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NTA JEE Main 12th April 2014 Online - Mathematics

For the following questions answer them individually

If $$\left(2 + \frac{x}{3}\right)^{55}$$ is expanded in the ascending powers of x and the coefficients of powers of x in two consecutive terms of the expansion are equal, then these terms are:

Let p, q, r denote arbitrary statements. Then the logically equivalent of the statement $$p \Rightarrow (q \vee r)$$ is:

Let $$\bar{X}$$ and M.D. be the mean and the mean deviation about $$\bar{X}$$ of n observations $$x_i$$, i = 1, 2, n. If each of the observations is increased by 5, then the new mean and the mean deviation about the new mean, respectively, are:

If $$\begin{vmatrix} a^2 & b^2 & c^2\\ (a+\lambda)^2 & (b+\lambda)^2 & (c+\lambda)^2 \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2  \end{vmatrix}$$ = $$k\lambda \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix}$$, $$\lambda \neq 0$$, then k is equal to:

If $$f(\theta) = \begin{vmatrix} 1 & \cos\theta & 1 \\ -\sin\theta & 1 & -\cos\theta \\ -1 & \sin\theta & 1 \end{vmatrix}$$ and A and B are respectively the maximum and the minimum values of $$f(\theta)$$, then (A, B) is equal to:

Statement I: The equation $$(\sin^{-1}x)^3 + (\cos^{-1}x)^3 - a\pi^3 = 0$$ has a solution for all $$a \geq \frac{1}{32}$$. 

Statement II: For any x $$\in$$ R, $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$ and $$0 \leq \left(\sin^{-1}x - \frac{\pi}{4}\right)^2 \leq \frac{9\pi^2}{16}$$.

Let f, g : R → R be two functions defined by $$f(x) = \begin{cases} x\sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$$ and $$g(x) = xf(x)$$. Statement I: f is a continuous function at x = 0. Statement II: g is a differentiable function at x = 0.

Let $$f$$ and $$g$$ be two differentiable functions on R such that $$f'(x) > 0$$ and $$g'(x) < 0$$ for all $$x \in R$$. Then for all x:

The general solution of the differential equation, $$\sin 2x\left(\frac{dy}{dx} - \sqrt{\tan x}\right) - y = 0$$, is:

A symmetrical form of the line of intersection of the planes $$x = ay + b$$ and $$z = cy + d$$ is: