Question

Find the values of a and b such that the function defined by
$straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell 5 comma end cell row cell ax plus straight b comma end cell row cell 21 comma end cell end table close table row cell if space straight x less or equal than 2 space space space space space space space space space space end cell row cell if space 2 less than straight x less than 10 end cell row cell if space straight x greater or equal than 10 space space space space space space space space end cell end table$
is a continous function.

Solution

straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell 5 comma end cell row cell ax plus straight b comma end cell row cell 21 comma end cell end table close table row cell if space straight x less or equal than 2 space space space space space space space space space space end cell row cell if space 2 less than straight x less than 10 end cell row cell if space straight x greater or equal than 10 space space space space space space space space end cell end table

∴ f is continuous function
∴ f is continuous at x = 2 and x = 10
∴ f is right continuous at x = 2 and left continuous at .x = 10.

therefore space space Lt with straight x rightwards arrow 2 to the power of plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 2 right parenthesis space rightwards double arrow space 2 straight a plus straight b equals 5 space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis space space space space space Lt with straight x rightwards arrow 10 to the power of minus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 10 right parenthesis space rightwards double arrow space 10 straight a plus straight b equals 21 space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

Subtracting (1) from (2), we get,
8 a = 16 or a = 2
∴ from (1), 4 + 6 = 5⇒ b = 1
∴ we have a = 2, b = 1

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