NTA JEE Main 12th April 2019 Shift 2

Let $$f(x) = 5 - |x - 2|$$ and $$g(x) = |x + 1|$$, $$x \in R$$. If $$f(x)$$ attains maximum value at $$\alpha$$ and $$g(x)$$ attains minimum value at $$\beta$$, then $$\lim_{x \to -\alpha\beta} \frac{(x - 1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$$ is equal to

Let $$\alpha \in \left(0, \frac{\pi}{2}\right)$$, be constant. If the integral $$\int \frac{\tan x + \tan\alpha}{\tan x - \tan\alpha} dx = A(x)\cos 2\alpha + B(x)\sin 2\alpha + C$$, where C is a constant of integration, then the functions A(x) and B(x) are respectively

A value of $$\alpha$$ such that $$\int_\alpha^{\alpha+1} \frac{dx}{(x + \alpha)(x + \alpha + 1)} = \log_e\left(\frac{9}{8}\right)$$ is

If the area (in sq. units) bounded by the parabola $$y^2 = 4\lambda x$$ and the line $$y = \lambda x$$, $$\lambda > 0$$, is $$\frac{1}{9}$$, then $$\lambda$$ is equal to

The general solution of the differential equation $$(y^2 - x^3)dx - xy\,dy = 0$$, $$(x \neq 0)$$ is (where c is a constant of integration)

Let $$\alpha \in R$$ and the three vectors $$\vec{a} = \alpha\hat{i} + \hat{j} + 3\hat{k}$$, $$\vec{b} = 2\hat{i} + \hat{j} - \alpha\hat{k}$$ and $$\vec{c} = \alpha\hat{i} - 2\hat{j} + 3\hat{k}$$. Then the set S = {$$\alpha$$: $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ are coplanar}

A plane which bisects the angle between the two given planes $$2x - y + 2z - 4 = 0$$ and $$x + 2y + 2z - 2 = 0$$, passes through the point

The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines $$\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$$ and $$\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - 2\hat{k})$$ is

A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins Rs 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is:

For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is $$\frac{4}{5}$$, then the probability that he is unable to solve less than two problems is