Join WhatsApp Icon JEE WhatsApp Group

NTA JEE Main 12th April 2019 Shift 1 - Mathematics

For the following questions answer them individually

If the normal to the ellipse $$3x^2 + 4y^2 = 12$$ at a point P on it is parallel to the line, $$2x + y = 4$$ and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to:

Let P be the point of intersection of the common tangents to the parabola $$y^2 = 12x$$ and the hyperbola $$8x^2 - y^2 = 8$$. If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio:

If A is a symmetric matrix and B is skew-symmetric matrix such that $$A + B = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix}$$, then AB is equal to:

For $$x \in \left(0, \frac{3}{2}\right)$$, let $$f(x) = \sqrt{x}$$, $$g(x) = \tan x$$ and $$h(x) = \frac{1 - x^2}{1 + x^2}$$. If $$\phi(x) = ((h \circ f) \circ g)(x)$$, then $$\phi\left(\frac{\pi}{3}\right)$$ is equal to:

A 2m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25 cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is:

If m is the minimum value of k for which the function $$f(x) = x\sqrt{kx - x^2}$$ is increasing in the interval [0, 3] and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to:

Consider the differential equation, $$y^2 dx + \left(x - \frac{1}{y}\right) dy = 0$$. If value of y is 1 when x = 1, then the value of x for which y = 2, is

Let $$\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$$ and $$\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$$ be two vectors. If a vector perpendicular to both the vectors $$\vec{a} + \vec{b}$$ and $$\vec{a} - \vec{b}$$ has the magnitude 12 then one such vector is: