A function f(x) is defined by f(x) = \begin{cases} \frac{x^2 + kx - 10}{x - 2}, & x \neq 2 \\ 7, & x = 2 \end{cases} If f(x) is continuous at x = 2, what is the numerical value of the constant k?

A function f(x) is defined by f(x) = \begin{cases} \frac{x^2 + kx - 10}{x - 2}, & x \neq 2… — JAMB UTME

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{x^2 + kx - 10}{x - 2}, & x \neq 2 \\ 7, & x = 2 \end{cases}

f(x)

is continuous at

x = 2

, what is the numerical value of the constant

k

Answer: 3

The numerical value of the constant

k

is 3.

By definition of continuity,

f(x)

is continuous at

x = 2

\lim_{x \to 2} f(x) = f(2) = 7

. As

x \to 2

, the denominator

x - 2

approaches

0

. For the quotient to have a finite limit, the numerator

x^2 + kx - 10

must also evaluate to

0

x = 2

, yielding

2^2 + 2k - 10 = 0

. Solving this gives

2k = 6

, so

k = 3

. Substituting

k = 3

gives

\lim_{x \to 2} \frac{(x-2)(x+5)}{x-2} = 7

, confirming that

k = 3

is correct.

Apply the definition of continuity at

x = 2

\lim_{x \to 2} f(x) = f(2) = 7

A function

f(x)

is continuous at

x = a

if and only if

\lim_{x \to a} f(x) = f(a)

Set the numerator to zero at the point of discontinuity

x = 2

2^2 + k(2) - 10 = 0

Because the denominator

(x - 2) \to 0

x \to 2

, the limit can only exist if the numerator also approaches

0

, forming an indeterminate form

\frac{0}{0}

that can be simplified.

Solve for the unknown parameter

k

4 + 2k - 10 = 0 \implies 2k - 6 = 0 \implies k = 3

Linear algebraic equation solving.

Verify that the simplified limit equals

f(2)

\lim_{x \to 2} \frac{x^2 + 3x - 10}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 5)}{x - 2} = \lim_{x \to 2} (x + 5) = 7

Canceling the common factor

(x - 2)

yields

7

, which matches

f(2) = 7

Continuity of a Piecewise Function and Limit Existence