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Midterm

# SolTest1Y2012.pdf

7 Pages
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School
University of Toronto Scarborough
Department
Mathematics
Course
MATA30H3
Professor
Sophie Chrysostomou
Semester
Winter

Description
MATA30: Calculus I - Solution of Midterm Test 1 1. (a) [ 4 marks] Give the deﬁnition of g(x) = arctanx and give the domain and its range. Solution: The deﬁnition of g(x) = arctanx is given by: arctanx = y ⇐⇒ tany = x ▯π π ▯ the domain of g is R and it’s range,is. 2 2 (b) [ 2 marks] Give the graph of g(x) = arctanx. Solution: 3 2 1 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 -1 -2 -3 (c) [ 4 marks] Find the exact value of the expression: ▯ ▯√ ▯ ▯ ▯▯ cos arctan 3 + arcsin √1 2 ▯ ▯ √ 1 Solution: First we evaluate arctan(nd arcsin√ . √ √ 2 arctan(3) = θ ⇐⇒ tanθ = 3 ⇐⇒ θ = π ▯ ▯ 3 1 1 π arcsin √ = α ⇐⇒ sinα = √ ⇐⇒ α = . 2 2 4 ▯ ▯√ ▯ ▯ ▯▯ ▯ ▯ cos arctan 3 + arcsin √1 = cos π + π 2 3 4 π π π π = cos cos − sin sin 3 4 3 4 √ 1 1 3 1 = ▯√ − ▯ √ 2 2 2 2 √ = 1 √ 3 2 2 1 −log (x)+log (3) 2. (a) [ 4 marks] Find all x satisfying: 4 9 = 3 Solution: −lo4 (x)9log (3l4g () lo9 (3) 3 = 4 = 4 4 −1 2 loa b 1 = x ▯ 4 since a = b and log9(3) =2 2 = x 2 2 So 3 =x therefore x =3 sin(x) + sin(2x) (b) [ 6 marks] Prove the identity: = tanx. 1 + cos(x) + cos(2x) Solution sinx + sin(2x) LHS = 1 + cos(x) + cos(2x) sin(x) + 2sin(x)cos(x) = 1 + cos(x) + 2cos (x) − 1 = sin(x)[1 + 2cos(x)] cos(x) + 2cos (x) sin(x)[1 + 2cos(x)] = cos(x)[1 + 2cos(x)] sin(x) = = tan(x) = RHS cos(x) 3. [5 marks each; 25 marks total] Evaluate each of the following limits, else explain why the limit does not exist. Justify your answer. Do notˆspital’s rule. x − 9 (a) lim√ √ x→3 3 − x Solution: x − 9 lim√ √ = lim (x√− 3)√x + 3) x→3 3 − x x→3 3 − x √ √ √ √ ( x − 3)( x + 3)(x + 3) = lim √ √ x→3 3 − x √ √ √ √ = lx→3−( x + 3)(x + 3) = −(2 3)(3 + 3) = −12 3 ▯ ▯ ▯ ▯ 2 1 1 (b) lim x sin cos x→0 x x Solution: Since −1▯≤ ▯inx ≤ 1 and −1 ≤ c▯ ▯ ≤ 1 for all x, then: 1 1 −1 ≤ sin ≤ 1 and −1 ≤ cos ≤ 1 for all x ▯= 0. x x ▯ ▯ ▯ ▯ 1 1 Thus: −1 ≤ sin x cos x ≤ 1 for all x ▯= 0. Since x ≥ 0 for all x ▯= 0, then, ▯ 1▯ ▯ 1 −x ≤ x sin cos ≤ x2 for all x ▯= 0. x x 2 2 Sincex→0m −x = 0 and x→0 x = 0, therefore by the Squeeze theorem ▯ 1 ▯1▯ lim x sin cos = 0 x→0 x x |x + 5| (c) lim x→−52x + 10 2 Solution: We take the left and right limits: If x → −5− then x < −5 therefore x + 5 < 0 and |x + 5| = −(x + 5). Thus, lim |x + 5|= lim −(x + 5) = lim −1 = − 1 x→−5− 2x + 10 x→−5− 2(x + 5) x→−5 − 2 2 + If x → −5 then x > −5 therefore x + 5 > 0 and |x + 5| = (x + 5). Thus, lim |x + 5|= lim (x + 5) = lim 1= 1 x→−5+ 2x + 10 x→−5 +2(x + 5) x→−5+ 2 2 |x + 5| |x + 5| |x + 5| Sincex→−5− 2x + 10 ▯=x→−5+ 2x + 10, thenx→−5 2x + 10does not exist. 1 − cosx (d) lim x→0 xtanx Solution: 1 − cosx (1 − cosx) (1 + cosx) lim = lim ▯ x→0 xtanx x→0 xtanx (1 + cosx) 2 = lim (1 − cos x) x→0xtan(x)(1 + cosx) (sin x) = lim sin(x) x→0x cos(x) + cosx) sin (x)cos(x) = lim x→0xsin(x)(1 + cosx) sin(x)cos(x) = lim x→0x(1 + cosx) sin(x) cos 1 1 = lix→0 x ▯(1 + cosx) = 1 ▯1 + 1= 2 √ 2 (e) lim x + 5 − (x + 1) x→2 2 − x Solution: √ √ √ x2 + 5 − (x + 1) ( x2 + 5 − (x + 1))( x + 5 + (x + 1)) lim = lim ▯ √ 2 x→2 2 − x x→2 2 − x ( x + 5 + (x + 1)) 2 2 = lim (x +√5 − (x + 1) ) x→2 (2 − x)( x + 5 + (x + 1)) (x + 5 − (x + 2x + 1)) = lim √ 2 x→2 (2 − x)( x + 5 + (x + 1)) = lim √(4 − 2x) x→2 (2 − x)( x + 5 + (x + 1)) 2(2 − x) = lim √ 2 x→2 (2 − x)( x + 5 + (x + 1)) 2 = lim √ x→2 ( x + 5 + (x + 1)) 2 2 1 = √ = 6 = 3 ( 4 + 5 + (2 + 1)) 3 √ x + 4 4. [6 marks] Let f(x) = . Find all the horizontal and vertical asymptotes of f. x − 6 Solution: For horizontal asymptotes: √ √x +4 x + 4 x √ lim = lim 6 since x → ∞, then x > 0, and x = |x| x2 x→∞ x − 6 x→∞ 1 − x √ 2 √ +4 = lim x2 x→∞ 1 − 6 x ▯ 2 x 24 = lim x x→∞ 1 − 6 x ▯ √ 1 + 2 = lim x = 1 = 1 x→∞ 1 − 6 1 x Thus y = 1 is a horizontal asymptote (as x → ∞). √ √ x + 4 x +4 √ lim = lim x since x → −∞, then x < 0, and x = −|x| = −x2 x→−∞ x − 6 x→−∞ 1 − 6 x √ x√+4 − x2 = lim 6 x→−∞ 1 − x ▯ x +4 − x2 = lim
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