JEE MAIN - Mathematics (2019 - 10th January Morning Slot - No. 17)
Let $${\rm I} = \int\limits_a^b {\left( {{x^4} - 2{x^2}} \right)} dx.$$ If I is minimum then the ordered pair (a, b) is -
$$\left( {\sqrt 2 , - \sqrt 2 } \right)$$
$$\left( {0,\sqrt 2 } \right)$$
$$\left( { - \sqrt 2 ,\sqrt 2 } \right)$$
$$\left( { - \sqrt 2 ,0} \right)$$
Explanation
Let f(x) = x2(x2 $$-$$ 2)
_10th_January_Morning_Slot_en_17_1.png)
As long as f(x) lie below the x-axis, define integral will remain negative,
so correct value of (a, b) is ($$-$$ $$\sqrt 2 $$, $$\sqrt 2 $$) for minimum of I
As long as f(x) lie below the x-axis, define integral will remain negative,
so correct value of (a, b) is ($$-$$ $$\sqrt 2 $$, $$\sqrt 2 $$) for minimum of I
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