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Mathematics: Post your doubts here!

Suchal Riaz

Suchal Riaz

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Thought blocker said:
GCE As and a level
4i)
(x/k)+k = x^2/4
=>4((x/k)+k) = x^2
=>4x/k=x^2-4k
=>4x = kx^2 - 4k^2
=>kx^2-4x-4k^2 = 0
Use b^2 - 4ac = 0
16 +16k^3 = 0
k = -1 :)

4ii)
kx^2-4x-4k^2 = 0 put k as -1 and get your x value as -2
now as we have y = x^(2)/4 put 2 as x, you get y as 1
so P is (-2,1) :)

6ii)
from part i) you got 2sin^(2)x + 3sinx -2 = 0
By solving this quadratic eqn, you get 2 values of sin theta i.e is 0.5 and -2 well sin inverse is not defined at -2 so consider sin theta to be 0.5 ; Now, sin inverse 0.5 is 30. Here we have 2y instead of x so for one value of y it will be 30/2 = 15 and for 2nd value of y will be (180-30)/ 2 = 75 . :)

7ii)
For AB to be unit vector it should equal to 1
hence,
(k-1)^2 + K^2 + (2k - 2)^2 = 1
6k^2 -10k + 4 = 0
now you got k as 1 or 0.67

8i)
Use ar^(n-1) formula here,
so for second term , ar^(2-1) = 24 and for fourth term, ar^(4-1) = 13.5
Now equate each other and get r as 3/4 and first term (a) as 32 :)

8ii)
use a/(1-r) formula, you get answer as 128 :)

8b)
ZaqZainab Suchal Riaz Idk this one ask 'em.


10ii)
When d2y/dx^2 is > 0 it has minimum value. Here it is > 0 i.e is minimum point :)


11)
Suchal Riaz ZaqZainab

Actually I am tired of typing. :/ Help him now, I have to do organic chemistry :/
Click to expand...
where are question?
 
A

Ayeshak93

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Can anyone tell me why in the last part the answer is -2.5《x《0....I get the -2.5 part but not the zero part...
 

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Thought blocker

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