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102-2008&2009-3-M10-July2009(1)

102-2008&2009-3-M10-July2009(1) - Kuwait University...

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Unformatted text preview: Kuwait University Department of Mathematics and Computer Science Math 102 : Calculus II Summer Semester 2009 In- Term Test 1 Saturday 18 July 2009 Duration: 90 minutes Justify all your answers 1. Consider fir] = |r|tan_133 for —oo < :r <: 00. (a) Show that f is one-to—one on (—00, 00). (b) Determine the domain of f‘l. (c) Find (firs/4)- 2. Prove that ln(1/:c] = — Inn: for all 3: > 0. 3. Solve logfi(4 — a") = 1 — logfil:.r + 3). 4. Use logarithmic differentiation to find dy/dn: when _ lefll y _ {€33 —i—1)[sin2 :rh/m—k—l 5. Evaluate the following. 5—1 1 (a) L4 tlnt d]- (b) 1 + ln 3: d3. 2—3:rln3: 23+1 823—].- 6. Find the value of cos_1[sin(51rf4]). Total marks: 25 [2 marks] [1 mark] [2 marks] [2 marks] [2 marks] [2 marks] [3 marks each] [1 mark] 7. Prove that cosh 2.1: = cosh2 :r+sinh2 1', using the definition of hyperbolic functions. [1 mark] 00 CO Kuwait University Department of Mathematics and Computer Science Math 102 : Calculus II Summer Semester 2009 In-Term Test 1 ANSWERS 1. (a) fix) = :carctansr and f’(.r) = arctanr + :r/(l+.1"2) when .r > 0, while firs) = —:carctan:r and f"(:c] = —arctan:c — :r/(l—l—Ifl) when :r < 0. So, flat} > 0 and f’(:c) > 0 for :r > 0, f(0) = 0, and, fix] < 0 and f’(3:) > 0 for :r < 0. Hence, no horizontal line intersects the graph of y = fir) more than once. (13) As :6 —> :Izoo, arctanzt' —> :lz’iT/Q, and hence fix) —> :lzoo, respectively. Since f is continuous on (—00, 00), this implies that the range of f is [—00, 00}. Therefore the domain of f‘1 is (—oo,oo). (c) When :3 =1, firs): 113/4. So f_1(1r/4}= 1 and —.r _ 1 _ 1 _ 1 U 1W“ ‘ f’lf—lm/MJ_f’{1}_arctan1+1/(1+12) 4 1 fif4+1f2 _ 7r+2' 2. By definition, lf'z flit ln(1/.r) =/1 t Substituting t = 1/21, so dt = (—1/u2} (in, gives ln(1/:r) = —/ E = —ln:r. 1 1i- 3. log5(4—.r) = 1—logfi(:c+3] => logfi{(4—:r](:r+3)} = 1 (=> (4—m)(:c+3) = 6 4:» 32—3—6=0 4:» {so 3)(;r:2)—0 <—> :L‘-3or.r=—2. Since, 4 — :t' > 0 and 3: + 3 :> 0 when :r: = 3 or :c = —2, both are valid solutions. ln |y| = ln |:c| +32 — ln[e32 + 1) — 21n |sin5c| — élnm + 1) => ldy 1 3833: 1 — 2 2 t — ydzc 3+ 3 e3$+1 cor 2(sr+1) :> 3:: 2 fl: 2$+—r+2 — 3e —2cot:r: —:rex-p(23:) . d1- 23:(:r:+1) 832+1 (e3z+1](sm I) r+1 OO 99 5. (a) Substitute u = Int. 80 (in = (1ft) cit. Then 1 s— —1 / irit=f ldu=ln|u| 5—2 t Int _2 u —1 =ln|—1|—ln|—2|= —ln2. —2 (b) Substituteu=2— 331nm. Sc- du= —3[ln:r+1)d:r. Then 1+ln3: _ _ _1 = —%ln|2—3xln:r| +C. (c) /233+1 / 23 1/ 1 432+1d$ 432+1d3+4 32+(112Fd‘r = ilnhlar2 +1)+%aretan(2:r) + C. ((1) Substitute x = —111 11.. So doc = (—lfu] (in. Then du = arecesu + ( ‘fu—izéfidubfni—s = arceos[e s‘z} +C. An alternative answer is — arcsin u+C = — arcsin(e_$) +C. Alternative substitutio are :r = lnu leading to see—11H— C or —cse_1u + C with u = ex, or, :r = élnfir2 + leading to arctan “it + C or —c-:)t_1 n + C with u = 82’ — 1. 6. arccos(sin(5'rrf4)] = arecos(—1fv/§) = 311,14. Alternatively, arccos(sin(5'rrf4)] = 1*er — arcsin(sin(51r/4)) = '1er — aresin(sin(—Trf4)) = 1*er — [—zrrfdl] = 31TH. 7. cosh2 3: + sinh2 x ll H’H—h‘fi. m a w‘i‘ LT) :13. U to I‘d—“N rt- N to I m I H U to — €22+2+e—22 €21_2+8—2:— 22+ —2:: _ 4 + 4 _ 2 = cesh23: O. CO ...
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