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UPSC NDA I 2015 Mathematics Paper

Question Numbers: 67-68

Given a function

f(x)=\begin{cases} -1 & if & x\le 0 \\ ax+b & if & 0 < x<1 \\ 1 & if & x\ge 1 \end{cases}

; Where a, b are constants. The function continuous everywhere.

Question : 67 of 120

Marks: +1, -0

What is the value of a?

Solution:

Concept:

A function

f(x)

is continuous at a point if its left-hand limit, right-hand limit, and the function value at that point are all equal.

Explanation:

The given piecewise function is

f(x) = \begin{cases} -1 & x \le 0 \\ ax+b & 0 < x < 1 \\ 1 & x \ge 1 \end{cases}

Since

f

is continuous everywhere, we check the boundaries

x=0

and

x=1

x=0

\lim_{x \to 0^-} f(x) = -1

and

\lim_{x \to 0^+} f(x) = a(0)+b = b

Equating gives

b = -1

x=1

\lim_{x \to 1^-} f(x) = a(1)+b = a+b

and

\lim_{x \to 1^+} f(x) = 1

Equating gives

a+b = 1

Substitute

b = -1

into

a+b = 1

a - 1 = 1

, so

a = 2

Answer:

a = 2

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