1.

If \(f(x) = \begin{cases} ax^2+b &,0≤x<{1}\\ 4 &, x=1\\x+3&,1<x≤2\end{cases} \) , then the value of (a, b) for which f(x) cannot be continuous at x = 1, isA. (2, 2) B. (3, 1) C. (4, 0) D. (5, 2)

Answer»

Option : (D)

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 :-

\(f(x) = \begin{cases} ax^2+b &,0≤x<{1}\\ 4 &, x=1\\x+3&,1<x≤2\end{cases} \) 

\(\lim\limits_{x \to 1^-}f(x)\) = \(\lim\limits_{h\to0}f(1-h)\)2+ b

= a + b

Function f(x) is discontinuous at x = 1

\(\lim\limits_{x \to 1^-}f(x)\) ≠ f(1)

⇒ a + b ≠ 4

Checking option,

(5,2) is answer.



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