(D) Given inequations are $x^2-1 \leq 0$ and $x^2-x-2 \geq 0$. For $x^2-1 \leq 0$: $(x-1)(x+1) \leq 0$ This implies $x \in [-1, 1]$. For $x^2-x-2 \geq

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(D) Given inequations are $x^2-1 \leq 0$ and $x^2-x-2 \geq 0$. For $x^2-1 \leq 0$: $(x-1)(x+1) \leq 0$ This implies $x \in [-1, 1]$. For $x^2-x-2 \geq 0$: $(x-2)(x+1) \geq 0$ This implies $x \in (-\infty, -1] \cup [2, \infty)$. The intersection of the two sets $x \in [-1, 1]$ and $x \in (-\infty, -1] \cup [2, \infty)$ is the single point $\{-1\}$. Thus,the set of all values of $x$ is $\{-1\}$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic inequations and Newton Formula

Medium

The set of all values of $x$ which satisfy both the inequations $x^2-1 \leq 0$ and $x^2-x-2 \geq 0$ simultaneously is

TS EAMCET 2023 All TS EAMCET

A
$(-1, 2)$
B
$(-1, 1)$
C
$(-2, -1)$
D
$\{-1\}$

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4-2.Quadratic Equations and Inequations

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Solution of quadratic inequations and Newton Formula

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