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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 : 60 of 120

Marks: +1, -0

What is the value of b?

Solution:

Concept:

For a function to be continuous at a point, the left-hand limit, right-hand limit, and the function value at that point must all be equal.

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}

is continuous at

x = 3

For

x < 3

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

Thus left-hand limit:

\lim_{x \to 3^{-}} f(x) = \lim_{x \to 3^{-}} \left( \frac{x-3}{-(x-3)} + a \right) = -1 + a

For

x > 3

|x-3| = (x-3)

Thus right-hand limit:

\lim_{x \to 3^{+}} f(x) = \lim_{x \to 3^{+}} \left( \frac{x-3}{x-3} + b \right) = 1 + b

x = 3

f(3) = a - b

Continuity gives:

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

From

-1 + a = a - b

we get

-1 = -b

, so

b = 1

(Check:

1 + b = a - b

gives

1 + 1 = a - 1

a = 3

, consistent.)

Answer:

b = 1

(Option B).

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