -6</p> <p>z = 2</p> <p>Therefore, AD = 3i - 6j + 2k</p> <p>The area of parallelogram ABCD is given by</p> <p>Area of ABCD = |AD x AB|</p> <p>=|(3i - 6j + 2k) x (2i - 4j + 5k)|</p> <p>=|(14i + 11j + 2k)|</p> <p>= √(14² + 11² + 2²)</p> <p>= √321</p> <p>Therefore, the area of parallelogram ABCD is √321 square units.</p> Signup and view all the answers

<p>A relation R on a set A is reflexive if (a, a) ∈ R for all a ∈ A. It is symmetric if (a, b) ∈ R implies (b, a) ∈ R. It is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R.</p> <p>Let's analyze if the relation R meets these properties:</p> <p><strong>Reflexivity:</strong></p> <p>We can see that <strong>R is not reflexive</strong> . For example, (1, 1) does not belong to R because |1² - 1²| = 0, which is not less than 8.</p> <p><strong>Symmetry</strong>:</p> <p>Let's take a look at some cases:</p> <p>(2, 1) ∈ R because |2² - 1²| = 3 &lt; 8.</p> <p>(1, 2) ∈ R because |1² - 2²| = 3 &lt; 8.</p> <p>Similarly, all pairs (a, b) and (b, a) with a and b belonging to the set A, will fulfill the condition, therefore, <strong>R is symmetric</strong>.</p> <p>We can see the symmetry by looking at the definition: the condition for a pair (x, y) to belong to R depends on the absolute value of x² - y², which is independent of the order of x and y.</p> <p><strong>Transitivity</strong>:</p> <p>We can see that <strong>R is not transitive</strong>. For example, (1, 3) ∈ R because |1² - 3² | = 8 &lt; 8. Also, (3, 4) ∈ R because |3² - 4² | = 7 &lt; 8. However, (1, 4) does not belong to R because |1² - 4² | = 15 is not less than 8.</p> <p>Therefore, the relation R is not reflexive, is symmetric, and is not transitive.</p> Signup and view all the answers

<p>Given that f(x) = ax + b, and f(1) = 1 and f(2) = 3, we can set up a system of equations:</p> <p>a + b = 1</p> <p>2a + b = 3</p> <p>Subtracting the first equation from the second, we get</p> <p>a = 2</p> <p>Substituting a = 2 into the first equation, we get</p> <p>2 + b = 1 b = -1</p> <p>Therefore, the function f(x) is:</p> <p>f(x) = 2x - 1</p> <p>To check if the function is one-to-one, we need to see if different values of x produce different values of f(x). If we assume that f(x₁) = f(x₂), then</p> <p>2x₁ - 1 = 2x₂ - 1</p> <p>2x₁ = 2x₂ x₁ = x₂</p> <p>Therefore, the function is <strong>one-to-one</strong>. This means that no two different values of x produce the same output value of f(x) in the range of the function.</p> <p>A function is onto if every element in the codomain Y is an image of some element in the domain X. In other words, the range of the function is equal to its codomain. The codomain of the function is R. The output values of the expression 2x - 1 cover the entire R. Therefore, the function is <strong>onto</strong>. This means that the range of the function is equal to its codomain, which is R.</p> <p>Therefore, the function f(x) = 2x - 1 is both one-to-one and onto.</p> Signup and view all the answers

