Let f x be a differentiable function such that f - 0 - 1- and the sequence an is defined as a1-2 and an-limx-x2fan-1x-f02 for n- 2- If the value of -i-110ai-2k- then the value of k is

A_n=lim_x→∞x^2(f(a_n 1x) f(0))^2 Putting x=a_n 1h If x→∞, then h→0 nbsp; a_n=lim_h→ 0(a_n 1h)^2(f(h) f(0))^2 =(a_n 1)^2lim_h→ 0(f(h) f(0)h)^2 ⇒ a_n=[a_n 1· ...

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Byju's Answer

Standard XII

Mathematics

Integration of Piecewise Continuous Functions

Let f x be a ...

Question

Let $f (x)$ be a differentiable function such that $f^{'} (0) = 1,$ and the sequence ${a_{n}}$ is defined as $a_{1} = 2$ and $a_{n} = lim x \to \infty x^{2} {(f (\frac{a_{n - 1}}{x}) - f (0))}^{2}$ for $n \geq 2.$ If the value of $10 \prod i = 1 a_{i} = 2^{k},$ then the value of $k$ is

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Solution

$a_{n} = lim x \to \infty x^{2} {(f (\frac{a_{n - 1}}{x}) - f (0))}^{2}$
Putting $x = \frac{a_{n - 1}}{h}$
If $x \to \infty,$ then $h \to 0$
$a_{n} = lim h \to 0 {(\frac{a_{n - 1}}{h})}^{2} {(f (h) - f (0))}^{2} = (a_{n - 1})^{2} lim h \to 0 {(\frac{f (h) - f (0)}{h})}^{2} \Rightarrow a_{n} = [a_{n - 1} \cdot f^{'} (0)]^{2} \Rightarrow a_{n} = (a_{n - 1})^{2} (∵ f^{'} (0) = 1)$

Now, $a_{n} = (a_{n - 1})^{2}$
Hence, $a_{2} = a_{1}^{2} = 2^{2};$
$a_{3} = a_{2}^{2} = 2^{4};$
$a_{4} = a_{3}^{2} = 2^{8}$ and so on.
$∴ 10 \prod i = 1 a_{i} = 2^{(1 + 2 + 2^{2} + 2^{3} + \dots + 2^{9})} = 2^{⎛ ⎜ ⎝ \frac{2^{10} - 1}{2 - 1} ⎞ ⎟ ⎠} = 2^{1023}$
$∴ k = 1023$

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Let f(x) be a differentiable function such that f′(0)=1, and the sequence {an} is defined as a1=2 and an=limx→∞x2(f(an−1x)−f(0))2 for n≥2. If the value of 10∏i=1ai=2k, then the value of k is

Let $f (x)$ be a differentiable function such that $f^{'} (0) = 1,$ and the sequence ${a_{n}}$ is defined as $a_{1} = 2$ and $a_{n} = lim x \to \infty x^{2} {(f (\frac{a_{n - 1}}{x}) - f (0))}^{2}$ for $n \geq 2.$ If the value of $10 \prod i = 1 a_{i} = 2^{k},$ then the value of $k$ is