Q1 .If

f(x)=\( \begin{array}{ccc} \cos x & 1 & 0 \\ 1 &2\cos x &1 \\ 0 &1 &2\cos x \end{array}\)

then \int^{\frac{\pi}{2}}_0 \ f(x)\ dx =

1) -1\3
2) 1\4
3) 1\2
4) none of these

Q2.If

f(x)=\( \begin{array}{ccc} \cos x & 1 & 0 \\ 1 &\cos x &1 \\ 0 &1 &\cos x \end{array}\)

then f'\(\frac{\pi}{3}\) =

1) \frac{11\sqrt{3}}{8}

2) \frac{5\sqrt{3}}{8}

3) \frac{-5\sqrt3}{8}

4) none of these.

Please give your answer with correct explanation

Q3. If [a] denotes the greatest integer less than or equal to a and -1\leq x<0, 0\leq y<1, 1\leq z<2,then

\(\begin{array}{ccc}[x]+1 & [y] & [z]\\
[x] & [y]+1 & [z] \\
[x] & [y] & [z]+1\end{array}\) =

1) [x]
2) [y]
3) [z]
4) none of these3

Please give your answer with correct explanation..

Q4. If A+B+C=\pi then value of determinant

\(\begin{array}{ccc}\sin^2A & \cos A\sin A & \cos^2A \\ \sin^2B & \cos B\sin B & \cos^2B\\ \sin^2C & \cos C \sin C &\cos^2C\end{array}\) =

1) 2sin(A+B)cos(B-C)sin(C-A)
2) 2sin(A-B)cos(B-C)cos(C-A)
3) sin(A-B)sin(B-C)sin(-C+A)
4) cos(A-B)cos(B-C)cos(C-A)

Please give your answer with correct explanation

Q5. If

A= \[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\]

is a square root of I2 , then alfa,beta and amma will satisfy the relation

1) 1 + \alpha^2 + \beta\gamma = 0
2) 1 - \alpha^2 + \beta\gamma = 0
3) 1 + \alpha^2 - \beta\gamma = 0
4) -1 + \alpha^2 + \beta\gamma = 0

Please give your answer with correct explanation...

ModEDIT: Threads merged and edited. Please, post your attempt first and if something is not correct in this way, say so because some of them are vague.

Did you read the note posted by the moderator?

"ModEDIT: Threads merged and edited. Please, post your attempt first and if something is not correct in this way, say so because some of them are vague."

Read this first:

Expectations for the Homework Help board

Do not simply retype or paste a question from your book or study material

We won't do your homework questions for you.
You were given the assignment for you to learn.

If you come up with your own answer and post it for us to critique that is within reason.

If you have some SPECIFIC questions that you couldn't find or didn't understand, we may help with that.
But this is your assignment, so show us you have at least attempted to complete it on your own.

Thank you.

You meant that first I will type my attempt and than you will correct it
I think you are mad
I don't have any mathematical key board
Therefore fore it will take a lot of time to type it

You could always write your attempts with a pencil and scan your work as a picture (if you have access to a scanner or digital camera).

Anyway, these problems are generally just a combination of two different things that you already know how to do. They might be a little tedious, but I'm guessing you can easily calculate

\int_0^{2\pi}\cos x \; dx

and other such integrals. Likewise, I'm sure you know how to calculate the derivative of such functions. So all you have to do for the first two questions is integrate or differentiate all of the elements of the matrices individually, and then find the determinant afterwards.

Finding a determinant is tedious, but it's simple math you've probably known how to do since you were 8 years old! Just addition, subtraction, and multiplication.

For Q3, you'll need to figure out the values of [x], [y], and[z]. It's pretty straightforward; if -1 <= x < 0, then what's the greatest integer less than or equal to x? -1 right? You can apply the same logic to the other two. Then you just have to evaluate each of the elements of the matrix, then find the determinant afterwards.

Q4 will require you to find the determinant, then use some trig identities to simplify the results.

