Consider the function f: \mathbbR arrow \mathbbR defined by f(x)=\frac2 x\sqrt1+9 x^2 . If the composition of f, \underbrace(f \circ f \circ f \circ \cdots \circ f)_10 times (x)=\frac2^10 x\sqrt1+9 \alpha x^2 , then the value of \sqrt3 \alpha+1 is equal to \underline\hspace2cm .

Functions PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q56 with Solution

View written solutionFree

Correct answer: 1024

Let $f(x)=\frac{2x}{\sqrt{1+9x^2}}.$ We need the 10-times composition: $f^{(10)}(x)=\underbrace{(f\circ f\circ \cdots \circ f)}_{10\text{ times}}(x)=\frac{2^{10}x}{\sqrt{1+9\alpha x^2}}.$ We must find $\sqrt{3\alpha+1}$ .
First, find the effect of one more composition in general. Assume $f^{(n)}(x)=\frac{2^n x}{\sqrt{1+9A_n x^2}}$ for some coefficient $A_n$ .

Then $f^{(n+1)}(x)=f\big(f^{(n)}(x)\big)=\frac{2\,f^{(n)}(x)}{\sqrt{1+9\left(f^{(n)}(x)\right)^2}}.$

Substitute: $f^{(n+1)}(x)=\frac{2\cdot \dfrac{2^n x}{\sqrt{1+9A_nx^2}}}{\sqrt{1+9\left(\dfrac{2^n x}{\sqrt{1+9A_nx^2}}\right)^2}}.$

Simplify the denominator:

=1+\frac{9\cdot 4^n x^2}{1+9A_nx^2} =\frac{1+9A_nx^2+9\cdot 4^n x^2}{1+9A_nx^2}.$$ Hence $$\sqrt{1+9\left(f^{(n)}(x)\right)^2}= \frac{\sqrt{1+9(A_n+4^n)x^2}}{\sqrt{1+9A_nx^2}}.$$ Therefore, $$f^{(n+1)}(x)=\frac{2^{n+1}x}{\sqrt{1+9(A_n+4^n)x^2}}.$$ So the recurrence is $$A_{n+1}=A_n+4^n.$$ 3. Initial value: For $n=1$, $$f^{(1)}(x)=\frac{2x}{\sqrt{1+9x^2}},$$ so $$A_1=1.$$ Thus, $$A_{10}=1+4+4^2+\cdots+4^9.$$ This is a geometric series: $$A_{10}=\frac{4^{10}-1}{4-1}=\frac{4^{10}-1}{3}.$$ Since in the question $A_{10}=\alpha$, we get $$\alpha=\frac{4^{10}-1}{3}.$$ 4. Now compute: $$3\alpha+1=3\cdot \frac{4^{10}-1}{3}+1=4^{10}.$$ Therefore, $$\sqrt{3\alpha+1}=\sqrt{4^{10}}=4^5=2^{10}=1024.$$ 5. Final answer: $$\boxed{1024}$$

View written solutionFree

Correct answer: 1024

Let $f(x)=\frac{2x}{\sqrt{1+9x^2}}.$ We need the 10-times composition: $f^{(10)}(x)=\underbrace{(f\circ f\circ \cdots \circ f)}_{10\text{ times}}(x)=\frac{2^{10}x}{\sqrt{1+9\alpha x^2}}.$ We must find $\sqrt{3\alpha+1}$ .
First, find the effect of one more composition in general. Assume $f^{(n)}(x)=\frac{2^n x}{\sqrt{1+9A_n x^2}}$ for some coefficient $A_n$ .

Then $f^{(n+1)}(x)=f\big(f^{(n)}(x)\big)=\frac{2\,f^{(n)}(x)}{\sqrt{1+9\left(f^{(n)}(x)\right)^2}}.$

Substitute: $f^{(n+1)}(x)=\frac{2\cdot \dfrac{2^n x}{\sqrt{1+9A_nx^2}}}{\sqrt{1+9\left(\dfrac{2^n x}{\sqrt{1+9A_nx^2}}\right)^2}}.$

Simplify the denominator:

=1+\frac{9\cdot 4^n x^2}{1+9A_nx^2} =\frac{1+9A_nx^2+9\cdot 4^n x^2}{1+9A_nx^2}.$$ Hence $$\sqrt{1+9\left(f^{(n)}(x)\right)^2}= \frac{\sqrt{1+9(A_n+4^n)x^2}}{\sqrt{1+9A_nx^2}}.$$ Therefore, $$f^{(n+1)}(x)=\frac{2^{n+1}x}{\sqrt{1+9(A_n+4^n)x^2}}.$$ So the recurrence is $$A_{n+1}=A_n+4^n.$$ 3. Initial value: For $n=1$, $$f^{(1)}(x)=\frac{2x}{\sqrt{1+9x^2}},$$ so $$A_1=1.$$ Thus, $$A_{10}=1+4+4^2+\cdots+4^9.$$ This is a geometric series: $$A_{10}=\frac{4^{10}-1}{4-1}=\frac{4^{10}-1}{3}.$$ Since in the question $A_{10}=\alpha$, we get $$\alpha=\frac{4^{10}-1}{3}.$$ 4. Now compute: $$3\alpha+1=3\cdot \frac{4^{10}-1}{3}+1=4^{10}.$$ Therefore, $$\sqrt{3\alpha+1}=\sqrt{4^{10}}=4^5=2^{10}=1024.$$ 5. Final answer: $$\boxed{1024}$$