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

Last updated at March 22, 2023 by

The smallest value of the polynomial x 3 – 18x 2 + 96x in [0, 9] is

(A)126Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  (B) 0

(C) 135Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  (D) 160

Β 

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

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Transcript

Question 24 The smallest value of the polynomial x3 – 18x2 + 96x in [0, 9] is 126 (B) 0 (C) 135 (D) 160 𝒇(π‘₯)=π‘₯^3βˆ’18π‘₯^2+96π‘₯ Finding 𝒇’(x) 𝒇′(𝒙)=γ€–3π‘₯γ€—^2βˆ’36π‘₯+96 𝑓′(π‘₯)=πŸ‘(𝒙^πŸβˆ’πŸπŸπ’™+πŸ‘πŸ) Putting 𝒇’(𝒙)=𝟎 3(π‘₯^2βˆ’12π‘₯+32)=0 π‘₯^2βˆ’12π‘₯+32 = 0 π‘₯^2βˆ’8π‘₯βˆ’4π‘₯+32=0 π‘₯(π‘₯βˆ’8)βˆ’4(π‘₯βˆ’8)=0 (π‘₯βˆ’4)(π‘₯βˆ’8)=0 So, 𝒙=πŸ’, πŸ– Since, 𝒙 ∈ [𝟎 , πŸ—] Hence , calculating 𝒇(𝒙) at 𝒙=𝟎 , πŸ’ , πŸ– , πŸ— 𝒇(πŸ’) =(4)^3βˆ’18(4)^2+96(4) = 64 – 18 Γ— 16 + 96 Γ— 4 = 160 𝑓(8) =(8)^3βˆ’18(8)^2+96(8) = 512 – 18 Γ— 64 + 768 =128 𝒇(πŸ—) =(9)^3βˆ’18(9)^2+96(9) = 729 – 18 Γ— 81 + 864 =135 Hence, Minimum value of 𝑓(π‘₯) is 0 at 𝒙 = 0 So, the correct answer is (B)

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