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S1B - SOLUTIONS TO EXAM 1B 1 1[10 Marks Find the largest...

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Unformatted text preview: SOLUTIONS TO EXAM 1B 1 1. [10 Marks] Find the largest domain of the function f(1:) 2 —.. 7r — 6 arcs1nzc Express your answer using interval notation. Solution The domain of arcsina: is [71, 1]. The domain of f are the numbers in this interval for which 71' the denominator of f($) is nonzero. Note 7r — Garcsinx 2 0 when arcsinzr 2 3’ i.e. when :6 2 %. Thus the domain of f is {—16} U 6,1]. 2. [15 Marks] f is the function with domain [0, 00) defined by flax) 2 (a) Show f is one—to—one. Solution Assume f(a:1) 2 f(a:2). Then 1+;t2' 1 1 1+x§ : 1+x§ Hag? = Has; 3:? = mg Hence .171 2 332 since both 331 and 332 are greater than or equal to zero. Thus f is one2t02one. (b) Find the domain of f ‘1. Express your answer in interval notation. Solution The domain of f’1 is the range of f which consists for all positive numbers less than or equal to f(0) 2 1. Thus the domain of f’1 is (0,1]. (c) Find the formula for f ‘1. Solution .r 2 f_1(y) if y 2 flm) 2 1 Then l+x2' 1 1+ac2 2 — y 1 1— y y 1—9; fly = ac: —. () y Note that we chose the positive squareroot because at is in the domain of f which consists of numbers greater than or equal to zero. ,2 3. [15 Marks] Find tan arcsin “L . 1 + .702 Solution Let 6 2 arcsin 11:2. Then sin6 2 11:2. In the right triangle below with angle (9, let the opposite side have length .752 and let the hypotenuse have length 1 + 12. By the Pythagorean Theorem7 the length of the adjacent side is (1 + x2)? — (:52)‘2 = ym = m Hence .2 ,.2 . .1, 1, tan (arcsm 1 > 2 tan9 2 'l+2x2 4. JUSTIFY YOUR ANSWERS. 2 o x/ 5 — «13 — , ~ (a) [15 Marks] Evaluate lim # x~>72 .T —]— 2 Solution To evaluate this limit, rationalize the numerator. . x/zr2 [ 5 13 m2 , \/:r2 [ 5 \/13 $2 \/.702 I 5 l \/13 1:2 11m — 2 11m x—>—2 a:+2 m—>—2 :1?+2 .x/x2+5+\/13—x2 ( x2 +5)2 a (x/13—m2)2 (3:2 +5) a (13am?) = $EQ2WZ£BW = lim 299—4: lim M “1%72 <x+2><m+m) 1KH72($__2)(\/m+m) : lim 2(a: 2) _ 2(4) _ 8_ 4 x—>—2\/x2+5+\/13—1"2 ¢9+¢9 ‘7 3 COSQZC (b) [10 Marks] Evaluate lim Solution Since —1 g cosw g 1, it follows that 0 g cos2 1' g 1. Hence , 0 , c0s21: _ 1 hm — S hm S hm —. 2 The first and third limits above are zero. By the Pinching Theorem7 T111130 CO: ac = 0 too. (c) [10 Marks] Evaluate $131) tan(39) c0t(59). Solution First replace tan 39 and cot 59 by their definitions. . _ sin 39 cos 59 . cos 59 . sin 39 ($135 tan(39) COt<50> — 913]) cos 39 sin 59 — ($135 cos 39 ($135 sin 59 = 35135 3:335:22???- 56 5. (a) [10 Marks] State the Intermediate Value Theorem. Solution Let f be a continuous function with domain an interval I. If A and B are in the range of f, with A < B7 then the range of f contains the entire interval [14, B]. (b) [15 Marks] Does the Intermediate Value Theorem apply to the function f (r) = g for or 75 O. J USTIFY YOUR ANSWER. Solution The domain of f consists of all nozero numbers which is not an interval. Therefore, the Intermediate Value Theorem does not apply to f. 6. [15 Marks] Determine the values of as for which the function below is continuous, has a removable discontinuity or has an essential discontinuity. J UST IFY YOUR ANSWERS. 1le if ac # 0 and a? is rational f(;,;): 1 if$:0 x2 if ac is irrational Solution Note all equals +1 if a: > 0 and equals —1 if :v < 0. See the graph off below. Therefore the leftehand limit at a: : 0 and the rightehand limit at at : 0 do not exist. Hence f has an essential discontinuity at a: = 0. When 00 < 0, note 902 > 0 while I:—1 : 71. Hence lim f (m) DNE for every 0 < 0 and f has an essential m—m discontinuity at every (2 < 0. When :1: > 0, note :1;2 2 % 2 +1 irand only 11:1: : 1. Therefore, lim f(:l;) 2 1 2 f0) and f is (continuous x—d at a: : 1. In addition7 lim f (:c) DNE7 and has an essential discontinuity when 0 > 0 and 0 7E 1. 77—)(1. ...
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