MCQ
If $\int_0^k {\frac{{dx}}{{2 + 8{x^2}}}} = \frac{\pi }{{16}}\,,$ then $k = $
- A$1$
- ✓$\frac{1}{2}$
- C$\frac{1}{4}$
- DNone of these
$ = \frac{1}{4}|{\tan ^{ - 1}}t|_0^{2k} = \frac{1}{4}{\tan ^{ - 1}}2k$.
Comparing it with the given value, we get
${\tan ^{ - 1}}2k = \frac{\pi }{4} $
$\Rightarrow 2k = 1$
$ \Rightarrow k = \frac{1}{2}$.
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$f_n(x)=\sum_{j=1}^n \tan ^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right) \text { for all } x \in(0, \infty)$
(Here, the inverse trigonometric function $\tan ^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ )
Then, which of the following statement(s) is (are) TRUE?
$(A)$ $\sum_{ j =1}^5 \tan ^2\left( f _{ j }(0)\right)=55$
$(B)$ $\sum_{ j =1}^{10}\left(1+ f _{ j }^{\prime}(0)\right) \sec ^2\left( f _{ j }(0)\right)=10$
$(C)$ For any fixed positive integer $n$, $\lim _{x \rightarrow \infty} \tan \left(f_n(x)\right)=\frac{1}{n}$
$(D)$ For any fixed positive integer $n, \lim _{x \rightarrow \infty} \sec ^2\left(f_n(x)\right)=1$