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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 59-60

Consider the following for the next items that follow:

Let

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

and f(x) be continuous at x = 3.

Question : 59 of 120

Marks: +1, -0

What is the value of a ?

Solution:

Concept:

For a function to be continuous at

x=3

, the left-hand limit must equal the right-hand limit and the function value at that point.

Explanation:

Given

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

and

f

is continuous at

x=3

For

x<3

|x-3|=-(x-3)

, so

\frac{x-3}{|x-3|}=-1

. Hence

\text{LHL} = -1 + a

For

x>3

|x-3|=x-3

, so

\frac{x-3}{|x-3|}=1

. Hence

\text{RHL} = 1 + b

x=3

f(3)=a-b

Continuity gives

\text{LHL} = \text{RHL} = f(3)

Thus

-1+a = 1+b = a-b

From

-1+a = a-b

, we get

-1 = -b

, so

b=1

Substitute into

1+b = a-b

1+1 = a-1

⇒

2 = a-1

⇒

a=3

Answer:

a=3

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