JEE Mains · Maths · STD 12 - 1. relation and function
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 \(f, \underbrace{(f \circ f \circ f \circ \ldots \circ f)}_{10 \text { times }}(x)=\frac{2^{10} x}{\sqrt{1+9 \alpha x^2}}\), then the value of \(\sqrt{3 \alpha+1}\) is equal to ...........
- A \(1044\)
- B \(1075\)
- C \(1056\)
- D \(1024\)
Answer & Solution
Correct Answer
(D) \(1024\)
Step-by-step Solution
Detailed explanation
\( \mathrm{f}(\mathrm{f}(\mathrm{x}))=\frac{2 \mathrm{f}(\mathrm{x})}{\sqrt{1+9 \mathrm{f}^2(\mathrm{x})}}=\frac{4 \mathrm{x}}{\sqrt{1+9 \mathrm{x}^2+9.2^2 \mathrm{x}^2}} \)…
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