Limits, Continuity & Differentiability
Continuity And Differentiability
nta_abhyas_2025
Grade 12
Question:
If $f(x) = \begin{cases} \sqrt{t+x^2} & x > 0 \\ a & x = 0 \text{ is continuous at } x = 0 \text{ for some constants } a, b \text{ and } c, \text{ then the value of } \frac{abc}{b^2} \text{ is equal to} \\ \frac{c^2+x}{b^2+x^2} & x < 0 \end{cases}$
Step-by-Step Solution
Key Concept: Rationalization of denominators combined with limit evaluation and continuity conditions
We rationalize the denominator: $\lim_{x \to 0} \frac{\sqrt{x}}{\sqrt{1+x^2}-1} \cdot \frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}+1} = \lim_{x \to 0} \frac{\sqrt{x}(\sqrt{1+x^2}+1)}{1+x^2-1} = \lim_{x \to 0} \frac{\sqrt{x}(\sqrt{1+x^2}+1)}{x^2}$. Simplifying: $\lim_{x \to 0} \frac{\sqrt{1+x^2}+1}{x^{3/2}} = \lim_{x \to 0} \frac{\sqrt{1+x^2}+2}{x^{3/2}} = 4$. Also, $\lim_{f(x)} = \lim_{\frac{a}{x^{a-1}}} = 4$, which gives $a = 2$.
Correct Answer: 2