A function is one-one (injective) if distinct inputs produce distinct outputs.


A function is onto (surjective) if every element in the codomain has at least one corresponding input in the domain.


Let's analyze the function \(f: Z \rightarrow Z\).


Option (A): \(f(x) = 2x - 1\)

This function is one-one because if \(f(x_1) = f(x_2)\), then \(2x_1 - 1 = 2x_2 - 1\), which implies \(x_1 = x_2\).


However, it is not onto. The output is always an odd integer. An even integer like 2 in the codomain Z has no pre-image in the domain Z, since \(2x - 1 = 2\) gives \(x = 3/2\), which is not an integer.


Option (B): \(f(x) = 3x^2 + 5\)

This function is not one-one because different inputs can produce the same output. For example, \(f(1) = 3(1)^2 + 5 = 8\) and \(f(-1) = 3(-1)^2 + 5 = 8\). Since \(f(1) = f(-1)\) but \(1 \neq -1\), it's not one-one.


Option (C): \(f(x) = x + 5\)

This function is one-one because if \(f(x_1) = f(x_2)\), then \(x_1 + 5 = x_2 + 5\), which implies \(x_1 = x_2\).


This function is also onto. For any integer \(y\) in the codomain, we can find an integer \(x\) in the domain such that \(f(x) = y\). Solving \(x + 5 = y\) gives \(x = y - 5\). Since \(y\) is an integer, \(y-5\) is also an integer. Thus, it is onto.


Since it is both one-one and onto, this is the correct answer.


Option (D): \(f(x) = 5x^3\)

This function is one-one because if \(f(x_1) = f(x_2)\), then \(5x_1^3 = 5x_2^3\), which implies \(x_1 = x_2\).


However, it is not onto. For an integer \(y=1\) in the codomain, we have \(5x^3 = 1\), so \(x = \sqrt[3]{1/5}\), which is not an integer.
Quick Tip: For a linear function \(f(x) = ax + b\) to be a bijection from Z to Z (both one-one and onto), the coefficient 'a' must be either 1 or -1.

Let the given expression be denoted by \(E\).

\(E = 4 \cos\left[\frac{1}{2}\cos^{-1}\left(\frac{1}{8}\right)\right]\)


Let \(\theta = \cos^{-1}\left(\frac{1}{8}\right)\). This implies that \(\cos(\theta) = \frac{1}{8}\).


The expression simplifies to \(E = 4 \cos\left(\frac{\theta}{2}\right)\).


We use the half-angle identity for cosine: \(\cos(\theta) = 2\cos^2\left(\frac{\theta}{2}\right) - 1\).


We can rewrite this as \(2\cos^2\left(\frac{\theta}{2}\right) = 1 + \cos(\theta)\).


Substituting the value of \(\cos(\theta)\):

\(2\cos^2\left(\frac{\theta}{2}\right) = 1 + \frac{1}{8} = \frac{9}{8}\).

\(\cos^2\left(\frac{\theta}{2}\right) = \frac{9}{16}\).

\(\cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{9}{16}} = \pm\frac{3}{4}\).


The range of the principal value of \(\cos^{-1}(x)\) is \([0, \pi]\).


So, \(0 \le \theta \le \pi\), which implies \(0 \le \frac{\theta}{2} \le \frac{\pi}{2}\).


In this interval (the first quadrant), the value of cosine is positive.


Therefore, we take the positive value: \(\cos\left(\frac{\theta}{2}\right) = \frac{3}{4}\).


Now, substitute this back into the expression for E:

\(E = 4 \times \frac{3}{4} = 3\).
Quick Tip: Master the trigonometric identities, especially the half-angle formulas like \(\cos(2x) = 2\cos^2(x) - 1\). They are essential for simplifying expressions involving inverse trigonometric functions.

We are given the matrix equation \(A = 6I\).


