Test Index

UPSC NDA I 2015 Mathematics Solved Paper

For the next two (2) items that follow :

Given function

f(x)=\begin{cases} -1, & \text{if } x \leq 0 \\ ax+b, & \text{if } 0 < x < 1 \\ 1, & \text{if } x \geq 1 \end{cases}

where a, b are constants.The function is continuous everywhere.

[NDA I 2015]

Question : 67 of 120

Marks: +1, -0

Solution:

For the given function to be continuous on R, left hand limit should be equal to right hand limit at 0 and 1.

Therefore

\lim\limits_{x\to 0^+} f(x)

\lim\limits_{x\to 0^+} (ax+b)

=b

\lim\limits_{x\to 0^+} f(x)

=\lim\limits_{x\to 0^+} (-1)

=-1

LHL = RHL

b=-1

Similarly applying for x = 1 we get

\lim\limits_{x\to 1^-} f(x)

\lim\limits_{x\to 1^+} (ax+b)

=a+b

\lim\limits_{x\to 1^+} f(x)

=1

LHL = RHL

a+b=1

a+(-1)=1

a=2

Go to Question:

Prev Question

Next Question