Q
Numerical
Functions
Let the domain of the function
$f(x) = \log_3\log_5\!\bigl(7-\log_2(x^2-10x+85)\bigr) + \sin^{-1}\!\left(\left|\dfrac{3x-7}{17-x}\right|\right)$
be $(\alpha,\beta]$. Then $\alpha+\beta$ is equal to :
(A) 12 (B) 8 (C) 10 (D) 9
1. Domain of nested logarithms: working inside-out
When a function contains nested (composite) logarithms like $\log_a(\log_b(\log_c(\cdots)))$, the domain is found by unwrapping from the outermost layer inward. The outermost function $\log_a$ requires its argument to be positive. That argument is the next logarithm $\log_b$, and $\log_b(A) > 0$ iff $A > 1$ (when $b > 1$). So you impose $A > 1$, which generates an inequality on the next layer. This is repeated until you reach the actual variable $x$. In this question, $\log_3(\log_5(A)) > 0$ is NOT required — only that $\log_5(A)$ is positive. Because $\log_3$ is defined for any positive argument, but $\log_3(y)$ being real only needs $y > 0$. Therefore the condition chain is: $7 – \log_2(x^2-10x+85) > 1$, which upon unwrapping gives $(x-3)(x-7) < 0$.
2. Solving $|f(x)| \le |g(x)|$ by squaring
When both sides of an inequality involve absolute values, squaring is the cleanest method. If $|A| \le |B|$, then squaring gives $A^2 \le B^2$, i.e. $A^2 – B^2 \le 0$, i.e. $(A-B)(A+B) \le 0$. This avoids splitting into multiple cases based on signs of $A$ and $B$ separately. In this problem, $|3x-7| \le |17-x|$ becomes $(3x-7)^2 \le (17-x)^2$. Expanding and simplifying yields $8x^2 – 8x – 240 \le 0$, or $x^2 – x – 30 \le 0$, which factors as $(x-6)(x+5) \le 0$, giving $-5 \le x \le 6$. The boundary values $x = -5$ and $x = 6$ are included because equality holds there. Always verify that $x = 17$ (where denominator vanishes) is not in the final domain.
3. Domain of $\sin^{-1}(f(x))$: argument must lie in $[-1,1]$
The inverse sine function $\sin^{-1}(t)$ is defined only when $-1 \le t \le 1$. When the argument involves an absolute value, the condition simplifies to $0 \le |t| \le 1$, i.e. $|t| \le 1$. This is because absolute values are always non-negative. So the effective condition is just $|t| \le 1$. For a ratio, you set $\left|\frac{3x-7}{17-x}\right| \le 1$ and solve as shown above. The result includes the equality case, meaning the boundary $x = 6$ (where the ratio equals 1) is included in the domain. Always check that the denominator is non-zero; here $17 – x \ne 0$ requires $x \ne 17$, which does not affect the domain in this problem.
4. Intersection of domains and interval notation
When a function is a sum of two terms, the overall domain is the intersection of the individual domains. Here the logarithm part requires $x \in (3,7)$ and the arcsin part requires $x \in [-5,6]$. The intersection is all $x$ that satisfy both simultaneously: $x \in (3,7) \cap [-5,6] = (3,6]$. Note that 3 is open (excluded from both domains’ reasoning) and 6 is closed (included in the arcsin domain via equality, and satisfies the log condition strictly). The domain is written as $(\alpha,\beta] = (3,6]$ in the question’s notation. Endpoint verification is always essential in domain problems: test $x = \alpha$ and $x = \beta$ explicitly to confirm which is included and which is excluded.
Q
Why is $x^2-10x+85$ always positive?
⌄
Because $(x-5)^2+60 \ge 60 > 0$ for all real $x$. The discriminant is negative.
Q
Why does $\log_3(\log_5(A))$ require $A > 1$, not just $A > 0$?
⌄
$\log_3$ needs its argument $\log_5(A) > 0$. Since $\log_5(A) > 0$ iff $A > 1$, the overall condition is $A > 1$.
Q
How does $7-\log_2(x^2-10x+85) > 1$ give $(3,7)$?
⌄
It gives $x^2-10x+85 < 64$, i.e. $x^2-10x+21 < 0$, i.e. $(x-3)(x-7) < 0$, so $x \in (3,7)$.
Q
How to handle $\left|\frac{3x-7}{17-x}\right| \le 1$?
⌄
Square both sides: $(3x-7)^2 \le (17-x)^2$, giving $x^2-x-30 \le 0$, so $-5 \le x \le 6$.
Q
What is the domain from the log part alone?
⌄
$x \in (3, 7)$ — open at both ends.
Q
What is the domain from the arcsin part alone?
⌄
$x \in [-5, 6]$ — closed at both ends.
Q
What is the final combined domain?
⌄
$(3,7) \cap [-5,6] = (3,6]$. So domain $= (\alpha,\beta] = (3,6]$.
At $x=3$, $\log_5(1) = 0$ and $\log_3(0)$ is undefined. So $x=3$ is not in the domain.
At $x=6$, the arcsin argument equals 1 (valid), and the log part is defined. So $x=6$ is included.
Q
What is $\alpha + \beta$?
⌄
$\alpha = 3$, $\beta = 6$. So $\alpha + \beta = 9$. Option (D).