Question

Let \(f:R→R\) be a function defined by
\(f(x) =   \begin{cases}     3e^x       & \quad \text {if}\  {x<0}\\     x^2+3x+3  & \quad \text {if}\  0≤x<1   \\ x^2-3x-3 & \quad \text {if} \ x≥1\end{cases}\)

• A. f is continuous on R
• B. f is not continuous on R
• C. f is continuous on R\{0}
• D. f is continuous on R\{1}
• E. f is not continuous on R\{0,1}

Answer 1

The Correct Option is A

Given that:
\(f(x) =   \begin{cases}     3e^x       & \quad \text {if}\  {x<0}\\     x^2+3x+3  & \quad \text {if}\  0≤x<1   \\ x^2-3x-3 & \quad \text {if} \ x≥1\end{cases}\)
To determine the continuity of the function f(x), we need to examine its behavior at the points where it changes definition, i.e., at 0 x=0 and t 1 x=1.
Let's analyze the function f(x) at these points:
Then , 1.at \(x=0 ,\) LHS and RHS respectively are,
\(limx→0−​f(x)=limx→0−​3e^x=3e^0=3\)
\(limx→0+​f(x)=limx→0+​(x^2+3x+3)=02+3⋅0+3=3\)
Now, let's evaluate f(0):
\(f(0)=0^2+3⋅0+3=3\)
Since the limit of \(f(x)\)as \(x\) approaches \(0\) and \(f(0)\) are the same, the function is continuous at \(x=0\).
At \(x=1\): LHS  and RHS respectively are,
\(limx→1−​f(x)=limx→1−​(x2−3x−3)=12−3⋅1−3=−5\)
\(limx→1+​f(x)=limx→1+​(x2−3x−3)=12−3⋅1−3=−5\)
Now, let's evaluate f(1):
\(f(1)=1^2−3⋅1−3=−5\)
Based on the above analysis, the function f(x) is continuous on its entire domain R.

So, the correct option is (A): f is continuous on R.