Mat135_PT_T1_solutions - www.prep101.com Mat135 Practice...

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Unformatted text preview: www.prep101.com Mat135 Practice Test 1 Solutions Part A 1. Solution We want to find linear factors that are shared by the numerator and denominator. Notice that: 23 − 22 − 8 ⋅ 2 + 12 = 0 = 23 − 12 ⋅ 2 + 16 So we can divide by ( x − 2) on the top and bottom. Using long division: (x − 2)(x − 2)(x + 3) = x + 3 . x 3 − x 2 − 8 x + 12 ( x − 2) x 2 + x − 6 = = 3 2 (x − 2) x + 2 x − 8 (x − 2)(x − 2)(x + 4) x + 4 x − 12 x + 16 ( ( ) ) x 3 − x 2 − 8 x + 12 2 + 3 5 = =. x → 2 x 3 − 12 x + 16 2+4 6 Hence, lim Therefore, the answer is 2. 5 and (c). 6 Solution Notice that 10 = eln10 , ⇒ y = 102 x+1 = (eln10 )2 x+1 = eln10(2 x+1) Apply the Chain Rule and exponential differentiation: dy d = eln10(2 x+1) (ln10(2 x + 1)) = 2ln10(eln10 ) 2 x+1 = (2ln10)102 x+1. dx dx Therefore, the answer is (b). 3. Solution f(x) = ln(x − 2) ⇒ y = ln (x − 2) First, interchange x and y as follows: x = ln (y − 2) Now solve for y in terms of x: x = ln (y − 2) ⇒ ex = y − 2 ⇒ y = 2 + ex. www.prep101.com −1 x Hence, f (x) = 2 + e . Thus, the answer is (d). 4. Solution We will go about this by constructing the graph in question. Recall the following: To shift horizontally, we replace x in the equation with (x+d) where d is our horizontal shift (d is positive for a shift to the left and negative for a shift to the right.) We add a vertical shift to the “end” of the equation (positive for a shift in the upward direction and negative for a shift in the downward direction.) 3( x + 4 ) f(x) = 2e f(x) = 2e 3 x +12 left by 4 −4+5 +1 upward by 5 Therefore, the answer is (a). 5. Solution We’ll look at the exponent first: −1 1 Considering only the exponent, as x → 0+, − 2 → −∞. I.e. lim+ 2 = −∞ . x →0 x x 1 Now, we’ll let − 2 = t, and re-write the given expression as follows: x 2 lim e −1 / x = lim e t = 0 . x →0+ t → −∞ Therefore, the answer is (a). 6. lim x →∞ Solution x 4x + 2 x = lim x →∞ 4x + 2 x →∞ = lim Therefore, the answer is (b). 7. Solution What is the slope of the line y = 3x − 5? 1 x/ x = lim = (4 x + 2) / x x → ∞ 4 + 2 / x 11 =. 42 www.prep101.com The slope of the line is given by the first derivative, and hence y’(x) = 3. Now we must find a point x where the tangent line to the curve y = x + ln x has a slope of 3. To that end, we first find the derivative function of the curve: 1 y = x + ln x ⇒ y’(x) = 1 + . x The slope of the tangent line to the curve is given by the above function, so we can now solve the point at which this new function is equal to 3: 1 1+ =3 x 1 ⇒ =2 x 1 ⇒x= . 2 Therefore, the answer is (b). 8. Solution Find the slope of the tangent line (i.e. the derivative of the function): y′ y = 7 x ⇒ ln y = ln(7 x ) ⇒ y = x ln 7 ⇒ = ln 7 ⇒ y ′ = y ln 7 = 7 x ln 7 y Now find where this derivative equals ln 49: ln 49 y’(x) = 7x(ln7) = ln 49 ⇒ 7x = = log749 = 2. ln 7 ln 2 Hence, x = log 7 2 = . ln 7 Therefore, the answer is (d). 