JEE Main 2024 4 April Evening Shift
Exam JEE Main
Year 2024
Date 4 April
Shift Evening Shift
Subject Mathematics
Chapter Functions JEE Main 2024 · 4 April · Evening Shift · Mathematics · Functions
Consider the function f: ℝ arrow ℝ defined by f(x)= 2 x 1+9 x^2 ....
Consider the function
f : R → f: \mathbb{R} \rightarrow f : R → R \mathbb{R} R defined by
f ( x ) = 2 x 1 + 9 x 2 f(x)=\frac{2 x}{\sqrt{1+9 x^2}} f ( x ) = 1 + 9 x 2 2 x . If the composition of
f , ( f ∘ f ∘ f ∘ ⋯ ∘ f ) ⏟ 10 times ( x ) = 2 10 x 1 + 9 α x 2 f, \underbrace{(f \circ f \circ f \circ \cdots \circ f)}_{10 \text{ times }}(x)=\frac{2^{10} x}{\sqrt{1+9 \alpha x^2}} f , 10 times ( f ∘ f ∘ f ∘ ⋯ ∘ f ) ( x ) = 1 + 9 α x 2 2 10 x , then the value of
3 α + 1 \sqrt{3 \alpha+1} 3 α + 1 is equal to ______
Step-by-step solution
f ( f ( x ) ) = 2 f ( x ) 1 + 9 f 2 ( x ) = 4 x 1 + 9 x 2 + 9. 2 2 x 2 f ( f ( f ( x ) ) ) = 2 3 x / 1 + 9 x 2 1 + 9 ( 1 + 2 2 ) 2 2 x 2 1 + 9 x 2 = 2 3 x 1 + 9 x 2 ( 1 + 2 2 + 2 4 ) \begin{aligned}
& 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)}}
\end{aligned} f ( f ( x )) = 1 + 9 f 2 ( x ) 2 f ( x ) = 1 + 9 x 2 + 9. 2 2 x 2 4 x f ( f ( f ( x ))) = 1 + 9 ( 1 + 2 2 ) 1 + 9 x 2 2 2 x 2 2 3 x / 1 + 9 x 2 = 1 + 9 x 2 ( 1 + 2 2 + 2 4 ) 2 3 x \therefore
B y o b s e r v a t i o n By observation B yo b ser v a t i o n α = 1 + 2 2 + 2 4 + … . + 2 18 = 1 ( ( 2 2 ) 10 − 1 2 2 − 1 ) = 2 20 − 1 3 3 α + 1 = 2 20 → 3 α + 1 = 2 10 = 1024 \begin{aligned}
& \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
\end{aligned} α = 1 + 2 2 + 2 4 + … . + 2 18 = 1 ( 2 2 − 1 ( 2 2 ) 10 − 1 ) = 3 2 20 − 1 3 α + 1 = 2 20 → 3 α + 1 = 2 10 = 1024 $
Practice Functions on Quantrex Academy →
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