InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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, thenA. a=0, b=1B. a-0, b=0C. a=1, b=0D. none of these |
|
Answer» Correct Answer - B 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. |
|