∫exlogsinx + cotxdx is equal to from Mathemat

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

31.

If a and b are unit vectors such that a b a × b = 14, then angle between a and b is :

  • π3

  • π4

  • π6

  • π2


32.

The order and degree of the differential equation sinxdx + dy = cosxdx - dy is :

  • (1, 2)

  • (2, 2)

  • (1, 1)

  • (2, 1)


33.

0πcosxdx is equal to :

  • 12

  • - 2

  • 1

  • - 1


34.

sin2xsin3xsin5xdx is equal to :

  • 15logesin5x - 13logesin3x + c

  • 13logesin3x - 15logesin5x

  • 13logesin3x + 15logesin5x

  • - 12cos2x + 13logesin3x


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35.

exlogsinx + cotxdx is equal to

  • excot(x) + c

  • exlog(sin(x)) + c

  • exlog(sin(x)) + tan(x) + c

  • ex + sin(x) + c


B.

exlog(sin(x)) + c

We know exfx + f'xdx = exfx +c If f(x) = logsinx f'(x) = 1sinxcosx = cotx exlogsinx + cotxdx = exlogsinx + c


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36.

- 1010loga + xa - xdx is equal to :

  • 0

  • - 2log(a + 10)

  • 2loga + 10a - 10

  • 2log(a + 10)


37.

The point of intersection of the line r = 7i^ + 10j^ + 13k^ + s2i^ + 3j^ + 4k^ and r = 3i^ + 5j^ + 7k^ + si^ + 2j^ + 3k^ is :

  • i^ + j^ - k^

  • 2i^ - j^ + 4k^

  • i^ - j^ + k^

  • i^ + j^ + k^


38.

Define f(x) = 0xsintdt, x  0, Then :

  • f is increasing only in the interval 0, π2

  • f is decreasing in the interval 0, π

  • f attains maximum at x = π2

  • f attains minimum at x = π


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39.

Let f(x) = sin2πx1 + π2. Then, fx + f- xdx is equal to :

  • 0

  • x + c

  • x2 - cosπx2π + c

  • x2 - sin2πx4π + c


40.

The interior angles of a polygon are AP. The smallest angle is 120° and the common difference is 5°. The number of sides of the polygon is :

  • 9

  • 10

  • 16

  • 5


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