<p>To prove the given result, we will differentiate the given equation implicitly with respect to x:</p> <p>`(√1-x² + √1-y²)’ = [a(x - y)]’</p> <p>(1/2√1-x²)(-2x) + (1/2√1-y²)(-2y)(dy/dx) = a(1 - dy/dx)`</p> <p>Simplifying:</p> <p><code>-x/√1-x² - y(dy/dx)/√1-y² = a - a(dy/dx)</code></p> <p>Rearranging the terms to separate dy/dx:</p> <p><code>y(dy/dx)/√1-y² + a(dy/dx) = a + x/√1-x²</code></p> <p>Factoring out dy/dx:</p> <p><code>dy/dx [y/√1-y² + a] = a + x/√1-x²</code></p> <p>Finally, solving for dy/dx:</p> <p><code>dy/dx = (a + x/√1-x²) / (y/√1-y² + a)</code></p> <p>To simplify the result, we can multiply the numerator and denominator by √1-y²:</p> <p><code>dy/dx = [(a + x/√1-x²) * √1-y²] / [(y/√1-y² + a) *√1-y²]</code></p> <p><code>dy/dx = (a√1-y² + (x√1-y²)/√1-x²) / (y + a√1-y²)</code></p> <p><code>dy/dx = (√1-y² * (a√1-x² + x)) / (√1-x² *(y + a√1-y²))</code></p> <p>Now, we can see that the numerator and denominator contain common factors. Simplifying, we get</p> <p><code>dy/dx = √1-y² / √1-x² = (1 - y²)/(1 - x²)</code></p> <p>Therefore, the result <code>dy/dx = (1 - y²)/(1 - x²)</code> is verified.</p> Signup and view all the answers

<p>To solve this problem, we need perform the following steps:</p> <ol> <li> <p>Convert the equation to a standard form.</p> </li> <li> <p>Find the integrating factor.</p> </li> <li> <p>Multiply the equation by the integrating factor.</p> </li> <li> <p>Integrate both sides and solve for y.</p> </li> <li> <p>Use the initial condition to find the value of the constant of integration.</p> </li> <li> <p>Convert the equation to a standard form. This involves isolating the terms related to y and its derivatives on one side of the equation and the terms related to x on the other:</p> </li> </ol> <p>x² dy/dx - xy = x² cos²(y/2x)</p> <p>Dividing both sides by x²:</p> <p>dy/dx - (y/x) = cos²(y/2x)</p> <p>Now the equation is in the standard form:</p> <p>dy/dx + P(x)y = Q(x)</p> <p>where P(x) = -1/x and Q(x) = cos²(y/2x).</p> <ol start="2"> <li>Find the integrating factor. The integrating factor (IF) is given by:</li> </ol> <p>IF = e^(∫P(x)dx) = e^(∫(-1/x)dx) = e^(-ln|x|) = e^(ln(1/|x|))</p> <p>Since x is positive, we can simplify it to:</p> <p>IF = 1/x</p> <ol start="3"> <li>Multiply the equation by the integrating factor.</li> </ol> <p>Multiplying the equation <code>dy/dx - (y/x) = cos²(y/2x)</code> by 1/x, we get</p> <p>(1/x) * dy/dx - (y/x²)= (1/x)cos²(y/2x)</p> <ol start="4"> <li>Integrate both sides and solve for y. The left side of the equation can be written as a derivative of a product:</li> </ol> <p><code>d/dx(y/x) = (1/x)cos²(y/2x)</code></p> <p>Integrating both sides with respect to x:</p> <p>y/x = ∫(1/x)cos²(y/2x) dx + C`</p> <p>to solve the remaining integral on the right side, we can use the substitution method, setting u = y/2x. This yields du = (1/2x)(dy/dx) - (y/2x²)dx. Rearranging this, we can express dy/dx as:</p> <p>dy/dx = 2xdu + (y/x) = 2xdu + (y/x²)</p> <p>Substituting into the integral and simplifying:</p> <p>y/x = ∫cos²(y/2x)(1/x)(2xdu + (y/x²)) dx + C</p> <p>y/x = ∫2cos²(u)du + C = sin(2u) + C</p> <p>Substituting back u = y/2x and simplifying: y/x = sin(y/x) + C</p> <p>Now, we have an implicit solution for y in terms of x.</p> <ol start="5"> <li>Use the initial condition to find the constant of integration. Let's use the initial condition x = 1 and y = π/2. Substituting into the implicit solution:</li> </ol> <p>π/2 = sin(π/2) + C</p> <p>Simplifying, we get</p> <p>π/2 = 1 + C</p> <p>Therefore, C = π/2 - 1</p> <p>Substituting the value of C back into the implicit solution:</p> <p>y/x = sin(y/x) + π/2 - 1</p> <p>This is the particular solution to the differential equation.</p> Signup and view all the answers