Let the tangent to the curve y = x3 - 2x2 - x + 5 at the points P and Q be parallel to the line y = -2x + 3. Find the coordinates of the points P and Q.

Victor, let's knock this out. Here are the solutions to the questions from the image: Question 1: Let the tangent to the curve y = x^3 - 2x^2 - x + 5 at the points P and Q be parallel to the line y = -2x + 3. Find the coordinates of the points P and Q. Step 1: Find the derivative of the curve to get the slope of the tangent. (dy)/(dx) = (d)/(dx)(x^3 - 2x^2 - x + 5) = 3x^2 - 4x - 1 Step 2: The tangent lines are parallel to y = -2x + 3, so their slope is -2. Set the derivative equal to -2. 3x^2 - 4x - 1 = -2 3x^2 - 4x + 1 = 0 Step 3: Solve the quadratic equation for x. We can factor the quadratic: (3x-1)(x-1) = 0. So, 3x-1=0 x = (1)/(3) or x-1=0 x = 1. Step 4: Find the corresponding y-coordinates for each x value using the original curve equation y = x^3 - 2x^2 - x + 5. For x = 1: y = (1)^3 - 2(1)^2 - (1) + 5 = 1 - 2 - 1 + 5 = 3 So, one point is (1, 3). For x = (1)/(3): y = ((1)/(3))^3 - 2((1)/(3))^2 - ((1)/(3)) + 5 y = (1)/(27) - 2((1)/(9)) - (1)/(3) + 5 y = (1)/(27) - (2)/(9) - (1)/(3) + 5 To combine these, find a common denominator, which is 27: y = (1)/(27) - (6)/(27) - (9)/(27) + (135)/(27) y = (1 - 6 - 9 + 135)/(27) = (121)/(27) So, the other point is ((1)/(3), (121)/(27)). The coordinates of the points P and Q are ((1)/(3), (121)/(27)) and (1, 3). Comparing with the options, this matches option (B). The final answer is (B) P((1)/(3), (121)/(27)), Q(1, 3). Question 2: The radius of a circle is increasing at a rate of 2.0 mm/sec. At what rate is the area increasing when radius is 100 mm? Step 1: Write down the formula for the area of a circle. A = r^2 Step 2: Differentiate the area A with respect to time t using the chain rule. (dA)/(dt) = (d)/(dt)( r^2) = 2 r (dr)/(dt) Step 3: Substitute the given values: (dr)/(dt) = 2.0 mm/sec and r = 100 mm. (dA)/(dt) = 2 (100 mm) (2.0 mm/sec) (dA)/(dt) = 400 mm^2/sec Comparing with the options, this matches option (B). The final answer is (B) 400 mm^2/sec. Question 3: Suppose the distance S in centimetre travelled by a body after time t in seconds is given by S = t^2 - 3t + 7. Find the time at which the body comes to rest? Step 1: The body comes to rest when its velocity is zero. Velocity v is the derivative of distance S with respect to time t. v = (dS)/(dt) = (d)/(dt)(t^2 - 3t + 7) = 2t - 3 Step 2: Set the velocity to zero and solve for t. 2t - 3 = 0 2t = 3 t = (3)/(2) = 1.5 sec Comparing with the options, this matches option (B). The final answer is (B) 1.5 sec. Question 4: The turning point of the function y = x(x-1)^2, at x=1. Step 1: Expand the function. y = x(x^2 - 2x + 1) = x^3 - 2x^2 + x Step 2: Find the first derivative (dy)/(dx). (dy)/(dx) = 3x^2 - 4x + 1 Step 3: Evaluate the first derivative at x=1. (dy)/(dx)|_x=1 = 3(1)^2 - 4(1) + 1 = 3 - 4 + 1 = 0 Since (dy)/(dx) = 0 at x=1, it is a critical point. Step 4: Find the second derivative (d^2y)/(dx^2). (d^2y)/(dx^2) = (d)/(dx)(3x^2 - 4x + 1) = 6x - 4 Step 5: Evaluate the second derivative at x=1. (d^2y)/(dx^2)|_x=1 = 6(1) - 4 = 2 Since (d^2y)/(dx^2) > 0 at x=1, the turning point is a minimum point. Comparing with the options, this matches option (A). The final answer is (A) is a minimum point. Question 5: If 3x^2 - 5y^2 + 2xy = 7, find (dy)/(dx)? Step 1: Differentiate both sides of the equation with respect to x using implicit differentiation. (d)/(dx)(3x^2) - (d)/(dx)(5y^2) + (d)/(dx)(2xy) = (d)/(dx)(7) 6x - 10y