Math Honors Calculus I Final Examination, Fall Semester, 2013

Transcription

1 Math 2 - Honors Calculus I Final Eamination, Fall Semester, 2 Time Allowed: 2.5 Hours Total Marks:. (2 Marks) Find the following: ( (a) 2 ) sin 2. (b) + (ln 2)/(+ln ). (c) The 2-th Taylor polynomial centered at a = for the function (d) f (2) (), where f() = e 2. f() = (6 Marks) Sketch the curve y = + by accomplishing the following steps: (a) Determine the domain. (b) Find the intercepts. (c) Determine symmetry of the function, if any. (d) Compute its asymptotes. (e) Find the intervals where the function is increasing or decreasing. (f) Determine maimum and minimum values, if any. (g) Study the concavity and determine the points of inflection. (h) Sketch the curve.. ( Marks) Find the value of a such that + ( a + a 4. (4 Marks) Suppose a function y is defined by Show that ) = e. y = 2 y +, y() =. [ yn + y n y + y ] =, n y n y n y 4 y where y i = y (i) (), i =,, 2,. (continue to the net page...)

2 5. (4 Marks) If a, b, and c are constants such that find the value of 4a + b c. a 2 + b(e ) + sin(c) =, 6. (4 Marks) Find the point on the parabola y = 2 at which the tangent line cuts from the first quadrant the triangle with the smallest area. y P 2

3 . (2 Marks) Find the following: ( (a) 2 ) sin 2. (b) Solution (ln 2)/(+ln ). Solutions ( 2 ) sin 2 2 sin 2 2 sin 2 Solution Let y = (ln 2)/(+ln ). Then sin sin 2 H 2 sin cos 2 4 sin H 2 cos H 4 sin 2 24 = =. (ln 2) ln ln y + ln = 4 2 sin 2 2 H (ln 2) = ln 2. Thus, by the continuity of the eponential function, we have (ln 2)/(+ln ) y eln y = e ln y = e ln 2 = 2. (c) The 2-th Taylor polynomial centered at a = for the function f() = Solution Denote P 2 () = The function P 2 is a polynomial of degree 2 centered at a =, and f() P 2 () = 24. Thus, P 2 is the 2-th Taylor polynomial centered at a = for the given function f.

4 (d) f (2) (), where f() = e 2. Solution For f() = e 2, we have f() = [ + ( 2 ) + ( 2 ) ( 2 ) 6 ] + 2! 6! = + 5 2! + 2 6! +. Thus, by Taylor s epansion, the coefficient of 2 is f (2) () 2! = 6!, so that f (2) () = 2! 6!. 2. (4 Marks) Sketch the curve y = + by accomplishing the following steps: (a) Determine the domain. (b) Find the intercepts. (c) Determine symmetry of the function, if any. (d) Compute its asymptotes. (e) Find the intervals where the function is increasing or decreasing. (f) Determine maimum and minimum values, if any. (g) Study the concavity and determine the points of inflection. (h) Sketch the curve. Solution The given function is y = f() = + = ( + )( 2 + ). (a) The function f is defined for all since the radical is an odd-degree root, so the domain of f is (, ). (b) The discriminant of the quadratic function 2 + is negative, so it has no real roots. By solving the equation f() = ( + )( 2 + ) =, we get only -intercept at =. The y-intercept is f() =. (c) f is neither even/odd, nor periodic. 4

5 (d) Since f() = ± ± + = + / = ±, ± f does not have any horizontal asymptote. Since f() is bounded in any bounded interval, it does not have any vertical asymptote. However, f has a slant asymptote. In fact, since and [f() ] f() + + / = + =, ( ) ( + ) / ( ) ( ( + ) / + ) 2/ + ( + ) / + 2 ( + ) 2/ + ( + ) / + 2 ( + ) ( + ) 2/ + ( + ) / [( + / ) 2/ + ( + / ) / + =, so y = is a slant asymptote as. Similarly, we can have [f() ] =, so y = is a slant asymptote as. (e) The derivative of f is f () = ( + ) 2/ 2 = 2 ( + ) 2/. Thus, the function f > on R, ecept at = and =. continuous at these two points, we know that f is increasing on R. Since f is (f) By Part (e), f does not change its sign on R, by the First Derivative Test, the function f has no local etrema. 5

