NTA JEE Mains 05th April 2024 Shift 2

The number of real solutions of the equation $$x|x + 5| + 2|x + 7| - 2 = 0$$ is _________

If $$1 + \frac{\sqrt{3} - \sqrt{2}}{2\sqrt{3}} + \frac{5 - 2\sqrt{6}}{18} + \frac{9\sqrt{3} - 11\sqrt{2}}{36\sqrt{3}} + \frac{49 - 20\sqrt{6}}{180} + \ldots$$ upto $$\infty = 2 + \left(\sqrt{\frac{b}{a}} + 1\right)\log_e\left(\frac{a}{b}\right)$$, where $$a$$ and $$b$$ are integers with $$\gcd(a, b) = 1$$, then $$11a + 18b$$ is equal to ______

The number of solutions of $$\sin^2 x + (2 + 2x - x^2)\sin x - 3(x - 1)^2 = 0$$, where $$-\pi \leq x \leq \pi$$, is ________

Let a line perpendicular to the line $$2x - y = 10$$ touch the parabola $$y^2 = 4(x - 9)$$ at the point $$P$$. The distance of the point $$P$$ from the centre of the circle $$x^2 + y^2 - 14x - 8y + 56 = 0$$ is __________

Let $$a > 0$$ be a root of the equation $$2x^2 + x - 2 = 0$$. If $$\lim_{x \to \frac{1}{a}} \frac{16(1 - \cos(2 + x - 2x^2))}{(1 - ax)^2} = \alpha + \beta\sqrt{17}$$, where $$\alpha, \beta \in Z$$, then $$\alpha + \beta$$ is equal to ______

Let the mean and the standard deviation of the probability distribution $$\begin{array}{|c|c|c|c|c|} \hline X & \alpha & 1 & 0 & -3 \\ \hline P(X) & \frac{1}{3} & K & \frac{1}{6} & \frac{1}{4} \\ \hline \end{array}$$ be $$\mu$$ and $$\sigma$$, respectively. If $$\sigma - \mu = 2$$, then $$\sigma + \mu$$ is equal to ________

Let the maximum and minimum values of $$\left(\sqrt{8x - x^2 - 12} - 4\right)^2 + (x - 7)^2$$, $$x \in \mathbb{R}$$ be $$M$$ and $$m$$, respectively. Then $$M^2 - m^2$$ is equal to _________

If $$f(t) = \int_0^{\pi} \frac{2x \, dx}{1 - \cos^2 t \sin^2 x}$$, $$0 < t < \pi$$, then the value of $$\int_0^{\frac{\pi}{2}} \frac{\pi^2 dt}{f(t)}$$ equals _________

Let $$y = y(x)$$ be the solution of the differential equation $$\frac{dy}{dx} + \frac{2x}{(1+x^2)^2} y = xe^{\frac{1}{(1+x^2)}}$$; $$y(0) = 0$$. Then the area enclosed by the curve $$f(x) = y(x)e^{-\frac{1}{(1+x^2)}}$$ and the line $$y - x = 4$$ is __________

Let the point $$(-1, \alpha, \beta)$$ lie on the line of the shortest distance between the lines $$\frac{x+2}{-3} = \frac{y-2}{4} = \frac{z-5}{2}$$ and $$\frac{x+2}{-1} = \frac{y+6}{2} = \frac{z-1}{0}$$. Then $$(\alpha - \beta)^2$$ is equal to __________