Question

Let $f ( x )$ satisfy the requirements of Lagrange's mean value theorem in $[0,2]$. If $f (0)=0$ and $f '( x ) \leq \frac{1}{2}$ for all $x$ in $[0, 2]$, then

• A. $| f ( x )| \leq 2$
• B. $f ( x ) \leq 1$
• C. $f(x)=2 x$
• D. $f(x)=3$ for atleast one $x$ in $[0,2]$

Answer 1

Answer: (B)

Applying Lagrange's mean value theorem to $f ( x )$
in $[0, x], 0 <\,x \leq 2$, we get
$\frac{ f ( x )- f (0)}{ x -0}= f '( c )$ for some $c \in(0,2)$
$\Rightarrow \frac{ f ( x )}{ x } \equiv f '( c ) \leq \frac{1}{2} $
$\Rightarrow f ( x ) \leq \frac{1}{2} x \leq \frac{1}{2} .2 \,\,\,\,\,\,\,(\because x \leq 2)$
$\Rightarrow f ( x ) \leq 1$