Question 24 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at Dec. 4, 2021 by Teachoo

The smallest value of the polynomial x
^{
3
}
– 18x
^{
2
}
+ 96x in [0, 9] is
(A)126 (B) 0
(C) 135 (D) 160

This question is similar to
Ex 6.5, 7 - Chapter 6 Class 12
- Application of Derivatives

Transcript

Question 24
The smallest value of the polynomial x3 โ 18x2 + 96x in [0, 9] is
126 (B) 0
(C) 135 (D) 160
๐(๐ฅ)=๐ฅ^3โ18๐ฅ^2+96๐ฅ
Finding ๐โ(x)
๐โฒ(๐)=ใ3๐ฅใ^2โ36๐ฅ+96
๐โฒ(๐ฅ)=๐(๐^๐โ๐๐๐+๐๐)
Putting ๐โ(๐)=๐
3(๐ฅ^2โ12๐ฅ+32)=0
๐ฅ^2โ12๐ฅ+32 = 0
๐ฅ^2โ8๐ฅโ4๐ฅ+32=0
๐ฅ(๐ฅโ8)โ4(๐ฅโ8)=0
(๐ฅโ4)(๐ฅโ8)=0
So, ๐=๐, ๐
Since, ๐ โ [๐ , ๐]
Hence , calculating ๐(๐) at ๐=๐ , ๐ , ๐ , ๐
๐(๐) =(4)^3โ18(4)^2+96(4)
= 64 โ 18 ร 16 + 96 ร 4
= 160
๐(8) =(8)^3โ18(8)^2+96(8)
= 512 โ 18 ร 64 + 768
=128
๐(๐) =(9)^3โ18(9)^2+96(9)
= 729 โ 18 ร 81 + 864
=135
Hence, Minimum value of ๐(๐ฅ) is 0 at ๐ = 0
So, the correct answer is (B)

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