calculus-solutions - SMT 2013 Calculus Test Solutions...

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SMT 2013 Calculus Test Solutions February 2, 2013 1. Compute lim x 3 x 2 + 2 x - 15 x 2 - 4 x + 3 . Answer: 4 Solution: Note that x 2 + 2 x - 15 x 2 - 4 x + 3 = ( x - 3)( x + 5) ( x - 3)( x - 1) = x + 5 x - 1 . Then lim x 3 x + 5 x - 1 = 3 + 5 3 - 1 = 4 . 2. Compute all real values of b such that, for f ( x ) = x 2 + bx - 17, f (4) = f 0 (4). Answer: 3 Solution: We have that f (4) = 4 b - 1 and f 0 (4) = 2(4) + b = b + 8. Setting these equal to each other, we see that b = 3 . 3. Suppose a and b are real numbers such that lim x 0 sin 2 x e ax - bx - 1 = 1 2 . Determine all possible ordered pairs ( a, b ). Answer: (2 , 2) and ( - 2 , - 2) Solution: Since this is in an indeterminate form, we can use L’Hˆ opital’s Rule to obtain lim x 0 sin 2 x ae ax - b = 1 2 . However, the numerator goes to zero, so the denominator must also go to zero to give us another indeterminate form. This implies that a = b . Using L’Hˆ opital’s Rule again, we have that lim x 0 2 cos 2 x a 2 e ax = 1 2 . The numerator goes to 2, so the denominator must go to 4. Therefore, a = b = ± 2, giving us ( a, b ) = (2 , 2) and ( - 2 , - 2) . 4. Evaluate Z 4 0 e x dx . Answer: 2 e 2 + 2 Let w = x so that w 2 = x and dx = 2 w dw . Then the integral becomes 2 Z 2 0 we w dw . To find this integral, use integration by parts: u = w du = dw ; dv = e w dw v = e w Z we w dw = uv - Z v du = we w - Z e w dw = ( w - 1) e w . Evaluating 2( w - 1) e w at our limits of integration yields 2 e 2 + 2 .
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SMT 2013 Calculus Test Solutions February 2, 2013 5. Evaluate lim x 0 sin 2 (5 x ) tan 3 (4 x ) (log(2 x + 1)) 5 . Answer: 50 Solution 1: For any function f with f (0) = 0, we know that lim x 0 f ( x ) x = lim x 0 f ( x ) - f (0) x = f 0 (0) . sin(5 x ), tan(4 x ), and log(2 x + 1) are all 0 at x = 0, and their derivatives at 0 are 5, 4, and 2, respectively. So, divide numerator and denominator by x 5 and re-arrange to get lim x 0 sin 2 (5 x ) tan 3 (4 x ) (log(2 x + 1)) 5 = lim x 0 sin(5 x ) x 2 · tan(4 x ) x 3 log(2 x +1) x 5 = 5 2 · 4 3 2 5 = 50 . Solution 2: Recall from Taylor series that if f (0) = 0, then f ( x ) f 0 (0) x when x is small. This allows us to write lim x 0 sin 2 (5 x ) tan 3 (4 x ) (log(2 x + 1)) 5 = lim x 0 (5 x ) 2 (4 x ) 3 (2 x ) 5 = 50 . 6. Compute X k =0 Z π 3 0 sin 2 k x dx. Answer: 3 Bring the sum into the integral, so we have Z π 3 0 X k =0 sin 2 k x dx. The integrand is a geometric series, so the answer is Z π 3 0 1 1 - sin 2 x dx = Z π 3 0 sec 2 x dx = tan π 3 - tan(0) = 3 . 7. The function f ( x ) has the property that, for some real positive constant C , the expression f ( n ) ( x ) n + x + C is independent of n for all nonnegative integers n , provided that n + x + C 6 = 0. Given that f 0 (0) = 1 and Z 1 0 f ( x ) dx = C + ( e - 2), determine the value of C . Note: f ( n ) ( x ) is the n -th derivative of f ( x ), and f (0) ( x ) is defined to be f ( x ). Answer: 3 - e Solution: Since f ( n ) ( x ) / ( n + x + C ) is independent of n , we can say that it is equal to g ( x ). Multiplying by ( n + x + C ), we have that f ( n ) ( x ) = ( n + x + C ) g ( x ) .
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SMT 2013 Calculus Test Solutions February 2, 2013 Taking a derivative with respect to x , we obtain f ( n +1) ( x ) = ( n + x + C ) g 0 ( x ) + g ( x ) .
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