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JEE Mains · Maths · STD 11 - 12. limits

मान लीजिए \(\mathrm{a}>0\) समीकरण \(2 \mathrm{x}^2+\mathrm{x}-2=0\) का एक मूल है। यदि \(\lim _{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{\left(1-a x^2\right)}=\alpha+\beta \sqrt{17}\), जहाँ \(\alpha, \beta \in Z\) तो \(\alpha+\beta\) = ...........

  1. A \(195\)
  2. B \(170\)
  3. C \(149\)
  4. D \(315\)
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Answer & Solution

Correct Answer

(B) \(170\)

Step-by-step Solution

Detailed explanation

\(\lim _{x \rightarrow \frac{1}{a}} 16 \cdot \frac{\left(1-\cos 2\left(x-\frac{1}{a}\right)\left(x-\frac{1}{b}\right)\right)}{4\left(x-\frac{1}{b}\right)^2} \times \frac{4\left(x-\frac{1}{b}\right)^2}{a^2\left(x-\frac{1}{a}\right)^2}\)…
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