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UPSC NDA I 2016 Mathematics Solved Paper

For the next five items that follow :

Consider the function

f(x)=\left|x-1\right|+x^2

where

x\in\mathbb{R}

[NDA I 2016]

Question : 20 of 120

Marks: +1, -0

Solution:

f(x)

has local minimum at one point only

(-\infty,\infty)

f'(x)=\begin{cases}2x-1,&x<1\\2x+1,&x\ge1\end{cases}

Clearly, for

(x>1);f'(x)>0\ge\&\;\text{for}\;(x<1)

x=\frac{1}{2}

is the point of local minima

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