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) $2<a<5$

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Answer:

Correct Answer: 4. (a)

Solution:

  1. 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 $(-1,0) \cup(0,1)$



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