Tiêu đề 1. Solution to Refresher Exam 3 - MSTEC Part.pdf ✅

SITUATION: Given the partial fraction below: x 2 − 7x + 4 (x + 2)(x 2 − x − 12) = A (x + 2) + B (x − 4) + C (x + 3) 1. Determine the value of A. A. -11/3 B. -4/21 C. 34/7 D. -7/2 2. Determine the value of B. A. -11/3 B. -4/21 C. 34/7 D. -7/2 3. Determine the value of C. A. -11/3 B. -4/21 C. 34/7 D. -7/2 Solution: SOLVE A: (x + 2) [ x 2 − 7x + 4 (x + 2)(x 2 − x − 12) = A (x + 2) + B (x − 4) + C (x + 3) ] x 2 − 7x + 4 (x 2 − x − 12) = A + B(x + 2) (x − 4) + C(x + 2) (x + 3) Let x = -2 so that x + 2 = 0 Evaluate: x 2 − 7x + 4 (x 2 − x − 12) when x = −2 − 11 3 = A + 0 + 0 A = − 11 3 SOLVE B: (x − 4) [ x 2 − 7x + 4 (x + 2)(x 2 − x − 12) = A (x + 2) + B (x − 4) + C (x + 3) ] x 2 − 7x + 4 (x + 2)(x + 3) = A(x − 4) (x + 2) + B + C(x − 4) x + 3 Let x = 4 so that x – 4 = 0 Evaluate: x 2 − 7x + 4 (x + 2)(x + 3) when x = 4 − 4 21 = 0 + B + 0 B = − 4 21
SOLVE C: (x + 3) [ x 2 − 7x + 4 (x + 2)(x 2 − x − 12) = A (x + 2) + B (x − 4) + C (x + 3) ] x 2 − 7x + 4 (x + 2)(x − 4) = A(x + 3) (x + 2) + B(x + 3) (x − 4) + C Let x = -3 so that x + 3 = 0 Evaluate: x 2 − 7x + 4 (x + 2)(x − 4) when x = −3 34 7 = C 4. Solve for the inequality: |2x − 4| < 10 A. All real numbers B. Null set C. (-3,7) D. [-3,7] Solution: −10 < 2x − 4 < 10 −6 < 2x < 14 −3 < x < 7 5. What is the sum of the coefficients of the expansion of (x + y –z)8? A. 0 B. 1 C. 2 D. 3 Sum of coeffcients = (1 + 1 − 1) 8 = 1 6. Of what quadrant is A, if sec(A) is negative and csc(A) is positive? A. I B. II C. III D. IV Secant functions are negative on quadrants II and III, while cosecant functions are positive on quadrants I and II. Therefore, the answer is quadrant II 7. A force F = 1000kN passes through the point (1, 2, 5) to (7, 8, 10). Determine the magnitude of the moment of the force about the origin. A. 1233.78 B. 3305.16 C. 3490 D. 2313.90 vf =< (7 − 1) (8 − 2) (10 − 5) > = < 6 6 5 > λ = < 6 6 5 > √6 2 + 6 2 + 5 2 =< 0.609 0.609 0.508 > F = 1000 < 0.609 0.609 0.508 > = < 609 609 508 > r =< (1 − 0) (2 − 0) (5 − 0) > = < 1 2 5 > M = rxF M = < 1 2 5 > x < 609 609 508 >

10. The sum of all edges of a rectangular prism is 80cm. If the length of the longest rod that fits inside the prism is 12cm, determine the surface area. A. 179 B. 256 C. 203 D. 288 4(L + W + H) = 80 L + W + H = 20 L 2 + W2 + H 2 = 122 = 144 (L + W + H) 2 = L 2 + W2 + H 2 + 2(LW + LH + WH) 202 = 122 + SA SA = 256cm2 11. Find the Taylor Series for f(x) = x3 – 10x2 + 6 about x = 3. A. -57 – 13(x-3) – (x-3)2 + (x-3)3 C. -57 – 23(x-3) – (x-3)2 + 2(x-3)3 B. -57 – 33(x-3) – (x-3)2 + 3(x-3)3 D. -57 – 33(x-3) – (x-3)2 + 4(x-3)3 To find the Taylor series for the function f(x) = x 3 − 10x 2 + 6 about x = 3, we first need to compute the derivatives of the function at x = 3 and then use the Taylor series formula. The Taylor series expansion of a function f(x) about “x = a” is given by: f(x) = f(a) + f ′(a)(x − a) + f ′′(a) 2! (x − a) 2 + f ′′′(a) 3! (x − a) 3 ... Step 1: Calculate f(3) f(3) = 3 3 − 10(3) 2 + 6 = −57 Step 2: Calculate first derivative of f(x) substitute x = 3 f ′(x) = 3x 2 − 20x f ′(3) = 3(3) 2 − 20(3) = −33 Step 3: Calculate second derivative of f(x) substitute x = 3 f ′ ′(x) = 6x − 20 f ′′(3) = 6(3) − 20 = −2 Step 4: Calculate third derivative of f(x) substitute x = 3 f′′ ′ (x) = 6 f ′′ ′(3) = 6 Step 5: Calculate fourth derivative of f(x) substitute x = 3 f′′′′(x) = 0 f′′′′(3) = 0 Step 6: As can be seen from the pattern, the fourth derivative onwards will always be zero.