6 (g) The second derivative of f is f () = 2 ( 2 )( + ) 5/ 2 + ( + ) 2/ 2 2 = ( + ) 5/, which changes sign at =,. Since f () is positive for large negative, the last epression shows that f () >, if < < or < < ; f () <, if < <. By the Concavity Test, the function f is concave upward on (, ) and (, ), and concave downward on (, ). There are two inflection points at (, f( )) = (, ), (, f()) = (, ). (h) Use the information in Parts (a)-(g), we have the following graph of f. y y =. ( Marks) Find the value of a such that + ( ) a = e. + a Solution The it is an indeterminate form of type. To find the it, let 6

7 ( ) a y =. Then, by applying l Hospital s Rule, + a ( ) a ln y ln a + H ln( a) ln( + a) a +a 2 ( + a) ( a) ( 2 ) ( a)( + a) 2a 2 a 2 ( 2 ) + 2a a 2 ( ) = 2a. /2 Thus, by the continuity of the eponential function, we have ( ) a + a y eln y = e ln y = e 2a. Solving the equation e 2a = e gives a = (4 Marks) Suppose a function y is defined by Show that where y i = y (i) (), i =,, 2,. y = 2 y +, y() =. [ yn + y n y + y ] =, n y n y n y 4 y Solution By the Leibniz formula, differentiating the equation y = 2 y + gives y (n) = [ 2 y + ] (n ) = 2 y (n ) + (n ) 2y (n 2) + Thus, or y (n) () = (n )(n 2)y (n ) (), y n y n = (n )(n 2) = n 2 n. (n )(n 2) 2 2y (n ). 7

8 Hence, so that = y n y n + y n y + y y n y 4 y ) + ( n 2 n = n, ( n n 2 ) ( + 2 ) ( + ) 2 [ yn + y n y + y ] =, n y n y n y 4 y 5. (4 Marks) If a, b, and c are constants such that find the value of 4a + b c. Solution By applying l Hospital s rule, we much have a 2 + b(e ) + sin c =, a 2 + b(e ) + sin c b + c =, H 2a + be + c cos c , otherwise the last it will be divergent. By applying l Hospital s rule again, we have = H 2a + be c 2 sin c = 2a + b, 4 so that = 2a + b, or 2a + b = 4. 4 Hence, 4a + b c = 2(4 b) + b ( b) = (4 Marks) Find the point on the parabola y = 2 at which the tangent line cuts from the first quadrant the triangle with the smallest area. y P 8

9 Solution For the given function y = 2, the derivative is y = 2. Thus, an equation of the tangent line to the curve at = a is y ( a 2 ) = 2a( a), or y = 2a + + a 2. The -intercept of the line is = ( + a 2 )/(2a); and the y-intercept is y = + a 2. Thus, the area cut off by the line in the from the first quadrant is A(a) = ( 2 2a + a 2) ( + a 2 ) = 4 (a + 2a + /a). We need to find the absolute minimum of A on the interval (a, ), since a >. The derivative of A is A = 4 (a2 + 2 /a 2 ) = 4 a 2 (a 4 + 2a 2 ) = 4 a 2 (a 2 + )(a 2 ) so A has one critical number a = on (, ). It follows that A (a) >, if < a < ; A (a) <, if < a <. Thus, the function A has a local minimum value [ /2 ] [ ] A( ) = 2 + ( )/2 + ( ) 2 = 4. Since A is increasing on (, ), and decreasing on (, ), the local minimum is also the absolute minimum. Hence, at the point (a, a 2 ) = (, 2 ), the tangent line of y = 2 cuts from the first quadrant the triangle with the smallest area. 9