Q5 I'm not sure about. I'm not familiar with "I2". What is it?

Thanks for the answer I had solved the third one but not able to solve the rest questions

ApoorvGoel, based on the multiple choice answers, I'd say I gave you some bad advice. You need to do the determinants BEFORE you do the integrals, derivatives, etc.

It was difficult to tell which order to do it in because the way the problems are written, there's no determinant operator (the parentheses around the matrix in Q1, for example, should look more like absolute value bars). Given that, you definitely need to do the determinant first.

I'll help you with Q1, since that's harder than Q2: the determinant is

f(x)=\cos x(4\cos^2 x -1)-2\cos x

Next you have to find the integral

\int_0^{\frac{\pi}{2}}\cos x(4\cos^2 x -1)-2\cos x\;dx = \int_0^{\frac{\pi}{2}}\cos x(4\cos^2 x -3);dx=\int_0^{\frac{\pi}{2}}\cos x(-4\sin^2 x +1)\;dx=-\frac43\sin^3 x+\sin x |_0^{\frac{\pi}{2}}=-\frac13

Thanks for your answer and now please help me in solving the 4th and 5th question

All right, question 5 is easy (now that I just realized that I2 probably means the 2x2 identity matrix).

One property of matrices is that the determinant of the square root of a matrix A is the same as the square root of the determinant of A. Written in math, this is

\begin{vmatrix}\sqrt {\bold{A}}\end{vmatrix}=\sqrt{\begin{vmatrix}\bold A \end{vmatrix}}

If A is a square root of the identity matrix, that means its determinant must be equal to the square root of the determinant of the identity matrix (which is 1).

So all you have to do is set the determinant of A to 1, then rearrange to find your answer.

I'm still working on question 4. I know the answer is #3, but I'm trying to prove why.

Okay, question #4 was a pain. Maybe there's an much simpler obvious way that I'm not seeing?

Anyway, you first have to go through the tedium of calculating the determinant. Let's call it D:

D=\begin{vmatrix}\sin^2A & \cos A\sin A & \cos^2A \\ \sin^2B & \cos B\sin B & \cos^2B\\ \sin^2C & \cos C \sin C &\cos^2C\end{vmatrix} = \sin^2 A \(\cos B \sin B \cos^2 C-\cos^2 B \cos C \sin C \)-\cos A \sin A \( \sin^2 B \cos^2 C - \cos^2 B \sin^2 C\) + \cos^2 A \( \sin^2 B \cos C \sin C - \cos B \sin B \sin^2 C \)

Now start factoring and simplifying using trig identities:

D= \sin^2 A \cos B \cos C \(\sin B \cos C-\cos B\sin C \)-\cos A \sin A \( \sin^2 B \cos^2 C - \cos^2 B \sin^2 C\) + \cos^2 A \sin B \sin C\( \sin B \cos C - \cos B \sin C \)

D= \sin^2 A \cos B \cos C \sin (B-C)-\cos A \sin A \( \sin B \cos C + \cos B \sin C\) \( \sin B \cos C - \cos B \sin C\) + \cos^2 A \sin B \sin C \sin (B-C)

D= \sin^2 A \cos B \cos C \sin (B-C)-\cos A \sin A \( \sin B \cos C + \cos B \sin C\)\sin (B-C) + \cos^2 A \sin B \sin C \sin (B-C)

D= \sin(B-C) \(\sin^2 A \cos B \cos C-\cos A \sin A \( \sin B \cos C + \cos B \sin C\) + \cos^2 A \sin B \sin C\)

D= \sin(B-C) \(\sin^2 A \cos B \cos C-\cos A \sin A \sin B \cos C -\cos A \sin A \cos B \sin C + \cos^2 A \sin B \sin C\)

D= \sin(B-C) \(\sin A \cos B -\cos A \sin B \) \(\sin A \cos C -\cos A \sin C \)

D= \sin(B-C) \sin (A-B) \sin (A-C)



Note that it doesn't matter that A+B+C=\pi.

Thanks a lot for your answer