Check sibling questions

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

Slide94.JPG

Slide95.JPG
Slide96.JPG Slide97.JPG

This video is only available for Teachoo black users

This video is only available for Teachoo black users

Get live Maths 1-on-1 Classs - Class 6 to 12


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)

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.