Let D be the domain of the function f(x) = sin⁻¹(log₃ₓ((6 + 2 log₃ x)/-5x)). If the range of the fun

Q: Wait, why do we reject the negative root or flip the inequality sign here?

Great question! This is the most common trap in JEE logarithmic inequalities. When you have an inequality like $\log_b(v) \le 1$, you exponentiate both sides to get $v \le b^1$. However, if the base $b$ is between 0 and 1 (which is the case here, as $3x < 1$), the function $f(t) = \log_b(t)$ is a decreasing function. Therefore, the inequality sign must reverse. Always check the base!

Q: Why is the greatest integer function $[x]$ equal to 0?

Look closely at the domain we derived. We found $x < \frac{1}{27}$. Since $\frac{1}{27}$ is approximately $0.037$, all values of $x$ in our domain $D$ are strictly between 0 and 1. By definition, for any $x \in (0, 1)$, the greatest integer $[x]$ is 0. This simplifies our function $g(x) = x - [x]$ to just $g(x) = x$.

Q: Why can we just ignore the $\alpha^2$ term in the final calculation?

In JEE Advanced, we often look for the most significant term when the options are integers. We calculated $\alpha^2 = 3^{-\frac{23}{3}}$, which is $3^{-7.66}$. This is a tiny fraction, approximately $0.0002$. When added to $135$, it doesn't change the integer value. In competitive exams, if you are confident in your integer part, you can safely treat such negligible terms as zero.

Question

Let $D$ be the domain of the function $f(x) = \sin^{-1} \left(\log_{3x} \left(\frac{6 + 2 \log_3 x}{-5x}ight)ight)$. If the range of the function $g : D 	o \mathbb{R}$ defined by $g(x) = x - [x]$, ($[x]$ is the greatest integer function), is $(\alpha, \beta)$, then $\alpha^2 + \frac{5}{\beta}$ is equal to

Options:

(A) 135

(B) 45

(C) 46

(D) 136

Sigma Percentile · Accepted Answer

### Analyzing the Setup The function provided is $f(x) = \sin^{-1} \left( \log_{3x} \left( \frac{6 + 2 \log_3 x}{-5x} ight) ight)$. To find its domain, we must peel back the layers of this composite function, starting from the outermost constraint. The domain of the inverse sine function, $\sin^{-1}(u)$, is strictly defined as $u \in [-1, 1]$. Therefore, the entire logarithmic expression must satisfy: $$-1 \le \log_{3x} \left( \frac{6 + 2 \log_3 x}{-5x} ight) \le 1$$ ### The Logarithmic Trap For the logarithm $\log_{3x}(v)$ to be defined, the base $3x$ must satisfy $3x > 0$ and $3x eq 1$. This implies $x > 0$ and $x eq \frac{1}{3}$. Furthermore, the argument $v = \frac{6 + 2 \log_3 x}{-5x}$ must be strictly positive. Since $x > 0$, the denominator $-5x$ is always negative. For the fraction to be positive, the numerator must also be negative: $$6 + 2 \log_3 x < 0 \Rightarrow \log_3 x < -3 \Rightarrow x < 3^{-3} = \frac{1}{27}$$ Thus, our initial domain constraint is $x \in (0, \frac{1}{27})$. ### The Inequality Dance We now solve $-1 \le \log_{3x}(v) \le 1$. Since $x < \frac{1}{27}$, the base $3x < \frac{1}{9}$. Because the base is between $0$ and $1$, the logarithmic function is strictly decreasing. When we remove the logarithm, the inequality signs must flip: $$(3x)^1 \le \frac{6 + 2 \log_3 x}{-5x} \le (3x)^{-1}$$ Focusing on the right-hand inequality: $$\frac{6 + 2 \log_3 x}{-5x} \le \frac{1}{3x}$$ Multiplying both sides by $3x$ (which is positive), we obtain: $$\frac{18 + 6 \log_3 x}{-5} \le 1 \Rightarrow 18 + 6 \log_3 x \ge -5$$ Solving for $x$: $$6 \log_3 x \ge -23 \Rightarrow \log_3 x \ge -\frac{23}{6} \Rightarrow x \ge 3^{-\frac{23}{6}}$$ ### The Fractional Part Revelation The domain of the function is $D = [3^{-\frac{23}{6}}, \frac{1}{27})$. We are asked to evaluate properties related to the fractional part function $g(x) = x - [x]$. Since all $x \in D$ satisfy $0 < x < \frac{1}{27}$, the greatest integer part $[x]$ is always $0$. Consequently, $g(x) = x$ for all $x$ in the domain. We identify the boundaries as $\alpha = 3^{-\frac{23}{6}}$ and $\beta = \frac{1}{27}$. ### Final Calculation We are tasked with calculating the value of $\alpha^2 + \frac{5}{\beta}$. Substituting our identified values: $$\alpha^2 = (3^{-\frac{23}{6}})^2 = 3^{-\frac{23}{3}}$$ $$\frac{5}{\beta} = 5 \div \frac{1}{27} = 5 imes 27 = 135$$ The final result is: $$\mathbf{135 + 3^{-\frac{23}{3}}}$$

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Visualized Solution

$C o n s t r ain t f or sin^{- 1}$

$L o g A r g u m e n tP os i t i v i t y$

$D e d u c in g t h e N u m er a t o r^{'} s S i g n$

$S o l v in g f or t h e U pp er B o u n d$

$A na l y z in g t h e L o g B a se$

$A ppl y in g t h e B a se P r o p er t y$

$S o l v in g t h e R i g h t I n e q u a l i t y$

$F in d in g t h e L o w er B o u n d$

$D e f inin g t h eD o main D$

$A na l y z in g t h e F u n c t i o n g (x)$

$R an g eo f g (x)$

$F ina l C a l c u l a t i o n S e t u p$

$A pp r o x ima t i o nan d F ina l A n s w er$

The Sigma Insight: Domain and Range of Inverse Trigonometric Functions

Analyzing the Setup

The Logarithmic Trap

The Inequality Dance

The Fractional Part Revelation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Constraintforsin−1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LogArgumentPositivity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DeducingtheNumerator′sSign

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SolvingfortheUpperBound

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AnalyzingtheLogBase

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ApplyingtheBaseProperty

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SolvingtheRightInequality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingtheLowerBound

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DefiningtheDomainD

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AnalyzingtheFunctiong(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rangeofg(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalCalculationSetup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ApproximationandFinalAnswer