![\"blog-maths-2018-maths-advanced-solutions-13-a-iii\"](\"https:\/\/www.matrix.edu.au\/wp-content\/uploads\/2018\/10\/blog-maths-2018-maths-advanced-solutions-13-a-iii.png\")
<\/p>\nHave you ever seen the 2018 HSC Arithmetic Superior Paper, but? On this put up, we\u2019ll work our approach by means of the 2018 HSC Maths Superior paper and provide the options, written by our main instructor Oak Ukrit and his group.<\/p>\n
<\/p>\n
Learn on to see the way to reply all the 2018 questions.<\/p>\n
<\/p>\n
<\/p>\n
1. (B) (7^{-1.3} = 0.07968 = 0.08\u00a0) (2 d.p.)<\/p>\n
2. (C) Use mid-point components, outcomes (Q(13, 7))<\/p>\n
3. (A) Let (y=0) provides (x=-6)<\/p>\n
4. (D) Given centre of the circle at (Q(3,-2)). Use perpendicular distance components to acquire radius of the circle, which is (4).From these data, we will write out the equation of the circle \u2013 which is choice D.<\/p>\n
5. (D) (frac{d}{dx}sin(ln{x}) = frac{1}{x} cos(ln{x})) utilizing chain rule.<\/p>\n
6. (C) P(Matching Pairs) = P(Any Footwear) (instances P(Matching Shoe) = 1timesfrac{1}{7}\u00a0 =\u00a0 frac{1}{7})<\/p>\n
7. (C) On condition that (int_0^4f(x),dx = 10) which takes into consideration unfavourable space between (x=3) and (x=4)
\nTherefore (int_0^3 f(x),dx = 10+3 = 13)
\nDue to this fact (int_{-1}^3 f(x),dx = 13-2 = 11)<\/p>\n
8. (D) Since (x^{2}=4ay), at (y=4, x=12) (through symmetry) Substituting values of (x) and (y) provides (a=9)<\/p>\n
9. (B) Level of inflexion happens when gradient of (f'(x)=0) and gradient of (f'(x)) adjustments signal. The one level satisfies each circumstances is level (x=b)<\/p>\n
10. (D) Think about space beneath the curve in every choice, (f(x)=cos{frac{x}{2}}) is the one choice that satisfies the given situation.<\/p>\n
11. (a)<\/p>\n
start{align*}
\nfrac{3}{3 + sqrt{2}} instances frac{3 \u2013 sqrt{2}}{3 \u2013 sqrt{2}} &= frac{9 \u2013 3sqrt{2}}{9 \u2013 2}
\n&= frac{9 \u2013 3sqrt{2}}{7}
\nfinish{align*}<\/p>\n
11. (b)<\/p>\n
start{align*}
\n1 \u2013 3x &> 10
\n-3x &> 9
\nx &< -3
\nfinish{align*}<\/p>\n
11. (c)<\/p>\n
start{align*}
\nfrac{8x^3 \u2013 27y^3}{2x \u2013 3y} &= frac{(2x \u2013 3y)(4x^2 + 6xy + 9y^2)}{2x \u2013 3y}
\n&= 4x^2 + 6xy + 9y^2
\nfinish{align*}<\/p>\n
11. (d) i.<\/p>\n
Given (T_3 = 8), (T_{20} = 59)
\nstart{align*}
\na + second &= 8 quad (1)
\na + 19d &= 59 quad (2)
\nfinish{align*}
\nAfter fixing concurrently, (d = 3)<\/p>\n
11. (d) ii.<\/p>\n
start{align*}
\nT_{50} &= (a + second) + 47d
\n&= 8 + 47(3) = 149
\nfinish{align*}<\/p>\n
11. (e)<\/p>\n
start{align*}
\nint_0^3 e^{5x},dx = left[frac{1}{5}e^{5x}right]_0^3
\n= frac{1}{5}e^{15} \u2013 frac{1}{5}e^0
