# Let $\dpi{100} f:R\rightarrow R$   be defined by$\dpi{100} f(x)=\left\{\begin{matrix} k-2x, & i\! fx\leq -1\\ 2x+3, & i\! fx> -1 \end{matrix}\right.$If $\dpi{100} f$ has a local minimum at $\dpi{100} x=-1$, then a possible value of $\dpi{100} k$ is Option 1) 1 Option 2) 0 Option 3) –1/2 Option 4) –1

S Sabhrant Ambastha

As we learnt in

Differentiability -

Let  f(x) be a real valued function defined on an open interval (a, b) and  $x\epsilon$ (a, b).Then  the function  f(x) is said to be differentiable at   $x_{\circ }$   if

$\lim_{h\rightarrow 0}\:\frac{f(x_{0}+h)-f(x_{0})}{(x_{0}+h)-x_{0}}$

$or\:\:\:\lim_{h\rightarrow 0}\:\frac{f(x)-f(x_{0})}{x-x_{0}}$

-

$f (x) = \left\{\begin{matrix} k-2x & x\leqslant -1\\ 2x+3 & x\leqslant -1 \end{matrix}\right.$

$\\ \therefore k -2 (-1) = 2 (-1)+3\\ k+2 = -2+3 \\k+2=1\\k=-1$

Since f (x) is differential so it must be continuous.

Option 1)

1

This option is incorrect.

Option 2)

0

This option is incorrect.

Option 3)

–1/2

This option is incorrect.

Option 4)

–1

This option is correct.

Exams
Articles
Questions