Let the domain of the function $f(x) = \log_3 \log_5 \log_7 (9x - x^2 - 13)$ be the interval $(m, n)$. Let the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ have eccentricity $\frac{n}{3}$ and the length of the latus rectum $\frac{8m}{3}$. Then $b^2 - a^2$ is equal to :
A) 7 B) 9 C) 11 D) 5
A) 7 B) 9 C) 11 D) 5
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Solution Steps
1
Determine the domain $(m, n)$
For $f(x)$ to be defined, the argument of the outermost log must be positive:
$\log_5 \log_7 (9x - x^2 - 13) > 0$
$\log_7 (9x - x^2 - 13) > 5^0 \implies \log_7 (9x - x^2 - 13) > 1$
$9x - x^2 - 13 > 7^1 \implies 9x - x^2 - 13 > 7$
$x^2 - 9x + 20 < 0 \implies (x-4)(x-5) < 0$
Thus, $x \in (4, 5)$, which gives $m = 4$ and $n = 5$.
2
Set up Hyperbola properties
Eccentricity $e = \frac{n}{3} = \frac{5}{3}$.
Length of latus rectum $L = \frac{8m}{3} = \frac{8 \times 4}{3} = \frac{32}{3}$.
3
Use Latus Rectum formula
Length of latus rectum $= \frac{2b^2}{a} = \frac{32}{3}$.
$b^2 = \frac{16a}{3} \quad \text{---(Equation 1)}$
4
Use Eccentricity formula
$e^2 = 1 + \frac{b^2}{a^2} \implies (\frac{5}{3})^2 = 1 + \frac{b^2}{a^2}$
$\frac{25}{9} = 1 + \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = \frac{16}{9}$
$b^2 = \frac{16a^2}{9} \quad \text{---(Equation 2)}$
5
Solve for $a$ and $b$
Equating $b^2$ from (1) and (2):
$\frac{16a}{3} = \frac{16a^2}{9} \implies \frac{1}{3} = \frac{a}{9} \implies a = 3$.
Substitute $a=3$ into Eq 1: $b^2 = \frac{16 \times 3}{3} = 16 \implies b = 4$.
6
Final Calculation
$b^2 - a^2 = 16 - 3^2 = 16 - 9 = 7$.
Final Answer: 7 (Option A)
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Theory
1. Domain of Logarithmic Functions
The domain of a function $y = \log_b(g(x))$ consists of all values of $x$ for which $g(x) > 0$. When logarithms are nested, like $f(x) = \log(\log(\log(u)))$, the condition must be applied starting from the outermost log and working inwards. Each successive argument must be greater than zero. For a base $b > 1$, the inequality $\log_b(u) > k$ is equivalent to $u > b^k$. In this problem, we sequentially solve three layers of inequalities to find the range $(m, n)$.
2. Standard Hyperbola Properties
A standard hyperbola with its transverse axis along the x-axis is represented by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Here, $2a$ is the length of the transverse axis and $2b$ is the length of the conjugate axis. Key parameters include the coordinates of the foci $(\pm ae, 0)$ and the vertices $(\pm a, 0)$. The latus rectum is a chord perpendicular to the transverse axis passing through the foci, with a total length given by $2b^2/a$.
3. Hyperbola Eccentricity
Eccentricity ($e$) measures how "open" the hyperbola is. For a hyperbola, $e$ is always greater than 1. It represents the ratio of the distance from the center to a focus to the distance from the center to a vertex ($e = c/a$). The relationship between the semi-axes and eccentricity is expressed as $b^2 = a^2(e^2 - 1)$, which can be rearranged to $e = \sqrt{1 + (b/a)^2}$. Solving for $a$ and $b$ often requires using this relation in conjunction with the latus rectum length.
4. Quadratic Inequalities
Finding the domain often leads to a quadratic inequality of the form $ax^2 + bx + c < 0$. To solve this, one finds the roots of the corresponding quadratic equation $ax^2 + bx + c = 0$. If the roots are $\alpha$ and $\beta$ (with $\alpha < \beta$), the solution for a "less than zero" inequality with a positive $x^2$ coefficient is the interval $(\alpha, \beta)$. This "between the roots" principle is a fundamental tool in JEE calculus and coordinate geometry crossover problems.
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FAQs
1
Why did we only solve the outermost log inequality?
⌄
Actually, solving $\log_5 \log_7 (u) > 0$ automatically implies $\log_7(u) > 1$ (which is > 0) and $u > 7$ (which is > 0). The most restrictive condition usually comes from the outermost layer when the base is greater than 1.
2
What if the base of the log was less than 1?
⌄
If the base $0 < b < 1$, the inequality sign flips. For example, $\log_{0.5}(x) > 1 \implies x < (0.5)^1$.
3
What is the physical meaning of the latus rectum?
⌄
It is the chord passing through the focus and perpendicular to the major (or transverse) axis. Its length helps define the curvature of the conic section.
4
How did we get m=4 and n=5?
⌄
By solving $(x-4)(x-5) < 0$, we found the interval to be (4, 5). Comparing this with the given (m, n) identifies m as 4 and n as 5.
5
Is eccentricity always greater than 1 for a hyperbola?
⌄
Yes. For a circle $e=0$, for an ellipse $0 < e < 1$, for a parabola $e=1$, and for a hyperbola $e > 1$.
6
Can b² be less than a² in a hyperbola?
⌄
Yes, unlike an ellipse where $a^2$ must be the largest, in a hyperbola $b^2$ can be smaller, equal to, or larger than $a^2$.
7
What is a rectangular hyperbola?
⌄
It is a hyperbola where $a = b$. In that case, the eccentricity is always $\sqrt{2}$.
8
What if the hyperbola was conjugate?
⌄
If the transverse axis was along the y-axis, the equation would be $y^2/b^2 - x^2/a^2 = 1$ and the eccentricity would be $e = \sqrt{1 + a^2/b^2}$.
9
Why did we ignore the outermost log₃?
⌄
The domain of $\log_3(U)$ is $U > 0$. Here $U = \log_5 \log_7 (...)$. That's why we set that entire inner expression to be greater than 0.
10
Are m and n always integers in these problems?
⌄
Not necessarily, but in competitive exams like JEE, they are often integers or simple fractions to make the following coordinate geometry parts solvable.
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