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Functions

Section: Mathematics

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Question : 1 of 33

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\lfloor t \rfloor

denotes the greatest integer function and

\lfloor t - m \rfloor = \lfloor t \rfloor - m

m \in \mathbb{Z}

k = 2\lfloor 2x - 1 \rfloor - 1

3\lfloor 2x - 2 \rfloor + 1 = 2\lfloor 2x - 1 \rfloor - 1

then the range of

f(x) = \lfloor k + 5x \rfloor

[AP EAPCET 23 May 2025 S2]

$\{7,8,9\}$
$\{4,5,6\}$
$\{5,6,7\}$
$\{6,7,8\}$

Solution:

\because \lfloor 2x - 2 \rfloor = \lfloor (2x - 1) - 1 \rfloor

= \lfloor 2x - 1 \rfloor - 1

\text{ let, } \lfloor 2x - 1 \rfloor = a

\Rightarrow \lfloor 2x - 2 \rfloor = a - 1

\text{ so, } 3(a-1) + 1 = 2a - 1

\Rightarrow 3a - 3 + 1 = 2a - 1 \Rightarrow a = 1

thus

\lfloor 2x - 1 \rfloor = 1

\Rightarrow 1 \leq (2x - 1) < 2 \Rightarrow 2 \leq 2x < 3

\Rightarrow 1 \leq x < 1.5

so,

x \in [1, 1.5)

Now,

k = 2\lfloor 2x - 1 \rfloor - 1 = 2(1) - 1 = 1

for

x \in [1, 1.5)

f(x) = \lfloor k + 5x \rfloor = \lfloor 1 + 5x \rfloor

\text{ at } x=1, f(x) = \lfloor 6 \rfloor = 6 \quad (\text{ minimum })

\text{ at } x=1.5, f(x) = \lfloor 1 + 5 \times 1.5 \rfloor

= \lfloor 8.5 \rfloor = 8 \quad \text{ (maximum) }

so,

f(x)

can be

6, 7, 8

Hence, the range of

f(x)=\{6,7,8\}

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