$S=\cot ^{-1}\left(\frac{1+2 \times 6}{4}\right)+\cot ^{-1}\left(\frac{1+3 \times 8}{10}\right)+\cot ^{-1} \left(\frac{1+4 \times 10}{18}\right) \ldots \ldots \ldots$ upto $\infty$ terms is equal to
$\tan ^{-1} 2$
$\cot ^{-1} 2$
$\frac{\pi}{2}-\cot ^{-1} 2$
$-\tan ^{-1} 2$
The area of the region $\left\{(x, y): x y<8,1 \leq y \leq x^2\right\}$ is
$8 \ln 2-\frac{14}{3}$
$16 \ln 2-\frac{14}{3}$
$16 \ln 2-6$
$8 \ln 2-\frac{7}{3}$
If $\alpha$ is a root of $x^4=1$ with negative principal argument, then the principle argument of $\Delta(A)$, where $\Delta(A)=\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha^n & \alpha^{n+1} & \alpha^{n+3} \\ \frac{1}{\alpha^{n+1}} & \frac{1}{\alpha^n} & 0\end{array}\right|$
$\frac{5 \pi}{4}$
$\frac{\pi}{4}$
$-\frac{3 \pi}{4}$
$-\frac{\pi}{4}$
Let $f(x)=\cos ^{-1}\left(2 x \sqrt{1-x^2}\right)$, then $f^{\prime}(0.6)$ equals to
$\frac{5}{2}$
$-\frac{5}{2}$
$\frac{2}{5}$
$-\frac{2}{5}$
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