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Functions: Let Represent Sequence Formed Solutions Equation Represent S

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

Question
f:[0,1]→Rf:\left[0,1\right]\to\mathbb R, f(x)=4x(1−x)f(x) = 4x\left(1-x\right), fn(x)=f(fn−1(x))f_n(x) = f\left(f_{n-1}(x)\right) ∀ n≥1\forall\,n \ge 1 and f0(x)=xf_0(x) = x. Let {bk(n)}\left\{b_k(n)\right\} represent the sequence formed by the solutions of the equation fn(x)=f0(x)f_n(x) = f_0(x) and B(n)B(n) represent the sum of all terms of the sequence {bk(n)}\left\{b_k(n)\right\}.
ANumber of terms in the sequence {bk(4)}\left\{b_k(4)\right\} is 1616correct
BThe value of B(2)B(3)\dfrac{B(2)}{B(3)} is 147310\dfrac{147}{310}
CNumber of non-zero terms in {bk(3)}\left\{b_k(3)\right\} is 77correct
DThe value of B(2)B(3)=121240\dfrac{B(2)}{B(3)} = \dfrac{121}{240}
Solution
Step 1: ff is quadratic, so f2=f∘ff_2 = f\circ f has degree 44, f3f_3 has degree 88, and in general
deg⁡fn=2n\deg f_n = 2^n
 Hence fn(x)−xf_n(x)-x is a polynomial of degree 2n2^n and has at most 2n2^n roots. Step 2: Substitute x=sin⁡2θx = \sin^2\theta with θ∈[0,π2]\theta \in \left[0,\tfrac\pi2\right]:
f(sin⁡2θ)=4sin⁡2θcos⁡2θ=sin⁡22θf\left(\sin^2\theta\right) = 4\sin^2\theta\cos^2\theta = \sin^2 2\theta
 so by induction fn(sin⁡2θ)=sin⁡2(2nθ)f_n\left(\sin^2\theta\right) = \sin^2\left(2^n\theta\right). Therefore
fn(x)=x  ⟺  sin⁡2(2nθ)=sin⁡2θ  ⟺  2nθ=±θ+mπf_n(x) = x \iff \sin^2\left(2^n\theta\right) = \sin^2\theta \iff 2^n\theta = \pm\theta+m\pi
θ=mπ2n−1orθ=mπ2n+1\theta = \frac{m\pi}{2^n-1} \quad\text{or}\quad \theta = \frac{m\pi}{2^n+1}
 Step 3: Count these θ\theta in [0,π2]\left[0,\tfrac\pi2\right]. The first family needs 0≤m≤2n−120 \le m \le \tfrac{2^n-1}2, i.e. m=0,1,…,2n−1−1m = 0,1,\dots,2^{n-1}-1, which is 2n−12^{n-1} values; the second needs 0≤m≤2n+120 \le m \le \tfrac{2^n+1}2, i.e. m=0,1,…,2n−1m = 0,1,\dots,2^{n-1}, which is 2n−1+12^{n-1}+1 values. The two families share only θ=0\theta = 0, because m2n−1=m′2n+1\dfrac m{2^n-1} = \dfrac{m'}{2^n+1} gives m(2n+1)=m′(2n−1)m\left(2^n+1\right) = m'\left(2^n-1\right), and 2n−12^n-1, 2n+12^n+1 are odd and differ by 22, so their H.C.F. is 11 and (2n−1)\left(2^n-1\right) divides mm, which with m<2n−1m<2^n-1 gives m=0m = 0. So the count is
2n−1+(2n−1+1)−1=2n2^{n-1}+\left(2^{n-1}+1\right)-1 = 2^n
 sin⁡2θ\sin^2\theta is strictly increasing on [0,π2]\left[0,\tfrac\pi2\right], so these are 2n2^n distinct numbers in [0,1]\left[0,1\right]; the degree is also 2n2^n, so all the roots are real, simple and lie in [0,1]\left[0,1\right]. Step 4:
#{bk(4)}=24=16.TRUE\#\left\{b_k(4)\right\} = 2^4 = 16 . \quad\textbf{TRUE}
#{bk(3)}=23=8\#\left\{b_k(3)\right\} = 2^3 = 8
 Here x=0x = 0 is one of them, since f(0)=0f(0) = 0, and the 88 roots are distinct, so
#{non-zero}=7.TRUE\#\left\{\text{non-zero}\right\} = 7 . \quad\textbf{TRUE}
 Step 5:
f2(x)=4(4x−4x2)(1−4x+4x2)=16x−80x2+128x3−64x4f_2(x) = 4\left(4x-4x^2\right)\left(1-4x+4x^2\right) = 16x-80x^2+128x^3-64x^4
f2(x)−x=−64x4+128x3−80x2+15xf_2(x)-x = -64x^4+128x^3-80x^2+15x
B(2)=−128−64=2B(2) = -\frac{128}{-64} = 2
 Expanding f3=f∘f2f_3 = f\circ f_2 gives a degree-88 polynomial whose two leading terms are
f3(x)−x=−16384 x8+65536 x7−⋯f_3(x)-x = -16384\,x^8+65536\,x^7-\cdots
 so
B(3)=−65536−16384=4B(3) = -\frac{65536}{-16384} = 4
 (In general, if fn−1f_{n-1} has leading terms axd+bxd−1ax^d+bx^{d-1} then fn=4fn−1−4fn−12f_n = 4f_{n-1}-4f_{n-1}^2 has leading terms −4a2x2d−8abx2d−1-4a^2x^{2d}-8abx^{2d-1}, and subtracting xx changes neither of these when d≥2d \ge 2, so B(n)=−−8ab−4a2=−2ba=2B(n−1)B(n) = -\tfrac{-8ab}{-4a^2} = -\tfrac{2b}a = 2B\left(n-1\right) and B(n)=2 n−1B(n) = 2^{\,n-1}.) Step 6:
B(2)B(3)=24=12=0.5\frac{B(2)}{B(3)} = \frac24 = \frac12 = 0.5
 while 147310=0.474194\dfrac{147}{310} = 0.474194 and 121240=0.504167\dfrac{121}{240} = 0.504167. Both (2) and (4) are FALSE. Answer: (1), (3)
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