NDA 2024 Maths · Mathematics · Continuity and Differentiability · easy

The function f(x)=begincases(x^2-9)/(x-3),&x≠3\\[4pt] k,&x=3endcases is continuous at x=3.

Single correct option

The function

f(x)=\begin{cases}\dfrac{x^2-9}{x-3},&x\ne3\\[4pt] k,&x=3\end{cases}

is continuous at

x=3

. The value of

k

Show answer and solution

Answer: (B) $6$

For continuity

k=\lim_{x\to3}f(x)

. For

x\ne3

\frac{x^2-9}{x-3}=x+3

, whose limit at

x=3

is 6. Trap: substituting

x=3

directly gives

\frac00

, and cancelling incorrectly can lead to 3 or 9. The removable discontinuity is repaired by

k=6

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