NTA JEE Main 9th April 2019 Shift 1

If the tangent to the curve, $$y = x^3 + ax - b$$ at the point $$(1, -5)$$ is perpendicular to the line, $$-x + y + 4 = 0$$, then which one of the following points lies on the curve?

Let $$S$$ be the set of all values of $$x$$ for which the tangent to the curve $$y = f(x) = x^3 - x^2 - 2x$$ at $$(x, y)$$ is parallel to the line segment joining the points $$(1, f(1))$$ and $$(-1, f(-1))$$, then $$S$$ is equal to:

$$\int \sec^2 x \cdot \cot^{4/3} x \, dx$$ is equal to:

The value of $$\int_0^{\pi/2} \frac{\sin^3 x}{\sin x + \cos x} dx$$ is:

The area (in sq. units) of the region $$A = \{(x, y) : x^2 \le y \le x + 2\}$$ is:

The solution of the differential equation $$x\frac{dy}{dx} + 2y = x^2$$, $$(x \neq 0)$$ with $$y(1) = 1$$, is:

Let $$\vec{\alpha} = 3\hat{i} + \hat{j}$$ and $$\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}$$. If $$\vec{\beta} = \vec{\beta_1} - \vec{\beta_2}$$, where $$\vec{\beta_1}$$ is parallel to $$\vec{\alpha}$$ and $$\vec{\beta_2}$$ is perpendicular to $$\vec{\alpha}$$, then $$\vec{\beta_1} \times \vec{\beta_2}$$ is equal to:

A plane passing through the points $$(0, -1, 0)$$ and $$(0, 0, 1)$$ and making an angle $$\frac{\pi}{4}$$ with the plane $$y - z + 5 = 0$$, also passes through the point:

If the line, $$\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$$ meets the plane, $$x + 2y + 3z = 15$$ at a point P, then the distance of P from the origin is:

Four persons can hit a target correctly with probabilities $$\frac{1}{2}$$, $$\frac{1}{3}$$, $$\frac{1}{4}$$ and $$\frac{1}{8}$$ respectively. If all hit at the target independently, then the probability that the target would be hit, is: