Mathematics: Post your doubts here!

[sma786]

they already said leave the answer in surds..so when u do the equation x= -b ±√(b^2-4ac) / 2a u will end up with x =1/4 (-20 + √40) or 1/4(-20-√40) u can simplify it..did u get it now ?

I have rationalized the denominator, it is although not necessary!

thanks both got the answerr..

 

[SHAHBAZ NAZIR]

plz tell me the names of books for alevel maths .p1,p3,m1,s1

 

[Silver Wing]

dude tht is for stationary point of a curve.... no?

Thats right, but the stationary point in a Quadratic curve gives you either max or min point on it and this is where our range starts.
Hope I have helped you.

 

[TERMINATOR]

Salaaaam. Can you please help me differentiate this?

 

[falcon678]

Thats right, but the stationary point in a Quadratic curve gives you either max or min point on it and this is where our range starts.
Hope I have helped you.

ooh i see.... got it...
thanks...!

 

[VelaneDeBeaute]

Salaaaam. Can you please help me differentiate this?

View attachment 18414

Double differentiation?
Bring the power -1/2 to the start, multiply it with the constant already there. Less 1 from the power and quote the exact bracket with the power -3/2. Finally differentiate the contents of the brackets, i.e. the base, and it will be two. Now simplify!

 

[TERMINATOR]

Double differentiation?
Bring the power -1/2 to the start, multiply it with the constant already there. Less 1 from the power and quote the exact bracket with the power -3/2. Finally differentiate the contents of the brackets, i.e. the base, and it will be two. Now simplify!

Thank you, I highly appreciate it.

 

[M.Kabir Nawaz]

y=9x+1/x
coordinates where tangent is horizontal.
From Addmaths Ch 15.1 Q14b

 

[whitecorp]

y=9x+1/x
coordinates where tangent is horizontal.
From Addmaths Ch 15.1 Q14b

Differentiating y=9x+1/x wrt to x on both sides gives
dy/dx = 9 -1/x^2

When the tangent is horizontal,
dy/dx=0 =====> 9 -1/x^2 =0
x^2 =1/9
x= 1/3 or -1/3
When x=1/3, y= 6
When x= -1/3, y= -6
Hence, the required coordinates are (1/3, 6) and (-1/3, -6) (shown)

Hope this helps. Peace.

 

[cool Asviva]

please help me in this question.

Q: find the point where the normal at (2, 4) to y=x^2 cuts the curve again.
my answer is ( -7/4 , 49/16).
but it is (-9/4 , 81/16) in book answers.

well...thnx in advance.

 

[M.Kabir Nawaz]

If we plot the derived function on graph, then would it give gradient?
Differentiation.

 

[whitecorp]

please help me in this question.

Q: find the point where the normal at (2, 4) to y=x^2 cuts the curve again.
my answer is ( -7/4 , 49/16).
but it is (-9/4 , 81/16) in book answers.

well...thnx in advance.

The book answer is correct.

dy/dx =2x

At the point (2, 4), gradient of tangent =2(2)=4, and hence gradient of normal = -1/4

Equation of normal is y-4 = -1/4 ( x-2) = -1/4 x +1/2

ie y = -1/4x +9/2

When this normal cuts the curve again, we set x^2 = -1/4x +9/2

Solving gives x=2 (rejected because this is the original point) or x=-9/4

When x= -9/4, y =81/16

So the required set of coordinates is (-9/4. 81/16 ) . (shown)

Hope this helps. Peace.

 

[whitecorp]

If we plot the derived function on graph, then would it give gradient?
Differentiation.

You mean the derivative of the function?

 

[M.Kabir Nawaz]

You mean the derivative of the function?

yes

 

[SilverCrest]

f(x) is 2x+5
g(x) is 8/x-3

given that the equation fg(x) is 5-kx where k isa constant,has no solutions find the set of possible values for k

Can someone solve this for me?

 

[whitecorp]

yes

Then yes, it would.

Hope this helps. Peace.

 

[cool Asviva]

whitecorp
ohh...thnxx!!

 

[Minato112]

f(x) is 2x+5
g(x) is 8/x-3

given that the equation fg(x) is 5-kx where k isa constant,has no solutions find the set of possible values for k

Can someone solve this for me?

f(x) is 2x+5
g(x) is 8/x-3

fg(x) = 2(8/x-3) + 5
= 16/x - 6 + 5
= 16/x - 1

Now,

fg(x) = 5-kx
fg(x) = 16/x - 1

5-kx = 16/x - 1
5-kx = (16 - x) / x
5x - kx^2 = 16 - x
kx^2 - 6x + 16 = 0

Condition for no real solution: b^2 - 4ac < 0

a = k, b = -6, c = 16

(-6)^2 - 4 (k) (16) < 0
64k > 36
k > 9/16

Please note that I did not use any calculator and did it mentally. So please check the calculations and let me know.

 

[asexamskillme111]

2011 Paper 31.
Q7.

how do you integrate (2x-2)ln(x). I know you have to use the formula but I can't seem to integrate (x^2-2x)(1/x)

 

[iKhaled]

2011 Paper 31.
Q7.

how do you integrate (2x-2)ln(x). I know you have to use the formula but I can't seem to integrate (x^2-2x)(1/x)

man u serious ? u r an A level student and u dont know how to integrate (x^2-2x)(1/x)..

(x^2-2x)/x = x(x-2)/x cancel the x's!! then u will have x-2 and thats pretty easy to integrate eh..i guess u were a bit confused right ?