The given matrix is \(A = \begin{bmatrix} x & 0 & m
y & z & 0
0 & 0 & 6 \end{bmatrix}\).

\(I\) is the 3x3 identity matrix, \(I = \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix}\).


First, we calculate \(6I\) by scalar multiplication:

\(6I = 6 \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0
0 & 6 & 0
0 & 0 & 6 \end{bmatrix}\).


Now, we equate the matrices \(A\) and \(6I\):

\(\begin{bmatrix} x & 0 & m
y & z & 0
0 & 0 & 6 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0
0 & 6 & 0
0 & 0 & 6 \end{bmatrix}\).


For two matrices to be equal, their corresponding elements must be equal.


By comparing the elements, we get:

\(x = 6\)

\(m = 0\)

\(y = 0\)

\(z = 6\)


We need to find the value of \(x + y + z + m\).

\(x + y + z + m = 6 + 0 + 6 + 0 = 12\).
Quick Tip: The equation \(A = kI\), where \(A\) is a square matrix and \(k\) is a scalar, implies that \(A\) is a scalar matrix. In a scalar matrix, all diagonal elements are equal to \(k\) and all off-diagonal elements are zero.

The order of a matrix is given by the number of its rows and the number of its columns.


The format is (number of rows) × (number of columns).


The given matrix is \(B = \begin{bmatrix} 31 & 42
53 & 64
75 & 86 \end{bmatrix}\).


Let's count the number of rows (horizontal lines of numbers).


Row 1: [31 42]


Row 2: [53 64]


Row 3: [75 86]


There are 3 rows.


Now let's count the number of columns (vertical lines of numbers).


Column 1: \(\begin{bmatrix} 31
53
75 \end{bmatrix}\)


Column 2: \(\begin{bmatrix} 42
64
86 \end{bmatrix}\)


There are 2 columns.


Therefore, the order of the matrix B is 3 × 2.
Quick Tip: A simple way to remember the order is "Row by Column" or "RC Cola". Always state the number of rows first, followed by the number of columns.

To find \((A - B)^2\), we multiply the matrix \((A - B)\) by itself.

\((A - B)^2 = (A - B)(A - B)\).


We use the distributive property of matrix multiplication.

\(= A(A - B) - B(A - B)\)


Now, distribute again for each term:

\(= (A \cdot A - A \cdot B) - (B \cdot A - B \cdot B)\)

\(= A^2 - AB - BA + B^2\).


It is crucial to remember that matrix multiplication is not commutative in general, which means that \(AB\) is not necessarily equal to \(BA\).


Therefore, we cannot combine the terms \(-AB\) and \(-BA\) into \(-2AB\).


The correct expansion remains \(A^2 - AB - BA + B^2\).
Quick Tip: Never apply standard algebraic identities like \((a-b)^2 = a^2 - 2ab + b^2\) directly to matrices. These identities depend on the commutative property of multiplication, which does not hold for matrices. Always expand matrix products using the distributive law.

We are given the determinant equation and need to solve for \(x\).

\(\begin{vmatrix} 5 & 3 & -1
-7 & x & 2
9 & 6 & -2 \end{vmatrix} = 0\)


We expand the determinant along the first row (\(R_1\)).

\(5 \begin{vmatrix} x & 2
6 & -2 \end{vmatrix} - 3 \begin{vmatrix} -7 & 2
9 & -2 \end{vmatrix} + (-1) \begin{vmatrix} -7 & x
9 & 6 \end{vmatrix} = 0\)


Now, we calculate the 2x2 determinants.

\(5(x(-2) - 2(6)) - 3((-7)(-2) - 2(9)) - 1((-7)(6) - x(9)) = 0\)

\(5(-2x - 12) - 3(14 - 18) - (-42 - 9x) = 0\)

\(-10x - 60 - 3(-4) + 42 + 9x = 0\)

\(-10x - 60 + 12 + 42 + 9x = 0\)


Combine the terms with \(x\) and the constant terms.

