Mathematics: Post your doubts here!

[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

 

[Thought blocker]

If you guys want there is a whats app group for AS and A level students, you can get yourself added there and help people to clarify their doubts and get your self helped by posting your doubts within few minutes. If you wanna get added, leave a whats app message on: +919426116018 saying 'Add me to the A level group.' Thank yiu.

 

[AAnsarii]

Two lines have equations y=m1x+c1 and y=m2x+c2, and m1m2= -1. Prove that the lines are perpendicular.
(PS for 22, I couldn't type the 1s and 2s in subscript, sorry.)

 

[Holmes]

studyingrobot457
Finally I got a reply. Thanks.

 

[Deeksha Tamang]

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.

but why can't accept 5/4? the answer is only -1/2. BUT it helped me a lot thanks

 

[Holmes]

Postion vector of Reflection???
m/j/2017/32

 

[Thought blocker]

View attachment 63199
Postion vector of Reflection???
m/j/2017/32

Foot of perpendicular from A to l = 3/2 i + 3/2 j + 3k
Thus reflection of A in l = 2(Foot of perpendicular) - OA = (3i + 3j + 6k) - ( i + 2j + 4k) = 2i + j + 2k

 

[Holmes]

Foot of perpendicular from A to l = 3/2 i + 3/2 j + 3k
Thus reflection of A in l = 2(Foot of perpendicular) - OA = (3i + 3j + 6k) - ( i + 2j + 4k) = 2i + j + 2k

Oh Nice

 

[Deeksha Tamang]

View attachment 63199
Postion vector of Reflection?
m/j/2017/32

can to tell me how to find foot ofthe perpendicular???plzzz

 

[Thought blocker]

can to tell me how to find foot ofthe perpendicular???plzzz

Dis explains it well. (Y)

 

[syed1995]

View attachment 63162 can anyone plz some this

2-3x < |x-3|
2-3x < x-3 or 2-3x < -(x-3)
2-4x < -3
-4x < -5
x < 5/4

or

2-3x < -(x-3)
2-3x < -x +3
-2x < 1
x < -1/2

since the intersection of both is x<-1/2 that's the answer.

 

[Holmes]

Help me out in both questions!

 

[Deeksha Tamang]

Dis explains it well. (Y)

thanks a lot

 

[Thought blocker]

View attachment 63261
View attachment 63262
View attachment 63263
Help me out in both questions!

Differentiate the eqn given to u and equate it to 0. solve for x and get your y substituting the eqn you get for x in the main eqn.

Draw cosec(1/2 x) = y and 1/3 (x) + 1 = y on same axis and show that the graph interstcts at one point in the given domain.

for next part, do this:
Let f(x) = coesec(1/2 x) - x/3 - 1
now substitute values of x that is 1.4 and 1.6 and see if there is a sign change...
by that i mean if u sub value of x as 1.4 u will get some number either +ve or -ve and same way u get a + or - number when u sub 1.6... so if u see a sign chenge then it means that solution lies between 1.4 and 1.6

If you want detailed answer contact me on whats app : +919426116018

 

[Thought blocker]

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Hope to see you all soon. Good luck.

 

[Holmes]

Differentiate the eqn given to u and equate it to 0. solve for x and get your y substituting the eqn you get for x in the main eqn.

Draw cosec(1/2 x) = y and 1/3 (x) + 1 = y on same axis and show that the graph interstcts at one point in the given domain.

for next part, do this:
Let f(x) = coesec(1/2 x) - x/3 - 1
now substitute values of x that is 1.4 and 1.6 and see if there is a sign change...
by that i mean if u sub value of x as 1.4 u will get some number either +ve or -ve and same way u get a + or - number when u sub 1.6... so if u see a sign chenge then it means that solution lies between 1.4 and 1.6

If you want detailed answer contact me on whats app : +919426116018

Kindly help me out in Q4 too .
Thanks.

 

[Thought blocker]

Kindly help me out in Q4 too .
Thanks.

first line corresponds to d solution of 4th question

 

[Holmes]

first line corresponds to d solution of 4th question

If I was able to do correct differentiation, would i be asking you???