f(x)=x3-18x2+96xf(x)=3x2-36x+96f’(x)=03x2-36x+96=0x2-12x+32=0x2-8x-4x+32=0x(x-8)-4(x-8)=0Now we have f(0)=0,f(4)=160,f(9)=135Hence in the given interval least value is f(0)=0Option(D)

Q. 13

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Class 12th RD Sharma - Volume 1 18. Maxima and Minima

Answer :

f(x)=x³-18x²+96x

f(x)=3x²-36x+96

f’(x)=0

3x²-36x+96=0

x²-12x+32=0

x²-8x-4x+32=0

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

Now we have f(0)=0,f(4)=160,f(9)=135

Hence in the given interval least value is f(0)=0

Option(D)

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