question_answer

  Let\[f:R\to R\]be defined by \[f(x)=\left\{ \begin{matrix}    k-2x, & if & x\le -1  \\    2x+3, & if & x>-1  \\ \end{matrix} \right.\] If\[f\]has a local minimum at\[x=1,\]then a possible value of k is       AIEEE  Solved  Paper-2010

A) 1

B)                                       0                             

C)        \[-\frac{1}{2}\]                 

D)        \[-1\]

Correct Answer: D

Solution :

\[f:R\to Rf(x)=\left\{ \begin{matrix}    k-2x & x\le -1  \\    2x+3 & x\ge -1  \\ \end{matrix} \right\}\] \[f'(x)=\left\{ \begin{matrix}    -2 & x1  \\ \end{matrix} \right\}\] \[k-2x=+1\] \[k=-1\]