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Mathematics: Post your doubts here!

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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! :)
 
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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.
 
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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.:eek:

well...thnx in advance.:)
 
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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.:eek:

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.
 
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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?
 
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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. :)
 
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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)
 
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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 :p ?
 
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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 :p ?

LOL. No idea what I was thinking. Must've had a dumb moment.
 
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If y = tan^-1 x, show that dy/dx = (1/x^2 + 1)?
Should use the identity (dy/dx) * (dx/dy) = 1.
In this case x = tan y. then dx/dy = sec^2 y

hence dy/dx = cos^2 y now substitute y = tan^(-1) x:

dy/dx = cos^2( tan^(-1) x).
denote the angle tan^(-1) x as B and find cos^2 B if tan B = x;
using identity sec^2(B) = 1 + tan^2 B, obtain

dy/dx = 1/(1+x^2)
 
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