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AQA Core 1 C1 Unofficial Mark Scheme 2017

Q8b 9/4 <t < 8/3

EDIT: Below are the answers to this paper provided by @RDKGames


QUESTION 1

(a) 17−6717-6\sqrt{7}

(b) x=43x=\frac{4}{3}



QUESTION 2

(a) The other stationary point is (43,57527)(\frac{4}{3}, \frac{575}{27})

(b) d2ydx2(−2)=22⇒\frac{d^2y}{dx^2}(-2)=22 \Rightarrow M is a minimum.

(c) Sketch the curve

Spoiler





QUESTION 3

(a) x+2x+2 is a factor of p(x)p(x) so p(−2)=0⇒(−2)3+b(−2)2+c(−2)+24=0p(-2)=0 \Rightarrow (-2)^3+b(-2)^2+c(-2)+24=0 so 2b−c+8=02b-c+8=0 as required.

(b) x−3x-3 leaves a remainder of -30 so p(3)=−30⇒3b+c+27=0p(3)=-30 \Rightarrow 3b+c+27=0

(c) b=−7b=-7 and c=−6c=-6



QUESTION 4

(a) Line AB has equation 10y+11x=2810y+11x=28

(b) ABC is a right-angle so k=−4k=-4 or k=112k=\frac{11}{2}



QUESTION 5

(a) Normal to the curve at A is y=117x+15417y=\frac{1}{17}x+\frac{154}{17}

(b) Area under the curve between -1 and 2 is 27920\frac{279}{20}

(c) The area of the shaded region is 35120\frac{351}{20}



QUESTION 6

(a) (x+10)2+(y−7)2=102(x+10)^2+(y-7)^2=10^2

(b) The radius of the circle is 10 and horizontally it is displaced by 10 units from the centre. Therefore the circle is tangent to the y-axis. Similarly, it is displaced by 7 vertically from the centre and since 7<10 we have 2 distinct intersections with the x-axis.

(c) Intersection with y=kx+2y=kx+2 is given by (x−10)2+(kx−5)2=102(x-10)^2+(kx-5)^2=10^2 so expanding it gives (1+k2)x2+10(2−k)x+25=0(1+k^2)x^2+10(2-k)x+25=0

(d) Tangent line implies 1 intersection, which implies 1 solution to the above, which further implies the discriminant of the above quadratic is 0. Thus b2−4ac=100(2−k)2−4(1+k2)(25)=0b^2-4ac=100(2-k)^2-4(1+k^2)(25)=0 which yields k=34k=\frac{3}{4}



QUESTION 7

(a) i. y=115−2xy=\frac{1}{15}-2x

(a) ii. Shaded region is S=(x+y)2−y2=x2+2xyS=(x+y)^2-y^2=x^2+2xy and subbing in y=y=115−2xy=y=\frac{1}{15}-2x yields S=3(5x−x2)S= 3(5x-x^2)

(b) i. 5x−x2=254−(x−52)25x-x^2=\frac{25}{4}-(x-\frac{5}{2})^2

(b) ii. Max S is achieved by maximising 5x−x25x-x^2. Clearly, the max of this is 254\frac{25}{4} so max S = 754\frac{75}{4}



QUESTION 8

(a) dhdt(3)=3cm3h−1\frac{dh}{dt}(3) = 3 \text{cm}^3\text{h}^{-1}

(b) h decreasing means dhdt≤0⇒94≤t≤83\frac{dh}{dt} \leq 0 \Rightarrow \frac{9}{4} \leq t \leq \frac{8}{3}
(edited 6 years ago)

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Reply 1
Avatar for M451
M451
14
Yes, one question right ;-;

First exam where I really can't remember Q numbers but anyone get a value of k as 3/4?
Reply 2
Avatar for Workedmisty
Workedmisty
OP
3
Original post by M451
Yes, one question right ;-;

First exam where I really can't remember Q numbers but anyone get a value of k as 3/4?

I got k as 11/2 and -4
Reply 3
Avatar for aliyaahh
aliyaahh
4
What did you get for the integration question ?
k was 4/3 for one question from what i remember, integration was 351/20
Reply 5
Avatar for Workedmisty
Workedmisty
OP
3
Original post by aliyaahh
What did you get for the integration question ?

279/20
I got that!
What about the right angles question, I got k as 2 and 10.
Reply 8
Avatar for Workedmisty
Workedmisty
OP
3
Original post by C1waseasyaf
Swear question 9b was the hardest?? What did you guys get?

9/4<t<8/3
Integration was 279/20, (shaded region was 351/20) and k was 4/3


Posted from TSR Mobile
Reply 10
Avatar for dan11h
dan11h
1
Original post by C1waseasyaf
Swear question 9b was the hardest?? What did you guys get?


there was no ****ing q9...!!!!!!!!!!!
Original post by Legend15
Integration was 279/20, (shaded region was 351/20) and k was 4/3


Posted from TSR Mobile


Acc k might have been 3/4. It was the inverse of Q1b


Posted from TSR Mobile
what will grade boundaries be like?
Original post by BenMurphy11
what will grade boundaries be like?


Low i hope
what did people get for that question where you had to find the equation of normal to curve but in the form y=mx+c. I got like y=1/17+something?
Reply 15
Avatar for M451
M451
14
Original post by DanDanDidAnExam
what did people get for that question where you had to find the equation of normal to curve but in the form y=mx+c. I got like y=1/17+something?


Think I got 1/17 x + 154/17 which didn't exactly fill me with confidence
Original post by M451
Think I got 1/17 x + 154/17 which didn't exactly fill me with confidence


Yeah same. But i was thinking 153/17 gets an interger so i was like wtf about it & thought i did it wrong
Original post by M451
Think I got 1/17 x + 154/17 which didn't exactly fill me with confidence


I got that too, I'm pretty sure it's right
Original post by DanDanDidAnExam
what did people get for that question where you had to find the equation of normal to curve but in the form y=mx+c. I got like y=1/17+something?


Crap i forgot and kept 17y on the left. Would i still get the mark 😞🙏🏽🙏🏽🙏🏽🙏🏽
Original post by MKaur18
Crap i forgot and kept 17y on the left. Would i still get the mark 😞🙏🏽🙏🏽🙏🏽🙏🏽
youd drop one mark as you didn't fully simplify and put it in the form y=mx+c

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