Question:
The number of integers that satisfy both the inequalities
\[
x^2-2x+8>0
\]
and
\[
x^2-3x+2\le 0
\]
is:
The number of integers that satisfy both the inequalities
\[
x^2-2x+8>0
\]
and
\[
x^2-3x+2\le 0
\]
is:
Show Hint
If a quadratic has a negative discriminant and a positive coefficient of \(x^2\), then it remains positive for all real values of \(x\). This observation can save considerable calculation time in inequality problems.
Updated On: Jun 10, 2026
- \(1\)
- \(2\)
- \(4\)
- Infinite
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The Correct Option is B
Solution and Explanation
Concept:
To find the number of integers satisfying two inequalities simultaneously, we first solve each inequality separately and then find their common solution set. Finally, we count the integers lying in that intersection.
Step 1: Solve the inequality \(x^2-2x+8>0\) Consider \[ x^2-2x+8. \] The discriminant is \[ D=b^2-4ac =(-2)^2-4(1)(8) =4-32 =-28. \] Since \[ D<0 \] and the coefficient of \(x^2\) is positive, the quadratic never touches the \(x\)-axis and remains positive for all real values of \(x\). Therefore, \[ x^2-2x+8>0 \] is satisfied for every real number \(x\). \[ \boxed{\text{All real numbers satisfy the first inequality}} \]
Step 2: Solve the inequality \(x^2-3x+2\le0\) Factorising, \[ x^2-3x+2=(x-1)(x-2). \] Thus, \[ (x-1)(x-2)\le0. \] A quadratic with positive leading coefficient is non-positive between its roots. Hence, \[ 1\le x\le2. \] \[ \boxed{x\in[1,2]} \]
Step 3: Find the common solution The first inequality is true for all real numbers. The second inequality gives \[ x\in[1,2]. \] Therefore the common solution set is \[ [1,2]. \]
Step 4: Count the integers The integers in \[ [1,2] \] are \[ 1,\;2. \] Total number of integers \[ =2. \] Hence, \[ \boxed{2} \]
Step 1: Solve the inequality \(x^2-2x+8>0\) Consider \[ x^2-2x+8. \] The discriminant is \[ D=b^2-4ac =(-2)^2-4(1)(8) =4-32 =-28. \] Since \[ D<0 \] and the coefficient of \(x^2\) is positive, the quadratic never touches the \(x\)-axis and remains positive for all real values of \(x\). Therefore, \[ x^2-2x+8>0 \] is satisfied for every real number \(x\). \[ \boxed{\text{All real numbers satisfy the first inequality}} \]
Step 2: Solve the inequality \(x^2-3x+2\le0\) Factorising, \[ x^2-3x+2=(x-1)(x-2). \] Thus, \[ (x-1)(x-2)\le0. \] A quadratic with positive leading coefficient is non-positive between its roots. Hence, \[ 1\le x\le2. \] \[ \boxed{x\in[1,2]} \]
Step 3: Find the common solution The first inequality is true for all real numbers. The second inequality gives \[ x\in[1,2]. \] Therefore the common solution set is \[ [1,2]. \]
Step 4: Count the integers The integers in \[ [1,2] \] are \[ 1,\;2. \] Total number of integers \[ =2. \] Hence, \[ \boxed{2} \]
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