- Home
- Standard 11
- Mathematics
જો $a \in R$ હોય અને સમીકરણ $ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$ ને પૂર્ણાંક ઉકેલ ન હોય તો $a$ શકય કિંમતો . . . અંતરાલમાં હોય . .
$\left( { - 1,0} \right) \cup \left( {0,1} \right)$
$\left( {1,2} \right)$
$\left( { - 2, - 1} \right)$
$\left( { - \infty , - 2} \right) \cup \left( {2,\infty } \right)$
Solution
Here, $a \in R$ and equation is
$-3\{x-[x]\}^{2}+2\{x-[x]\}+a^{2}=0$
$\text { Let } \quad t=x-[x]$
$\therefore-3 t^{2}+2 t+a^{2}=0$
$\Rightarrow t=\frac{1 \pm \sqrt{1+3 a^{2}}}{3}$
$\because \quad t=x-[x]=\{x\} \quad[\text { fractional part }]$
$\therefore \quad 0 \leq t \leq 1$
$\Rightarrow 0 \leq \frac{1 \pm \sqrt{1+3 a^{2}}}{3} \leq 1$
Taking positive sign
$\because \quad[\{x\}>0]$
$\therefore 0 \leq \frac{1 \pm \sqrt{1+3 a^{2}}}{3}<1$
$\Rightarrow \sqrt{1+3 a^{2}}<2$
$\Rightarrow \quad 1+3 a^{2}<4$
$\Rightarrow a^{2}-1<0$
$\Rightarrow \quad(a+1)(a-1)<0$
For no integral solution of a we consider the interval $(-1,0) \cup(0,1)$
