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Interval of a for which a^x = log_a(x) has exactly 3 solutions | JEE Main

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Question
If the interval of aa for which ax=log⁡axa^x = \log_ax has exactly 33 solutions is (0, b−b)\left(0,\ b^{-b}\right), then [b]\left[b\right] is, where [ ⋅ ]\left[\ \cdot\ \right] is G.I.F.
Solution
Answer: 2
Step 1: log⁡ax\log_a x needs a>0, a≠1a > 0,\ a \ne 1. The given interval (0,b−b)⊂(0,1)(0, b^{-b}) \subset (0,1), so 0<a<10 < a < 1. Step 2: y=axy = a^x and y=log⁡axy = \log_a x are inverse of each other, so their graphs are reflections of each other in the line y=xy = x. Hence if (p,q)(p, q) is a point of intersection, so is (q,p)(q, p), i.e. every intersection off the line y=xy = x occurs in a pair. Step 3: On y=xy = x: ax=xa^x = x. For 0<a<10 < a < 1, axa^x is decreasing and xx is increasing, so this has exactly one root. Let it be x=tx = t, so
at=t,0<t<1...(1)a^t = t, \qquad 0 < t < 1 \quad ...(1)
 Total number of solutions is therefore 11 or 33, and it is 33 exactly when the pair off the line exists. Question attachment Step 4: The pair exists when y=axy = a^x cuts y=xy = x at x=tx = t with slope less than −1-1.
ddx(ax)∣x=t=atln⁡a=tln⁡a\frac{d}{dx}\left(a^x\right)\Big|_{x=t} = a^t \ln a = t \ln a
 From (1), ln⁡t=tln⁡a\ln t = t \ln a, so the condition tln⁡a<−1t \ln a < -1 becomes
ln⁡t<−1  ⟹  t<1e\ln t < -1 \implies t < \frac{1}{e}
 Step 5: From (1), ln⁡a=ln⁡tt\ln a = \dfrac{\ln t}{t}. Let g(t)=ln⁡ttg(t) = \dfrac{\ln t}{t}, 0<t<10 < t < 1.
g′(t)=1−ln⁡tt2>0(∵ln⁡t<0<1)g'(t) = \frac{1 - \ln t}{t^2} > 0 \quad (\because \ln t < 0 < 1)
 So ln⁡a\ln a is strictly increasing in tt, i.e. tt is strictly increasing in aa. Step 6: g(1e)=−11/e=−eg\left(\dfrac{1}{e}\right) = \dfrac{-1}{1/e} = -e, so
t<1e  ⟺  ln⁡a<−e  ⟺  a<e−et < \frac{1}{e} \iff \ln a < -e \iff a < e^{-e}
 ∴\therefore the required interval is (0, e−e)\left(0,\ e^{-e}\right). Step 7:
b−b=e−e  ⟹  bln⁡b=eb^{-b} = e^{-e} \implies b \ln b = e
 bln⁡b<0b \ln b < 0 for 0<b<10 < b < 1, and for b≥1b \ge 1, ddb(bln⁡b)=1+ln⁡b>0\dfrac{d}{db}(b \ln b) = 1 + \ln b > 0, so bln⁡bb \ln b is strictly increasing there. Hence b=eb = e is the only root. Step 8:
[b]=[e]=[2.718…]=2[b] = [e] = [2.718\ldots] = 2
 Answer: 22.
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