If \(\rm f(x)= \frac{x^2+x-6}{x^2 +kx-3}\) is not continuous at x = 3, then find the value of k ?

Concept:

For a function say f, \(\mathop {\lim }\limits_{x \to a} f(x)\) exists

\( ⇒ \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = l = \;\mathop {\lim }\limits_{x \to a} f(x)\), where l is a finite value.

Any function say f is said to be continuous at point say a if and only if \(\mathop {\lim }\limits_{x \to a} f(x) = l = f\left( a \right)\), where l is a finite value.

Calculation:

 Given: \(\rm f(x)= \frac{x^2+x-6}{x^2 +kx-3}\) is not continuous at x = 3.

So, if any function is not continuous at x = a then \(\mathop {\lim }\limits_{x \to a} f(x) = l \neq f\left( a \right)\)

So, for the function f(x) if denominator is 0 at x = 3 then we can say that f(3) is infinite and limit cannot exist.

Let's find the value of k for which the denominator of f(x) is 0 for x = 3.

So, substitute x = 3 in x2 + kx - 3 = 0

⇒ 3² + 3k - 3 = 0.

⇒ 6 + 3k = 0.

⇒ k = - 2.

Hence, option 1 is correct.