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Range of abc when f(a) = f(b) = f(c) for |log_3 x - 1| | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
Suppose
f(x)={∣log⁡3x−1∣,0<x≤94−x,x>9f(x) = \begin{cases}\left|\log_3 x-1\right|, & 0<x \le 9 \\ 4-\sqrt x, & x>9\end{cases}
 defined on the positive set of real numbers; provided that aa, bb, c∈Rc \in \mathbb{R} which are not equal to each other satisfying f(a)=f(b)=f(c)f\left(a\right) = f\left(b\right) = f\left(c\right), then abcabc can be
A8181
B8585correct
C8888correct
D143143correct
Solution
Question attachment Step 1: For 0<x<30<x<3: log⁡3x<1\log_3x<1, so f=1−log⁡3xf = 1-\log_3x, decreasing from +∞+\infty to 00. For 3≤x≤93 \le x \le 9: f=log⁡3x−1f = \log_3x-1, increasing from 00 to 11. For x>9x>9: f=4−xf = 4-\sqrt x, decreasing from 4−3=14-3 = 1 downwards. ff is continuous at x=3x = 3 and at x=9x = 9 (both formulas give 11). Step 2: A horizontal line y=ky = k meets each branch at most once, so three distinct solutions need it to meet all three. The three branches have ranges (0,∞)\left(0,\infty\right), [0,1]\left[0,1\right] and (−∞,1)\left(-\infty,1\right), whose common part is
0<k<10<k<1
 (At k=0k = 0 the first two branches meet at the single point x=3x = 3; at k=1k = 1 the last two meet at the single point x=9x = 9.) Step 3:
1−log⁡3a=k⟹a=3 1−k1-\log_3a = k \Longrightarrow a = 3^{\,1-k}
log⁡3b−1=k⟹b=3 1+k\log_3b-1 = k \Longrightarrow b = 3^{\,1+k}
4−c=k⟹c=(4−k)24-\sqrt c = k \Longrightarrow c = \left(4-k\right)^2
 For 0<k<10<k<1 these give a∈(1,3)a \in \left(1,3\right), b∈(3,9)b \in \left(3,9\right), c∈(9,16)c \in \left(9,16\right), so each lies on its own branch and aa, bb, cc are distinct, as the question requires. Step 4:
abc=3 1−k⋅3 1+k⋅(4−k)2=32(4−k)2=9(4−k)2abc = 3^{\,1-k}\cdot3^{\,1+k}\cdot\left(4-k\right)^2 = 3^2\left(4-k\right)^2 = 9\left(4-k\right)^2
 Step 5: 4−k∈(3,4)4-k \in \left(3,4\right), so (4−k)2∈(9,16)\left(4-k\right)^2 \in \left(9,16\right) and
abc∈(81, 144)abc \in \left(81,\ 144\right)
 an open interval, its endpoints coming from k=1k = 1 and k=0k = 0, where two of the three points coincide. So
81∉(81,144)  (it is the excluded endpoint)81 \notin \left(81,144\right) \ \text{ (it is the excluded endpoint)}
85, 88, 143∈(81,144)85,\ 88,\ 143 \in \left(81,144\right)
 Step 6: abc=85abc = 85 needs (4−k)2=859\left(4-k\right)^2 = \tfrac{85}9, i.e. k=0.9268k = 0.9268; then a=1.0837a = 1.0837, b=8.3046b = 8.3046, c=9.4446c = 9.4446, all distinct, with f(a)=f(b)=f(c)=0.9268f\left(a\right) = f\left(b\right) = f\left(c\right) = 0.9268 and abc=85.00abc = 85.00 For abc=81abc = 81 one needs k=1k = 1, giving b=c=9b = c = 9, which is not allowed. Answer: (2), (3) and (4)
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