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Answer 5) y = ln (2/(3-e^(2x)))
but ln is there if content of the bracket is not negative or zero
hence i wrote the range also otherwise i said take the abstract value of them to be effective at upper ranges
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Answer 5) y = ln (2/(3-e^(2x)))
Good thinking, should've done that. That wasn't required though, wouldn't lose anything there.but ln is there if content of the bracket is not negative or zero
hence i wrote the range also otherwise i said take the abstract value of them to be effective at upper ranges![]()
I got a copy of the paper from my exam centre, so I have it with me. I can either email you the exam paper OR tell you the questions via PM if you wish. The cosec2theta question is number 4 btw
can u solve the question here..plzI got 1-x+1/2x^2 and I'm pretty sure it's correct. I will have all the correct answers in about 2 hours though
Sure. Credit to prisonbreak94hey aaditya memonif you have the paper could you send it to me at [email protected]. i would really appreciate it
Was there a need to find the terms in x^2 (like the -1/8x^2 and 3/8x^2) because (1-0.5x)(1-0.5x) will give you the terms in x and x^2.(1-x)^0.5 * (1+x)^-0.5
(1-x)^0.5 = 1-0.5x + (0.5)(-0.5)/2 x^2 =1-0.5x-(1/8)x^2
(1+x)^-0.5 = 1-0.5x + (-0.5)(-1.5)/2 x^2 = 1-0.5x+(3/8)x^2
Multiply them and you get 1-x+0.5x^2
Sure. Credit to prisonbreak94
what was question number 8 and tell it's total marks too?Answer 1) 2.30
Answer 2) 1.32
Answer 3) 1 - x + 0.5x^2
Answer 4) 201.47 degrees and 338.53 degrees.
Answer 5) y = ln (2/(3-e^(2x)))
Answer 6) pi/12 and 5pi/12 , maximum point
Answer 7) -0.5 + 0.5i
Answer 8) Pretty easy, if someone wants the working, i'll be more than happy to post it here.
Answer 9) y = x-1 and pi/4(e^2 -1)
Answer 10)The two points are (7,4,5) and (3,2,1). Distance between them = 6.
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