# HSC Maths Superior Examination Paper Options

Have you ever seen the 2018 HSC Arithmetic Superior Paper, but? On this put up, we’ll 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.

1. (B) (7^{-1.3} = 0.07968 = 0.08 ) (2 d.p.)

2. (C) Use mid-point components, outcomes (Q(13, 7))

3. (A) Let (y=0) provides (x=-6)

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 – which is choice D.

5. (D) (frac{d}{dx}sin(ln{x}) = frac{1}{x} cos(ln{x})) utilizing chain rule.

6. (C) P(Matching Pairs) = P(Any Footwear) (instances P(Matching Shoe) = 1timesfrac{1}{7}  =  frac{1}{7})

7. (C) On condition that (int_0^4f(x),dx = 10) which takes into consideration unfavourable space between (x=3) and (x=4)
Therefore (int_0^3 f(x),dx = 10+3 = 13)
Due to this fact (int_{-1}^3 f(x),dx = 13-2 = 11)

8. (D) Since (x^{2}=4ay), at (y=4, x=12) (through symmetry) Substituting values of (x) and (y) provides (a=9)

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)

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.

## Written Response

11. (a)

start{align*}
frac{3}{3 + sqrt{2}} instances frac{3 – sqrt{2}}{3 – sqrt{2}} &= frac{9 – 3sqrt{2}}{9 – 2}
&= frac{9 – 3sqrt{2}}{7}
finish{align*}

11. (b)

start{align*}
1 – 3x &> 10
-3x &> 9
x &< -3
finish{align*}

11. (c)

start{align*}
frac{8x^3 – 27y^3}{2x – 3y} &= frac{(2x – 3y)(4x^2 + 6xy + 9y^2)}{2x – 3y}
&= 4x^2 + 6xy + 9y^2
finish{align*}

11. (d) i.

Given (T_3 = 8), (T_{20} = 59)
start{align*}
a + second &= 8 quad (1)
a + 19d &= 59 quad (2)
finish{align*}
After fixing concurrently, (d = 3)

11. (d) ii.

start{align*}
T_{50} &= (a + second) + 47d
&= 8 + 47(3) = 149
finish{align*}

11. (e)

start{align*}
int_0^3 e^{5x},dx = left[frac{1}{5}e^{5x}right]_0^3
= frac{1}{5}e^{15} – frac{1}{5}e^0
= frac{1}{5}(e^{15} – 1)
finish{align*}

11. (f)

start{align*}
frac{d}{dx}(x^2tan x) &= 2xtan x + x^2sec^2x quad (textual content{product rule})
finish{align*}

11. (g)

start{align*}
frac{d}{dx}left(frac{e^x}{x+1}proper) &= frac{(x+1)e^x – e^x}{(x+1)^2} quad textual content{(quotient rule)}
= frac{xe^x}{(x+1)^2}
finish{align*}

12. (a) i.

start{collect*}
angle ABN = 130°
angle ABC + 130° + 120° = 360°
textual content{due to this fact}   angle ABC = 110°
finish{collect*}

12. (a) ii.

Utilizing cosine rule,
start{align*}
AC^2 &= AB^2 + BC^2 – 2ABcdot BCcdot cos 110°
&= 320^2 + 190^2 – 2(320)(190)cos 110°
textual content{due to this fact} AC &approx 420text{km} quad textual content{(nearest 10km)}
finish{align*}

12. (b)

Given (y = cos 2x),
start{collect*}
frac{dy}{dx} = -2sin(2x)
textual content{at } x = frac{pi}{6}, quad frac{dy}{dx} = -sqrt{3}
textual content{due to this fact}   y – frac{1}{2} = -sqrt{3}left(x – frac{pi}{6}proper)
finish{collect*}
Rearrange,
\$\$y = -sqrt{3}x + frac{pisqrt{3} + 3}{6}\$\$

12. (c) i.

start{collect*}
textual content{In } triangle ADF, triangle ABE
angle ADF = angle ABE quad textual content{(angles of sq. is 90°)}
textual content{observe: } EC = FC, DC = BC
DF = DC – FC = BE
textual content{due to this fact} triangle ADF equiv triangle ABE,,(SAS)
finish{collect*}

12. (c) ii.

start{align*}
A_{ACEF} &= 14^2 – 2left(frac{1}{2} instances 10times 14right)
= 196 – 140
= 56 textual content{cm}^2
finish{align*}

12. (d) i.

start{collect*}
x = frac{t^3}{3} – 2t^2 + 3t
v = frac{dx}{dt} = t^2 – 4t + 3
textual content{at } t = 0, v = 3 textual content{ms}^{-1}
finish{collect*}

12. (d) ii.

start{collect*}
textual content{stationary when } v= 0
t^2 – 4t + 3 = (t-3)(t-1) = 0
textual content{due to this fact} textual content{at } t= 1,3,,s textual content{ particle is stationary}
finish{collect*}

12. (d) iii. Acceleration is zero at (v_text{max})
start{align*}
t &= -frac{b}{2a} textual content{ from quadratic } t^2 – 4t + 3
&= frac{4}{2}
&=2,,s
finish{align*}
\$\$textual content{due to this fact} textual content{at } t = 2, x = frac{8}{3} – 8 + 6 = frac{2}{3},,m\$\$

13. (a) i.

