Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202416 May 2024Evening ShiftMathematicsLimitsActual

_ y 0 1+ 1+y^4 - 2 y^4 =

Options

  1. A0
  2. B1 2 2
  3. C1 4 2
  4. D1 2 2 ( 2 +1)

Correct answer

C. 1 4 2

Step-by-step solution

By rationalising, we get aligned & _ y 0 1+ 1+y^4 - 2 y^4 & = _ y 0 1+ 1+y^4 -2 y^4 ( 1+ 1+y^4 + 2 ) & = _ y 0 1+y^4 -1 y^4 ( 1+ 1+y^4 + 2 ) & = _ y 0 1+y^4 -1 y^4 ( 1+ 1+y^4 + 2 ) 1+y^4 +1 1+y^4 +1 & = _ y 0 y^4 y^4 ( 1+ 1+y^4 + 2 ) ( 1+y^4 +1 ) & = 1 ( 1+ 1+0 + 2 )( 1+0 +1) = 1 4 2 aligned

Practice Limits on Quantrex Academy →

More from Limits

If _ x 0 ( p 2x + 1 - 2x x + x ) = 1 then the value of ' p ' is 2026_ x 2 ( 1 - x x ) is equal to 2026The value of _ x 3 [ 1 x-3 + 9x 27-x^3 ] is: 2026_ x 0 (1 - 2x)(3 + x) x 4x is equal to: 2026If _ x 3 ( x^2 - ax - 3b x - 3 ) = 5 , then a + b = 2026If f(x) = cases x^2 - 1 & if x 2 x + 1 & if x < 2 cases , then _ x 1 f(x) + _ x 2 f(x) = 2026_ x 4 2 2 -( x+ x)^3 1- 2 x = 2025Let [x] denote the greatest integer less than or equal to x . Then _ x 2⁺ ( [x]^3 3 - [ x 3 ]^3 )= 2025 Full Limits list All MHT CET PYQs