Question:
Let \(\alpha\) and \(\beta\) be the roots of \(ax^2 + bx + c = 0\), then \(\lim_{x \to \alpha} \frac{1 – \cos(ax^2 + bx + c)}{(x – \alpha)^2}\) is equal to:
मान लीजिए \(\alpha\) और \(\beta\) समीकरण \(ax^2 + bx + c = 0\) के मूल हैं, तो \(\lim_{x \to \alpha} \frac{1 – \cos(ax^2 + bx + c)}{(x – \alpha)^2}\) किसके बराबर है?
Options:
- 0
- \(\frac{1}{2}(\alpha – \beta)^2\)
- \(\frac{a^2}{2}(\alpha – \beta)^2\)
- \(-\frac{a^2}{2}(\alpha – \beta)^2\)
- 0
- \(\frac{1}{2}(\alpha – \beta)^2\)
- \(\frac{a^2}{2}(\alpha – \beta)^2\)
- \(-\frac{a^2}{2}(\alpha – \beta)^2\)
Correct Answer: c)
Explanation:
Express \(ax^2 + bx + c = a(x – \alpha)(x – \beta)\).
As \(x \to \alpha\), \(\theta = ax^2 + bx + c \to 0\).
Using the standard limit \(\lim_{\theta \to 0} \frac{1 – \cos\theta}{\theta^2} = \frac{1}{2}\):
\[\lim_{x \to \alpha} \frac{1 – \cos(ax^2 + bx + c)}{(x – \alpha)^2} = \lim_{x \to \alpha} \frac{1 – \cos(ax^2 + bx + c)}{(ax^2 + bx + c)^2} \cdot \frac{(ax^2 + bx + c)^2}{(x – \alpha)^2}\]
\[= \frac{1}{2} \cdot \lim_{x \to \alpha} \left[ \frac{a(x – \alpha)(x – \beta)}{x – \alpha} \right]^2 = \frac{1}{2} \cdot [a(\alpha – \beta)]^2 = \frac{a^2}{2}(\alpha – \beta)^2\]
Hence, the correct option is (c).
\(ax^2 + bx + c = a(x – \alpha)(x – \beta)\) लिखने पर।
मानक सीमा \(\lim_{\theta \to 0} \frac{1 – \cos\theta}{\theta^2} = \frac{1}{2}\) का प्रयोग करने पर:
\[\lim_{x \to \alpha} \frac{1 – \cos(ax^2 + bx + c)}{(x – \alpha)^2} = \frac{1}{2} \cdot \left[ \lim_{x \to \alpha} a(x – \beta) \right]^2 = \frac{a^2}{2}(\alpha – \beta)^2\]
अतः सही विकल्प (c) है।