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NTA JEE Main 26th August 2021 Shift 1 - Mathematics

For the following questions answer them individually

The sum of the series $$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{2^2}{x^4+1} + \ldots + \frac{2^{100}}{x^{2^{100}}+1}$$ when $$x = 2$$ is:

Let $$ABC$$ be a triangle with $$A(-3, 1)$$ and $$\angle ACB = \theta$$, $$0 < \theta < \frac{\pi}{2}$$. If the equation of the median through B is $$2x + y - 3 = 0$$ and the equation of angle bisector of C is $$7x - 4y - 1 = 0$$, then $$\tan \theta$$ is equal to:

On the ellipse $$\frac{x^2}{8} + \frac{y^2}{4} = 1$$, let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line $$x + 2y = 0$$. Let S and S' be the foci of the ellipse and $$e$$ be its eccentricity. If A is the area of the triangle SPS', then the value of $$(5 - e^2) \cdot A$$ is

The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If $$\alpha$$ and $$\sqrt{\beta}$$ are the mean and standard deviation respectively for correct data, then $$(\alpha, \beta)$$ is:

Out of all the patients in a hospital 89% are found to be suffering from heart ailment and 98% are suffering from lungs infection. If $$K$$% of them are suffering from both ailments, then $$K$$ can not belong to the set:

If $$A = \begin{bmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{bmatrix}$$, $$B = \begin{bmatrix} 1 & 0 \\ i & 1 \end{bmatrix}$$, $$i = \sqrt{-1}$$, and $$Q = A^T B A$$, then the inverse of the matrix $$AQ^{2021}A^T$$ is equal to:

Let $$\theta \in \left(0, \frac{\pi}{2}\right)$$. If the system of linear equations
$$(1 + \cos^2 \theta)x + \sin^2 \theta y + 4\sin 3\theta z = 0$$
$$\cos^2 \theta x + (1 + \sin^2 \theta)y + 4\sin 3\theta z = 0$$
$$\cos^2 \theta x + \sin^2 \theta y + (1 + 4\sin 3\theta)z = 0$$
has a non-trivial solution, then the value of $$\theta$$ is:

Let $$f(x) = \cos\left(2\tan^{-1}\sin\left(\cot^{-1}\sqrt{\frac{1-x}{x}}\right)\right)$$, $$0 \lt x \lt 1$$. Then:

Let $$y = y(x)$$ be a solution curve of the differential equation $$(y+1)\tan^2 x \, dx + \tan x \, dy + y \, dx = 0$$, $$x \in \left(0, \frac{\pi}{2}\right)$$. If $$\lim_{x \to 0^+} xy(x) = 1$$, then the value of $$y\left(\frac{\pi}{4}\right)$$ is:

A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is $$k$$ (meter), then $$\left(\frac{4}{\pi} + 1\right)k$$ is equal to _________

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