The values of x, which satisfy both the inequations x^2 - 1 = 0, lie in the interval

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q11 – AP-EAMCET 2026

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Home Maths Complex & QuadraticQuestion #63

The values of

x

, which satisfy both the inequations

x^{2} - 1 \leq 0

and

x^{2} - x - 2 \geq 0

, lie in the interval

Clue 1 of 3: Quadratic Inequalities

To find the values of $x$ that satisfy a quadratic inequality, first find the roots of the corresponding quadratic equation.

🧠 Full Solution Path

Step 1: Solve the first inequality✦ Active

The first inequality is $x^{2} - 1 \leq 0$ . We can factor this as a difference of squares:

(x - 1) (x + 1) \leq 0

The critical points are $x = - 1$ and $x = 1$ . For a quadratic inequality of the form $(x - a) (x - b) \leq 0$ where $a < b$ , the solution is the interval $[a, b]$ . Thus, the solution set for $x^{2} - 1 \leq 0$ is:

S_{1} = [- 1, 1]

💡 Teacher's Secret Hint

Remember that for a quadratic $a x^{2} + b x + c$ with $a > 0$ , if $a x^{2} + b x + c \leq 0$ , the solution lies between the roots. If $a x^{2} + b x + c \geq 0$ , the solution lies outside the roots.

Step 2: Solve the second inequality○ Expand

The second inequality is $x^{2} - x - 2 \geq 0$ . We can factor the quadratic expression:

(x - 2) (x + 1) \geq 0

The critical points are $x = - 1$ and $x = 2$ . For a quadratic inequality of the form $(x - a) (x - b) \geq 0$ where $a < b$ , the solution is $(- \infty, a] \cup [b, \infty)$ . Thus, the solution set for $x^{2} - x - 2 \geq 0$ is:

S_{2} = (- \infty, - 1] \cup [2, \infty)

💡 Teacher's Secret Hint

Always check the sign of the leading coefficient. If it's positive (as in this case for both inequalities), the parabola opens upwards. This helps determine if the solution is between or outside the roots.

Step 3: Find the intersection of the solution sets○ Expand

We need to find the values of $x$ that satisfy both inequalities, which means finding the intersection of $S_{1}$ and $S_{2}$ :

S_{1} \cap S_{2} = [- 1, 1] \cap ((- \infty, - 1] \cup [2, \infty))

Let's analyze the intersection:

The interval $S_{1} = [- 1, 1]$ includes all numbers from $- 1$ to $1$ , including the endpoints.

The interval $S_{2} = (- \infty, - 1] \cup [2, \infty)$ includes all numbers less than or equal to $- 1$ , or all numbers greater than or equal to $2$ .

The only value common to both sets is $x = - 1$ .

Therefore, the values of $x$ that satisfy both inequations lie in the set ${- 1}$ .

💡 Teacher's Secret Hint

A number line sketch can be very helpful for visualizing the intersection of intervals, especially for more complex unions and intersections.

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Maths Question 11 – AP-EAMCET 2026

Clue 1 of 3: Quadratic Inequalities

🧠 Full Solution Path