List of practice Questions

Find the interval in which the function \( f(x) = x^4 - 4x^3 + 10 \) is strictly decreasing.
If \( x^y = e^x - y \), prove that \( \frac{dy}{dx} = \frac{\log x (1 + \log x)^2}{1 + \log x} \).
If \( y = \cos^3(\sec^2 2t) \), find \( \frac{dy}{dt} \).
Find the principal value of \[ \tan^{-1}(1) + \cos^{-1} \left(-\frac{1}{2}\right) + \sin^{-1} \left(-\frac{1}{\sqrt{2}}\right). \]
Express \( \tan^-1 \left( \frac\cos x1 - \sin x \right) \), where \( -\frac\pi2<x<\frac\pi2 \), in the simplest form.

Assertion (A): The corner points of the bounded feasible region of a L.P.P. are shown below. The maximum value of \( Z = x + 2y \) occurs at infinite points. 
Reason (R): The optimal solution of a LPP having bounded feasible region must occur at corner points. 
 

Graph Content
The function \( f(x) = kx - \sin x \) is strictly increasing for:
The probability distribution of a random variable \( X \) is: \[ \begin{array}{c|c|c|c|c|c} \hline X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & 0.1 & k & 2k & k & 0.1 \\ \hline \end{array} \]
where \( k \) is some unknown constant. The probability that the random variable \( X \) takes the value 2 is:

The maximum value of \( Z = 4x + y \) for a L.P.P. whose feasible region is given below is: 
 

Graph for region
The lines \[ \frac{1 - x}{2} = \frac{y - 1}{3} = \frac{z}{1} \] and \[ \frac{2x - 3}{2p} = \frac{y - 1}{1} = \frac{z - 4}{7} \] are perpendicular to each other for \( p \) equal to:
The vector equation of a line passing through the point \( (1, -1, 0) \) and parallel to the \( y \)-axis is:
Let \( \vec{a} \) be any vector such that \( |\vec{a}| = a \). The value of \[ |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 \] is:
The vectors \[ \vec{a} = 2\hat{i} - \hat{j} + \hat{k}, \quad \vec{b} = \hat{i} - 3\hat{j} - 5\hat{k}, \quad \vec{c} = -3\hat{i} + 4\hat{j} + 4\hat{k} \] represent the sides of:
If \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( |\vec{a}| = 1 \), \( |\vec{b}| = 2 \), and \( \vec{a} \cdot \vec{b} = \sqrt{3} \), then the angle between \( 2\vec{a} \) and \( -\vec{b} \) is:
The integrating factor of the differential equation \[ (x + 2y^2) \frac{dy}{dx} = y, \quad (y > 0) \] is:
The general solution of the differential equation \( x \, dy + y \, dx = 0 \) is:
The value of the integral: \[ I = \int_0^3 \frac{dx}{\sqrt{9 - x^2}} \] is:
For what value of \( k \), the function given below is continuous at \( x = 0 \)?
\[ f(x) = \begin{cases} \frac{\sqrt{4} + x - 2x}{x}, & x \neq 0 \\ k, & x = 0 \end{cases} \]
Derivative of \( e^2x \) with respect to \( e^x \) is:
Given that \[ \begin{bmatrix} 1 & x \end{bmatrix} \begin{bmatrix} 4 & 0 \\ -2 & 0 \end{bmatrix} = 0 \] then the value of \( x \) is:
If \( A = \begin{bmatrix} -2 & 0 & 0 \\ 1 & 2 & 3 \\ 5 & 1 & -1 \end{bmatrix} \), then the value of \( |A \cdot \text{adj}(A)| \) is:
If \( A = \begin{bmatrix} 2 & 1 \\ -4 & -2 \end{bmatrix} \), then the value of \( I - A + A^2 - A^3 + \dots \) is:

Given that $ A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix} $, matrix $ A $ is:

If \[ \begin{bmatrix} a & c & 0 \\ b & d & 0 \\ 0 & 0 & 5 \end{bmatrix} \] is a scalar matrix, then the value of \[ a + 2b + 3c + 4d \] is:
State two consumer responsibilities that Ashish did not fulfil while buying the LCD