JEE MAIN - Mathematics (2023 - 8th April Evening Shift - No. 11)
If $$\alpha > \beta > 0$$ are the roots of the equation $$a x^{2}+b x+1=0$$, and $$\lim_\limits{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right), \text { then } \mathrm{k} \text { is equal to }$$ :
$$2 \beta$$
$$\beta$$
$$\alpha$$
$$2 \alpha$$
Explanation
Since, $\alpha, \beta$ are roots of $a x^2+b x+1=0$
Replace $x \rightarrow \frac{1}{x}$
$$ \frac{a}{x^2}+\frac{b}{x}+1=0 \Rightarrow x^2+b x+a=0 $$
So, $\frac{1}{\alpha}, \frac{1}{\beta}$ are the roots.
Now, $\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{1-\cos \left(x^2+b x+a\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}}$
$$ =\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2\left(\frac{x^2+b x+a}{2}\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}} $$
$$ \begin{aligned} & =\lim _{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2 \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{4 \times 2 \alpha^2 \frac{\left(x-\frac{1}{\alpha}\right)^2\left(x-\frac{1}{\beta}\right)^2}{4}\left(x-\frac{1}{\beta}\right)^2}\right]^{\frac{1}{2}} \\\\ & =\lim _{x \rightarrow \frac{1}{\alpha}}\left[ \pm \frac{1}{2} \frac{\sin \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{\alpha \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}\left(x-\frac{1}{\beta}\right)\right] \\\\ & =\frac{1}{2 \alpha}\left[\frac{-1}{\alpha}+\frac{1}{\beta}\right) \\\\ & \Rightarrow \frac{1}{k}\left[\frac{1}{\beta}-\frac{1}{\alpha}\right]=\frac{1}{2 \alpha}\left[\frac{1}{\beta}-\frac{1}{\alpha}\right] \\\\ & \Rightarrow k=2 \alpha \end{aligned} $$
Replace $x \rightarrow \frac{1}{x}$
$$ \frac{a}{x^2}+\frac{b}{x}+1=0 \Rightarrow x^2+b x+a=0 $$
So, $\frac{1}{\alpha}, \frac{1}{\beta}$ are the roots.
Now, $\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{1-\cos \left(x^2+b x+a\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}}$
$$ =\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2\left(\frac{x^2+b x+a}{2}\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}} $$
$$ \begin{aligned} & =\lim _{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2 \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{4 \times 2 \alpha^2 \frac{\left(x-\frac{1}{\alpha}\right)^2\left(x-\frac{1}{\beta}\right)^2}{4}\left(x-\frac{1}{\beta}\right)^2}\right]^{\frac{1}{2}} \\\\ & =\lim _{x \rightarrow \frac{1}{\alpha}}\left[ \pm \frac{1}{2} \frac{\sin \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{\alpha \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}\left(x-\frac{1}{\beta}\right)\right] \\\\ & =\frac{1}{2 \alpha}\left[\frac{-1}{\alpha}+\frac{1}{\beta}\right) \\\\ & \Rightarrow \frac{1}{k}\left[\frac{1}{\beta}-\frac{1}{\alpha}\right]=\frac{1}{2 \alpha}\left[\frac{1}{\beta}-\frac{1}{\alpha}\right] \\\\ & \Rightarrow k=2 \alpha \end{aligned} $$
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