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thanks
Assalamualaikum,
This question is from P3 ,P2 and P3 book , Exercise 18A
Q.2) Intregation;(x-2)/(x-4)^1/2.dx
Thanks in advance for cooperation!
Assalamualaikum,
This question is from P3 ,P2 and P3 book , Exercise 18A
Q.2) Intregation;(x-2)/(x-4)^1/2.dx
Thanks in advance for cooperation!
@mustehssun Iqbal: well, i need to do this... bcoz of unavailability of time, i wont be doin it.. i need to know what you will get when the lower part gets expanded.. anything it comes, i suggest you to do with partial fractions...by getting two individual terms or may be three..so that u can integrate easily..if u didnot get it, i will do it for you...
what ansers u get for the 2nd part?
last part?
i) higher than the top of the tower means higher than 25 m.http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s04_qp_4.pdf
Q.7 need help. kindly explain in detail whichever formulas and ways u use in this ques. Thanks.
First of all, you'll simultaneously solve the curve equation and the line equation.
y = x^2
y = x + 2
x^2 = x + 2
x^2 - x - 2 = 0
x^2 - 2x +1x -2 = 0
x ( x - 2) + 1 (x - 2) =0
x = 2, x = -1
Now we have the x-coordinates of the point at which the curve meets the line. Accordingly, we'll find their y-coordinates which when calculated are y = 4 and y = 1.
So now we have two points, (2 , 4) and (-1 , 1). Next, we'll differentiate the curve equation 'y = x^2' to obtain the gradient function which is '(dy/dx) = 2x'. We'll insert the value x=2 in this equation to obtain a value of the gradient. The value of gradient is '4'. Inserting the values of y=4, x=2 and m=4 in the equation 'y - y1 = m (x - x1)', we get the first equation of the tangent which's 'y = 4x -4'. Similarly, using the other point which's (-1 , 1), we'll insert the value x = -1 in the gradient function to again obtain a value of the gradient which, for this point, comes out to be 'm = -2'. Again, we'll substitute the values of x = -1, y = 1 and m = -2 in the equation 'y - y1 = m (x - x1)' to obtain a second equation of the tangent which's 'y = -2x - 1'. Lastly, we'll solve both the equations of the tangents simultaneously to find their points of intersection.
-2x-1 = 4x-4
3=6x
.5=x
y=4x-4
y=4(.5)-4
y=-2
Therefore, the coordinates of the point of intersection of the tangents to the graph y=x^2 are ( .5 , -2 ).
Hey again.
But the question seems a lil different.
Find the equations of the tangents of the graph of y=x^2 AT THE POINT AT WHICH IT MEETS THE LINE WITH THE EQUATION y=x+2.
So Here they're asking us to find the point where it meets the line with equation y=x+2
the answer given is the same as yours but it doesn't satisfy the line eqn thing.
Thankyou!Assalamoalaikum!
integrate => [g(x)]^2 - [f(x)]^2
square of equation of upper curve - square of lower curve...and integrate!
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