Question

If the function $straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell space space space 3 ax plus straight b space space if space straight x greater than 1 end cell row cell space space space space space space 11 space space space space space space space space if space straight x equals 1 end cell row cell space space 5 ax minus 2 straight b space space if space straight x less than 1. end cell end table close$
is continous at x=1, find the values of a and b.

Solution

We space have straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell space space space 3 ax plus straight b space space if space straight x greater than 1 end cell row cell space space space space space space 11 space space space space space space space space if space straight x equals 1 end cell row cell space space 5 ax minus 2 straight b space space if space straight x less than 1. end cell end table close therefore straight f left parenthesis 1 right parenthesis equals 11 space Lt with straight x rightwards arrow 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 1 to the power of minus below left parenthesis 5 ax minus 2 straight b right parenthesis space left square bracket Put space straight x equals 1 minus straight h comma space straight h greater than 0 space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 1 to the power of minus right square bracket space space space space space space space space space space space space space space space space equals Lt with straight x rightwards arrow 0 below open curly brackets 5 straight a left parenthesis 1 minus straight h right parenthesis minus 2 straight b close curly brackets equals 5 straight a left parenthesis 1 minus 0 right parenthesis minus 2 straight b equals 5 straight a minus 2 straight b Lt with straight x rightwards arrow 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 1 to the power of plus below open parentheses 3 ax plus straight b close parentheses space space left square bracket put space straight x equals 1 plus straight h comma space straight h greater than 0 space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 1 to the power of plus right square bracket space space space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below open curly brackets 3 straight a left parenthesis 1 plus straight h right parenthesis plus straight b close curly brackets equals 3 straight a left parenthesis 1 plus 0 right parenthesis plus straight b equals 3 straight a plus straight b Since space straight f left parenthesis straight x right parenthesis space is space continous space at space straight x equals 1 therefore Lt with straight x rightwards arrow 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 1 right parenthesis

∴ 5a – 2b = 3 a + b = 11
∴5a – 2b = 11 ...(1)
3a + b =11 ...(2)
Multiplying (1) by 1 and (2) by 2 , we get,
5a – 2b = 11
6a + 2b = 22
Adding these two equation, we get,
11 a = 33 ⇒ a = 3
∴ from (1), 15 – 2b = 11 ⇒ 2b = 4
∴ b = 2
∴ we have a = 3, b = 2.

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