Question

pleeease answer fast

Answer 1

Answer:

don't know the answer

Step-by-step explanation:

plz thank me

Answer 2

Required Answer :

Given :

$\tt \: f(x) = \begin{cases} \tt \: ax + b,& \tt \: x < 1 \\ \tt 4,& \tt \: x = 1 \\ \tt b - ax,& \tt \: x > 1\end {cases}$

Solution :

$\sharp \: \: \tt \lim_{x\to 1^{ + } } \: f(x) = \lim_{x\to 1^{ } } \: f(x) = \lim_{x\to 1^{ - } } \: f(x) \: \: \: \bf[ \: \because \: function \: \: is \: \: defined \: \: at \: \: x=1 \: ]\\ \\ \\ \\ \tt \lim_{x\to 1^{ + } } \: f(x) = \lim_{x\to 1^{ - } } \: f(x) = \lim_{x\to 1^{ } } \: f(x) \\ \\ \\ \implies \tt \lim_{x\to 1^{ + } } \: b - ax = \lim_{x\to 1^{ - } } \: ax + b = \lim_{x\to 1^{ } } \: f(x) \\ \\ \\ \tt \natural \: \: \: putting \: \: limiting \: \: value \\ \\ \\ \begin{array}{c |c} \implies \tt \: b - a = 4 \underline{ \: \: \: \: \: \: \: }(1) & \tt \implies a + b = 4 \underline{ \: \: \: \: \: \: }(2) \\ \end{array} \\ \\ \\ \tt \sharp \: \: Adding \: \: {eq}^{n} \: \: (1) \: \: \& \: \: (2) \\ \\ \\ \tt \: b - \cancel{a} + \cancel{a} + b = 4 + 4 \\ \\ \\ \large \bf \therefore \: \: \: b = 4 \\ \\ \& \\ \\ \bf \large\therefore \: \: \: \: a = 0 \\ \\ \\ \\ \LARGE \red{\mathfrak{ Happy \: \: New \: \: Year \: \: 2021} }\\ \\ \\ \small \tt \colorbox{aqua}{@StayHigh}$