NCERT Exemplar - MCQs

Chapter 6 Class 12 Application of Derivatives
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## (C) 135Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  (D) 160

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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 β 18x2 + 96x in [0, 9] is 126 (B) 0 (C) 135 (D) 160 π(π₯)=π₯^3β18π₯^2+96π₯ Finding πβ(x) πβ²(π)=γ3π₯γ^2β36π₯+96 πβ²(π₯)=π(π^πβπππ+ππ) Putting πβ(π)=π 3(π₯^2β12π₯+32)=0 π₯^2β12π₯+32 = 0 π₯^2β8π₯β4π₯+32=0 π₯(π₯β8)β4(π₯β8)=0 (π₯β4)(π₯β8)=0 So, π=π, π Since, π β [π , π] Hence , calculating π(π) at π=π , π , π , π π(π) =(4)^3β18(4)^2+96(4) = 64 β 18 Γ 16 + 96 Γ 4 = 160 π(8) =(8)^3β18(8)^2+96(8) = 512 β 18 Γ 64 + 768 =128 π(π) =(9)^3β18(9)^2+96(9) = 729 β 18 Γ 81 + 864 =135 Hence, Minimum value of π(π₯) is 0 at π = 0 So, the correct answer is (B)