Question 24 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Dec. 4, 2021 by Teachoo

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The smallest value of the polynomial x
^{
3
}
β 18x
^{
2
}
+ 96x in [0, 9] is

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Question 24
The smallest value of the polynomial x3 β 18x2 + 96x in [0, 9] is
126 (B) 0
(C) 135 (D) 160
π(π₯)=π₯^3β18π₯^2+96π₯
Finding πβ(x)
πβ²(π)=γ3π₯γ^2β36π₯+96
πβ²(π₯)=π(π^πβπππ+ππ)
Putting πβ(π)=π
3(π₯^2β12π₯+32)=0
π₯^2β12π₯+32 = 0
π₯^2β8π₯β4π₯+32=0
π₯(π₯β8)β4(π₯β8)=0
(π₯β4)(π₯β8)=0
So, π=π, π
Since, π β [π , π]
Hence , calculating π(π) at π=π , π , π , π
π(π) =(4)^3β18(4)^2+96(4)
= 64 β 18 Γ 16 + 96 Γ 4
= 160
π(8) =(8)^3β18(8)^2+96(8)
= 512 β 18 Γ 64 + 768
=128
π(π) =(9)^3β18(9)^2+96(9)
= 729 β 18 Γ 81 + 864
=135
Hence, Minimum value of π(π₯) is 0 at π = 0
So, the correct answer is (B)

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