If f(x) = x 1 - [ x ] ^2 then (where [.] denotes greatest integer function)

A. f(x) is increasing in x (0,1)

If f(x) = x 1 - [ x ] ^2 then (where [.] denotes greatest integer…

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MCQ

If $f(x)$ = $x\sqrt {1 - {{\left[ x \right]}^2}} $ then (where $[.]$ denotes greatest integer function)

✓
$f(x)$ is increasing in $x$ $ \in $ $(0,1)$
B
$x$ = $1$ is point of local maxima of $f(x)$
C
$f(x)$ is negative function
D
Rolle's theorem is applicable on $f (x)$ in $x$ $ \in $ $[0,1]$

✓

Answer

Correct option: A.

$f(x)$ is increasing in $x$ $ \in $ $(0,1)$

a
(1) For $x \in(0,1),[x]=0 \Rightarrow f(x)=x,$ which is increasing

(2) $\quad \because f\left(1^{+}\right)=0, f(1)=1, f(1)=0$

$\therefore f\left(1^{-}\right)>f(1)=f\left(1^{+}\right) \Rightarrow x=1$ is not a point of local maxima.

(3) Domain of $f(x)$ is given by $-1 \leq[x] \leq 1$

${ \Rightarrow x \in [ - 1,2)}$

${ \Rightarrow f(x) = \left\{ {\begin{array}{*{20}{l}}
{0,}&{ - 1 \le x < 0}\\
{x,}&{0 \le x < 1}\\
{0,}&{1 \le x < 2}
\end{array} \Rightarrow f(x) \ge 0} \right.}$

(4) $\quad \because \quad f(x)$ is discontinuous at $x=1.$

$\therefore $ Rolle's theorem is not applicable

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