\((-10x + 9x) + (-60 + 12 + 42) = 0\)

\(-x + (-6) = 0\)

\(-x = 6\)

\(x = -6\).
Quick Tip: When expanding a 3x3 determinant, be careful with the sign pattern (+, -, +) for the cofactors. A small sign error is a common mistake.

We are given the inverse of matrix A, \(A^{-1}\), and we need to find matrix A.


The property we use is that the inverse of an inverse matrix is the original matrix: \(A = (A^{-1})^{-1}\).


Let \(B = A^{-1} = \begin{bmatrix} 7 & 2
8 & 2 \end{bmatrix}\). We need to find \(B^{-1}\).


The formula for the inverse of a 2x2 matrix \(\begin{bmatrix} a & b
c & d \end{bmatrix}\) is \(\frac{1}{ad-bc} \begin{bmatrix} d & -b
-c & a \end{bmatrix}\).


First, calculate the determinant of B:

\(\det(B) = (7)(2) - (2)(8) = 14 - 16 = -2\).


Next, find the adjoint of B by swapping the diagonal elements and negating the off-diagonal elements:

\(adj(B) = \begin{bmatrix} 2 & -2
-8 & 7 \end{bmatrix}\).


Now, calculate the inverse of B:

\(A = B^{-1} = \frac{1}{\det(B)} adj(B) = \frac{1}{-2} \begin{bmatrix} 2 & -2
-8 & 7 \end{bmatrix}\).


Multiply each element of the adjoint matrix by the scalar \(\frac{1}{-2}\):

\(A = \begin{bmatrix} \frac{2}{-2} & \frac{-2}{-2}
\frac{-8}{-2} & \frac{7}{-2} \end{bmatrix} = \begin{bmatrix} -1 & 1
4 & -\frac{7}{2} \end{bmatrix}\).


This matches option (C).
Quick Tip: To find the original matrix from its inverse, simply take the inverse of the given inverse matrix. Remember the formula for a 2x2 inverse: swap the main diagonal elements, negate the others, and divide by the determinant.

We are given the equation \(\sqrt{x} + \sqrt{y} = \sqrt{a}\).


This can be written as \(x^{1/2} + y^{1/2} = a^{1/2}\).


We need to find \(\frac{dy}{dx}\), so we differentiate both sides of the equation with respect to \(x\).

\(\frac{d}{dx} (x^{1/2} + y^{1/2}) = \frac{d}{dx} (a^{1/2})\).


Using the power rule and the chain rule:

\(\frac{1}{2}x^{-1/2} + \frac{1}{2}y^{-1/2} \cdot \frac{dy}{dx} = 0\) (since \(a\) is a constant, its derivative is 0).

\(\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0\).


Now, we solve for \(\frac{dy}{dx}\).

\(\frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}\).


Multiply both sides by \(2\sqrt{y}\):

\(\frac{dy}{dx} = -\frac{2\sqrt{y}}{2\sqrt{x}}\).

\(\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}\).
Quick Tip: This is an example of implicit differentiation. When differentiating a term with 'y' with respect to 'x', always remember to multiply by \(\frac{dy}{dx}\) according to the chain rule.

We are given \(y = \tan^{-1}\left(\frac{1-\cos x}{\sin x}\right)\).


To simplify the differentiation, we first simplify the expression inside the \(\tan^{-1}\) using trigonometric half-angle identities.


The identities are:
\(1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)\)
\(\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\)


Substitute these into the expression:

\(\frac{1-\cos x}{\sin x} = \frac{2\sin^2\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}\).


Assuming \(\sin\left(\frac{x}{2}\right) \neq 0\), we can cancel terms:

\(= \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \tan\left(\frac{x}{2}\right)\).


Now substitute this simplified form back into the equation for \(y\):

\(y = \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right)\).


Assuming \(\frac{x}{2}\) lies in the principal value range of \(\tan^{-1}\), we have:

\(y = \frac{x}{2}\).


