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Mathematics Question on Limit and Continuity

Let f:R→Rf:R→R be a function defined by
f(x)={3exif x<0 x2+3x+3if 0≤x<1 x2−3x−3if x≥1f(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

f is continuous on R

Explanation

Solution

Given that:
f(x)={3exif x<0 x2+3x+3if 0≤x<1 x2−3x−3if x≥1f(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,x=0 , LHS and RHS respectively are,
limx→0−​f(x)=limx→0−​3ex=3e0=3limx→0−​f(x)=limx→0−​3e^x=3e^0=3
limx→0+​f(x)=limx→0+​(x2+3x+3)=02+3⋅0+3=3limx→0+​f(x)=limx→0+​(x^2+3x+3)=02+3⋅0+3=3
Now, let's evaluate f(0):
f(0)=02+3⋅0+3=3f(0)=0^2+3⋅0+3=3
Since the limit of f(x)f(x)as xx approaches 00 and f(0)f(0) are the same, the function is continuous at x=0x=0.
At x=1x=1: LHS and RHS respectively are,
limx→1−​f(x)=limx→1−​(x2−3x−3)=12−3⋅1−3=−5limx→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=−5limx→1+​f(x)=limx→1+​(x2−3x−3)=12−3⋅1−3=−5
Now, let's evaluate f(1):
f(1)=12−3⋅1−3=−5f(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.