Given (y = 6x^2 – x^3), (frac{dy}{dx} = 12x – 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,

minimal stationary level at ((0,0))
most stationary level at ((4,32))

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).

13. (a) iii

13. (b) i.

start{collect*}
textual content{In } triangle CBD, BC = CD quad textual content{(isosceles triangle)}
textual content{Simiarly, In } triangle ABC, AB = CD quad textual content{(isosceles triangle)}
textual content{Since } angle ABC textual content{ is frequent, } triangle CBD , ||| , triangle ABC ,,textual content{(Equiangular)}
finish{collect*}

13. (b) ii.

start{align*}
frac{BD}{BC} &= frac{DC}{AB} quad textual content{(matching sides in ratios of comparable triangles)}
frac{BD}{2} &= frac{2}{3}
BD &= frac{4}{3}
textual content{due to this fact} AD+ BD &= AB
&= 3 – frac{4}{3}
&= frac{5}{3} textual content{ items}
finish{align*}

13. (c) i.

start{align*}
P(50) & = 184 quad quad textual content{* items in thousands and thousands}
92e^{50k} & = 184
okay &= frac{1}{50}lnleft(frac{184}{92}proper)
&approx 0.0319 quad textual content{(4 d.p.)}
finish{align*}

13. (c) ii.

start{align*}
P(110) &= 92e^{0.0139(110)}
&= 424 textual content{ million} quad textual content{(nearest million)}
finish{align*}

14. (a) i.

Utilizing sine rule,
start{align*}
A_{triangle KLM} &= frac{1}{2}(3)(6)sin60°
&= frac{9sqrt{3}}{2} textual content{ unit}^2
finish{align*}

14. (a) ii.

start{align*}
A_{triangle KLN} + A_{triangle NLM} &= frac{9sqrt{3}}{2}
frac{1}{2}(3x)sin30° + frac{1}{2}(6x)sin30° &= frac{9sqrt{3}}{2}
frac{3x}{4} + frac{6x}{4} &= frac{9sqrt{3}}{2}
9x &= 18sqrt{3}
x &= 2sqrt{3} textual content{ items}
finish{align*}

14. (b)

Rearrange the equation to make (x) the topic.
start{align*}
x &= (y – 1)^{frac{1}{4}}
V &= piint_1^{10}(y – 1)^{frac{1}{2}},dy
&= 18pi textual content{ items}^3
finish{align*}

14. (c)

If (f(x)) has no stationary factors, (f'(x)) has no roots.

(f'(x) = 3x^2 + 2kx + 3)

No roots if (Delta < 0, B^2 – 4AC < 0). Due to this fact
\$\$-3 < okay < 3\$\$

14. (d) i.

DAY 1: (n = 1 Rightarrow 2^{1} + 1 = 3) downloads
DAY 3: (n = 3 Rightarrow 2^{3} + 3 = 11) downloads

14. (d) ii.

Think about AP & GP individually.
GP for (2^n):
start{collect*}
a = 2, r = 2, n = 20
S_{20} = frac{a(r^n-1)}{r-1} = 2097150
finish{collect*}
AP for (n):
\$\$ a = 1, d = 1, n = 20 \$\$
start{align*}
S_{20} &= frac{n}{2}(a+l)
&= frac{20}{2}(1 + 20)
&= 210
finish{align*}
Due to this fact, the Complete variety of downloads is (2097150 + 210 = 2097360).

14. (e) i.

start{align*}
textual content{P(a minimum of one fault)} &= 1 – textual content{P(no fault)}
&= 1 – (0.9 instances 0.95)
&= 0.145
finish{align*}

14. (e) ii.

start{align*}
textual 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)
&= frac{137}{160}
finish{align*}

15. (a) i.

let (t = 0), (L(0) = 12 + 2cos(0) = 14 textual content{hrs})

15. (a) ii.

As (-1 leq cosleft(frac{2pi t}{366}proper) leq 1), it’s least when (cosleft(frac{2pi t}{366}proper) = -1). Thus (min [L(t)] = 12 – 2 = 10 textual content{hrs}).

15. (a) iii.