Now, differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{d}{dx}\left(\frac{x}{2}\right) = \frac{1}{2}\).
Quick Tip: Before differentiating complex inverse trigonometric functions, always try to simplify the inner expression using trigonometric identities. This often reduces the problem to differentiating a much simpler algebraic function.

Let the function be \(y = f(x) = x^x\), where \(x > 0\).


To find the minimum value, we need to find the derivative of the function and set it to zero.


Since the function has a variable in the exponent, we use logarithmic differentiation.


Take the natural logarithm of both sides:

\(\ln y = \ln(x^x) = x \ln x\).


Differentiate both sides with respect to \(x\):

\(\frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln x)\).

\(\frac{1}{y} \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x}\) (using the product rule).

\(\frac{1}{y} \frac{dy}{dx} = \ln x + 1\).

\(\frac{dy}{dx} = y (\ln x + 1) = x^x (\ln x + 1)\).


To find critical points, set the derivative to zero:

\(x^x (\ln x + 1) = 0\).


Since \(x > 0\), \(x^x\) is always positive. Therefore, we must have:

\(\ln x + 1 = 0\).

\(\ln x = -1\).

\(x = e^{-1} = \frac{1}{e}\).


To confirm this is a minimum, we can use the second derivative test. The sign of \(\frac{dy}{dx}\) changes from negative to positive around \(x = 1/e\), indicating a minimum.


The minimum value of the function is the value of \(y\) at this critical point, \(x = 1/e\).


Minimum value = \(\left(\frac{1}{e}\right)^{\frac{1}{e}} = (e^{-1})^{\frac{1}{e}} = e^{-1/e}\).
Quick Tip: For functions of the form \(f(x)^{g(x)}\), logarithmic differentiation is the standard method to find the derivative. Set the derivative to zero to find critical points for potential maxima or minima.

Let the given integral be \(I\).

\(I = \int \frac{2x^3}{4+x^8} dx = \int \frac{2x^3}{2^2+(x^4)^2} dx\).


This integral is in a form that suggests substitution. Let's use the substitution method.


Let \(u = x^4\).


Then, differentiate \(u\) with respect to \(x\):

\(\frac{du}{dx} = 4x^3\), which implies \(du = 4x^3 dx\).


From this, we can write \(2x^3 dx = \frac{1}{2} du\).


Now substitute \(u\) and \(du\) into the integral:

\(I = \int \frac{\frac{1}{2} du}{2^2+u^2} = \frac{1}{2} \int \frac{1}{2^2+u^2} du\).


We use the standard integration formula \(\int \frac{1}{a^2+u^2} du = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C\).


In our case, \(a=2\).

\(I = \frac{1}{2} \left[ \frac{1}{2}\tan^{-1}\left(\frac{u}{2}\right) \right] + C\).

\(I = \frac{1}{4}\tan^{-1}\left(\frac{u}{2}\right) + C\).


Finally, substitute back \(u = x^4\):

\(I = \frac{1}{4}\tan^{-1}\left(\frac{x^4}{2}\right) + C\).
Quick Tip: Recognize standard integral forms. When you see a denominator like \(a^2 + u^2\), think of the inverse tangent function. The key is to find the right substitution for \(u\) such that its derivative (\(du\)) appears in the numerator.

Let the given integral be \(I\).

\(I = \int e^x \frac{x}{(1+x)^2} dx\).


This integral is in the special form \(\int e^x (f(x) + f'(x)) dx = e^x f(x) + C\).


We need to manipulate the fraction \(\frac{x}{(1+x)^2}\) to fit this form.


We can write the numerator \(x\) as \((1+x) - 1\).


So, \(\frac{x}{(1+x)^2} = \frac{(1+x) - 1}{(1+x)^2} = \frac{1+x}{(1+x)^2} - \frac{1}{(1+x)^2}\).


