Thanx Evaluate I = sin 2x / cos2 (2x) dx. The correct answer is 1 + 2 / root2 -r

Question

Thanx

Evaluate I = sin 2x / cos2 (2x) dx. The correct answer is 1 + 2 / root2 -root2 - 1 / 2 - (1 + root2 / 2) (1 + root2 / 2) None of the above. Write in sigma notation: 4 / 1 + 9 / 2 + 16 / 3 + 25 / 4 + 36 / 5. The correct answer is k2 / (k - 1) k2 / (k - 1) (k - 1)2 / k (k - 1) None of the above. Find the following limit limn rightarrow infinity ln n4 / n4. The correct alternative is 4 0 -1 / 6 1 / 4 None of the above. If f(x) = 1 / 1 - x, determine P6 (x) in the point 0. The correct answer is P6(x) = 1-x + x2 - x3 + x4 - x5 + x6 P6(x) = 1 + x + x2 /2 + x3 / 3 + x4 / 4 + x5 / 5 + x6 / 6 P6(x) = 1 + x + x2 / 2 + x3 / 3! + x4 / 4! + x5 / 5! P6(x) = 1 + x + x2 + x3 + x4 + x5 + x6 None of the above. Find the Taylor polynomial of order n generated by f(x) = ln 2x in the point alpha = 1. The correct answer is Pn,1 (x) = ln 2 + (x -1) - (x - 1)2 / 2 + (x - 1 )3 / 3 + + (-1)n-1 (x - 1)n / n Pn,1 (x) = ln 2 + (x - 1) - (x - 1)2 /2! + (x - 1)3 / 3! + + (-1)n-1 (x - 1)n / n! Pn,1 (x) = ln 2 + (x - 1) - (x - 1) 2 / 2! + (x - 1)3 / 3! + + (-1)n(x - 1)n / n! Pn,1 (x) = ln 2 - (x - 1) + (x - 1)2 / 2! - (x - 1)3 / 3! + + (-1)n (x - 1)n / n! None of the above.

Explanation / Answer

16The integral is -1/2cos (2x) evaluated from pi/8 to pi/2 = -1/2cos(pi) + 1/2cos(pi/4) = -1/2(-1) + 1/2(1/sqrt(2)) =

1/2 + sqrt(2)

4 is the answer

17 numerator is squared 1 greater sequence number; denominator is sequence number

It should be (k+1)^2/k

Note how in 2 the summation is from 2 to 6, which allows the transformation.

2

18 4 ln n/n^4 goes to 0

2

19

Think of the geometric progression with first term 1 and ratio x sums to 1/1-x

4

20 ln(2x) = ln(2) + ln(x)

Clearly, the solution is 4 This is the standard expression for ln(x) with ln(2) added.

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