If the function f(x) is given by `f(x)={{:(,2^(1//(x-1)),x lt 1),(,ax^(2)+bx,x ge 1):}` is everywhere differentiable, then

Options

A. a=0, b=1
B. a=0, b=0
C. a=1, b=0
D. none of these

Answer

B

Explanation

Clearly, f(x) is everywhere continuous and differentiable except possible at x=1.
At x=1, we have
`underset(x to 1^(-))lim f(x)=underset(x to 1^(-))lim 2^(1//(x-1))=underset(h to 0)lim 2^(-1//h)=0`
`underset(x to 1^(+))lim f(x)=underset(x to 1^(+))lim (ax^(2)+bx)=a+b`
and
f(1)=a+b
For f(x) to be differentiable at x=1, it must be continuous there at .
`therefore underset(x to 1^(-))lim f(x)=underset(x to 1^(+))lim f(x)=f(1)`
`Rightarrow 0=a+b........(i)`
Also,
`("LHD at x=1")=underset(x to 1^(-))lim (f(x)-f(1))/(x-1)a`
`Rightarrow ("LHD at x=1")=underset(h to 0)lim (f(1-h)-f(1))/(-h)`
`Rightarrow ("LHD at x=1")=underset(h to 0)lim (2^(-1//h)-(a+b))/(-h)a`
`Rightarrow ("LHD at x=1")=underset(h to 0)lim (2^(-1//h))/(-h)" "[therefore a+b=0]`
`Rightarrow ("LHD at x=1")=-underset(h to 0)lim (1//h)/(2^(1//h))" "[(oo)/(oo)"form"]`
`Rightarrow ("LHD at x=1")=-underset(h to 0)lim (-1//h^(2))/(2^(1//h)"log_(e) 2(-(1)/(h^(2))))`
`Rightarrow ("LHD at x=1")=-underset(h to 0)lim (2^(-1//h))/(log_(e)2)=(0)/(log_(e)2)=0`
and
`("RHD at x=1")=underset(x to 1^(+))lim (f(x)-f(1))/(x-1)`
`Rightarrow ("RHD at x=1")=underset(h to 0)lim (f(1+h)-f(1))/(h)`
`Rightarrow ("RHD at x=1")=underset(h to 0)lim (a(1+h)^(2)+b(1+h)-(a+b))/(h)`
`Rightarrow ("RHD at x=1")=underset(h to 0)lim (ah^(2)+2ah+bh)/(h)`
`Rightarrow ("RHD at x=1")=underset(h to 0)lim ah+2a+b=2a+b`
For f(x) to be differentiable at x=1, we must have (LHD at x=1)=(RHD at x=1)
`Rightarrow 0=2a+b......(ii)`
Solving (i) and (ii) we get a=b=0.