Math 27 Midterm Exam Answer Key PDF

MATH 27 Midterm Exam Answer Key I. (Multiple Choice) 1. D 6. D 2. B 7. C 3. A 8. B 4. C...

MATH 27 Midterm Exam Answer Key I. (Multiple Choice) 1. D 6. D 2. B 7. C 3. A 8. B 4. C 9. B 5. D 10. A II. (Problem Solving) x − 16 1. Without using L’Hopital’s Rule, evaluate lim √. (4 points) x→16 √x − 4 √ √   x − 16 0 x − 16 x+4 (x − 16)( x + 4) Solution. lim √ = lim √ ·√ = lim = lim ( x + 4) = 8 x→16 x−4 0 x→16 x−4 x + 4 x→16 x − 16 x→16 (1pt for each item after the given) ( π cos , x≤2 2. Let f be the function defined by f (x) = x √. Determine if f is continuous at x = 2. If x3 − 2x − 4, x > 2 the function is not continuous at x = 2, determine the type of discontinuity. (4 points) π Solution. Notice that f (2) = cos = 0. 2 π π √ Now, lim− f (x) = lim− cos = cos = 0, and lim+ f (x) = lim+ x3 − 2x − 4 = 2. x→2 x→2 x 2 x→2 x→2 Hence, lim f (x) does not exist. x→2 Therefore, f has an essential discontinuity at 2. (2pts each for the one-sided limits. 1pt for determining the non-existence of the limit. 1pt for the conclusion.) 3. Determine y ′ if √
3 3x2 + x (a) y = √. (4 points) 7− 3x √ √ √ 2 3(6x + 1) − ( 3(3x2 + x)(− 31 · x− 3 )
′ (7 − 3 x) Solution. y = √ (7 − 3 x)2 (2pts for the derivative of the numerator and 1pt for the derivative of denominator; 1pt for the proper use of quotient rule) (b) y = tan4 (5x6 ) + Arcsec (cos x) (4 points) ′ 3 6 2 6 5 1 Solution. y = (4 tan (5x ) sec (5x ))(30x ) + √ · − sin x cos x cos2 x − 1 (2pts each for the derivative of the addends) (c) y = 4csc x log5 (3x4 ) (4 points) 1 Solution. y ′ = 4csc x · 4 · 12x3 + log5 (3x4 ) · 4csc x ln 4 · (− csc x cot x) 3x · ln 5 (3pts for the derivative of the factors; 1pt for the proper use of product rule)  (d) y = x2 cot(x + 2y), where y is a differentiable function of x (5 points) Solution. Dx y = Dx [x2 cot(x + 2y)]   dy 2 2 dy = −x csc (x + 2y) 1 + 2 + cot(x + 2y)(2x) dx dx (3pts for proper implementation of product rule) dy dy + 2x2 csc2 (x + 2y) = 2x cot(x + 2y) − x2 csc2 (x + 2y) dx dx (1pt for distributive property and transposition) dy 2x cot(x + 2y) − x2 csc2 (x + 2y) = (1pt) dx 1 + 2x2 csc2 (x + 2y) dy x sin x 4. Suppose y is a function of x. Find if y =. (4 points) dx Arccot x x sin x Solution. |y| = Arccot x x sin x ln |y| = ln Arccot x ln |y| = ln |x| + ln | sin x| − ln |Arccot x| (1pt for proper implementation logarithmic properties) 1 dy 1 1 1 1 = + · (cos x) − ·− (2.5pts for proper differentiation) y dx x sinx Arccot x 1 + x2  dy x sin x 1 1 1 1 = + · (cos x) + · (0.5pt for the final answer) dx Arccot x x sin x Arccot x 1 + x2   1 5. Evaluate lim x sin. (4 points) x→+∞ x Solution.   1 lim x sin Form: 0 · ∞ x→+∞ x sin x1
0 = lim 1 Form: x→+∞ x 0 (1pt for algebraic manipulation and identification of form) 1 1

= lim cos x−x2 (2pts for correct application of L’Höpital’s rule) x→+∞ 1 − x2   1 = lim cos x→+∞ x = cos(0) = 1 (1pt) 1 6. Use L’Hopital’s Rule to evaluate lim x x. (5 points) x→∞ ln |x| Solution. Let y = x1/x. Then ln |y| =. (1pt) x ln |x|  ∞  Now, lim ln |y| = lim. x→∞ x→∞ x ∞ 1   1 x 1 Applying L’Hopital’s rule, we get lim = lim = 0 (3pts) x→∞ 1 x→∞ x ∞ lim ln |y| Finally, lim x1/x = ex→∞ = e0 = 1. (1pt) x→∞ End of exam Total: 60 points