Let `f(x)={(1-cos4x)/(x^2),\ \ \ if\ x<0a ,\ \ \ if\ x=0(sqrt(x))/(sqrt(16+sqrt(x))-4),\ \ \ if\ x >0` Determine the value of `a` so that `f(x)` is continuous at `x=0.`

Options

A. 2
B. 4
C. 6
D. 8

Answer

D

Explanation

Since, f(x) is continuous at x = 0
`therefore LHL=RHL =f(0) " " ` …(i)
Now, `lim_(x to 0^(-))f(x)=lim_(x to 0^(-))(1-cos 4x)/(x^(2))=lim_(x to 0^(-)) (2sin^(2)2x)/(x^(2))`
`LHL=lim_(x to 0^(-)) 8((sin2x)/(2x))^(2)=8(1)^(2)=8`
`RHL=lim_(x to 0^(+)) f(x) =lim_(x to 0^(+)) (sqrt(x))/(sqrt(16+sqrt(x))-4)`
`=lim_(x to 0^(+))(sqrt(x)(sqrt(16+sqrt(x))+4))/((sqrt(16)+sqrt(x))+4(sqrt(16 +sqrt(x))-4))`
`=lim_(x to 0^(+)) (sqrt(x)(sqrt(16+sqrt(x))+4))/(16+sqrt(x)-16)`
`=lim_(x to 0^(+))(sqrt(16+sqrt(x))+4)=8`
From Eq. (i),
`8=8=a`
`therefore a = 8`