If \( f(x)=\begin{cases}\displaystyle\frac{x^{2}+3x-10}{x-2}, & x\ne2\\ 2k+1, & x=2\end{cases} \) is continuous at \( x=2 \), then the value of \( k \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
For continuity at x = 2, the limit must equal \( f(2)=2k+1 \). Factorising the numerator,
\( \displaystyle \lim_{x\to2}\frac{x^{2}+3x-10}{x-2}=\lim_{x\to2}\frac{(x+5)(x-2)}{x-2} \)
\( \displaystyle =\lim_{x\to2}(x+5) \)
\( =7 \)
So
\( 2k+1=7 \)
\( k=3 \)
Why the other options are incorrect
Option (1) 7 — Incorrect. 7 is the value of the limit, that is of \( 2k+1 \), not of k.
Option (2) 4 — Incorrect. With k = 4, \( 2k+1=9 \), not 7.
Option (3) \( \displaystyle\frac{7}{2} \) — Incorrect. With \( k=\frac{7}{2} \), \( 2k+1=8 \), not 7.