Let [x] be the greatest integer less than or | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation $[x^2] = x + 1$. Since $[x^2]$ is an integer,$x + 1$ must be an integer,which implies $x$ is an integer. Let $x = n$,where $n \

Vedclass · Accepted Answer

no solution

Vedclass · Answer

(C) Given the equation $[x^2] = x + 1$. Since $[x^2]$ is an integer,$x + 1$ must be an integer,which implies $x$ is an integer. Let $x = n$,where $n \in \mathbb{Z}$. The equation becomes $[n^2] = n + 1$. Since $n^2$ is an integer,$[n^2] = n^2$. So,$n^2 = n + 1$,which implies $n^2 - n - 1 = 0$. The roots of this quadratic equation are $n = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$. Since $\frac{1 \pm \sqrt{5}}{2}$ are not integers,there is no integer $n$ that satisfies the equation. Therefore,the equation $[x^2] = x + 1$ has no solution.

Let $[x]$ be the greatest integer less than or equal to $x$ for a real number $x$. Then the equation $[x^2] = x + 1$ has:

Similar Questions

If the roots of the equation $x^2 + a^2 = 8x + 6a$ are real,then

$A$ value of $x$ satisfying $150 x \equiv 35 \pmod{31}$ is

If the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign,then $r = ......$

Let $a$ be an integer selected at random from the set $\{0, 1, 2, 3, ..., 9\}$. The probability that the equation $ax^2 - ax + 1 = 0$ has real roots is ...

Let $p(x) = x^2 - 5x + a$ and $q(x) = x^2 - 3x + b$,where $a$ and $b$ are positive integers. Suppose $\text{HCF}(p(x), q(x)) = x - 1$ and $k(x) = \text{LCM}(p(x), q(x))$. If the coefficient of the highest degree term of $k(x)$ is $1$,then the sum of the roots of $(x - 1) + k(x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module