\n= frac{1}{5}(e^{15} \u2013 1)
\nfinish{align*}<\/p>\n
11. (f)<\/p>\n
start{align*}
\nfrac{d}{dx}(x^2tan x) &= 2xtan x + x^2sec^2x quad (textual content{product rule})
\nfinish{align*}<\/p>\n
11. (g)<\/p>\n
start{align*}
\nfrac{d}{dx}left(frac{e^x}{x+1}proper) &= frac{(x+1)e^x \u2013 e^x}{(x+1)^2} quad textual content{(quotient rule)}
\n= frac{xe^x}{(x+1)^2}
\nfinish{align*}<\/p>\n
12. (a) i.<\/p>\n
start{collect*}
\nangle ABN = 130\u00b0
\nangle ABC + 130\u00b0 + 120\u00b0 = 360\u00b0
\ntextual content{due to this fact} \u00a0 angle ABC = 110\u00b0
\nfinish{collect*}<\/p>\n
12. (a) ii.<\/p>\n
Utilizing cosine rule,
\nstart{align*}
\nAC^2 &= AB^2 + BC^2 \u2013 2ABcdot BCcdot cos 110\u00b0
\n&= 320^2 + 190^2 \u2013 2(320)(190)cos 110\u00b0
\ntextual content{due to this fact} AC &approx 420text{km} quad textual content{(nearest 10km)}
\nfinish{align*}<\/p>\n
12. (b)<\/p>\n
Given (y = cos 2x),
\nstart{collect*}
\nfrac{dy}{dx} = -2sin(2x)
\ntextual content{at } x = frac{pi}{6}, quad frac{dy}{dx} = -sqrt{3}
\ntextual content{due to this fact} \u00a0 y \u2013 frac{1}{2} = -sqrt{3}left(x \u2013 frac{pi}{6}proper)
\nfinish{collect*}
\nRearrange,
\n$$y = -sqrt{3}x + frac{pisqrt{3} + 3}{6}$$<\/p>\n
12. (c) i.<\/p>\n
start{collect*}
\ntextual content{In } triangle ADF, triangle ABE
\nAD = AB quad textual content{(sides of sq. are equal)}
\nangle ADF = angle ABE quad textual content{(angles of sq. is 90\u00b0)}
\ntextual content{observe: } EC = FC, DC = BC
\nDF = DC \u2013 FC = BE
\ntextual content{due to this fact} triangle ADF equiv triangle ABE,,(SAS)
\nfinish{collect*}<\/p>\n
12. (c) ii.<\/p>\n
start{align*}
\nA_{ACEF} &= 14^2 \u2013 2left(frac{1}{2} instances 10times 14right)
\n= 196 \u2013 140
\n= 56 textual content{cm}^2
\nfinish{align*}<\/p>\n
12. (d) i.<\/p>\n
start{collect*}
\nx = frac{t^3}{3} \u2013 2t^2 + 3t
\nv = frac{dx}{dt} = t^2 \u2013 4t + 3
\ntextual content{at } t = 0, v = 3 textual content{ms}^{-1}
\nfinish{collect*}<\/p>\n
12. (d) ii.<\/p>\n
start{collect*}
\ntextual content{stationary when } v= 0
\nt^2 \u2013 4t + 3 = (t-3)(t-1) = 0
\ntextual content{due to this fact} textual content{at } t= 1,3,,s textual content{ particle is stationary}
\nfinish{collect*}<\/p>\n
12. (d) iii. Acceleration is zero at (v_text{max})
\nstart{align*}
\nt &= -frac{b}{2a} textual content{ from quadratic } t^2 \u2013 4t + 3
\n&= frac{4}{2}
\n&=2,,s
\nfinish{align*}
\n$$textual content{due to this fact} textual content{at } t = 2, x = frac{8}{3} \u2013 8 + 6 = frac{2}{3},,m$$<\/p>\n
13. (a) i.<\/p>\n
Given (y = 6x^2 \u2013 x^3), (frac{dy}{dx} = 12x \u2013 3x^2) Let (frac{dy}{dx} = 0), and clear up for (x). This offers (x = 0, 4). Discover related (y)-value and take a look at utilizing (frac{d^2y}{dx^2}). Thus now we have,<\/p>\n
