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Theory of Equations: Equation Denotes Gif Integral Solution Exactly Solution Lies

JEE Maths question with a full step-by-step solution.

Question
If a∈Ra \in \mathbb{R} and the equation (a−2)(x−[x])2+2(x−[x])+a2=0(a-2)(x-[x])^2 + 2(x-[x]) + a^2 = 0 (where [x][x] denotes GIF) has no integral solution and has exactly one solution in (2,3)(2, 3), then aa lies in the interval
A(0,1)(0, 1)
B(1,2)(1, 2)
C(−1,3)(-1, 3)
D(−1,0)(-1, 0)correct
Solution
Step 1: Let y=x−[x]y = x - [x], so y∈[0,1)y \in [0,1). The equation becomes:
f(y)=(a−2)y2+2y+a2=0f(y) = (a-2)y^2 + 2y + a^2 = 0
 Step 2: For 2<x<32 < x < 3: [x]=2[x] = 2, so y∈(0,1)y \in (0,1). For exactly one solution in (2,3)(2,3), we need exactly one root of f(y)=0f(y) = 0 in (0,1)(0,1), which requires f(0)⋅f(1)<0f(0) \cdot f(1) < 0. Step 3: f(0)=a2f(0) = a^2 and f(1)=(a−2)+2+a2=a2+a=a(a+1)f(1) = (a-2)+2+a^2 = a^2 + a = a(a+1).
f(0)⋅f(1)=a2⋅a(a+1)=a3(a+1)<0f(0) \cdot f(1) = a^2 \cdot a(a+1) = a^3(a+1) < 0
 Since a2≥0a^2 \geq 0, this reduces to a(a+1)<0a(a+1) < 0, i.e., −1<a<0-1 < a < 0. Step 4: For a∈(−1,0)a \in (-1, 0): f(0)=a2>0f(0) = a^2 > 0 (so y=0y = 0 is not a root, hence no integral solution at x=2x=2). ✓ Correct answer: (4)
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