JEE Main 2024 — 5 April, Shift 2 — Question 26

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

$=16 \times \frac{2}{\mathrm{a}^{2}}\left(\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}\right)^{2}$

$=\frac{32}{\mathrm{a}^{2}}\left(\frac{17}{4}\right)=\frac{17.8}{\mathrm{a}^{2}}=\frac{17 \times 8 \times 16}{(-1+\sqrt{117})^{2}}$ $=\frac{136.16}{18.2 \sqrt{7}} \times \frac{18+2 \sqrt{7}}{18+2 \sqrt{7}}$

$=\frac{136}{256}(18+2 \sqrt{7}) \cdot 16$

$=153+17 \sqrt{17}=\alpha+\beta \sqrt{17}$

$\alpha+\beta=153+17=170$