minimal stationary level at ((0,0))
\nmost stationary level at ((4,32))<\/p>\n
13. (a) ii. Present that (dfrac{d^2y}{dx^2}) at (x = 2) is zero, and (dfrac{d^2y}{dx^2}) adjustments signal at (x= 1) and (x = 3).<\/p>\n
13. (a) iii<\/p>\n
<\/p>\n
<\/p>\n
13. (b) i.<\/p>\n
start{collect*}
\ntextual content{In } triangle CBD, BC = CD quad textual content{(isosceles triangle)}
\ntextual content{Simiarly, In } triangle ABC, AB = CD quad textual content{(isosceles triangle)}
\ntextual content{Since } angle ABC textual content{ is frequent, } triangle CBD , ||| , triangle ABC ,,textual content{(Equiangular)}
\nfinish{collect*}<\/p>\n
13. (b) ii.<\/p>\n
start{align*}
\nfrac{BD}{BC} &= frac{DC}{AB} quad textual content{(matching sides in ratios of comparable triangles)}
\nfrac{BD}{2} &= frac{2}{3}
\nBD &= frac{4}{3}
\ntextual content{due to this fact} AD+ BD &= AB
\nAD &= AB \u2013 BD
\n&= 3 \u2013 frac{4}{3}
\n&= frac{5}{3} textual content{ items}
\nfinish{align*}<\/p>\n
13. (c) i.<\/p>\n
start{align*}
\nP(50) & = 184 quad quad textual content{* items in thousands and thousands}
\n92e^{50k} & = 184
\nokay &= frac{1}{50}lnleft(frac{184}{92}proper)
\n&approx 0.0319 quad textual content{(4 d.p.)}
\nfinish{align*}<\/p>\n
13. (c) ii.<\/p>\n
start{align*}
\nP(110) &= 92e^{0.0139(110)}
\n&= 424 textual content{ million} quad textual content{(nearest million)}
\nfinish{align*}<\/p>\n
14. (a) i.<\/p>\n
Utilizing sine rule,
\nstart{align*}
\nA_{triangle KLM} &= frac{1}{2}(3)(6)sin60\u00b0
\n&= frac{9sqrt{3}}{2} textual content{ unit}^2
\nfinish{align*}<\/p>\n
14. (a) ii.<\/p>\n
start{align*}
\nA_{triangle KLN} + A_{triangle NLM} &= frac{9sqrt{3}}{2}
\nfrac{1}{2}(3x)sin30\u00b0 + frac{1}{2}(6x)sin30\u00b0 &= frac{9sqrt{3}}{2}
\nfrac{3x}{4} + frac{6x}{4} &= frac{9sqrt{3}}{2}
\n9x &= 18sqrt{3}
\nx &= 2sqrt{3} textual content{ items}
\nfinish{align*}<\/p>\n
14. (b)<\/p>\n
Rearrange the equation to make (x) the topic.
\nstart{align*}
\nx &= (y \u2013 1)^{frac{1}{4}}
\nV &= piint_1^{10}(y \u2013 1)^{frac{1}{2}},dy
\n&= 18pi textual content{ items}^3
\nfinish{align*}<\/p>\n
14. (c)<\/p>\n
If (f(x)) has no stationary factors, (f'(x)) has no roots.<\/p>\n
(f'(x) = 3x^2 + 2kx + 3)<\/p>\n
No roots if (Delta < 0, B^2 \u2013 4AC < 0). Due to this fact
\n$$-3 < okay < 3$$<\/p>\n
14. (d) i.<\/p>\n
DAY 1: (n = 1 Rightarrow 2^{1} + 1 = 3) downloads
\nDAY 2: (n = 2 Rightarrow 2^{2}+ 2 = 6) downloads
\nDAY 3: (n = 3 Rightarrow 2^{3} + 3 = 11) downloads<\/p>\n
<\/p>\n
14. (d) ii.<\/p>\n
Think about AP & GP individually.
