HSC Grove 2u Ex6.4 Q13 Solution

Hey sir, I was wondering i you could do a solution for this question

thanks

Grove HSC 3u book Ex 8.6 Q13,15

Hey sir could you blog this qu from 8.6 q13!! thankyou!!

Q13

Q15

Thanks Grace :-)
I hope the tests went well. :-) JH

Thank you!

Jan,
You projected into our lives and dramatically increased the gradient of our learning. Thank you so much for being the best brumby artiste, hovercrafter and person who met a brain surgeon (and maths teacher) we could've asked for. You have been such an amazing teacher who has inspired all of us. Thank you for the immense dedication and commitment to our class, it hasn't gone unnoticed.

We wish we all the best in the future, good luck being the CEO of NASA!!

All the best
Year 11 accelerated.

Hollynomial, camilimetre, sarabola, f'(ern), stephstitution, domintegrate, decibel, geogebra, e(^mily), annalyse, madisum, issymetric, jessintegrate

:)

Grove HSC 3u ex6.1 Q11

2015 Math HSC 16b

MIF 2U HSC TY6 Q10

we dont integrate sec(x).

I cant find the question, please screenshot it. :-)


Dear Mr Hansen,
The question requires you to integrate sec(x), we only have a formula to integrate sec^2(x).
how would I do this question, could you send me your working?

Thanks, Anna

HSC exam, 2014, 15 (c)

HSC exam, 2014, 15 (c)​

Exp log, solve for x

Series AP question arithmetic

Exp log question

QUESTION: Given that the sum of the first 4 terms of an AP equals 34, the next 4 terms sum to 146, find $T_9+T_{10}$

QUESTION: Given that the sum of the first 4 terms of an AP equals 34, the next 4 terms sum to 146, find $T_9+T_{10}$

Grove HSC 3U TY8, series AP

Question 6 pls Trigonometric Functions

SOLUTION
Put $\sin x =\cos x$ to find where the two curves intersect..

Divide both sides by $\cos x$ to get  $\tan x=1$ and this gives solution $x=\pi/4$.

When $x=\pi/4$, $y=\sin \pi/4 = 1/\sqrt{2}$ as required.

JH 

Grove HSC 3U EX 4.8 Q10

- Thanks Mr Garben for this solution :-)

POST FORMAT PLEASE :

WHICH BOOK, e.g.           Grove HSC 3U
WHICH EXERCISE, e.g.   Ex 6.11
WHICH QUESTION
SNAPSHOT PREFERRED too please if possible

Derivative/Integral of $a^x$

Grove 4.8 Question 11

Sent from my iPhoneWhy in the answers is it pi/2?

Grove hsc 3u ex ,?? Q14 pls

which set is this from ?

You have to use $3^B=e^{B\ln 3}$

Sent from my iPhone

Grove HSC 3U chapter 4.2 Question 2 - Differentiating an exponential type fn.

Help!  .

Question 2. =============

$y=(e^{2x}+1)^7$
$y'=7.(e^{2x}+1)^6\times 2e^{2x}$ by chain rule.
$y'=14(e^{2x}+1)^6e^{2x}$
$y'=14 e^{2x} (e^{2x}+1)^6$

Now use product rule too for $y''$

$u=14e^{2x}$, $v=(e^{2x}+1)^6$
$u'=28e^{2x}$, $v'=6(e^{2x}+1)^5 2e^{2x}=12 e^{2x}(e^{2x}+1)^5$

$y'=14e^{2x}12 e^{2x}(e^{2x}+1)^5 + 28e^{2x}(e^{2x}+1)^6$

Factorising

$y'=28e^{2x}(e^{2x}+1)^5(7e^{2x}+1)$

Grove HSC 3U - Challenge Exercise 4

Grove HSC 3U - Challenge Exercise 4

Challenge exercise 5 q15

Help please !!

SOLUTION

Trig graph grove HSC ty5 q4a

Graph of $\Large y=\tan 2x - 2$

The best way to draw trigonometric graphs like these, is to first draw the basic trig graph, here it is $y=\tan x$. 
Then build it up from there, here we multiply by 2 (stretch), then subtract 2 (lower by 2)

Graph of $\Large y=\sin(2x+{\pi\over 3})$

The best way to draw trigonometric graphs like these, is to first draw the basic trig graph, here it is $y=\sin x$. 
Then build it up from there, here we multiply by 2 (stretch), then shift to left by ${\pi \over 6}$ 

Sketch $\Large y=3\sin x +1$

The best way to draw trigonometric graphs like these, is to first draw the basic trig graph, here it is $y=\sin x$. 
Then build it up from there, here we multiply by 3 (stretch), then add 1 (lift by 1)

AP SERIES QUESTION FROM JAN ACADEMY :-) Thanks Emily/Issie

Thanks to Emily, Issie, and the yr11 class for this solution :-)

Grove HSC 2U Ex 4.2 Q2o

Question= 

Answer =

How did they get that?

Thanks, Anna

SOLUTION

$u=e^{2x+1}$, $u'=2e^{2x+1}$
$v=2x+5$, $v'=2$

$y'= \displaystyle { vu'-uv' \over v^2}$
$y'= \displaystyle { (2x+5)2e^{2x+1}-e^{2x+1}\times 2 \over (2x+5)^2}$
$y'= \displaystyle {  2e^{2x+1}(2x+5-1)\over (2x+5)^2}$
$y'= \displaystyle {  2e^{2x+1}(2x+4)\over (2x+5)^2}$

$y'= \displaystyle {  2\times 2e^{2x+1}(x+2)\over (2x+5)^2}$
$y'= \displaystyle {  4e^{2x+1}(x+2)\over (2x+5)^2}$

:-)

Brought to you by Anna and the Jan Academy :-) :-)

Grove HSC 3U Ex 5.9 Q12

SOLUTION

Sent from my iPhoneQ12 pls!

