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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