UPSC NDA Math Model Paper 7

Consider the function $f(x)=3x^{4}-2x^{3}-6x^{2}+6x+1$ and differentiate it with respect to $x$ $f'(x)=12x^{3}-6x^{2}-12x+6$ Equating it to 0 we get, $12x^{3}-6x^{2}-12x+6=0$ $2x^{3}-x^{2}-2x+1=0$ $(x-1)(2x^{2}+x-1)=0$ $(x-1)(2x-1)(x+1)=0$ Therefore, the critical points are $x=1, x=\frac{1}{2}$ and $x=-1$. Now double derivative is given by: $f''(x)=6x^{2}-2x-2$ Now the function will have local maxima when $f''(x) < 0$ At $x=1$ the value of double derivative is: $f''(1)=2$ At $x=-1$ the value of double derivative is: $f''(-1)=6$ At $x=\frac{1}{2}$ the value of double derivative is: $f''\left(\frac{1}{2}\right)=-\frac{3}{2}$ Therefore, at $x=\frac{1}{2}$ the function has maxima.