Mathematics: Post your doubts here!

[The Sarcastic Retard]

How to find greatest and least values?

 

[qwertypoiu]

View attachment 59014
How to find greatest and least values?

You can change the given expression to:
10 + Rcos(blabla)

Since cosx is always between +1 and -1, the max value and min value of the expression will be 10+R, and 10-R

 

[The Sarcastic Retard]

You can change the given expression to:
10 + Rcos(blabla)

Since cosx is always between +1 and -1, the max value and min value of the expression will be 10+R, and 10-R

Thanks.

 

[holoholo]

Take both the roots you obtained in part (i), and cube it, if it evaluates to -1, that's proof that it satisfies the given equation in the question. Hope that makes sense.

yes i tried that but i did not get the right answer

 

[The Sarcastic Retard]

1(ii) Seeking for some help.

 

[nehaoscar]

View attachment 59023
1(ii) Seeking for some help.

log(base2)of20^-5 = -21.6 = roughly -21
log(base2)of20^5 = 21.6 = roughly 21

so n = 21 (from 1 to 21) + 21 (from -1 to -21) + 1 (the number 0)
n = 43

 

[The Sarcastic Retard]

4(i)

 

[The Sarcastic Retard]

log(base2)of20^-5 = -21.6 = roughly -21
log(base2)of20^5 = 21.6 = roughly 21

so n = 21 (from 1 to 21) + 21 (from -1 to -21) + 1 (the number 0)
n = 43

thanks.

 

[Rizwan Javed]

View attachment 59023
1(ii) Seeking for some help.

In the previous part, you got x = 21.6.

So if you replace the power of '5' in the previous part with the -5, you'll get x = - 21.6

In the second part he's talking about integers 'n' which satisfy the inequality, therefore, n must -21.6 < n < 21.6

Now just find the number of integers possible = 21+21+1 = 43 (21 integers on the negative side of number line, 21 on positive side, and zero. )

 

[The Sarcastic Retard]

In the previous part, you got x = 21.6.

So if you replace the power of '5' in the previous part with the -5, you'll get x = - 21.6

In the second part he's talking about integers 'n' which satisfy the inequality, therefore, n must -21.6 < n < 21.6

Now just find the number of integers possible = 21+21+1 = 43 (21 integers on the negative side of number line, 21 on positive side, and zero. )

Thanks.

 

[qwertypoiu]

Integrate : sqrt(25 - x^2)
lower bound 0 and uper bound is 5

(25 -x^2)^(1/2)
((25 - x^2)^(1/2 + 1))/(1/2+1 * -2x)
-((25-x^2)^(3/2))/3x

put the limits in and calculate.

This is not a correct method of integrating this function. This is a common misunderstanding.
When you integrate something within brackets raised to a certain power, you can integrate it by adding 1 to the power and dividing by this new power along with "derivative of inside the bracket", IF AND ONLY IF THE TERMS IN THE BRACKETS ARE LINEAR. ie they must be in the form (ax+b)^n

If there is a x² or x³ involved, or anything other than simple linear ax+b, the above rule CANNOT be applied.

In your particular case, the appropriate technique to be used is a special type of substitution known as trigonometric substitution.
You may substitute x=5sin@ :

∫ √(25-x²)dx = ∫ √(25-25sin²@)dx = ∫ √25(1-sin²@)dx = ∫5cos@dx

dx must also be changed.
x = 5sin@
dx/d@ = 5cos@
dx = 5cos@d@

So
∫5cos@dx = ∫5cos@(5cos@)d@ = ∫25cos²@d@
=25∫cos²@d@
=25 ∫ (½)(1+cos2@) d@
= 25/2 * (@ + ½sin2@)

We can covert above into terms that have x instead of @, but that's just extra work. We can just convert the lower bound (x=0) and upper bound (x=5) into @ Bounds:

x=5sin@
@ = arcsin(x/5)

Substitute x=0, @ = 0
Substitute x=5, @ = π/2

Substitute lower and upper bound to above integral thingy:

25/2 * (π/2 + ½sin(π) ) - 25/2 * (0 + ½sin(0))
= 25π/4

 

[nehaoscar]

Integrate : sqrt(25 - x^2)
lower bound 0 and uper bound is 5

Which paper is this from?

 

[zahra azam]

Is there any easy way or formula to find product-Moment correlation coefficient?

 

[shahzaib9291]

please explain in part 3 why BE is not equal to AC

 

[Mr.Physics]

View attachment 59068 please explain in part 3 why BE is not equal to AC

Point E will be somewhere above D. This makes ABCE a parallelogram. And in parallelogram, the two diagonals aren't equal. One is longer than the other and here BE is longer than AC.

 

[Mr.Physics]

Is there any easy way or formula to find product-Moment correlation coefficient?

This is from a guide

 

[The Sarcastic Retard]

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9709 -Mathematics/9709_w07_qp_2.pdf
Q2(ii) How can we give exact value?? I get in decimals also how to derive eqn? Is it just that we substitute Xn as x?
Q7(i) I did like this, temme what to do next;
cos^2(x) + 6sin(x)cos(x) + 9sin^(x)
(1-sin^2(x)) + 9sin^2(x) +3sin2(x)
1 + 8sin^2(x) + 3sin2(x)
I don't know what to do next.

Q8(i)I got -2 but answer is 2.
(ii) I dont know what ms says, is the equation y = e^-1(x)
Anybody?

 

[Rizwan Javed]

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9709 -Mathematics/9709_w07_qp_2.pdf
Q2(ii) How can we give exact value?? I get in decimals also how to derive eqn? Is it just that we substitute Xn as x?
Q7(i) I did like this, temme what to do next;
cos^2(x) + 6sin(x)cos(x) + 9sin^(x)
(1-sin^2(x)) + 9sin^2(x) +3sin2(x)
1 + 8sin^2(x) + 3sin2(x)
I don't know what to do next.

Q8(i)I got -2 but answer is 2.
(ii) I dont know what ms says, is the equation y = e^-1(x)
Anybody?

Q8 :

y = x^2 * e^-x

Use product rule to differentiate,

dy/dx = 2x e^-x + x^2 (-1)(e^-x)
= 2xe^-x -x^2 e^-x

For maximum values, dy/dx = 0, so,

0 = 2xe^-x -x^2 e^-x
0 = e^-x ( 2x - x^2)

Either:

e^-x = 0
N.A

Or:

2x - x^2 = 0
solving it you get,
x = 0 or x = 2

At x = 0, the graph is minimum as can be seen from the diagram, so we'll neglect that.

So the x - coordinate of M will 2.

 

[The Sarcastic Retard]

Q8 :

y = x^2 * e^-x

Use product rule to differentiate,

dy/dx = 2x e^-x + x^2 (-1)(e^-x)
= 2xe^-x -x^2 e^-x

For maximum values, dy/dx = 0, so,

0 = 2xe^-x -x^2 e^-x
0 = e^-x ( 2x - x^2)

Either:

e^-x = 0
N.A

Or:

2x - x^2 = 0
solving it you get,
x = 0 or x = 2

At x = 0, the graph is minimum as can be seen from the diagram, so we'll neglect that.

So the x - coordinate of M will 2.

Thanks. I forgot to take -ve sign in differentiating e^(-x) Can u explain my other doubts?

 

[Rizwan Javed]

Thanks. I forgot to take -ve sign in differentiating e^(-x) Can u explain my other doubts?

I haven't studied the other topics. So may be someone else might help.