JEE Main 2014 · 6 Apr · Shift 1 · Mathematics Q63

← JEE Main 2014 · 6 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 63 of 90

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

(-2,-1)

(-\infty,-2)\cup(2,\infty)

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

(1,2)

Answer and worked solution

Official answer: C

Put $t=x-[x]\in[0,1)$ , so $3t^2-2t-a^2=0$ . An integer $x$ means $t=0$ , which needs $a=0$ ; so $a\ne0$ .

For a solution to exist at all, the root $t=\dfrac{1+\sqrt{1+3a^2}}3$ must be less than 1, i.e. $\sqrt{1+3a^2}<2$ , i.e. $a^2<1$ .

Hence $a\in(-1,0)\cup(0,1)$ .

Topics:Location of roots Graphs of functions

← Previous question Next question →

JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 63