Theory of Equations 4 Question 4
4. If $a \in R$ and the equation $-3(x-[x])^{2}+2(x-[x])$ $+a^{2}=0$ (where, $[x]$ denotes the greatest integer $\leq x$ ) has no integral solution, then all possible values of $a$ lie in the interval
(2014 Main)
(a) $(-1,0) \cup(0,1)$
(b) $(1,2)$
(c) $(-2,-1)$
(d) $(-\infty,-2) \cup(2, \infty)$
5 For all ’ $x$ ‘, $x^{2}+2 a x+(10-3 a)>0$, then the interval in which ’ $a$ ’ lies is
(2004, 1M)
(a) $a<-5$
(b) $-5<a<2$
(c) $a>5$
(d) $2<a<5$
Show Answer
Solution:
- Put $t=x-[x]={X}$, which is a fractional part function and lie between $0 \leq{X}<1$ and then solve it.
Given, $a \in R$ and equation is
$$ -3{x-[x]}^{2}+2{x-[x]}+a^{2}=0 $$
Let $t=x-[x]$, then equation is
$$ \begin{array}{rlrl} & & -3 t^{2}+2 t+a^{2}=0 \\ \Rightarrow & & t & =\frac{1 \pm \sqrt{1+3 a^{2}}}{3} \\ \because & & t & =x-[x]={X} \quad \text { [fractional part] } \\ \therefore & 0 & \leq t \leq 1 \\ & & 0 & \leq \frac{1 \pm \sqrt{1+3 a^{2}}}{3} \leq 1 \end{array} $$
Taking positive sign, we get
$$ \begin{aligned} 0 & \leq \frac{1+\sqrt{1+3 a^{2}}}{3}<1 \\ \Rightarrow \quad \sqrt{1+3 a^{2}}<2 & \Rightarrow \quad 1+3 a^{2}<4 \\ \Rightarrow \quad a^{2}-1<0 & \Rightarrow \quad(a+1)(a-1)<0 \end{aligned} $$
$\therefore a \in(-1,1)$, for no integer solution of $a$, we consider $(-1,0) \cup(0,1)$
