Question:
Let \(I=\displaystyle\int_a^b\left(x^4-2x^2\right)dx\). If \(I\) is minimum then the ordered pair \((a,b)\) is -
Let \(I=\displaystyle\int_a^b\left(x^4-2x^2\right)dx\). If \(I\) is minimum then the ordered pair \((a,b)\) is -
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The integral is smallest when the interval covers exactly the region where the integrand \(x^2(x^2-2)\) is negative.
Updated On: Nov 5, 2026
- \(\left(\sqrt2,-\sqrt2\right)\)
- \(\left(0,\sqrt2\right)\)
- \(\left(-\sqrt2,\sqrt2\right)\)
- \(\left(-\sqrt2,0\right)\)
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The Correct Option is C
Solution and Explanation
\(f(x)=x^4-2x^2=x^2(x^2-2)\) is negative exactly for \(0<|x|<\sqrt2\) and positive for \(|x|>\sqrt2\).
\(I=\int_a^bf\,dx\) is minimum when we integrate over the whole negative region and nothing else: \(a=-\sqrt2,\ b=\sqrt2\).
Then \(I=2\left(\frac{(\sqrt2)^5}{5}-\frac{2(\sqrt2)^3}{3}\right)<0\) is the least value.
Answer: (3) \(\left(-\sqrt2,\sqrt2\right)\)
\(I=\int_a^bf\,dx\) is minimum when we integrate over the whole negative region and nothing else: \(a=-\sqrt2,\ b=\sqrt2\).
Then \(I=2\left(\frac{(\sqrt2)^5}{5}-\frac{2(\sqrt2)^3}{3}\right)<0\) is the least value.
Answer: (3) \(\left(-\sqrt2,\sqrt2\right)\)
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