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UPSC NDA II 2023 Mathematics Solved Paper

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Consider the following for the next two (02) items that follow :

Let

f(x)=\begin{cases} \frac{x-3}{|x-3|}+a, & x<3 \\ a-b, & x=3 \text{ and } \\ \frac{x-3}{|x-3|}+b, & x>3 \end{cases}.

f(x)

be continuous at

x=3

.

[NDA II 2023]

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Question : 60 of 120

Marks: +1, -0

What is the value of

b

-1
1
2
3

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