Question

If $f\left(x\right)=\{\begin{array}{c}x;x\le 1\\ x^2+bx+c;x>1\end{array}$, and $f{}^{\prime }\left(x\right)$ exists finitely for all $x\in R$ then

• A. $b=-1,c\in R$
• B. $c=1,b\in R$
• C. $b=1,c=-1$
• D. $b=-1,c=1$

Answer 1

The correct option is D $b=-1,c=1$

$f\left(x\right)=xforx<1=x^2+bx+cforx>1Asf\left(x\right)isdifferentiable\therefore f\left(1^{+}\right)=f\left(1^{-}\right)\therefore 1=1+b+c\Rightarrow b+c=0\therefore f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)\therefore 2+b=1\therefore b=-1\Rightarrow c=1$
Hence answer is D