Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET2013MathematicsLimits

_ x 0 ^3 x- ^3 x x^5 is equal to

Options

  1. A5 2
  2. B3 2
  3. C3 5
  4. D2 5

Correct answer

B. 3 2

Step-by-step solution

aligned & _ x 0 ^3 x- ^3 x x^5 & = _ x 0 [ array c (x+ x^3 3 + 2 15 x^5+ )^3 - (x- x^3 3 ! + x^5 5 ! + )^3 array ] x^5 & = _ x 0 [ array c (1+ x^2 2 + 2 15 x^4+ )^3 - (1- x^2 3 ! + x^4 5 ! + )^3 array ] x^2 & (1+ x^2 3 + 2 15 x^4+ )- (1- x^2 3 ! + x^4 5 ! + ) & = _ x 0 array c (1+ x^2 3 + )^2+ (1- x^2 3 ! + )^2 + (1+ x^2 3 + ) (1- x^2 3 ! + ) array x^2 & . [using (a^3-b^3 )=(a-b) (a^2+b^2+a b ) ] & = _ x 0 ( x^2 2 + x^4 8 + ) array c . (1+ x^2 3 + )^2+ (1- x^2 3 ! + )^2 + (1+ x^2 3 + ) (1- x^2 3 + ) array x^2 & = _

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 AP EAMCET PYQs