This simplifies to \(\frac{1}{1+x} - \frac{1}{(1+x)^2}\).


Now, let's check if this is in the form \(f(x) + f'(x)\).


Let \(f(x) = \frac{1}{1+x} = (1+x)^{-1}\).


Then, the derivative \(f'(x)\) is:

\(f'(x) = -1 \cdot (1+x)^{-2} \cdot \frac{d}{dx}(1+x) = -1(1+x)^{-2} = -\frac{1}{(1+x)^2}\).


So, the expression is indeed \(f(x) + f'(x)\).


The integral becomes \(I = \int e^x \left[ \frac{1}{1+x} + \left(-\frac{1}{(1+x)^2}\right) \right] dx\).


Using the formula, the result is:

\(I = e^x f(x) + C = e^x \frac{1}{1+x} + C\).
Quick Tip: When you see an integral involving \(e^x\) multiplied by a function, always try to check if the function can be expressed as the sum of another function and its derivative, i.e., \(f(x) + f'(x)\).

The area of the region bounded by a curve \(y = f(x)\), the x-axis, and the vertical lines \(x=a\) and \(x=b\) is given by the definite integral \(A = \int_{a}^{b} y \,dx\).


Here, the curve is the line \(y = x+1\).


The boundaries are given by \(x=1\) (lower limit, \(a=1\)) and \(x=3\) (upper limit, \(b=3\)).


The function \(y=x+1\) is positive in the interval \([1, 3]\).


So, the area is:

\(A = \int_{1}^{3} (x+1) dx\).


Now, we evaluate the integral:

\(A = \left[ \frac{x^2}{2} + x \right]_{1}^{3}\).


Substitute the upper and lower limits:

\(A = \left( \frac{3^2}{2} + 3 \right) - \left( \frac{1^2}{2} + 1 \right)\).

\(A = \left( \frac{9}{2} + \frac{6}{2} \right) - \left( \frac{1}{2} + \frac{2}{2} \right)\).

\(A = \frac{15}{2} - \frac{3}{2}\).

\(A = \frac{12}{2} = 6\).


The area is 6 square units.
Quick Tip: For simple geometric shapes like the trapezoid formed in this problem, you can also use geometry to verify your answer. The vertices are (1,0), (3,0), (3,4), and (1,2). The area of a trapezoid is \(\frac{1}{2} \times (sum of parallel sides) \times height = \frac{1}{2} \times (2+4) \times (3-1) = 6\).

The given differential equation is \(x \frac{dy}{dx} - y = 2x^2\).


To find the integrating factor, we must first write the equation in the standard linear form: \(\frac{dy}{dx} + P(x)y = Q(x)\).


Divide the entire equation by \(x\) (assuming \(x \neq 0\)):

\(\frac{dy}{dx} - \frac{1}{x}y = 2x\).


Now, comparing this with the standard form, we have:

\(P(x) = -\frac{1}{x}\).


The integrating factor (I.F.) is given by the formula \(I.F. = e^{\int P(x) dx}\).

\(I.F. = e^{\int -\frac{1}{x} dx}\).

\(I.F. = e^{-\int \frac{1}{x} dx} = e^{-\ln|x|}\).


Using the property of logarithms, \(m \ln a = \ln a^m\):

\(I.F. = e^{\ln|x|^{-1}} = e^{\ln\left(\frac{1}{|x|}\right)}\).


Since \(e^{\ln z} = z\), we have:

\(I.F. = \frac{1}{|x|}\).


For simplicity in solving, we can choose the integrating factor as \(\frac{1}{x}\).
Quick Tip: Always ensure a linear differential equation is in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\) before identifying \(P(x)\) to calculate the integrating factor. The coefficient of \(\frac{dy}{dx}\) must be 1.

A vector that is perpendicular to two given non-collinear vectors \(\vec{a}\) and \(\vec{b}\) is parallel to their cross product, \(\vec{a} \times \vec{b}\).


