If a in R and the equation - 3{left( {x - le | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Here, $a \in R$ and equation is

$-3\{x-[x]\}^{2}+2\{x-[x]\}+a^{2}=0$

$	ext { Let } \quad t=x-[x]$

$	herefore-3 t^{2}+2 t+a^{2}=0$

$\Righ

Vedclass · Accepted Answer

$\left( { - 1,0} \right) \cup \left( {0,1} \right)$

Vedclass · Answer

Here, $a \in R$ and equation is

$-3\{x-[x]\}^{2}+2\{x-[x]\}+a^{2}=0$

$	ext { Let } \quad t=x-[x]$

$	herefore-3 t^{2}+2 t+a^{2}=0$

$\Rightarrow t=\frac{1 \pm \sqrt{1+3 a^{2}}}{3}$

$\because \quad t=x-[x]=\{x\} \quad[	ext { fractional part }]$

$	herefore \quad 0 \leq t \leq 1$

$\Rightarrow 0 \leq \frac{1 \pm \sqrt{1+3 a^{2}}}{3} \leq 1$

Taking positive sign

$\because \quad[\{x\}>0]$

$	herefore 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 $(-1,0) \cup(0,1)$

If $a \in R$ and the equation $ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$ (where $[x]$ denotes the greatest integer $\leq\,x$)has no integral solution ,then all possible values of $a$ lie in the interval

Similar Questions

The maximum value $M$ of $3^x+5^x-9^x+15^x-25^x$, as $x$ varies over reals, satisfies

Equation $\frac{3}{{x - {a^3}}} + \frac{5}{{x - {a^5}}} + \frac{7}{{x - {a^7}}} = 0,a > 1$ has

Let $a, b, c, d$ be real numbers between $-5$ and $5$ such that $|a|=\sqrt{4-\sqrt{5-a}},|b|=\sqrt{4+\sqrt{5-b}},|c|=\sqrt{4-\sqrt{5+c}}$ $|d|=\sqrt{4+\sqrt{5+d}}$ Then, the product $a b c d$ is

The real roots of the equation ${x^2} + 5|x| + \,\,4 = 0$ are