Question

let $straight f left parenthesis straight x right parenthesis space equals space fraction numerator 1 minus space tan space straight x over denominator 4 straight x minus straight pi end fraction space comma space straight x not equal to space straight pi over 4 space straight x space element of space open square brackets 0 comma space straight pi over 2 close square brackets$ . If f(x) is continuous in $open square brackets 0 space comma space straight pi over 2 close square brackets comma$ then f (π/4) is

1

-1/2

-1

1/2

Solution

-1/2

straight f left parenthesis straight x right parenthesis space space equals space fraction numerator 1 minus space tan space straight x over denominator 4 straight x space minus space straight pi end fraction limit as straight x rightwards arrow space straight pi divided by 4 of space straight f left parenthesis straight x right parenthesis space equals stack space lim with straight x rightwards arrow straight pi divided by 4 below space open parentheses fraction numerator 1 minus tan space straight x over denominator 4 space straight x minus straight pi end fraction close parentheses By space straight L apostrophe space Hospital space rule limit as straight x rightwards arrow space straight pi divided by 4 of space open parentheses fraction numerator negative sec squared space straight x over denominator 4 end fraction close parentheses space equals space fraction numerator negative sec squared space straight pi divided by 4 over denominator 4 end fraction space equals space minus 2 over 4 rightwards double arrow space limit as straight x rightwards arrow straight pi divided by 4 of space straight f left parenthesis straight x right parenthesis space equals space minus space 1 divided by 2

Also f(x) is continuous in [0, π/ 2] ,
so f(x) will be continuous at π / 4 .
∴ Value of function = Value of limit ⇒ f(π/ 4) = −1/ 2

For Daily Free Study Material Join wiredfaculty