A function f(n) is defined from N→R (i.e., set of natural numbers to real numbers) such that f(n)=∑r=1n(sin(rπn)+cos(rπn));r∈N&n≥1, then which of the following option(s) is/are correct?

f(n)=sin(nπ2n)in(π2n)[sin(π+πn2)+cos(π+πn2)] =(cosπ2n)−sin(π2n)(sinπ2n)cot(π2n)−1=2 Range: {−1,0,3−1,2,.....}

A function f(n) is defined from N→R (i.e., set of natural numbers to real numbers) such that f(n)=∑r=1n(sin(rπn)+cos(rπn));r∈N&n≥1, then which of the following option(s) is/are correct?

f(n)=sin(nπ2n)in(π2n)[sin(π+πn2)+cos(π+πn2)] =(cosπ2n)−sin(π2n)(sinπ2n)cot(π2n)−1=2 Range: {−1,0,3−1,2,.....}

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A function $f (n)$ is defined from $N \to R$ (i.e., set of natural numbers to real numbers) such that $f (n) = \sum_{r = 1}^{n} (\sin (\frac{r π}{n}) + \cos (\frac{r π}{n})); r \in N & n \geq 1$ , then which of the following option(s) is/are correct?

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$f (n)$ contains only positive integers in its range

Number of real values of $x$ for which $2 f (x) + 1 = 0$ is $" 0 "$

$f (n)$ contains exactly one non-positive integer inits range

$f (4) = \sqrt{2}$

answer is B, D.

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Detailed Solution

$f (n) = \frac{\sin (\frac{n π}{2 n})}{i n (\frac{π}{2 n})} [\sin (\frac{\frac{π + π}{n}}{2}) + \cos (\frac{\frac{π + π}{n}}{2})] = \frac{(\cos \frac{π}{2 n}) - \sin (\frac{π}{2 n})}{(\sin \frac{π}{2 n})} \cot (\frac{π}{2 n}) - 1 = \sqrt{2} R a n g e : {- 1, 0, \sqrt{3} - 1, \sqrt{2}, .....}$