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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Limits

NTA Abhyas 2020 · Mathematics · Limits

If _ x → 0 ( 1 + px + qx^ 2 ) ^ cosecx = e^ 5 , then

If lim⁡x→0(1+px+qx2)cosecx=e5\lim_{x \to 0} ( 1 + px + qx^{2}{)}^{cosecx} = e^{5}, then

Options

  1. Ap=5,q∈Rp = 5, q \in \mathbb{R}
  2. Bp=5,q>0p = 5, q > 0
  3. Cq=5,p∈Rq = 5, p \in \mathbb{R}
  4. Dq=5,p=0q = 5, p = 0

Correct answer

A. p=5,q∈Rp = 5, q \in \mathbb{R}

Step-by-step solution

∵∵ given limit is of the form 1∞{1}^{\infty}
∴elim⁡x→0(cosecx){1+px+qx2−1}=e5∴ e^{\lim_{x \to 0}\left(cosecx\right)\left\{1 + px + qx^{2} - 1\right\}} = e^{5}
lim⁡x→0(px+qx2sinx)=5\lim_{x \to 0}\left(\dfrac{px + qx^{2}}{sin x}\right) = 5
lim⁡x→0(px+qx2x)⋅(xsinx)=5\lim_{x \to 0}\left(\dfrac{px + qx^{2}}{x}\right) ⋅ \left(\dfrac{x}{sin x}\right) = 5
lim⁡x→0(p+qx)(1)=5⇒p=5\lim_{x \to 0}\left(p + qx\right)\left(1\right) = 5⇒ p = 5
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