9. Solution y = xcos x ⇒ ln y = ln(xcos x) = cos xln x Differentiate to get: 1 ⎛1⎞ y’ = (cos x) ⎜ ⎟ + (ln x)(−sin x) y ⎝ x⎠ ⎤ ⎛ cos x ⎞ ⎡ cos x − sin x ln x ⎟ − sin x ln x ⎥ = xcos x ⎜ Thus, y’ = y ⎢ ⎦ ⎝x ⎠ ⎣x Therefore, the answer is (a). www.prep101.com 10. Solution ( ) Let the point of tangency be P a, a − 1 . , we find (0,-3), and (a, a − 1) , the slope of this line is The derivative ⇒ 1 dy = . If the tangent line passes through dx 2 x − 1 a −1 − 0 a −1 = . a − (− 3) a+3 dy must equal this slope at P: dx dy a −1 1 (a) = = dx 2 a −1 a + 3 ⇒ 2 ( a − 1) = a + 3 . Hence a = 5. 1 1 =, 2 5 −1 4 1 Then the equation of this line is given by y = ( x + 3). 4 Therefore, the slope of the tangent line is y y = x −1 Therefore, the answer is (e). −3 1a x www.prep101.com Part B 11. Solution f (x + h ) − f (x ) 3(x + h ) − 3x 3 = lim h →0 h h 3 f ′(x) = lim h→0 = lim h →0 3 ( x 3 + 3 x 2 h + 3 xh 2 + h 3 ) − 3 x 3 h = lim 9 x + 9 xh + 3h 2 = 9 x 2 h→0 12. ( 2 ) h ( 9 x 2 + 9 xh + 3h 2 ) 9 x 2 h + 9 xh 2 + 3h3 = lim = lim h →0 h →0 h h Solution (a) ln y = ln x arcsin x = arcsin x ln x ⎛1⎞ 1 dy ⎛1⎞ ⎟ = (arcsin x )⎜ ⎟ + (ln x )⎜ ⎜ 2⎟ y dx ⎝ x⎠ ⎝ 1− x ⎠ ⎛ arcsin x ⎛ dy ln x ⎞ ⎟ = x arcsin x ⎜ arcsin x + ln x = y⎜ + Thus, ⎜ ⎟ ⎜ dx x x 1 − x2 ⎠ 1 − x2 ⎝ ⎝ Then (b) Apply Product Rule and Chain rule: 2 2 2 f’(x) = 2x e x + x2 e x (2x) = e x (2x +2x3). (c) Apply Quotient rule: 1 3x 2 ln x − x3 3x 2 ln x − x 2 x f’(x) = = . (ln x )2 (ln x )2 (d) Let u = x2 + π + x. Then f(u) = tan u and f’(x) = f’(u) 13. du = 2x + 1. dx du = (sec2 u)(2x + 1) = (2x + 1)sec2( x2 + π + x). dx Solution ⎞ ⎟. ⎟ ⎠ www.prep101.com The function is continuous everywhere except at x = 3. By the definition of continuity, for f to be continuous at 3 ⇒ lim− f ( x ) = lim+ f (x ) = f (3) . ( ) ⇒ lim− 2 x + c = lim+ (cx + c + 2 ) = 3c + c + 2 x →3 2 x →3 x →3 x →3 ⇒ 6 + c = 4c + 2 ⇒ c 2 − 4c + 4 = 0 2 ⇒ ( c − 2 ) = 0 ⇒ c = 2. 2 14. Solution Let (x0, y0) be the point on y = ex at which the line is tangent and let L be the tangent line. The slope of L is y0 − 0 e x0 and also the slope of the line L is e x0 since y’ = ex. = x0 − a x0 − a e x0 = e x0 ⇒ x0 − a = 1 ⇒ x0 = a + 1. Hence, the slope of the line is ea + 1. x0 − a The equation of the line L: y − 0 = ea + 1(x − a) ⇒ y = ea + 1(x − a). Hence, 15. Solution Differentiate implicitly, to get cos(x − y)(1 − y’) = y’ + 0 ⇒ cos(x − y) − y’cos(x − y) = y’ ⇒ cos(x − y) = y’ + y’cos(x − y) = y’[1 + cos(x − y)]. Thus, y’ = cos( x − y ) . 1 + cos( x − y ) Hence, y” = [1 + cos(x − y )][− sin (x − y )](1 − y′) − cos(x − y )[− sin (x − y )](1 − y′) [1 + cos(x − y )]2 cos π / 2 ⎛π ⎞ = 0. Note that at ⎜ ,0 ⎟ , y’= 1 + cos π / 2 ⎝2 ⎠ ⎛π ⎞ Thus, at ⎜ ,0 ⎟ : ⎝2 ⎠ [1 + cos π / 2][− sin π / 2](1 − 0) − cos π / 2[− sin π / 2](1 − 0) = = 1(− 1)(1) − 0(− 1)(1) = −1 . y” = (1 + cos π / 2)2 (1 + 0)2 16. Solution We need to get the factor of “(3-x)” out of the denominator, as follows: www.prep101.com 7 − 10 − x ( 7 + 10 − x ) 7 − (10 − x) lim ⋅ = lim x →3 x →3 3− x ( 7 + 10 − x ) (3 − x)( 7 + 10 − x ) −(3 − x) = lim x →3 (3 − x )( 7 + 10 − x ) −1 = lim x →3 ( 7 + 10 − x ) −1 = . 27 Note: This question is marked very strictly. You’ll get zero unless your solution is completely correct. ...
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