1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise
  3. NCERT Exemplar - MCQs

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

Last updated at Dec. 4, 2021 by

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

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  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

    3. NCERT Exemplar - MCQs

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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’‡(๐‘ฅ)=๐‘ฅ^3โˆ’18๐‘ฅ^2+96๐‘ฅ Finding ๐’‡โ€™(x) ๐’‡โ€ฒ(๐’™)=ใ€–3๐‘ฅใ€—^2โˆ’36๐‘ฅ+96 ๐‘“โ€ฒ(๐‘ฅ)=๐Ÿ‘(๐’™^๐Ÿโˆ’๐Ÿ๐Ÿ๐’™+๐Ÿ‘๐Ÿ) Putting ๐’‡โ€™(๐’™)=๐ŸŽ 3(๐‘ฅ^2โˆ’12๐‘ฅ+32)=0 ๐‘ฅ^2โˆ’12๐‘ฅ+32 = 0 ๐‘ฅ^2โˆ’8๐‘ฅโˆ’4๐‘ฅ+32=0 ๐‘ฅ(๐‘ฅโˆ’8)โˆ’4(๐‘ฅโˆ’8)=0 (๐‘ฅโˆ’4)(๐‘ฅโˆ’8)=0 So, ๐’™=๐Ÿ’, ๐Ÿ– Since, ๐’™ โˆˆ [๐ŸŽ , ๐Ÿ—] Hence , calculating ๐’‡(๐’™) at ๐’™=๐ŸŽ , ๐Ÿ’ , ๐Ÿ– , ๐Ÿ— ๐’‡(๐Ÿ’) =(4)^3โˆ’18(4)^2+96(4) = 64 โ€“ 18 ร— 16 + 96 ร— 4 = 160 ๐‘“(8) =(8)^3โˆ’18(8)^2+96(8) = 512 โ€“ 18 ร— 64 + 768 =128 ๐’‡(๐Ÿ—) =(9)^3โˆ’18(9)^2+96(9) = 729 โ€“ 18 ร— 81 + 864 =135 Hence, Minimum value of ๐‘“(๐‘ฅ) is 0 at ๐’™ = 0 So, the correct answer is (B)

Chapter 6 Class 12 Application of Derivatives (Term 1)
Serial order wise

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