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  • Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise Multiple choice question, question 13.

Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise Multiple choice question, question 13.

Answers (1)

Answer: option(d) 0

Hint: For local maxima or minima, we must have f'(x)=0.

Given: f(x)=x^3-18x^2+96x

Solution:

We have,

f(x)=x^3-18x^2+96x

\Rightarrow f'(x)=3x^2-36x+96                                   

For maxima and minima f'(x)=0

\Rightarrow 3x^2-36x+96=0

\Rightarrow x^2-12x+32=0

\Rightarrow (x-4)(x-8)=0

\Rightarrow x=4,8

At x = 4, f(4)=(4)^3-18(4)^2+96(4)=64-288+384=160

At x = 8, f(4)=(8)^3-18(8)^2+96(8)=512-1152+768=128

Now at the extreme points of [0, 9]

f(0)=(0)^3-18(0)^2+96(0)=0

f(9)=(9)^3-18(9)^2+96(9)=729-1458=864=135

Hence, 0 is the minimum value in the range [0, 9].

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