MCQ
If $x = \sqrt {1 + \sqrt {1 + \sqrt {1 + .......\infty} } }$, then $x =$
- ✓$\frac{{1 + \sqrt 5 }}{2}$
- B$\frac{{1 - \sqrt 5 }}{2}$
- C$\frac{{1 \pm \sqrt 5 }}{2}$
- DNone of these
We have $x = \sqrt {1 + x} $
==> ${x^2} = 1 + x\,\,\,\, \Rightarrow {x^2} - x - 1 = 0$
==> $x = \frac{{1 \pm \sqrt {1 + 4} }}{2} = \frac{{1 \pm \sqrt 5 }}{2}$
As $x > 0$, we get $x = \frac{{1 + \sqrt 5 }}{2}$
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Consider the two statements :
($I$) $\mathrm{R}$ is reflexive but not symmetric.
($II$) $\mathrm{R}$ is transitive
Then which one of the following is true?
$S_n(x)=\sum_{k=1}^n \cot ^{-1}\left(\frac{1+k(k+1) x^2}{x}\right)$
where for any $x \in R , \cot ^{-1} x \in(0, \pi)$ and $\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then which of the following
statements is (are) $TRUE$?
$(A)$ $S _{10}( x )=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1+11 x ^2}{10 x }\right)$, for all $x >0$
$(B)$ $\lim _{n \rightarrow \infty} \cot \left(S_n(x)\right)=x$, for all $x>0$
$(C)$ The equation $S_3(x)=\frac{\pi}{4}$ has a root in $(0, \infty)$
$(D)$ $\tan \left( S _{ n }( x )\right) \leq \frac{1}{2}$, for all $n \geq 1$ and $x >0$