Question:
If the function $f(x)=\begin{cases}\dfrac{2x^2+3x-5}{x-1}, & x \ne 1 \\ k, & x=1\end{cases}$ is continuous at $x=1$, then the value of $k$ is:
If the function $f(x)=\begin{cases}\dfrac{2x^2+3x-5}{x-1}, & x \ne 1 \\ k, & x=1\end{cases}$ is continuous at $x=1$, then the value of $k$ is:
Show Hint
Cancel common factors before applying limits in continuity problems.
Updated On: Apr 24, 2026
- $6$
- $8$
- $-6$
- $7$
- $-7$
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
• Continuity: $\lim_{x \to a} f(x) = f(a)$
Step 1: Factor numerator
\[ 2x^2+3x-5 = (2x+5)(x-1) \]
Step 2: Simplify
\[ f(x) = \frac{(2x+5)(x-1)}{x-1} = 2x+5 \quad (x \ne 1) \]
Step 3: Apply limit
\[ \lim_{x \to 1} f(x) = 2(1)+5 = 7 \]
Step 4: Continuity condition
\[ k = 7 \] Final Conclusion:
\[ k = 7 \]
• Continuity: $\lim_{x \to a} f(x) = f(a)$
Step 1: Factor numerator
\[ 2x^2+3x-5 = (2x+5)(x-1) \]
Step 2: Simplify
\[ f(x) = \frac{(2x+5)(x-1)}{x-1} = 2x+5 \quad (x \ne 1) \]
Step 3: Apply limit
\[ \lim_{x \to 1} f(x) = 2(1)+5 = 7 \]
Step 4: Continuity condition
\[ k = 7 \] Final Conclusion:
\[ k = 7 \]
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