{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}