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NTA JEE Main 9th January 2019 Shift 2 - Mathematics

For the following questions answer them individually

Let $$A(4, -4)$$ and $$B(9, 6)$$ be points on the parabola, $$y^2 = 4x$$. Let $$C$$ be chosen on the arc AOB of the parabola, where $$O$$ is the origin, such that the area of $$\triangle ACB$$ is maximum. Then, the area (in sq. units) of $$\triangle ACB$$, is:

If $$A = \begin{bmatrix} e^t & e^{-t}\cos t & e^{-t}\sin t \\ e^t & -e^{-t}\cos t - e^{-t}\sin t & -e^{-t}\sin t + e^{-t}\cos t \\ e^t & 2e^{-t}\sin t & -2e^{-t}\cos t \end{bmatrix}$$, then $$A$$ is:

Let $$\vec{a} = \hat{i} + \hat{j} + \sqrt{2}\hat{k}$$, $$\vec{b} = b_1\hat{i} + b_2\hat{j} + \sqrt{2}\hat{k}$$ and $$\vec{c} = 5\hat{i} + \hat{j} + \sqrt{2}\hat{k}$$ be three vectors such that the projection vector of $$\vec{b}$$ on $$\vec{a}$$ is $$|\vec{a}|$$. If $$\vec{a} + \vec{b}$$ is perpendicular to $$\vec{c}$$, then $$|\vec{b}|$$ is equal to:

The equation of the plane containing the straight line $$\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$$ and perpendicular to the plane containing the straight lines $$\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$$ and $$\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$$ is:

An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is: