Question

If α > β > 0 are the roots of the equation ax2 + bx + 1 = 0, and \(\lim\limits_{x \to \frac{1}{\alpha}}\left(\frac{1-cos(x^2+bx+a)}{2(1-ax)^2}\right)^{1/2}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right),\) then k is equal to

• A. α
• B. β
• C. 2α
• D. 2β

Answer 1

The Correct Option is C

Given:
\[ ax^2 + bx + 1 = a(x - \alpha)(x - \beta) \quad \Rightarrow \quad \alpha \beta = \frac{1}{a} \] \[ x^2 + bx + a = a(1 - \alpha x)(1 - \beta x) \] 

Now, we need to evaluate the limit: \[ \lim_{x \to \frac{1}{\alpha}} \frac{1 - \cos(x^2 + bx + a)}{2(1 - \alpha x)^2} \] 

Rewriting the limit expression: \[ = \lim_{x \to \frac{1}{2}} \frac{1 - \cos(a(1 - \alpha x)(1 - \beta x))}{2a^2(1 - \beta x)^2} \] 

Simplifying: \[ = \left[\frac{1}{2} \cdot \frac{1}{2} a^2 \left( \frac{1 - \beta}{\alpha} \right)^2 \right]^{\frac{1}{2}} \] 

Simplifying further: \[ = \frac{1}{2} \cdot \frac{1}{2 \alpha \beta} \left( \frac{1 - \beta}{\alpha} \right) = \frac{1}{2} \left( \frac{1}{\alpha \beta} - \frac{1}{\alpha^2} \right) \] 

Therefore, we get: \[ k = 2\alpha \] 

So, the final answer is \( k = 2\alpha \).