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Can u tel me the position vector of p?
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ohh..yah i too got the value of lamda 3...Are you sure of the lambda thing? I think it was 3? And i'm not sure about the other value of lambda?
Does anyone remember the last question? NOT THE ANSWER, THE QUESTION?
Yes, coz they too give you an answer i x^2 (when multiplied by 1)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.
can you plz tell me what was the second part of vectores qsIn the 1st question, the value of x was 2.30 and in the vectors ques ii) part, the lamda was 2. correct me if i am wrong.
uhm weren't there 3 stationary points in q 6, as in pie/2, pie/12 and 5pie/12?
But doesn't lne get cancelled?
I got y= x -ln(3/2)
ohh..yah i too got the value of lamda 3...
Your binomial expansion is wrong. I do not remember the answer though
Hey guys, here are some of the answers my tutor got. Please remember that even a math tutor could make a mistake, so the answers below may not be 100% correct. Good luck to all of us!
1. x=2.3
2.ii. The root is 1.32
3. 1 - x + (1/2)x^2
4. pi/2, 7pi/6 ---Note: not 100% sure about this one because we did it really fast, since I didn't solve it completely
5. expression for differential is -y=ln[(-1/2)e^2x+(3/2)] -> y= -ln[-1/2 e^2x + 3/2]
6. i. angles are pi/2, pi/12 and 5pi/12 (3 angles)
ii. the point is a maximum point
7. u = -1/2+1/2i
8. ---
9. i.equation of tangent is y= x-1
ii. pi[(1/4)-(e^2/4)]
10. ii. position vector for point of intersection is 5,3,3
we did not solve the last part with the distance for the vector because I didn't attempt the question.
His binomial expansion is correct.
But doesn't lne get cancelled?
I got y= x -ln(3/2)
The question was dy/dx=e^(2x+y) right?
dy/dx=e(2x+y)
dy/dx=e^(2x) x e^( y )
[1/e^( y )] dy = e^(2x) dx
e^(-y) dy = e^(2x) dx
Integrate both the sides.
-e^(-y) = [e^(2x)/2] + c
Put x=0 and y=0 to find the value of 'c'.
-e^(-y) = [e^(2x)/2] + c
-e^(-0) = [e^(0)/2] + c
-1 = (1/2) + c
-1 - (1/2) = c
-3/2 = c
Put back this value of 'c' in the integrated solution.
-e^(-y) = [e^(2x)/2] + c
-e^(-y) = [e^(2x)/2] - (3/2)
Multiply the equation by a '-' sign.
- {-e^(-y) = [e^(2x)/2] - (3/2)}
e^(-y) = (3/2) - [e^(2x)/2]
Put 'ln' on both the sides.
ln e^(-y) = ln {(3/2) - [e^(2x)/2]}
-y = ln {(3/2) - [e^(2x)/2]}
y = - ln {(3/2) - [e^(2x)/2]}
Therefore, the answer was 'y = - ln {(3/2) - [e^(2x)/2]}'.
wouldn't e^2x/2 get cancelled when we take ln?
wouldn't e^2x/2 get cancelled when we take ln?
Exactly!!!! Ln always cancels an "e"!!!!
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