Consider the function \(f:\mathbb{R}\to\mathbb{R}\) defined by \(f(x)=\dfrac{2x}{\sqrt{1+9x^{2}}}\). If the composition of \(f\), \((f\circ f\circ f\circ\ldots\circ f)(x)\) (10 times) \(=\dfrac{2^{10}x}{\sqrt{1+9\alpha x^{2}}}\), then the value of \(\sqrt{3\alpha+1}\) is equal to ____.

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Sets Relations and Functions

Consider the function

f : R \to R

defined by

f (x) = \frac{2 x}{1 + 9 x ^{2}}

. If the composition of

f

(f \circ f \circ f \circ \dots \circ f) (x)

(10 times)

= \frac{2 ^{10} x}{1 + 9 α x ^{2}}

, then the value of

3 α + 1

is equal to ____.

Answer: 1024

f (x) = \frac{2 x}{1 + 9 x ^{2}}

. Writing

\frac{1}{f ( x ) ^{2}} = \frac{1}{4 x ^{2}} + \frac{9}{4}

, each application of

f

adds

\frac{9}{4}

\frac{1}{4 x ^{2}}

scaled appropriately; tracking the pattern, after

n

applications:

f^{(n)} (x) = \frac{2 ^{n} x}{1 + 9 x ^{2} ( \frac{2 ^{2 n} - 1}{2 ^{2} - 1} )} = \frac{2 ^{n} x}{1 + 9 x ^{2} \cdot \frac{4 ^{n} - 1}{3}}

For

n = 10

: coefficient of

x^{2}

inside the root is

9 \cdot \frac{4 ^{10} - 1}{3} = 3 (4^{10} - 1)

. Matching to the given form

\frac{2 ^{10} x}{1 + 9 α x ^{2}}

9 α = 3 (4^{10} - 1) \Rightarrow α = \frac{4 ^{10} - 1}{3}

3 α + 1 = 4^{10} - 1 + 1 = 4^{10} \Rightarrow 3 α + 1 = 4^{5} = 1024

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Integer answer