(D) Given equation: $x^2-4x+[x]+3=x[x]$Rearranging the terms: $x^2-4x+3=x[x]-[x]$Factorizing the left side: $(x-1)(x-3)=[x](x-1)$This implies: $(x-1)(

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a unique solution in $(-\infty, \infty)$

(D) Given equation: $x^2-4x+[x]+3=x[x]$Rearranging the terms: $x^2-4x+3=x[x]-[x]$Factorizing the left side: $(x-1)(x-3)=[x](x-1)$This implies: $(x-1)(x-3) - [x](x-1) = 0$$(x-1)(x-3-[x]) = 0$So,$x=1$ or $x-3=[x]$For $x-3=[x]$,we have $x-[x]=3$,which means the fractional part $\{x\}=3$.Since the fractional part $\{x\}$ must satisfy $0 \le \{x\} Thus,the only solution is $x=1$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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The equation $x^2-4x+[x]+3=x[x]$,where $[x]$ denotes the greatest integer function,has:

JEE MAIN 2023 All JEE MAIN

A
exactly two solutions in $(-\infty, \infty)$
B
no solution
C
a unique solution in $(-\infty, 1)$
D
a unique solution in $(-\infty, \infty)$

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4-2.Quadratic Equations and Inequations

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Solution of quadratic equations and Nature of roots

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The equation $x^2-4x+[x]+3=x[x]$,where $[x]$ denotes the greatest integer function,has:

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