Grove HSC 3U Ex 5.9 Question 4

Hello sir, how do you do question 4?

SOLUTION
Find $y'$, at $x=\pi/6$ and sub into $y-y_1=m(x-x_1)$

$y'=\cos(\pi-x)\times -1$
$y'=-\cos(\pi-x)$
So, tangent gradient is 
$m=y'(\pi/6)=-\cos (\pi-\pi/6)=-\cos(5\pi/6)=\cos(\pi/6)=\sqrt{3}/2$

Eqn of tangent

$y-0.5=\sqrt{3}/2(x-\pi/6)$
$2y-1=\sqrt{3}(x-\pi/6)$
$2y-1=\sqrt{3}x-\sqrt{3}\pi/6$
$12y-6=6\sqrt{3}x-\sqrt{3}\pi$
$6\sqrt{3}x-12y+6-\sqrt{3}\pi=0$

Thanks for the question.
Solution brought to you from the Jan Academy 2016, JSH :-)

Grove HSC 3U Ex 5.7 Question 1g and j

Hi sir, can you please help me with 1g and 1j

SOLUTION


The best way is to draw the basic graph of tan x and go from there..as shown below :-)

GROVE HSC 3u Ex 8.5 Q17

GROVE HSC 3U EX 8.5 Q11

GROVE HSC 3U EX 8.5 Q12

Grove HSC 3U EX 8.5 Q8

 

Formula is $S_n={n\over 2} (2a+(n-1)d)$

$a=5$

$d=4$

$S_n=152$

$ S_n= {n\over 2} (2a+(n-1)d)$

$ S_n= {n\over 2} (2(5)+4(n-1))$

$ S_n= {n\over 2}(10+4n-4)$

$S_n=2n^2+3n$

$\therefore 2n^2+3n=152$

$2n^2+3n-152=0$

$n={-3\pm \sqrt{3^2-4(2)(-152)} \over 2(2)}$

$n={-3\pm \sqrt{9+1216} \over 4}$

$n={-3\pm \sqrt{1225} \over 4}$

$n={-3\pm 35 \over 4}$

$n={-3+ 35 \over 4}, {-3- 35 \over 4}$

$n=8, -19/2$

Hence since $n$ is whole number 

$n=8$.

:-) JH 2016

Grove HSC 3U EX 8.4 Q11

Hi sir when you are doing arithmetic series how do you do it when you want to know if a number fits into a series? Like Q11 from 8.4?

Answer: I would find an expression for the nth Term $T_n$.

Is $0$ a term of the series $48 + 45 + 42 + ...$?

Method 1: write them all out (only works for small numbers!)
48+45+42+39+36+33+30+27+24+21+18+15+12+9+6+3+0+...
So yes $0$ is a term.

Method 2: all the numbers are multiples of 3, and they go down by 3, so 0 must eventually be reached!! So again Yes.

Method 3: It's an AP with $a=48$, $d=-3$, 
$T_n=a+(n-1)d$
$T_n=48-3(n-1)$
$T_n=51-3n$

Solve $51-3n=0$

$51=3n$

$n=51/3=17$

$T_{17}=0$

So Yes. The $17$ th term equals $0$.

JH 2016 :-)

Grove HSC 2u Ex7.1 Q19

Please give me some of the question so i know which one it is as i am working from the 3U book. is it series?

GROVE HSC 3U EX 4.8 Q8

GROVE HSC 3U EX 4.8 Q8

Question about using $y''$ to determinenature of stat points

Can you use $y''$ to determine the nature of stationary points?  YES

• If $y''<0$ for some stat point then it is a MAX (Concave down).
• If $y''>0$ for some stat point then it is a MIN (Concave up).
• If $y''=0$ for some stat point then you MUST use the first derivative test box to see what it is :-)

EXAMPLE: $y=x^2-1$

$y'=2x$  stationary point is when $x=0$

Here, $y''=2>0$.

$(0,-1)$ is the stat point.

Since $y''(0)=2$ , so $y''>0$ and hence its a MIN.

You can verify this with the first derivative test.

JH :-)

Grove Prelim Book 2U 11.7 Question 1

$y=x^2/12$

$y'=x/6$

At x=6, $y'=6/6=1$

Normal gradient is $m=-1/1=-1$

$y-3=-1(x-6)$

$y-3=-x+6$

$y=-x+9$

Solve 
$y=-x+9$  (1)
$x^2=12y$   (2)

(1) x 12:
$12y=-12x+108$   (1)

so (1) subbed into (2) gives

$x^2=-12x+108$

$x^2+12x-108=0$                  

                  sandbox:  108=54 x 2 = 27 x 2^2 = 3^3 x 2^2
                                                                           3x2=6, 3^2 x 2=18!! yes

$(x+18)(x-6)=0$

$x=-18$ is where the normal cuts the parabola again.

$y=- -18 +9=27$

$P(-18,27)$

Primitive fn question

Grove 2U HSC EX5.3 Q10 A sector of a circle with radius 5cm and angle pie/3 subtended at the centre is cut out of card board. It is then curved around to make a cone. Find Exact SA and V.

Grove 2U HSC EX5.3 Q10 

A sector of a circle with radius 5cm and angle pie/3 subtended at the centre is cut out of card board. It is then curved around to make a cone. Find Exact SA and V.

Camb2uY12-EX1F-Q4a Integration

Please note that Cambridges diagram has the functions the wrong way around!!

Locus question

Grove Prelim 3U EX 11.8 Q5c