Functions – Domain and Range – JEE Main 22 Jan 2026 Shift 2

Solution:

For the first term, $\log_{3}\log_{5}(7-\log_{2}(x^{2}-10x+85))$ to be defined:
1. $x^2 – 10x + 85 > 0$ (Always true as $D < 0$)
2. $7 – \log_2(x^2 – 10x + 85) > 0 \Rightarrow \log_2(x^2 – 10x + 85) < 7$ $$ x^2 - 10x + 85 < 2^7 = 128 $$ $$ x^2 - 10x - 43 < 0 $$ Roots are $\frac{10 \pm \sqrt{100 + 172}}{2} = 5 \pm \sqrt{68}$. Approx $( -3.2, 13.2 )$. 3. $\log_5(7 - \log_2(x^2 - 10x + 85)) > 0$
$$ 7 – \log_2(x^2 – 10x + 85) > 5^0 = 1 $$
$$ \log_2(x^2 – 10x + 85) < 6 $$ $$ x^2 - 10x + 85 < 64 $$ $$ x^2 - 10x + 21 < 0 $$ $$ (x-3)(x-7) < 0 \Rightarrow x \in (3, 7) $$
For the second term, $\sin^{-1}(|\frac{3x-7}{17-x}|)$ to be defined:
$$ -1 \le \frac{3x-7}{17-x} \le 1 $$
Since we are looking for intersection with $(3, 7)$, $17-x$ is positive.
$$ |\frac{3x-7}{17-x}| \le 1 \Rightarrow |3x-7| \le |17-x| $$
Squaring both sides (or splitting cases):
$$ (3x-7)^2 \le (17-x)^2 $$
$$ 9x^2 – 42x + 49 \le 289 – 34x + x^2 $$
$$ 8x^2 – 8x – 240 \le 0 $$
$$ x^2 – x – 30 \le 0 $$
$$ (x-6)(x+5) \le 0 \Rightarrow x \in [-5, 6] $$

Intersection of $(3, 7)$ and $[-5, 6]$ is $(3, 6]$.
Thus, $\alpha = 3, \beta = 6$.
$$ \alpha + \beta = 3 + 6 = 9 $$

Ans. (3)