Let's first calculate the cross product \(\vec{c} = \vec{a} \times \vec{b}\).

\(\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 1 & 2
0 & 1 & 1 \end{vmatrix}\).

\(\vec{c} = \hat{i}(1 \cdot 1 - 2 \cdot 1) - \hat{j}(2 \cdot 1 - 2 \cdot 0) + \hat{k}(2 \cdot 1 - 1 \cdot 0)\).

\(\vec{c} = \hat{i}(1 - 2) - \hat{j}(2 - 0) + \hat{k}(2 - 0)\).

\(\vec{c} = -\hat{i} - 2\hat{j} + 2\hat{k}\).


This vector \(\vec{c}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\). To get a unit vector, we must divide \(\vec{c}\) by its magnitude.


The magnitude of \(\vec{c}\) is \(|\vec{c}| = \sqrt{(-1)^2 + (-2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\).


A unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is \(\hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{-\hat{i} - 2\hat{j} + 2\hat{k}}{3}\).


This can be written as \(\hat{c} = -\frac{1}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}\).


The vector in the exact opposite direction, \(-\hat{c}\), is also a unit vector and is also perpendicular to both \(\vec{a}\) and \(\vec{b}\).

\(-\hat{c} = - \left( -\frac{1}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} \right) = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k}\).


Thus, there are exactly two unit vectors perpendicular to the given vectors.
Quick Tip: For any two non-zero, non-parallel vectors, there are always exactly two unit vectors perpendicular to both of them. They are \(\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\) and its negative.

This question relates to the fundamental concepts of Linear Programming Problems (LPP).


The feasible region is the set of all points that satisfy all the given constraints of the LPP.


The objective function is the linear function \(Z = ax + by\) which is to be maximized or minimized.


According to the Fundamental Theorem of Linear Programming (also known as the Corner Point Theorem), if an optimal solution (maximum or minimum value) for the objective function exists, it must occur at one of the vertices (or corner points) of the feasible region.


If the optimal value occurs at two adjacent corner points, then it also occurs at every point on the line segment connecting them. However, the search for the optimal solution is always focused on the corner points.


Therefore, the point yielding the maximum or minimum value lies at the corner points of the feasible region.
Quick Tip: To solve a linear programming problem graphically, the standard procedure is: 1. Graph the constraints to find the feasible region. 2. Identify the coordinates of all corner points. 3. Evaluate the objective function at each corner point. 4. The largest/smallest value is the maximum/minimum.

Let's analyze each inequality to determine the common region.


1. \(x \ge 0\): This inequality represents all points on and to the right of the y-axis. This includes the I and IV quadrants.


2. \(y \ge 0\): This inequality represents all points on and above the x-axis. This includes the I and II quadrants.


The combination of \(x \ge 0\) and \(y \ge 0\) restricts the solution to the first quadrant (I Quadrant).


3. \(x + y \le 1\): To understand this region, we first consider the line \(x + y = 1\).

This line has an x-intercept of 1 (when y=0) and a y-intercept of 1 (when x=0).

The inequality \(x+y \le 1\) represents all points on this line and on the side of the line that includes the origin (since \(0+0 \le 1\) is true).


The common region must satisfy all three conditions.


Since the region must already be in the I Quadrant due to the first two inequalities, and it is further bounded by the line \(x+y=1\), the entire resulting feasible region (a triangle with vertices at (0,0), (1,0), and (0,1)) lies completely within the I Quadrant.
Quick Tip: The inequalities \(x \ge 0\) and \(y \ge 0\) are known as non-negativity constraints in Linear Programming and they always restrict the feasible region to the first quadrant.

Let A be the event that A is selected, and B be the event that B is selected.


We are given the probabilities of selection:
\(P(A) = \frac{1}{5}\)
\(P(B) = \frac{1}{3}\)


We are interested in the event that none of them is selected. This corresponds to the event that A is NOT selected AND B is NOT selected.


