For what value of k, `f(x)={((2^(x+2)-16)/(4^(x)-16)",", x ne 2),(k",", x =2):}` is continuous at x = 2?

Options

A. 1
B. `(3)/(2)`
C. 2
D. `(1)/(2)`

Answer

D

Explanation

Given, `f(x) = {((2^(x+z))/(4^(x)-16) ",", x ne 2),( k",", x = 2):}`
`rArr f(x) ={( (4(2^(x)-4))/((2^(x))^(2)-4^(2))",", x ne 2),(k",", x =2):}`
`therefore f(x) = { ((4)/(2^(x)+4)",",x ne 2),( k",", x =2):}`
We have, `f(2) =k`
Case I When ` x lt 2`,
`lim_( x to 2^(-)) f(x) = lim_( x to 2^(-)) ((4)/(2^(x)+4)) = lim_(h to 0) ((4)/( 2^((2-h))+4))`
`=((4)/(4+4))=(1)/(2)`
Case II When `x gt 2, `
`lim_( x to 2^(+)) f(x)= lim_(x to 2^(+))((4)/(2^(x)+4))=lim_( h to 0)[ (4)/( 2^((2+h))+4)]`
`=((4)/(4+4))=(1)/(2)`
Now, function is continuous at x = 2, if
`lim_(x to 2^(+))f(x)=lim_(x to 2^(-)) f(x) =f(2) rArr (1)/(2) =(1)/(2) =k`
`therefore k =(1)/(2)`