The points of discontinuity of the function f(x) = {[1/5(2x^2+3),x≤1][6-5x,1<x<3][x-3,x≥3] is (are) : A. x = 1 B. x = 3 ​

Question

The points of discontinuity of the function\(f(x) = \begin{cases}\frac{1}{5}(2x^2+3)&,x ≤1\\ 6-5x &,1<x<3\\x-3&,x≥3\end{cases} \) is (are) :

A. x = 1 

B. x = 3 

C. x = 1, 3 

D. none of these

Answer 1

Option : (B)

Formula :-

(i) A function f(x) is said to be continuous at a point x = a of its domain, if
\(\lim\limits_{x \to a}f(x)\) = f(a)
\(\lim\limits_{x \to a^+}f(a+h)\) = \(\lim\limits_{x \to a^-}f(a-h)\) = f(a) 

Given :-

\(f(x) = \begin{cases}\frac{1}{5}(2x^2+3)&,x ≤1\\ 6-5x &,1<x<3\\x-3&,x≥3\end{cases} \) 

Now at x = 1

Therefore discontinuous at x = 3.