If f(x) = begin{cases} ax^2 - b, & 0 le x

Question

(D) For a function $f(x)$ to be continuous at $x = 1$,the left-hand limit,right-hand limit,and the value of the function at $x = 1$ must be equal.$1$.

Vedclass · Accepted Answer

All the above

Vedclass · Answer

(D) For a function $f(x)$ to be continuous at $x = 1$,the left-hand limit,right-hand limit,and the value of the function at $x = 1$ must be equal.$1$. Left-hand limit: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} (ax^2 - b) = a(1)^2 - b = a - b$.$2$. Right-hand limit: $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} (x + 1) = 1 + 1 = 2$.$3$. Value at $x = 1$: $f(1) = 2$.For continuity,$a - b = 2$.Checking the options:- For $A$: $a = 2, b = 0 \Rightarrow 2 - 0 = 2$ (Correct).- For $B$: $a = 1, b = -1 \Rightarrow 1 - (-1) = 2$ (Correct).- For $C$: $a = 4, b = 2 \Rightarrow 4 - 2 = 2$ (Correct).Since all options satisfy the condition $a - b = 2$,the correct answer is $D$.

If $f(x) = \begin{cases} ax^2 - b, & 0 \le x < 1 \\ 2, & x = 1 \\ x + 1, & 1 < x \le 2 \end{cases}$ is continuous at $x = 1$,then the most suitable values of $a$ and $b$ are:

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