\nGP for (2^n):
\nstart{collect*}
\na = 2, r = 2, n = 20
\nS_{20} = frac{a(r^n-1)}{r-1} = 2097150
\nfinish{collect*}
\nAP for (n):
\n$$ a = 1, d = 1, n = 20 $$
\nstart{align*}
\nS_{20} &= frac{n}{2}(a+l)
\n&= frac{20}{2}(1 + 20)
\n&= 210
\nfinish{align*}
\nDue to this fact, the Complete variety of downloads is (2097150 + 210 = 2097360).<\/p>\n
14. (e) i.<\/p>\n
start{align*}
\ntextual content{P(a minimum of one fault)} &= 1 \u2013 textual content{P(no fault)}
\n&= 1 \u2013 (0.9 instances 0.95)
\n&= 0.145
\nfinish{align*}<\/p>\n
14. (e) ii.<\/p>\n
start{align*}
\ntextual content{P(A, NF, NF)} + textual content{P(B, NF, NF)} &= frac{1}{2}left(frac{9}{10}proper)left(frac{9}{10}proper) + frac{1}{2}left(frac{19}{10}proper)left(frac{19}{10}proper)
\n&= frac{137}{160}
\nfinish{align*}<\/p>\n
15. (a) i.<\/p>\n
let (t = 0), (L(0) = 12 + 2cos(0) = 14 textual content{hrs})<\/p>\n
15. (a) ii.<\/p>\n
As (-1 leq cosleft(frac{2pi t}{366}proper) leq 1), it\u2019s least when (cosleft(frac{2pi t}{366}proper) = -1). Thus (min [L(t)] = 12 \u2013 2 = 10 textual content{hrs}).<\/p>\n
15. (a) iii.<\/p>\n
Let (L(t) = 11)<\/p>\n
start{align*}
\n11 &= 12 + 2cosleft(frac{2pi t}{366}proper)
\n-frac{1}{2} &= cosleft(frac{2pi t}{2}proper)
\nfrac{2pi t}{366} &= frac{2pi}{3},frac{4pi}{3}
\nt &= 122, 244
\nfinish{align*}<\/p>\n
15. (b)<\/p>\n
start{align*}
\nint_{0}^kfrac{1}{x+3},dx &= int_{okay}^{45} frac{1}{x+3},dx
\n[ln(x+3)]_0^okay &= [ln(x+3)]_k^{45}
\nln(okay+3) \u2013 ln 3 &= ln 48 \u2013 ln(okay+3)
\n2ln(okay+3) &= ln 144
\nokay+3 &= 12 quad (textual content{as a result of} okay > 0)
\nokay &= 9
\nfinish{align*}<\/p>\n
15. (c) i.<\/p>\n
start{align*}
\ntextual content{Space} &= -int_0^3(x^3 \u2013 7x),dx + int_0^32x,dx
\n&= int_0^3(2x + 7x \u2013 x^3),dx = int_0^3(9x- x^3),dx
\n&= left[frac{9}{2}x^2 \u2013 frac{1}{4}x^4right]^3_0
\n&= frac{81}{2} \u2013 frac{81}{4} = frac{81}{4} textual content{items}^2
\nfinish{align*}<\/p>\n
15. (c) ii.<\/p>\n
start{collect*}
\ntextual content{let } f(x) = 2x \u2013 (x^3 \u2013 7x) = 9x \u2013 x^3
\nf(0) = 0, f(1.5) = 10.125 = frac{81}{8}, f(3) = 0
\nh = frac{b-a}{2} = 1.5
\nfinish{collect*}
\nstart{align*}
\ntextual content{Space} &= frac{h}{3}left(1times 0 + frac{81}{8}instances 4 + 1 instances 0right)
\n&= frac{1}{2} instances 4 instances frac{81}{4} = frac{81}{4} textual content{items}^2
\nfinish{align*}<\/p>\n
15. (c) iii.<\/p>\n
Discovering co-ordinate of (P)
\nstart{align*}
\ny&=x^{3}-7x
\ny\u2019&=3x^{2}-7
\ntextual content{Let},,y\u2019&=2,Rightarrow x=sqrt{3},,,y=-4sqrt{3}
\ntextual content{due to this fact} P(sqrt{3},-4sqrt{3})
\nfinish{align*}<\/p>\n
15. (c) iv.<\/p>\n
start{align*}
\nd&=frac{|2sqrt{3}-(-4sqrt{3})|}{sqrt{2^{2}+1^{2}}}
\n&=frac{6sqrt{3}}{sqrt{5}}
\ntextual content{Therefore OA},&=sqrt{3^{2}+6^{2}}
\n&=sqrt{45}
\n&=3sqrt{5},,textual content{unit}
\ntextual content{due to this fact} ,textual content{Space}Delta,OAP,&=frac{1}{2}timesfrac{6sqrt{3}}{sqrt{5}}times3sqrt{5}
