InterviewSolution
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InterviewSolution
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If `f(x)-{x^2}-({x})^2, `where (x) denotes the fractional part of x, thenA. `f (x) ` is continuous at ` x = 2 ` but not at ` x = - 2 `B. `f (x) ` is continuous at ` x = - 2 ` but not at ` x = 2 `C. ` f (x)` is continuous at ` x = 2 and x = - 2 `D. `f (x) ` is discontinuous at ` x = 2 and x = - 2 ` |
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Answer» Correct Answer - A We have, ` f (x) = {x^(2)} - ( {x}) ^(2)` ` therefore lim_( x to 2 ^(-)) f (x) = lim_(h to 0 ) f ( 2 - h ) ` ` rArr lim_( x to 2^(-)) f (x) = lim _( h to 0 ){ (2 - h ) ^(2)} - ({ 2- h }) ^(2)` ` rArr lim _( x to 2 ^(-)) f (x) = lim_( h to 0 )[ ( 2 - h ) ^(2) - [(2 - h ) ^(2)] - (( 2 - h ) - [(2 - h)]) ^(2)]` ` rArr lim_( x to 2 ^(-)) f (x) = lim_( h to 0 )[ ( 2 - h) ^(2) - 3 - (2 - h - 1) ^(2)] ` ` rArr lim_( x to 2 ^(-)) f (x) = lim_( h to 0 ) [ 4 - 4h + h ^(2) - 3 - (1 - 2h + h ^(2))] = 0 ` ` lim_( x to 2 ^(+)) f (x) = lim_( h to 0 ) f ( 2 + h )` `rArr lim_( x to 2 ^(+)) f (x) = lim_( h to 0 ) [ {(2 + h ^(2)) } - { 2+ h } ^(2)]` ` rArr lim_(x to 2 ^(+)) f (x) = lim_( x to 0 ) [ ( (2 + h ) ^(2) - [ (2 + h ) ^(2)] ) - ((2 + h ) - [ 2 + h ] ) ^(2)]` ` rArr lim_( x to 2 ^(+)) f (x) =lim_( h to 0 ) [ ( 4 + 4h + h ^(2) - 4) - ( 2 + h - 2 )^(2)]` ` rArr lim_(x to 2^(+)) f (x) = lim_( h to 0 )4h =0 ` and, ` f ( 2 ) = { 2 ^(2)} - ( {2}) ^(2) = 0 - 0 = 0 ` ` therefore lim_( x to 2^(-)) f (x) = lim_( x to 2 ^(+)) f (x) = f (2) ` So, f (x) is continuous at ` x = 2`. Now, ` " " lim_( x to - 2^( - )) f (x) = lim_( h to 0 ) f (-2 - h )` ` rArr lim _( x to - 2^( -)) f (x) = lim_( h to 0) [{(-2 - h )^(2) } - ({( - 2 - h )}) ^(2) ] ` ` rArr lim_(x to - 2 ^(-))f (x) = lim_( h to 0 ) [ (( - 2 - h ) ^(2) - [ (-2 - h ) ^(2)] ) - (( - 2 - h ) - [ - 2 - h ] ) ^(2)]` ` rArr lim_( x to - 2 ^(-)) f (x) = lim _( h to 0) [ (-2 - h )^(2) - 4 - ( - 2 - h + 3 ) ^(2)]` `rArr lim_( x to - 2 ^( - )) f (x) = lim _( h to 0) [ 4h + h ^(2) - ( 1- h ) ^(2)] = - 1 ` ` " " lim_( x to - 2 ^(+)) f (x) = lim_( h to 0) f (-2 + h )` `rArr lim_( x to - 2 ^(+)) f (x) = lim_( h to 0 ) [{( - 2 + h ) ^(2)} - { ( - 2 + h )} ^(2) ] ` ` rArr lim_( x to - 2 ^( + ) ) f (x) =lim _( hto 0 ) [(( - 2 + h ) ^(2) - [ (-2 + h ) ^(2) ] )- (-2 + h - [ - 2 + h ] ) ^(2) ]` ` rArr lim _( x to - 2 ^( + )) f (x) = lim _( h to 0 ) [ (( - 2 + h ) ^(2) - 3 ) - ( - 2 + h + 2) ^(2)]` ` rArr lim _( x to - 2 ^(+)) f (x) = lim _(h to 0) [(-2 + h ) ^(2) - 3 - h ^(2)] = 1 ` ` therefore lim _( x to - 2^( - ))f (x) ne lim_( x to - 2 ^( + )) f (x) ` So, f (x) is continuous at ` x = - 2 ` |
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