Let (L(t) = 11)

start{align*}
11 &= 12 + 2cosleft(frac{2pi t}{366}proper)
-frac{1}{2} &= cosleft(frac{2pi t}{2}proper)
frac{2pi t}{366} &= frac{2pi}{3},frac{4pi}{3}
t &= 122, 244
finish{align*}

15. (b)

start{align*}
int_{0}^kfrac{1}{x+3},dx &= int_{okay}^{45} frac{1}{x+3},dx
[ln(x+3)]_0^okay &= [ln(x+3)]_k^{45}
ln(okay+3) – ln 3 &= ln 48 – ln(okay+3)
2ln(okay+3) &= ln 144
okay+3 &= 12 quad (textual content{as a result of} okay > 0)
okay &= 9
finish{align*}

15. (c) i.

start{align*}
textual content{Space} &= -int_0^3(x^3 – 7x),dx + int_0^32x,dx
&= int_0^3(2x + 7x – x^3),dx = int_0^3(9x- x^3),dx
&= left[frac{9}{2}x^2 – frac{1}{4}x^4right]^3_0
&= frac{81}{2} – frac{81}{4} = frac{81}{4} textual content{items}^2
finish{align*}

15. (c) ii.

start{collect*}
textual content{let } f(x) = 2x – (x^3 – 7x) = 9x – x^3
f(0) = 0, f(1.5) = 10.125 = frac{81}{8}, f(3) = 0
h = frac{b-a}{2} = 1.5
finish{collect*}
start{align*}
textual content{Space} &= frac{h}{3}left(1times 0 + frac{81}{8}instances 4 + 1 instances 0right)
&= frac{1}{2} instances 4 instances frac{81}{4} = frac{81}{4} textual content{items}^2
finish{align*}

15. (c) iii.

Discovering co-ordinate of (P)
start{align*}
y&=x^{3}-7x
y’&=3x^{2}-7
textual content{Let},,y’&=2,Rightarrow x=sqrt{3},,,y=-4sqrt{3}
textual content{due to this fact} P(sqrt{3},-4sqrt{3})
finish{align*}

15. (c) iv.

start{align*}
d&=frac{|2sqrt{3}-(-4sqrt{3})|}{sqrt{2^{2}+1^{2}}}
&=frac{6sqrt{3}}{sqrt{5}}
textual content{Therefore OA},&=sqrt{3^{2}+6^{2}}
&=sqrt{45}
&=3sqrt{5},,textual content{unit}
textual content{due to this fact} ,textual content{Space}Delta,OAP,&=frac{1}{2}timesfrac{6sqrt{3}}{sqrt{5}}times3sqrt{5}
&=9sqrt{3},,u^{2}
finish{align*}

16. (a) i.

start{align*}
V&= frac{1}{3}pi^{2}h,
however,, h^{2}+x^{2} &= 100
textual content{due to this fact} h&=sqrt{100-x^{2}}
V&=frac{1}{3}pi x^{2}sqrt{100-x^{2}}
finish{align*}

16. (a) ii.

start{align*}
frac{dV}{dx}&=frac{2}{3}pi xsqrt{100-x^{2}} – frac{2}{3}pi x^{3} frac{1}{sqrt{100-x^{2}}}timesfrac{1}{2}
&=frac{2pi x(100-x^{2})-pi x^{3}}{3sqrt{100-x^{2}}}
&=frac{pi x(200-2x^{2}-x^{2})}{3sqrt{100-x^{2}}}
&=frac{pi x(200-3x^{2})}{3sqrt{100-x^{2}}}
finish{align*}

16. (a) iii.

start{collect*}
Let frac{dV}{dx}=0
x=0,(Omit),,or,x=sqrt{frac{200}{3}}
textual content{Take a look at the character of the stationary level utilizing both desk methodology or Second by-product}
Rightarrow textual content{Max level at}, x=sqrt{frac{200}{3}}
textual content{Now}, 2pi x=10,theta ,,, textual content{By equating circumference}
textual content{due to this fact} theta=frac{2sqrt{2}pi}{sqrt{3}}
finish{collect*}

16. (b) i.

16. (b) ii.

\$\$Pr(Win) =frac{1}{36}instances(frac{2times20}{6})=frac{5}{27}\$\$

16 (c) i.

start{align*}
A_0&=300000
A_1&=300000times1.04-P
A_2&=A_1(1.04)-P(1.05)
&=300000(1.04)^{2}-P(1.04+1.05)
&=300000(1.04)^{2}-P(1.04+1.05),,,textual content{As Required}
finish{align*}

16. (c) ii.

start{align*}
A_3&=A_2(1.04)-P(1.05)^{2}
&=300000(1.04)^{3}-P((1.04)^{2}+1.05(1.04)+(1.05)^{2}),,,textual content{As Required}
finish{align*}

#### Related Posts

Platform Slot Gacor CMS Checker