Let $f: R \rightarrow R$ be defined by $f(x)=\left\{\begin{array}{ll}k-2 x, & \text { if } x \leq-1 \\ 2 x+3, & \text { if } x>-1\end{array}\right.$. If $f$ has a local minimum at $x=-1$, then a possible value of k is

JEE Main · 2010 · 11 Apr 2010 · Offline · Application of Derivatives

Let

f : R \to R

be defined by

f (x) = {k - 2 x, 2 x + 3, if x \leq - 1 if x > - 1

. If

f

has a local minimum at

x = - 1

, then a possible value of k is

Answer: (C)

For a local minimum at

x = - 1

the value there must not exceed the values just beside it. Just to the right the branch

2 x + 3

approaches

2 (- 1) + 3 = 1

and rises as

x

increases. The left branch

k - 2 x

falls as

x

increases, so coming up to

x = - 1

from the left it descends to

f (- 1) = k - 2 (- 1) = k + 2

. For

x = - 1

to be a minimum we need

f (- 1)

no larger than the right-hand neighbourhood,

k + 2 \leq 1

, that is

k \leq - 1

. The value that meets this is

k = - 1

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option