If `f(x)={(e^(x),xlt2),(a+bx,xge2):}` is differentiable for all `x epsilonR` then

Options

A. a+b=0
B. `a+2b=e^(2)`
C. `b=e^(2)`
D. all of these

Answer

D

Explanation

Clearly, f(x) is everywhere continuous and differentiable except possibly at x=2. At x=2 also at is given that f(x) is differentiable. So, it must be continuous there at.
Thus, we have
`underset(x to 2^(-))lim f(x)=underset(x to2^(+))lim f(2)` and (LHD at x=2)=(RHD at x=2)
`underset(x to 2^(-))lim e^(x)=underset(x to2^(+))lim a+bx=a+2b` and `{(d)/(dx)(e^(x))}_(x=2)={(d)/(dx)(a+bx)}_(x=2)`
`Rightarrow e^(2)=a+2b and e^(2)=b`
`Rightarrow a+2b=e^(2), b=e^(2) and a+b=0`