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8. sequence and series — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS8. sequence and series1 Mark
MCQ
For non-negative integers $n$, let

$f(n)=\frac{\sum_{k=0}^n \sin \left(\frac{k+1}{n+2} \pi\right) \sin \left(\frac{k+2}{n+2} \pi\right)}{\sum_{k=0}^n \sin ^2\left(\frac{k+1}{n+2} \pi\right)}$

Assuming $\cos ^{-1} x$ takes values in $[0, \pi]$, which of the following options is/are correct ?

$(1)$ $\sin \left(7 \cos ^{-1} f(5)\right)=0$

$(2)$ $f(4)=\frac{\sqrt{3}}{2}$

$(3)$ $\lim _{n \rightarrow \infty} f(n)=\frac{1}{2}$

$(4)$ If $\alpha=\tan \left(\cos ^{-1} f(6)\right)$, then $\alpha^2+2 \alpha-1=0$

  • A
    $1,2,3$
  • ✓
    $1,2,4$
  • C
    $1,2$
  • D
    $2,3$
✓

Answer

Correct option: B.
$1,2,4$
b
$\begin{array}{l}f(n)=\frac{\sum_{k=0}^n\left(\cos \left(\frac{\pi}{n+2}\right)-\cos \left(\frac{2 k+3}{n+2}\right) \pi\right)}{\sum_{k=0}^n\left(1-\cos \left(\frac{2 k+2}{n+2}\right) \pi\right)} \\ f(n)=\frac{(n+1) \cos \left(\frac{\pi}{n+2}\right)-\left(\sum_{k=0}^n \cos \left(\frac{2 k+3}{n+2}\right) \pi\right)}{(n+1)-\left(\sum_{k=0}^n \cos \left(\frac{2 k+2}{n+2}\right) \pi\right)}\end{array}$

$\begin{array}{r}f(n)=\frac{(n+1) \cos \frac{\pi}{n+2}-\left(\frac{\sin \left(\frac{(n+1) \pi}{n+2}\right)}{\sin \left(\frac{\pi}{n+2}\right)} \cdot \cos \left(\frac{n+3}{n+2}\right) \pi\right)}{(n+1)-\left(\frac{\sin \left(\frac{(n+1) \pi}{n+2}\right)}{\sin \left(\frac{\pi}{n+2}\right)} \cdot \cos \left(\frac{2(n+2) \pi}{2(n+2)}\right)\right)} \\ f(n)=\frac{(n+1) \cos \left(\frac{\pi}{n+2}\right)+\cos \left(\frac{\pi}{n+2}\right)}{(n+1)+1} \Rightarrow g(x)=\cos \left(\frac{\pi}{n+2}\right)\end{array}$

$(A)$ $\sin \left(7 \cos ^{-1} \cos \frac{\pi}{7}\right)=\sin \pi=0$

$(B)$ $f(4)=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$

$(C)$ $\lim _{n \rightarrow \infty} \cos \left(\frac{\pi}{n+2}\right)=1$

$(D)$ $\alpha=\tan \left(\cos ^{-1} \cos \frac{\pi}{8}\right)=\sqrt{2}-1 \Rightarrow \alpha+1=\sqrt{2}$

$\alpha^2+2 \alpha-1=0$

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