Question

For what value of k is the following function continuous at x = 2?
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell table row cell 2 straight x plus 1 comma space space space space straight x less than 2 end cell row cell space space space straight k space space space space comma space space space space space space straight x equals 2 end cell end table end cell row cell space 3 straight x minus 1 comma space space space space space straight x greater than 2 end cell end table close$

Solution

We space have straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell table row cell 2 straight x plus 1 comma space space space space straight x less than 2 end cell row cell space space space straight k space space space space comma space space space space space space straight x equals 2 end cell end table end cell row cell space 3 straight x minus 1 comma space space space space space straight x greater than 2 end cell end table close space Lt with straight x rightwards arrow 2 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 2 to the power of minus below left parenthesis 2 straight x plus 1 right parenthesis 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 space space space space space space space space space space space space space left square bracket Put space straight x equals 2 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 2 to the power of minus right square bracket equals Lt with straight h rightwards arrow 0 below open curly brackets 2 left parenthesis 2 minus straight h right parenthesis plus 1 close curly brackets equals Lt with straight h rightwards arrow 0 below left parenthesis 4 minus 2 straight h plus 1 right parenthesis equals 4 minus 0 plus 1 equals 5 space Lt with straight x rightwards arrow 2 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 2 to the power of plus below left parenthesis 3 straight x minus 1 right parenthesis 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 space space space space space space space space space space space space space left square bracket Put space straight x equals 2 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 2 to the power of plus right square bracket equals Lt with straight x rightwards arrow 0 below open curly brackets 3 left parenthesis 2 plus straight h right parenthesis minus 1 close curly brackets equals Lt with straight x rightwards arrow 0 below left parenthesis 6 plus 3 straight h minus 1 right parenthesis equals 6 plus 0 minus 1 equals 5 Since space straight f space is space continous space at space straight x equals 2.

∴ 5 = 5 = k ⇒ k = 5.

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