JEE Mains · Maths · STD 11 - 12. limits
Let \(\mathrm{a}>0\) be a root of the equation \(2 \mathrm{x}^2+\mathrm{x}-2=0\). If \(\lim _{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{\left(1-a x^2\right)}=\alpha+\beta \sqrt{17}\), where \(\alpha, \beta \in Z\) then \(\alpha+\beta\) is equal to ...........
- A \(195\)
- B \(170\)
- C \(149\)
- D \(315\)
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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