NTA JEE Main 15th April 2018 Shift 2

Tangents drawn from the point (-8, 0) to the parabola $$y^2 = 8x$$ touch the parabola at P and Q. If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to:

A normal to the hyperbola, $$4x^2 - 9y^2 = 36$$ meets the co-ordinate axes x and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the locus of P is:

$$\lim_{x \to 0} \frac{x\tan 2x - 2x\tan x}{(1 - \cos 2x)^2}$$ equals:

If the mean of the data: 7, 8, 9, 7, 8, 7, $$\lambda$$, 8 is 8, then the variance of this data is:

A tower T$$_1$$ of height 60 m is located exactly opposite to a tower T$$_2$$ of height 80 m on a straight road. From the top of T$$_1$$, if the angle of depression of the foot of T$$_2$$ is twice the angle of elevation of the top of T$$_2$$, then the width (in m) of the road between the feet of the towers T$$_1$$ and T$$_2$$ is:

Suppose A is any $$3 \times 3$$ non-singular matrix and $$(A - 3I)(A - 5I) = O$$, where $$I = I_3$$ and $$O = O_3$$. If $$\alpha A + \beta A^{-1} = 4I$$, then $$\alpha + \beta$$ is equal to:

If the system of linear equations
$$x + ay + z = 3$$
$$x + 2y + 2z = 6$$
$$x + 5y + 3z = b$$
has no solution, then:

Let $$f : A \to B$$ be a function defined as $$f(x) = \frac{x-1}{x-2}$$, where $$A = R - \{2\}$$ and $$B = R - \{1\}$$. Then f is:

Let $$f(x) = \begin{cases} (x-1)^{\frac{1}{2-x}}, & x > 1, x \neq 2 \\ k, & x = 2 \end{cases}$$. The value of k for which f is continuous at $$x = 2$$ is:

If $$f(x) = \sin^{-1}\left(\frac{2 \times 3^x}{1 + 9^x}\right)$$, then $$f'\left(-\frac{1}{2}\right)$$ equals: