InterviewSolution
Saved Bookmarks
| 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 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. |
|