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If f(x)={(2^(x+2)-16)/(4^x-16),ifx!=2k ...

If `f(x)={(2^(x+2)-16)/(4^x-16),ifx!=2k ,ifx=2i scon t inuou sa tx=2,fin dk`

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The correct Answer is:
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We have , `f(x) ={{:((2^(x+2)-16)/(4^(x)-16), if x ne 2 ),(k, if x = 2):}` ,`x = 2`.
Since, `f(x)` is continous at `x = 2`.
`:. LHL = RHL = f(2)`
At `x = 2, underset(xrarr2)lim (2^(x).2^(2)-2^(4))/(4^(x)-4^(2)) = underset(xrarr2)(lim)(4.(2^(x)-4))/((2^(x))^(2)-(4)^(2))`
`= underset(xrarr2)(lim)(4.(2^(x)-4))/((2^(x)-4)(2^(x)+4)) , [:' a^(2)-b^(2) = (a+b)(a-b)]`
`= underset(xrarr2)lim(4)/(2^(x)+4) = 4/8 = 1/2`
But `f(2) = k`
` :. k = 1/2`
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