Question:
The number of integral solution of \(\frac{x + 1}{x^2 + 2}>\frac{1}{4}\) is
The number of integral solution of \(\frac{x + 1}{x^2 + 2}>\frac{1}{4}\) is
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Check denominator positive: \(x^2 + 2>0\) always.
Updated On: Apr 7, 2026
- 1
- 2
- 5
- None of these
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Solve inequality.
Step 2: Detailed Explanation:
\(\frac{x + 1}{x^2 + 2}>\frac{1}{4}\)
\(4(x + 1)>x^2 + 2 \rightarrow 4x + 4>x^2 + 2 \rightarrow 0>x^2 - 4x - 2\)
\(x^2 - 4x - 2<0\)
Roots: \(2 \pm \sqrt{6} \approx 2 \pm 2.449 \rightarrow -0.449<x<4.449\)
Integers: 0, 1, 2, 3, 4 \(\rightarrow\) 5 solutions.
Step 3: Final Answer:
5 integral solutions.
Solve inequality.
Step 2: Detailed Explanation:
\(\frac{x + 1}{x^2 + 2}>\frac{1}{4}\)
\(4(x + 1)>x^2 + 2 \rightarrow 4x + 4>x^2 + 2 \rightarrow 0>x^2 - 4x - 2\)
\(x^2 - 4x - 2<0\)
Roots: \(2 \pm \sqrt{6} \approx 2 \pm 2.449 \rightarrow -0.449<x<4.449\)
Integers: 0, 1, 2, 3, 4 \(\rightarrow\) 5 solutions.
Step 3: Final Answer:
5 integral solutions.
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