Sum of the solutions of the equation [x^2] - | vedclass

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(D) Given the equation $[x^2] = 2x - 1$.Since $[x^2]$ is an integer,$2x - 1$ must be an integer,which implies $2x$ is an integer.By the definition of

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$3$

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(D) Given the equation $[x^2] = 2x - 1$.Since $[x^2]$ is an integer,$2x - 1$ must be an integer,which implies $2x$ is an integer.By the definition of the greatest integer function,$x^2 - 1 Substituting $[x^2] = 2x - 1$,we get $x^2 - 1 From $x^2 - 1 From $2x - 1 \le x^2$,we have $x^2 - 2x + 1 \ge 0$,which is $(x - 1)^2 \ge 0$,which is true for all $x$.Case $1$: $0 \le x^2 Case $2$: $1 \le x^2 Case $3$: $2 \le x^2 Case $4$: $3 \le x^2 The solutions are $x = 1/2, 1, 3/2$. The sum is $1/2 + 1 + 3/2 = 3$.

Sum of the solutions of the equation $[x^2] - 2x + 1 = 0$ is (where $[.]$ denotes the greatest integer function).

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