QJEEM24Mae551: JEE Main 2024 Mathematics - Class 12 JEEM24Mae551 Solution: Let be a root of the equation . If lim x → 1 a 16 ( 1 − cos ⁡ ( …

$\because 2 x^2+x-2=0$ has two roots where $a=\frac{\sqrt{17}-1}{4}$ and another root is $\frac{-\sqrt{17}-1}{4}$

And $2+x-2 x^2=-2\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)$

Now $$\underset{x\to \frac{1}{a}}{lim}\frac{16(1-\cos (2+x-2x^2))}{(1-ax{)}^2}$$

$\begin{aligned}& =\lim _{x \rightarrow \frac{1}{a}} \frac{32 \sin ^2\left(\frac{2+x-2 x^2}{2}\right)}{a^2\left(\frac{1}{a}-x\right)^2} \\& =\lim _{x \rightarrow \frac{1}{a}} \frac{\left(x+\frac{4}{\sqrt{17}+1}\right)^2 32 \cdot\left(\sin \left(\frac{1}{2} \cdot(-2)\right)\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)\right)^2}{a^2\left(\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)\right)^2} \\& =2 \cdot\left(\frac{1}{a}+\frac{4}{\sqrt{17}+1}\right)^2 \cdot\left(\frac{4}{\sqrt{17}-1}\right)^2 \\& =2\left(\frac{4}{\sqrt{17}-1}+\frac{4}{\sqrt{17}+1}\right)^2 \cdot\left(\frac{4}{\sqrt{17}-1}\right)^2 \\& =\frac{17 \times 4}{18-2 \sqrt{17}}=\frac{68}{9-\sqrt{17}} \\& =17(9+\sqrt{17}) \\& \alpha+\beta=170\end{aligned}$