If the area between y = mx2 and x = my2 (m > 0) is 1/4 sq unit

Subject

Mathematics

Class

JEE Class 12

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

21.

A unit vector perpendicular to both i + j + k and 2i + j + 3k is

  • 2i - j - k

  • 3i + j - 2k6


22.

The value of 04x - 1dx is

  • 52

  • 5

  • 4

  • 1


23.

If In0π4tannxdx, where where n is apositive integer, then I10 + I8 is

  • 19

  • 18

  • 17

  • 9


24.

If Inexsinx + cosx1 - sin2xdx is

  • ex . cscx +C

  • ex . cotx + C

  • ex . secx + C

  • ex . tanx + C


Advertisement
25.

When x > 0, then cos-11 - x21 +x2dx is

  • 2xtan-1x - log1 +x2 + C

  • 2xtan-1x + log1 +x2 + C

  • 2xtan-1x + log1 +x2 + C

  • 2xtan-1x - log1 +x2 + C


Advertisement

26.

If the area between y = mx2 and x = my2 (m > 0) is 1/4 sq units, then the value of m is

  • ± 32

  • ± 23

  • 2

  • 3


B.

± 23

Given curves; y = mx2 and y2m= x; m > 0Intersection point of both curves   x = mmx22 = m3x4     m3x4 - x = 0 xm3x3 - 1 = 0 xmx  - 1m2x2 + 1 + mx = 0 x = 0, x = 1/m and y = 0, y = 1/mWe take only the points       = (0, 0) and (1/m, 1/m)Now, the area of the curve= 01mxm - mx2dxGiven, 14 = 23m . x32 - m . x3301m      14 = 23m . 1m32 - m3 . 1m3     14 = 23m2 - 13m2    14 = 13m2  m2 = 43    m = ± 23


Advertisement
27.

If m and n are degree and order of 1 + y1223 = y2, then the value of m + nm - n is

  • 3

  • 4

  • 5

  • 2


28.

The general solution of dydx2 = 1 - x2 - y2 + x2y2 is

  • 2sin-1y = x1 - x2 + sin-1x + C

  • cos-1y = xcos-1x + C

  • sin-1y =12sin-1x + C

  • 2sin-1y = x1 - y2 + C


Advertisement
Advertisement