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Thanks!

^^just edited my earlier post. It is for the first one.

Now, the answer in the book says that the answer for question (1) is 1/2.

Is it right or there is a printing mistake?

**mttal24**- Replies: 24

I am trying to solve questions like:

1.

and

2.

How do I solve them?

for the first one, i get upto lim 1/n+2. Then what should I do?

If

t2/t1=t3/t2=t4/t3=.....=tn/t(n-1)=r (and 'r' also represents common ratio)

then the sequence is a GP.

Here,

1/2^4 divided by1/2 is equal to 1/2^7 divided by 1/2^4.

Thus, you can show that it is a gp

**mttal24**- Replies: 1

I have read all basics of trignometry and I have a question:

Find

I have read all the questions related to these on mathsisfun.com but all the questions there are like sin 330 is given and a related acute angle's value is given.

But how do I convert this into an acute angle?

Are you asking about proof by Mathematical Induction?

^^Thanks. Now I understood the concept.

benice wrote:

(1)

Notice that [x] ≤ x < [x]+1 for all x∈ R.

x ∈ R-Z

=> 0 < x-[x] < 1

=> 1/(x-[x]) > 1

=>f(x)= 1/sqrt(x-[x]) = sqrt(1/(x-[x]))> 1

Thanks. But what do you mean when you say that since x∈R-Z so the equation 0<x-[x]<1.

And thanks very much for the second question. I realised what I was doing wrong.

^^Thanks.

I have done it as you said.

x-[x] ≤0

=> x≤[x]

That is only possible when x is an element of Z(Integers)

=> Domain is (R-Z) [R is the set of real nos. and Z is that of integers].

Now for Range, what should I do?

**mttal24**- Replies: 6

I have been able to find the domain and ranges of questions but this one is not coming:

(1)

By [x] here, I mean Floor function or greatest integer function.

Also,

(2) This modulus function question, I am not able to understand how to get values for the conditions:

If f(x) be defined on

and is given byf(x) = {-1 for -2≤x≤0 and (x-1) for 0<x≤2 }

and g(x) = f(|x|) + |f(x)|. Find g(x). Here | | means Modulus.

Thanks.

Lets say, I begin answering by:

Now, I know sec/cos is negative between implies,

But, here we are taught by a different method where we see that the main identities that,

Similarly,

I tried understanding this in many books, which I wrote as

And, in some books I see that, suddenly out of the blues, a person adds

to and makes it

Now, I have understood that after reading the same question for the past 5 days.

But your method seems to be faster.

Thanks

And if

**mttal24**- Replies: 5

I know that a principal solution of a trignometric equation is 2 values lying between

But I don't know how to find them.

Lets take a question,

Find the principal solutions of

I know,

And, in some books I see that, suddenly out of the blues, a person adds to and makes it

I just know that values of sine and cos repeat after

Please give me full explanation on how to find this. Please explain with one more example so that it becomes clear to me :

That means that the fx is undefined for denominator=0 because anything/0=undefined

Can you explain how to solve

|x-2|/x-2 ≥ 0

My ans is:x ≥2

But book's ans is: x>2

Thanks. Finally I understood something...

In the book, the answer given is

soln. set of the inequation is (-infinity,-2) union (-1/2, +infinity).

But, you have given x<2 which would mean (-infinity, 2)

@bob bundy

I have two for you:

1. |x-1|/x+2<1

2. |x-1|+|x-2| ≥4

One thing more,

Can you show me a detailed procedure on how to solve inequations with the moduluses like | x-1 |

**mttal24**- Replies: 20

My question,

------------------------------------------------------------------------------------------------------

I understood that, all terms have to go on LHS, to solve in form:

1. But how to proceed ahead? Please explain me step-by-step.

------------------------------------------------------------------------------------------------------

2. When you reach the end, you get two values(here, fractions), say c and d. Then how do you decide whether:

or

Similarly,

or

**mttal24**- Replies: 1

Q. Show that for any sets A and B, A = (A ∩ B) ∪ (A B).

Please show me by the method of "let x∈A..."

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