\n&=9sqrt{3},,u^{2}
\nfinish{align*}<\/p>\n
16. (a) i.<\/p>\n
start{align*}
\nV&= frac{1}{3}pi^{2}h,
\nhowever,, h^{2}+x^{2} &= 100
\ntextual content{due to this fact} h&=sqrt{100-x^{2}}
\nV&=frac{1}{3}pi x^{2}sqrt{100-x^{2}}
\nfinish{align*}<\/p>\n
16. (a) ii.<\/p>\n
start{align*}
\nfrac{dV}{dx}&=frac{2}{3}pi xsqrt{100-x^{2}} \u2013 frac{2}{3}pi x^{3} frac{1}{sqrt{100-x^{2}}}timesfrac{1}{2}
\n&=frac{2pi x(100-x^{2})-pi x^{3}}{3sqrt{100-x^{2}}}
\n&=frac{pi x(200-2x^{2}-x^{2})}{3sqrt{100-x^{2}}}
\n&=frac{pi x(200-3x^{2})}{3sqrt{100-x^{2}}}
\nfinish{align*}<\/p>\n
16. (a) iii.<\/p>\n
start{collect*}
\nLet frac{dV}{dx}=0
\nx=0,(Omit),,or,x=sqrt{frac{200}{3}}
\ntextual content{Take a look at the character of the stationary level utilizing both desk methodology or Second by-product}
\nRightarrow textual content{Max level at}, x=sqrt{frac{200}{3}}
\ntextual content{Now}, 2pi x=10,theta ,,, textual content{By equating circumference}
\ntextual content{due to this fact} theta=frac{2sqrt{2}pi}{sqrt{3}}
\nfinish{collect*}<\/p>\n
16. (b) i.<\/p>\n
![\"blog-maths-2018-maths-advanced-solutions-16-b-iii\"](\"https:\/\/www.matrix.edu.au\/wp-content\/uploads\/2018\/10\/blog-maths-2018-maths-advanced-solutions-16-b-iii.png\")
<\/p>\n
16. (b) ii.<\/p>\n
$$Pr(Win) =frac{1}{36}instances(frac{2times20}{6})=frac{5}{27}$$<\/p>\n
16 (c) i.<\/p>\n
start{align*}
\nA_0&=300000
\nA_1&=300000times1.04-P
\nA_2&=A_1(1.04)-P(1.05)
\n&=300000(1.04)^{2}-P(1.04+1.05)
\n&=300000(1.04)^{2}-P(1.04+1.05),,,textual content{As Required}
\nfinish{align*}<\/p>\n
16. (c) ii.<\/p>\n
start{align*}
\nA_3&=A_2(1.04)-P(1.05)^{2}
\n&=300000(1.04)^{3}-P((1.04)^{2}+1.05(1.04)+(1.05)^{2}),,,textual content{As Required}
\nfinish{align*}<\/p>\n
<\/p>\n
Have you ever seen the 2018 HSC Arithmetic Superior Paper, but? On this put up, we\u2019ll work our approach by means of the 2018 HSC Maths Superior paper and provide the options, written by our main instructor Oak Ukrit and his group. Learn on to see the way to reply all the 2018 questions.<\/p>\n","protected":false},"author":1,"featured_media":377,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"class_list":["post-376","post","type-post","status-publish","format-standard","has-post-thumbnail","category-math"],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/posts\/376","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/comments?post=376"}],"version-history":[{"count":1,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/posts\/376\/revisions"}],"predecessor-version":[{"id":638,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/posts\/376\/revisions\/638"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/media\/377"}],"wp:attachment":[{"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/media?parent=376"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/categories?post=376"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/tags?post=376"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}