Given the function:
\[ f(x) = \begin{cases} \frac{(2x^2 - ax +1) - (ax^2 + 3bx + 2)}{x+1}, & \text{if } x \neq -1 \\ k, & \text{if } x = -1 \end{cases} \]
If \( a, b, k \in \mathbb{R} \) and \( f(x) \) is continuous for all \( x \), then the value of \( k \) is:
Given the function:
\[ f(x) = \begin{cases} \frac{(2x^2 - ax +1) - (ax^2 + 3bx + 2)}{x+1}, & \text{if } x \neq -1 \\ k, & \text{if } x = -1 \end{cases} \]
If \( a, b, k \in \mathbb{R} \) and \( f(x) \) is continuous for all \( x \), then the value of \( k \) is:
Show Hint
- \( -\frac{1}{3} \)
- \( 6 \)
- \( a - 2 \)
- \( a - 3 \)
The Correct Option is D
Solution and Explanation
Top Questions on Limits
- If \[ \lim_{x \to \infty} \left( \frac{e}{1 - e} \left( \frac{1}{e} - \frac{x}{1 + x} \right) \right)^x = \alpha, \] then the value of \[ \frac{\log_e \alpha}{1 + \log_e \alpha} \] equals:
- Let \[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \left( \frac{\tan \left( \frac{x}{2^{r+1}} \right) + \tan^3 \left( \frac{x}{2^{r+1}} \right)}{1 - \tan^2 \left( \frac{x}{2^{r+1}} \right)} \right) \] Then, \( \lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)} \) is equal to:
- If \[ \lim_{x \to \infty} \left( \frac{e}{1 - e} \left( \frac{1}{e} - \frac{x}{1 + x} \right) \right)^x = \alpha, \] then the value of \[ \frac{\log_e \alpha}{1 + \log_e \alpha} \] equals:
- \( \lim_{x \to 3} \frac{x^3 - 27}{x^2 - 9}. \)
If \( f(x) \) is given as: \( f(x) = \begin{cases} 3ax - 2b, & x<1 ax + b + 1, & x<1 \end{cases} \) and \( \lim_{x \to 1} f(x) \) exists, then the relation between \( a \) and \( b \) is:
Questions Asked in AP EAMCET exam
Given the function:
\[ f(x) = \begin{cases} \frac{2x e^{1/2x} - 3x e^{-1/2x}}{e^{1/2x} + 4e^{-1/2x}}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} \]
Determine the differentiability of \( f(x) \) at \( x = 0 \).
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