`f(x)={(|x|+3",","if", x le -3),(-2x",", "if", -3 lt x lt 3),(6x+2",","if",x ge 3):}` is

Options

A. continuous at x = -3 and discontinuous at x = 3
B. continuous at x = -3, 3
C. discontinuous at x = -3, 3
D. continuous at x = 3 and discontinuous at x = -3

Answer

A

Explanation

Here, `f(x)={(|x|+3",","if",x le -3),(-2x",","if",-3 lt x lt 3),(6x+2",","if", x ge 3):}`
At `x=-3, LHL =lim_(x to 3^(-)) f(x)=lim_(x to 3^(-))(|x|+3)`
Putting `x=(-3-h)" as " x to -3^(-), " then " h to 0.`
`therefore LHL = lim_(h to 0)[|-3-h|+2]=lim_(h to 0)(6+h)=6+0=6`
` RHL=lim_(x to 3^(+)) f(x)=lim_( x to 3^(+)) (-2x)`
Putting `x =-3+h" as " x to -3^(+), " then " h to 0`
`therefore RHL=lim_( h to 0) -2(-3+h)=lim_(h to 0) (6-2h)=6-2xx0=6`
Also, `f(-3)=|-3|+3=3+3=6[because f(x)=|x| +3]`
`therefore LHL=RHL=f(-3)`
Thus, f(x) is continuous at x = -3
At x = 3, `LHL=lim_(x to 3^(-)) f(x)=lim_(x to 3^(-))(-2x)` Putting ` x =3-h" as " x to 3^(-), " then " h to 0.`
`therefore LHL=lim_(h to 0) -2(3-h)`
`=lim_(h to 0) (-6+2h)=-6+2xx0= -6`
`RHL=lim_(x to 3^(+))f(x)=lim_(x to 3^(+))(6x+2)`
Putting `x =3+h" as " x to 3^(+), " then " h to 0.`
`therefore RHL=lim_(h to 0)[6(3+h)+2]=lim_(h to 0)(18+6h+2)`
`=lim_(h to 0) (20+6h)=20+6xx0=20`
`because LHL ne RHL`, therefore f(x) is discontinuous at x =3.