\texttt{find the domain of }
\sqrt{2\left\{x \right\}^2 -3\left\{x \right\}+1}
where
{x} is fractional part
dbt
2.given
\left[x \right]+ \left[x+\frac{1}{n} \right]+ \left[x+ \frac{2}{n} \right]+...\left[x+ \frac{n-1}{n} \right]= \left[nx \right]
find
\left[\frac{n+1}{2} \right]+\left[\frac{n+2}{4} \right]+\left[\frac{n+4}{8} \right]+...... = ?
3.
if
\left[x \right] =\left[\frac{x}{2} \right]+\left[\frac{x+1}{2} \right]
find
\left[\frac{n}{2} + \frac{2}{4}\right]+ \left[\frac{n}{4} + \frac{2}{4}\right]+\left[\frac{n}{8}+\frac{4}{8} \right]+.....
all above questions
[ ]
\texttt{is greatest integer function}
for practice
39Q1. f(x) = √2{x}² - 3{x} + 1
R1 : Square root term.
2{x}² - 3{x} + 1 ≥ 0
=> 2{x}² - 2{x} - {x} + 1 ≥ 0
=> 2{x}({x} - 1) - 1({x} - 1) ≥ 0
=> (2{x} - 1)({x} - 1) ≥ 0
=> ({x} - 1/2)({x} - 1) ≥ 0
{x} - 1 will always be negative as {x} → [0, 1)
So to make this non-negative,
{x} - 1/2 ≤ 0
=> {x} ≤ 1/2
=> {x} ≤ 0.5
But {x} ≥ 0
So {x} → [0, 0.5]
So as subho has found,
x ε [n, n + 0.5] where n ε Z or integer set.
493) [x]=[x2] + [x+12]=[x2] + [x2+12]
putting x=x2
[2x]=[x]+[x+12]
or, [x+12]=[2x]-[x]
given expression=[n2+12]+[n4+12]+[n8+12]+....=([n]-[n2])+([n2]-[n4])+(n4-n8)+...=[n][1][1][1]
