Mathematics: Post your doubts here!

[krishnapatelzz]

for your maths https://www.facebook.com/groups/302892556809834/

 

[NerdyBirdy]

Plz help me with this question
Find the value of the constant k for which the line y+2x=k is a tangent to the curve y=x^2-6x+14

 

[Thought blocker]

Plz help me with this question
Find the value of the constant k for which the line y+2x=k is a tangent to the curve y=x^2-6x+14

y = k - 2x
y = x^2 - 6x + 14

Equate them

k - 2x = x^2 - 6x + 14
x^2 - 4x + 14 - k = 0

As tangent meet once to the curve we can say that D = b^2 - 4ac = 0

16 - 4(14 - k) = 0
k = 10

P.s. If you wanna join As and A level whats app group, give me a message on my whats app number: +919426116018 and I also teach As and A level students on whats app, I teach in 3 ways. (i) Clear conceptual doubts (ii) Teach chapter wise (iii) Teach full course and all have the same rate of £5/hr..
There are many people taking benefits from this course, and some are there from the As and A level whats app group I mentioned above. You can persnoally ask them how they improved them selves or try yourself by just with one hour clearing concepts and then if you feel good you can continue with chapter wise and if you want you all chapters to be perfect as I teach you can also take the full course. The most benefit you get from this is YOU WILL GET 24 x 7 HELP from me, I will be solving your doubts from the chapter/concepts I taught you or had taught you before as soon as you send them apart from any technical issues I will be there to help all the time. And also solve the doubts from other chapters too but that will be solved when I will be having time. I hope to see you in my batch. Good luck.

P.s.s As and A level group is free, you can post your doubts there as much as you want admins and other people will try to help you as much as they can when they get time.

 

[Lenin Kumar Gandhi]

Anyone has a question bank for AS and A level Pure Mathematics P1 and P3. I had the question bank in the past for 0580-Extended Mathematics. It was called Q-Kit. It was really useful. You can make worksheets 'Topic-wise' in minutes... I hope to see something like that for AS and A-levels too.....
My e-mail ID: [email protected]

 

[sushil bista]

Does anyone has 8mundo's solved pastpapers.

 

[aqeelZamir]

Does anyone has 8mundo's solved pastpapers.

WHICH YEAR PAPER YOU NEED?

 

[aqeelZamir]

ANY ONE HAS 9709 2017-42 Q4 SOLUTION

 

[Holmes]

ANY ONE HAS 9709 2017-42 Q4 SOLUTION

2017 June or November????
Solution means .....Marking scheme? or you want explanation.

 

[sushil bista]

WHICH YEAR PAPER YOU NEED?

if u have all then please share it to me

 

[Holmes]

Help me out!

 

[Holmes]

Help me out Maths Geeks!

 

[abbas haider]

View attachment 63137
View attachment 63138

View attachment 63139

Help me out Maths Geeks!

We have 4T and 4B
all T stand next to each other and All B stand next T each other
which means we can treat 4T as 1 object = T ,and 4B as one object = B
now we have B and T (total arrangements right now are (between T and B)2 *4!(within B) * 4!(Within T ) = 2*4!*4!
but we know that 3B 's refuse to stand next to any T... and we know that there are two arrangements between groups( BT or TB)
which means that 1 B(the one who does'nt refuse = friendly person) must be facing to the T's always....thus it has 2 positions
and Rest 3Bs must be next to each other and so they will have arrangement withinthemselves ....which is 3!
so final answer
2(between groups)*4!(withinT) *3! (within B's that refused) = 2*3!*4! = 288

 

[Holmes]

We have 4T and 4B
all T stand next to each other and All B stand next T each other
which means we can treat 4T as 1 object = T ,and 4B as one object = B
now we have B and T (total arrangements right now are (between T and B)2 *4!(within B) * 4!(Within T ) = 2*4!*4!
but we know that 3B 's refuse to stand next to any T... and we know that there are two arrangements between groups( BT or TB)
which means that 1 B(the one who does'nt refuse = friendly person) must be facing to the T's always....thus it has 2 positions
and Rest 3Bs must be next to each other and so they will have arrangement withinthemselves ....which is 3!
so final answer
2(between groups)*4!(withinT) *3! (within B's that refused) = 2*3!*4! = 288

Thanks but I've got the explanation.

 

[Holmes]

View attachment 63075
Help me out!

anyone???

 

[Josephine Tan]

how do i get the answer for Q10 9709/31/on/11

 

[Deeksha Tamang]

can anyone plz some this

 

[Holmes]

Rule for Modulus: to remove Modulus take square on both sides.

4-6x+9x^2 < x^2-6x+9
simplify it you'll get
8x^2 -5 <0
8x^2<5
x^2<5/8
Take square root on both sides to get your required answer.

I Hope I helped a bit.

 

[Deeksha Tamang]

Rule for Modulus: to remove Modulus take square on both sides.

4-6x+9x^2 < x^2-6x+9
simplify it you'll get
8x^2 -5 <0
8x^2<5
x^2<5/8
Take square root on both sides to get your required answer.

I Hope I helped a bit.

but the answer is x>-1/2

 

[Holmes]

but the answer is x>-1/2

Deeksha Tamang thanks for making my correction! my mistake
Here you go with the right one:
Taking square on both sides
(2-3x)^2<(x-3)^2
you will get
8x^2-12x+9x^2<x^2-6x+4-9
Simplify it
8x^2-6x-5<0
Solve it with quadratic formula to get:
x=-1/2, 5/4

Hope I helped.

 

[studyingrobot457]

anyone???

Holmes here.
to expand tan(3x) you assume it is tan(2x+x)
then you use the rule tan(A+B)= (tan A + tan B)/(1-tanAtanB)
in this case (tan 2x + tan x)/(1-tan 2x tan x)
use the rule tan (2A) = 2tan A/1-tan^2 A
in this case it gets pretty much complicated

then this can be simplified