Test Index

Functions Part 2

Section: Mathematics

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Question : 58 of 73

Marks: +1, -0

Let

D

be the domain of the function

f(x)= \sin^{-1} \left( \log_{3x} \left( \frac{6+2\log_3 x}{-5x} \right) \right)

. If the range of the function

g: D \rightarrow R

g(x)=x-[x],([x]

is the greatest integer function

)

(\alpha, \beta)

\alpha^2+\frac{5}{\beta}

[12-Apr-2023 shift 1]

46
135
136
45

Solution:

\frac{6+2\log_3 x}{-5x} > 0 \& x > 0 \& x \neq \frac{1}{3}

\text{ this gives } x \in \left(0, \frac{1}{27}\right) \ldots (1)

-1 \leq \log_{3x} \left( \frac{6+2\log_3 x}{-5x} \right) \leq 1

3x \leq \frac{6+2\log_3 x}{-5x} \leq \frac{1}{3x}

15x^2+6+2\log_3 x \geq 0 \quad 6+2\log_3 x+\frac{5}{3} \geq 0

x \in \left(0, \frac{1}{27}\right) \ldots (2) \quad x \geq 3^{-\frac{23}{6}} \ldots (3)

\text{ form (1), (2) \& (3) }

x \in \left[ 3^{-\frac{23}{6}}, \frac{1}{27} \right)

\therefore \alpha

is small positive quantity

\& \beta = \frac{1}{27}

\therefore \alpha^2 + \frac{5}{\beta}

is just greater than 135

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