NCERT Exemplar - MCQs
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Last updated at August 8, 2026 by Teachoo
This question is similar to Ex 6.5, 7 - Chapter 6 Class 12 - Application of Derivatives
![MCQ - The smallest value of polynomial x3 – 18x2 + 96x in [0, 9] is - NCERT Exemplar - MCQs](https://cdn.teachoo.com/320ac995-5bf3-4f61-82a1-19eda2d30e0e/slide94.jpg)
Transcript
Question 11 The smallest value of the polynomial x3 ā 18x2 + 96x in [0, 9] is 126 (B) 0 (C) 135 (D) 160 š(š„)=š„^3ā18š„^2+96š„ Finding šā(x) šā²(š)=ć3š„ć^2ā36š„+96 šā²(š„)=š(š^šāššš+šš) Putting šā(š)=š 3(š„^2ā12š„+32)=0 š„^2ā12š„+32 = 0 š„^2ā8š„ā4š„+32=0 š„(š„ā8)ā4(š„ā8)=0 (š„ā4)(š„ā8)=0 So, š=š, š Since, š ā [š , š] Hence , calculating š(š) at š=š , š , š , š š(š) =(4)^3ā18(4)^2+96(4) = 64 ā 18 Ć 16 + 96 Ć 4 = 160 š(8) =(8)^3ā18(8)^2+96(8) = 512 ā 18 Ć 64 + 768 =128 š(š) =(9)^3ā18(9)^2+96(9) = 729 ā 18 Ć 81 + 864 =135 Hence, Minimum value of š(š„) is 0 at š = 0 So, the correct answer is (B)
