NTA Abhyas JEE Main2020MathematicsLimitsPractice

If l i m x → 0 ( 1 + p x + q x 2 ) c o s e c x = e 5 , then

Options

Ap = 5 , q ∈ R
Bp = 5 , q > 0
Cq = 5 , p ∈ R
Dq = 5 , p = 0

Correct answer

A. p = 5 , q ∈ R

Step-by-step solution

∵ given limit is of the form 1 ∞ ∴   e l i m x → 0 c o s e c x 1 + p x + q x 2 - 1 = e 5 l i m x → 0 p x + q x 2 sin ⁡ x = 5 l i m x → 0 p x + q x 2 x ⋅ x sin ⁡ x = 5 l i m x → 0 p + q x 1 = 5 ⇒ p = 5

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