Consider the function f: mathbb R rightarrow mathbb R defined by f(x)=frac 2 x sqrt 1+9 x 2 . If the composition of f , underbrace ( f of of o...o f ) 10 times ( x )=frac 2 10 x sqrt 1+9 α x 2 , then the value of √(3 α+1) is equal to .....

Mathematics · Functions

JEE Main 2024 — 4 April, Shift 2 — Question 24

Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x)=\frac{2 x}{\sqrt{1+9 x^{2}}}$ . If the composition of

$\mathrm{f}, \underbrace{(\text { f of of o...o } \mathrm{f})}_{10 \text { times }}(\mathrm{x})=\frac{2^{10} \mathrm{x}}{\sqrt{1+9 \alpha \mathrm{x}^{2}}}$ , then the value of $\sqrt{3 \alpha+1}$ is equal to .....

Answer: 1024

Numerical answer — enter this value.

Step-by-step solution

$f(f(x))=\frac{2 f(x)}{\sqrt{1+9 f^{2}(x)}}=\frac{4 x}{\sqrt{1+9 x^{2}+9.2^{2} x^{2}}}$

$f(f(f(x)))=\frac{2^{3} x / \sqrt{1+9 x^{2}}}{\sqrt{1+9\left(1+2^{2}\right) \frac{2^{2} x^{2}}{1+9 x^{2}}}}=\frac{2^{3} x}{\sqrt{1+9 x^{2}\left(1+2^{2}+2^{4}\right)}}$

$\therefore$ By observation $\alpha=1+2^{2}+2^{4}+\ldots+2^{18}=1\left(\frac{\left(2^{2}\right)^{10}-1}{2^{2}-1}\right)=\frac{2^{20}-1}{3}$

$3 \alpha+1=2^{20} \rightarrow \sqrt{3 \alpha+1}=2^{10}=1024$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Mathematics
Chapter: Functions
Topic: Composite Functions