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Gate 2016 EE (set 1) Q1

Gate 2016 EE (set 1) Q1

1. The maximum value attained by the function f(x)=x(x-1)(x-2) in the interval f[1, 2] is :--------

Solution:
 f(x)=x(x-1)(x-2)
\Rightarrow f(x)=x^3-3x^2+2
\Rightarrow f^1(x)=3x^2-6x+2=0
 \Rightarrow x=1\pm \frac{1}{\sqrt 3}
But
 x=1+\frac{1}{\sqrt 3} only lies on the interval [1, 2]
At  x=1+\frac{1}{\sqrt 3},
 f^{11}(x)=6x-6=6(1+\frac{1}{\sqrt 3})-6>0
 (1+\frac{1}{\sqrt 3}) is a point of minimum
 f(x)=x(x-1)(x-2)=0 at either ends x=1 \& x=2
Maximum value=0

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