JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 9)
If S1 and S2 are respectively the sets of local
minimum and local maximum points of the function,
ƒ(x) = 9x4 + 12x3 – 36x2 + 25, x $$ \in $$ R, then :
ƒ(x) = 9x4 + 12x3 – 36x2 + 25, x $$ \in $$ R, then :
S1 = {–1}; S2 = {0, 2}
S1 = {–2}; S2 = {0, 1}
S1 = {–2, 0}; S2 = {1}
S1 = {–2, 1}; S2 = {0}
Explanation
ƒ(x) = 9x4 + 12x3 – 36x2 + 25
ƒ'(x) = 36x3 + 36x2 – 72x
ƒ'(x) = 36x(x2 + x – 2)
ƒ'(x) = 36x(x + 2)(x - 1)_8th_April_Morning_Slot_en_9_2.png)
While moving left to right on x-axis whenever derivative changes sign from negative to positive, we get local minima, and whenever derivative changes sign from positive to negative, we get local maxima.
$$ \therefore $$ S1 = {–2, 1}
S2 = {0}
ƒ'(x) = 36x3 + 36x2 – 72x
ƒ'(x) = 36x(x2 + x – 2)
ƒ'(x) = 36x(x + 2)(x - 1)
While moving left to right on x-axis whenever derivative changes sign from negative to positive, we get local minima, and whenever derivative changes sign from positive to negative, we get local maxima.
$$ \therefore $$ S1 = {–2, 1}
S2 = {0}
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