Test Index

Functions Part 1

Section: Mathematics

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Question : 39 of 100

Marks: +1, -0

Let

\lfloor x \rfloor

denote the greatest integer

\leq x

x \in R

. If the domain of the real valued function

f(x)=\sqrt{\frac{\lvert \lfloor x \rfloor \rvert -2}{\lvert \lfloor x \rfloor \rvert -3}}

(-\infty, a) \cup [b, c) \cup [4, \infty), a < b < c

, then the value of

a+b+c

[20 Jul 2021 Shift 1]

8
1
$-2$
$-3$

Solution:

For domain,

\frac{\lvert \lfloor x \rfloor \rvert -2}{\lvert \lfloor x \rfloor \rvert -3} \geq 0

Case I : When

\lvert \lfloor x \rfloor \rvert - 2 \geq 0

and

\lvert \lfloor x \rfloor \rvert - 3 > 0

\therefore x \in (-\infty, -3) \cup [4, \infty)

........(1)

Case II : When

\lvert \lfloor x \rfloor \rvert - 2 \geq 0

and

\lvert \lfloor x \rfloor \rvert - 3 < 0

\therefore x \in [-2,3)

.......(2)

So, from (1) and (2)

we get

Domain of function

= (-\infty, -3) \cup [-2,3) \cup [4, \infty)

\therefore (a+b+c) = -3 + (-2) + 3

= -2 (a < b < c)

\Rightarrow

Option (3) is correct.

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