Q114 / 170 · Manipal (MET) 2018 · Mathematics

If $f(x)=\left\{\begin{array}{ll}\frac{x^2-9}{x-3}, & \text { if } x \neq 3 \\ 2 x+k, & \text { otherwise }\end{array}\right.$ is continuous at $x=3$, then $k$ is equal to

  • A
    3
  • Correct answer
    B
    0
  • C
    -6
  • D
    1/6

Solution

Since, $f(x)$ is continuous at $x=3$, then $\lim _{x \rightarrow 3} f(x)=f(3)$...(i) $\begin{aligned} & \therefore \lim _{x \rightarrow 3} \frac{x^2-9}{x-3}=\lim _{x \rightarrow 3} x+3=3+3=6 \\ & \text { and } \quad f(3)=2 \times 3+k=k+6 \end{aligned}$ $\therefore$ From Eq. (i), $6=k+6 \Rightarrow k=0$

Hence, the correct answer is option: 'B'.

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