1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
  3. NCERT Exemplar - MCQs
Check sibling questions

Question 11 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 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 11 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)
  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

    3. NCERT Exemplar - MCQs

    Tangents and Normals (using Differentiation) →
Tangents and Normals (using Differentiation) →
Chapter 6 Class 12 Application of Derivatives
Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo