If f(x) is continuous at x=3, then `f(x)=ax+1," for x "le3`
`=bx+3," for x "gt" 3 then "`

Options

A. `a+b=(-2)/(3)`
B. `a-b=(-2)/(3)`
C. `a-b=(2)/(3)`
D. `a+b=(2)/(3)`

Answer

C

Explanation

We have, f(x) is continuous at x=3.
where `f(x)={{:(ax+1," for "xle3),(bx+3," for "xgt3):}`
`therefore underset(x to 3^(-))(lim)f(x)=underset(x to 3^(+))(lim)f(x)=f(3)` . . . (i)
Now, `underset(x to 3^(-))(lim)f(x)=underset( hto 0)(lim)a(3-h)+1`
`=3a+1`
`underset(xto3^(+))(lim)f(x)=underset(hto0)(lim)f(3+h)` . . . (ii)
`=underset(h to 0)(lim)b(3+h)+3`
`=3b+3`
and `f(3)=3a+1`
Since, f(x) is continuous at x=3
`therefore`From Eqs. (i), (ii), (iii) and (iv)
we get `3a+1=3b+3`
`implies3a-3b=2`
`impliesa-b=(2)/(3)`