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$f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$ પ્રમાણે વ્યાખ્યાયિત વિધેય $f: \mathbb{R} \rightarrow \mathbb{R}$ ધ્યાને લો. જો $f$ નું સંયોજન $\underbrace{(f \circ f \circ f \circ \cdots \circ f)}_{1090 \cdots+1}(x)=\frac{2^{10} x}{\sqrt{1+9 \alpha x^2}}$ હોય, તો $\sqrt{3 \alpha+1}$ નું મૂલ્ચ .......... છે.
$1044$
$1075$
$1056$
$1024$
Solution
$ \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}} $
$ \mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{x})))=\frac{2^3 \mathrm{x} / \sqrt{1+9 \mathrm{x}^2}}{\sqrt{1+9\left(1+2^2\right) \frac{2^2 \mathrm{x}^2}{1+9 \mathrm{x}^2}}}=\frac{2^3 \mathrm{x}}{\sqrt{1+9 \mathrm{x}^2\left(1+2^2+2^4\right)}} $
$ \therefore \text { 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$
