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If f(x)={x2+3x−10x−2,x≠22k+1,x=2f(x) = \begin{cases} \frac{x^2+3x-10}{x-2}, & x \neq 2 \\ 2k+1, & x = 2 \end{cases} is continuous at x=2x = 2, then the value of kk is

Solution

✅ Correct Option: 4

For x≠2x\neq2, x2+3x−10x−2=(x+5)(x−2)x−2=x+5\frac{x^2+3x-10}{x-2}=\frac{(x+5)(x-2)}{x-2}=x+5, so the limit as x→2x\to2 is 7. Continuity requires 2k+1=72k+1=7, giving k=3k=3.

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