# 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