Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
KVPY2018MathematicsLimits

Suppose the limit L = lim n → ∞ n ∫ 0 1 1 1 + x 2 n d x exists and is larger than 1 2 . Then,

Options

  1. A1 2 < L < 2
  2. B2 < L < 3
  3. C3 < L < 3
  4. DL ≥ 4

Correct answer

A. 1 2 < L < 2

Step-by-step solution

We have, L = lim n → ∞ n ∫ 0 1 1 1 + x 2 n d x 1 + x 2 n > 1 + n x 2 ⇒ 1 1 + x 2 n < 1 1 + n x 2 ⇒ ∫ 0 1 d x 1 + x 2 n < ∫ 0 1 d x 1 + n x 2 ⇒ ∫ 0 1 d x 1 + x 2 n < 1 n tan - 1 n x 0 1 ⇒ ∫ 0 1 d x 1 + x 2 n < 1 n tan - 1 n ⇒ L = lim n → ∞ n ∫ 0 1 1 1 + x 2 n d x < lim n → ∞ n n tan - 1 n ⇒ L < tan - 1 ∞ = π 2 ∴ 1 2 < L < 2   ∵ π 2 < 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 KVPY PYQs