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Question 19 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Dec. 4, 2021 by

y = x (x – 3) 2 decreases for the values of x given by :

(A)1 < x < 3 Β 

(B) x < 0

(C) x > 0 Β 

(D) 0 < x < 3/2

Β 

This question is similar to Ex 6.2, 6 - Chapter 6 Class 12 - Application of Derivatives

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Transcript

Question 19 y = x (x – 3)2 decreases for the values of x given by : 1 < x < 3 (B) x < 0 (C) x > 0 (D) 0 < x < 3/2 𝑓(π‘₯)=π‘₯γ€–(π‘₯βˆ’3)γ€—^2 Calculating f’(𝒙) f’(π‘₯) = (π‘₯βˆ’3)^2+π‘₯Γ—2(π‘₯βˆ’3) f’(π‘₯) = (π‘₯βˆ’3) [(π‘₯βˆ’3)+2π‘₯] f’(π‘₯) = (π‘₯βˆ’3) (3π‘₯βˆ’3) f’(π‘₯) = (π‘₯βˆ’3) Γ—3(π‘₯βˆ’1) f’(𝒙) = πŸ‘(π’™βˆ’πŸ‘) (π’™βˆ’πŸ) Putting f’(𝒙) = 0 3(π‘₯βˆ’3) (π‘₯βˆ’1)=0 So, 𝒙 = 3 , 1 Plotting points on number line Hence, f is decreasing for the 𝟏<𝒙<πŸ‘. So, the correct answer is (A)

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