Question

If a function f(x) = \(\begin{Bmatrix} \rm 3x + 2, x \geq1\\ \rm 5, \;\;\;\;\;\;\;\;x<1 \end{Bmatrix}\), then the function at x = 1
1. Continuous and f(1) = 5
2. Continuous but f(1) cannot be determined
3. Not continuous 
4. None of the above

Answer 1

Correct Answer - Option 1 : Continuous and f(1) = 5

Concept:

The function is continuous at any point x when

Left hand limit  = Right hand limit = f(x)

Where LHL = \(\rm\lim_{α \rightarrow 0}\) f(x - α) and RHL = \(\rm\lim_{α \rightarrow 0}\) f(x + α)

Calculation:

Given f(x) = \(\begin{Bmatrix} \rm 3x + 2, x \geq1\\ \rm 5, \;\;\;\;\;\;\;\;x<1 \end{Bmatrix}\)

LHL = \(\rm\lim_{α \rightarrow 0}\)f(1 - α)

LHL = 5

f(1) = 3(1) + 2

f(1) = 5

RHL = \(\rm\lim_{α \rightarrow 0}\)f(1 + α)

RHL = \(\rm\lim_{α \rightarrow 0}\)[3(1 + α) + 2]

RLH = \(\rm\lim_{α \rightarrow 0}\)[5 + 3α]

RHL = 5

∵ LHL = RHL = f(x)

∴ f(x) is continuous and f(1) = 5