F n - r -0 n 1-4 r sin 42 r - then which of the following alternatives is-are correctA- f 2-4-1-2B- f 3-8-2-2-4C- f 43 -2-1D- f 5-0

The correct options are C f_4 ( 3π2 )=1 D f_5(π)=0Given : nbsp;f_n(θ)=∑_r=0^n14^rsin^4(2^rθ) ⇒ f_n(θ)=∑_r=0^n14^rsin^2(2^rθ)[1 cos^2(2^r θ)] ⇒ f_n(θ)=∑_r=0^n1 ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

f n θ=∑ r =0 ...

Question

If $f_{n} (θ) = n \sum r = 0 \frac{1}{4^{r}} {sin}^{4} (2^{r} θ)$ , then which of the following alternative(s) is/are correct?

f_{2} (\frac{π}{4}) = \frac{1}{\sqrt{2}}

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f_{3} (\frac{π}{8}) = \frac{2 + \sqrt{2}}{4}

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f_{4} (\frac{3 π}{2}) = 1

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f_{5} (π) = 0

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Solution

The correct options are
C $f_{4} (\frac{3 π}{2}) = 1$
D $f_{5} (π) = 0$
Given : $f_{n} (θ) = n \sum r = 0 \frac{1}{4^{r}} {sin}^{4} (2^{r} θ)$
$\Rightarrow f_{n} (θ) = n \sum r = 0 \frac{1}{4^{r}} {sin}^{2} (2^{r} θ) [1 - {cos}^{2} (2^{r} θ)] \Rightarrow f_{n} (θ) = n \sum r = 0 \frac{1}{4^{r}} [{sin}^{2} (2^{r} θ) - \frac{{sin}^{2} (2^{r + 1} θ)}{4}] \Rightarrow f_{n} (θ) = n \sum r = 0 [\frac{{sin}^{2} (2^{r} θ)}{4^{r}} - \frac{{sin}^{2} (2^{r + 1} θ)}{4^{r + 1}}] \Rightarrow f_{n} (θ) = [\frac{{sin}^{2} θ}{1} - \frac{{sin}^{2} (2 θ)}{4}] + [\frac{{sin}^{2} 2 θ}{4} - \frac{{sin}^{2} (4 θ)}{16}] + [\frac{{sin}^{2} 4 θ}{16} - \frac{{sin}^{2} (8 θ)}{64}] ⋮ ⋮ + [\frac{{sin}^{2} (2^{n} θ)}{4^{n}} - \frac{{sin}^{2} (2^{n + 1} θ)}{4^{n + 1}}] \Rightarrow f_{n} (θ) = {sin}^{2} θ - \frac{1}{4^{n + 1}} {sin}^{2} (2^{n + 1} θ)$

Now, checking the options, we get
$f_{2} (\frac{π}{4}) = {sin}^{2} \frac{π}{4} - \frac{1}{64} {sin}^{2} (8 \times \frac{π}{4}) \Rightarrow f_{2} (\frac{π}{4}) = \frac{1}{2}$

$f_{3} (\frac{π}{8}) = {sin}^{2} \frac{π}{8} - \frac{1}{256} {sin}^{2} (16 \times \frac{π}{8}) \Rightarrow f_{3} (\frac{π}{8}) = \frac{1 - cos \frac{π}{4}}{2} \Rightarrow f_{3} (\frac{π}{8}) = \frac{2 - \sqrt{2}}{4}$

$f_{4} (\frac{3 π}{2}) = {sin}^{2} \frac{3 π}{2} - \frac{1}{4^{5}} {sin}^{2} (32 \times \frac{3 π}{2}) \Rightarrow f_{4} (\frac{3 π}{2}) = 1$

$f_{5} (π) = {sin}^{2} π - \frac{1}{4^{6}} {sin}^{2} (64 π) = 0$

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If fn(θ)=n∑r=014rsin4(2rθ), then which of the following alternative(s) is/are correct?

If $f_{n} (θ) = n \sum r = 0 \frac{1}{4^{r}} {sin}^{4} (2^{r} θ)$ , then which of the following alternative(s) is/are correct?