Let \rm I = \int\limits_a^b ( x^4 - 2x^2 ) dx. If I is minimum then the ordered pair (a, b) is -

Definite Integration PYQ — JEE Main 2019 · 10 Jan · Shift 1 · Q36 with Solution

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Correct answer: C

We need to minimize $I=int_a^b (x^4-2x^2),dx.$ Let $f(x)=x^4-2x^2=x^2(x^2-2).$
First study the sign of the integrand. We have $f(x)<0 \iff x^2-2<0 \iff |x|<\sqrt{2},$ and $f(x)=0 \text{ at } x=0,\ \pm\sqrt{2}.$ Also, $f(x)>0 \text{ for } |x|>\sqrt{2}.$
To make the definite integral minimum, we should include the entire region where the function is negative and avoid any positive part. Hence the minimizing interval should be $[a,b]=[-\sqrt{2},\sqrt{2}].$
Verify by evaluating the integral. An antiderivative is $F(x)=\frac{x^5}{5}-\frac{2x^3}{3}.$ So $I=F(b)-F(a).$ For option C: $I=\int_{-\sqrt{2}}^{\sqrt{2}}(x^4-2x^2)dx.$ Since the integrand is even, $I=2\int_0^{\sqrt{2}}(x^4-2x^2)dx.$ Now, $\int (x^4-2x^2)dx=\frac{x^5}{5}-\frac{2x^3}{3}.$ Thus $I=2\left[\frac{x^5}{5}-\frac{2x^3}{3}\right]_0^{\sqrt{2}}.$ Using $(\sqrt{2})^3=2\sqrt{2},\qquad (\sqrt{2})^5=4\sqrt{2},$ we get

=2\cdot 4\sqrt{2}\left(\frac{1}{5}-\frac{1}{3}\right) =8\sqrt{2}\left(-\frac{2}{15}\right) =-\frac{16\sqrt{2}}{15}.$$ This is the most negative possible among the given intervals. 5. Check options briefly: - A: $$\int_{\sqrt2}^{-\sqrt2}f(x)dx=-\int_{-\sqrt2}^{\sqrt2}f(x)dx>0,$$ so not minimum. - B: interval covers only half of the negative region. - D: interval covers only half of the negative region. - C: covers the full negative region, hence gives the minimum value. Therefore, the ordered pair is $$\boxed{(-\sqrt2,\sqrt2)}.$$

View written solutionFree

Correct answer: C

We need to minimize $I=int_a^b (x^4-2x^2),dx.$ Let $f(x)=x^4-2x^2=x^2(x^2-2).$
First study the sign of the integrand. We have $f(x)<0 \iff x^2-2<0 \iff |x|<\sqrt{2},$ and $f(x)=0 \text{ at } x=0,\ \pm\sqrt{2}.$ Also, $f(x)>0 \text{ for } |x|>\sqrt{2}.$
To make the definite integral minimum, we should include the entire region where the function is negative and avoid any positive part. Hence the minimizing interval should be $[a,b]=[-\sqrt{2},\sqrt{2}].$
Verify by evaluating the integral. An antiderivative is $F(x)=\frac{x^5}{5}-\frac{2x^3}{3}.$ So $I=F(b)-F(a).$ For option C: $I=\int_{-\sqrt{2}}^{\sqrt{2}}(x^4-2x^2)dx.$ Since the integrand is even, $I=2\int_0^{\sqrt{2}}(x^4-2x^2)dx.$ Now, $\int (x^4-2x^2)dx=\frac{x^5}{5}-\frac{2x^3}{3}.$ Thus $I=2\left[\frac{x^5}{5}-\frac{2x^3}{3}\right]_0^{\sqrt{2}}.$ Using $(\sqrt{2})^3=2\sqrt{2},\qquad (\sqrt{2})^5=4\sqrt{2},$ we get

=2\cdot 4\sqrt{2}\left(\frac{1}{5}-\frac{1}{3}\right) =8\sqrt{2}\left(-\frac{2}{15}\right) =-\frac{16\sqrt{2}}{15}.$$ This is the most negative possible among the given intervals. 5. Check options briefly: - A: $$\int_{\sqrt2}^{-\sqrt2}f(x)dx=-\int_{-\sqrt2}^{\sqrt2}f(x)dx>0,$$ so not minimum. - B: interval covers only half of the negative region. - D: interval covers only half of the negative region. - C: covers the full negative region, hence gives the minimum value. Therefore, the ordered pair is $$\boxed{(-\sqrt2,\sqrt2)}.$$