NTA JEE Main 6th April 2014 Offline

If $$a \in R$$ and the equation $$-3(x - [x])^2 + 2(x - [x]) + a^2 = 0$$ (where $$[x]$$ denotes the greatest integer $$\leq x$$) has no integral solution, then all possible values of $$a$$ lie in the interval:

Let $$\alpha$$ and $$\beta$$ be the roots of equation $$px^2 + qx + r = 0$$, $$p \neq 0$$. If $$p$$, $$q$$, $$r$$ are in A.P. and $$\frac{1}{\alpha} + \frac{1}{\beta} = 4$$, then the value of $$|\alpha - \beta|$$ is:

If $$z$$ is a complex number such that $$|z| \geq 2$$, then the minimum value of $$\left|z + \frac{1}{2}\right|$$:

If $$(10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + \ldots + 10(11)^9 = k(10)^9$$, then $$k$$ is equal to:

Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is:

If the coefficients of $$x^3$$ and $$x^4$$ in the expansion of $$(1 + ax + bx^2)(1 - 2x)^{18}$$ in powers of $$x$$ are both zero, then $$(a, b)$$ is equal to:

Let $$f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$$ where $$x \in R$$ and $$k \geq 1$$. Then $$f_4(x) - f_6(x)$$ equals:

Let $$PS$$ be the median of the triangle with vertices $$P(2, 2)$$, $$Q(6, -1)$$ and $$R(7, 3)$$. The equation of the line passing through $$(1, -1)$$ and parallel to $$PS$$ is:

Let $$a$$, $$b$$, $$c$$ and $$d$$ be non-zero numbers. If the point of intersection of the lines $$4ax + 2ay + c = 0$$ and $$5bx + 2by + d = 0$$ lies in the fourth quadrant and is equidistant from the two axes then:

Let $$C$$ be the circle with center at $$(1, 1)$$ and radius = 1. If $$T$$ is the circle centered at $$(0, y)$$, passing through the origin and touching the circle $$C$$ externally, then the radius of $$T$$ is equal to: