SATHEE: Theory of Equations 4 Question 4

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) $(1, 2)$

(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) $- 5 < a < 2$

(d) $2 < a < 5$

Show Answer

Answer:

Correct Answer: 4. (a)

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

$\begin{aligned} - 3 t^{2} + 2 t + a^{2} = 0 \\ \Rightarrow & t & = \frac{1 \pm \sqrt{1 + 3 a^{2}}}{3} \\ ∵ & t & = x - [x] = X [fractional part] \\ ∴ & 0 & \leq t \leq 1 \\ 0 & \leq \frac{1 \pm \sqrt{1 + 3 a^{2}}}{3} \leq 1 \end{aligned}$

Taking positive sign, we get

$\begin{aligned} 0 & \leq \frac{1 + \sqrt{1 + 3 a^{2}}}{3} < 1 \\ \Rightarrow \sqrt{1 + 3 a^{2}} < 2 & \Rightarrow 1 + 3 a^{2} < 4 \\ \Rightarrow a^{2} - 1 < 0 & \Rightarrow (a + 1) (a - 1) < 0 \end{aligned}$

$∴ a \in (- 1, 1)$ , for no integer solution of $a$ , we consider $(- 1, 0) \cup (0, 1)$

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

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Theory of Equations 4 Question 4

4. If a∈R and the equation −3(x−[x])2+2(x−[x]) +a2=0 (where, [x] denotes the greatest integer ≤x ) has no integral solution, then all possible values of a lie in the interval

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

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