Let A' be the event that A is not selected, and B' be the event that B is not selected.


Using the rule of complementary events, \(P(A') = 1 - P(A)\).
\(P(A') = 1 - \frac{1}{5} = \frac{4}{5}\).


Similarly, \(P(B') = 1 - P(B)\).
\(P(B') = 1 - \frac{1}{3} = \frac{2}{3}\).


Since the selection of A and B are independent events, the probability of both A' and B' occurring is the product of their individual probabilities.

\(P(none is selected) = P(A' \cap B') = P(A') \times P(B')\).

\(P(none is selected) = \frac{4}{5} \times \frac{2}{3} = \frac{8}{15}\).
Quick Tip: For problems involving "at least one" or "none", it's often useful to think about complementary events. The probability of an event happening is 1 minus the probability of the event not happening.

First, let's evaluate the Reason (R).


Reason (R) states that two vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular if their dot product \(\vec{a} \cdot \vec{b} = 0\). This is the definition of orthogonality for non-zero vectors. So, Reason (R) is a true statement.


Next, let's evaluate the Assertion (A) using the condition from Reason (R).


Assertion (A) claims that \(\vec{a} = 4\hat{i} + \hat{j} - \hat{k}\) and \(\vec{b} = -2\hat{i} + 3\hat{j} - 5\hat{k}\) are mutually perpendicular.


Let's calculate their dot product:

\(\vec{a} \cdot \vec{b} = (4)(-2) + (1)(3) + (-1)(-5)\).

\(\vec{a} \cdot \vec{b} = -8 + 3 + 5\).

\(\vec{a} \cdot \vec{b} = -8 + 8 = 0\).


Since the dot product of the two vectors is 0, they are indeed mutually perpendicular. Therefore, Assertion (A) is also a true statement.


Finally, we check the relationship between (A) and (R). Assertion (A) is true because the condition stated in Reason (R) is satisfied. Thus, Reason (R) is the correct explanation for Assertion (A).


Hence, option (A) is the correct choice.
Quick Tip: For Assertion-Reason questions, follow a three-step process: 1. Determine if the Assertion is true or false. 2. Determine if the Reason is true or false. 3. If both are true, determine if the Reason correctly explains the Assertion.

First, we evaluate Reason (R).


Reason (R) gives the definition of a first-order homogeneous differential equation. An equation that can be expressed in the form \(\frac{dy}{dx} = F\left(\frac{y}{x}\right)\) is, by definition, homogeneous. Thus, Reason (R) is a true statement.


Next, we evaluate Assertion (A).


The given differential equation is \(x^2dy = (2xy + y^2)dx\).


To check if it's homogeneous, we rearrange it into the form \(\frac{dy}{dx}\).

\(\frac{dy}{dx} = \frac{2xy + y^2}{x^2}\).


Now, we try to express the right-hand side as a function of \(\frac{y}{x}\). We can do this by dividing both the numerator and the denominator by \(x^2\) (assuming \(x \neq 0\)).

\(\frac{dy}{dx} = \frac{\frac{2xy}{x^2} + \frac{y^2}{x^2}}{\frac{x^2}{x^2}} = \frac{2\left(\frac{y}{x}\right) + \left(\frac{y}{x}\right)^2}{1}\).


The right-hand side is a function of \(\frac{y}{x}\), i.e., \(F\left(\frac{y}{x}\right) = 2\left(\frac{y}{x}\right) + \left(\frac{y}{x}\right)^2\).


Since the equation can be written in the form stated in Reason (R), the Assertion (A) is a true statement.


Finally, Reason (R) provides the exact definition that we used to prove that Assertion (A) is true. Therefore, Reason (R) is the correct explanation for Assertion (A).


Hence, option (A) is the correct choice.
Quick Tip: An alternative way to check for homogeneity is to see if all terms in the differential equation have the same total degree. In \(x^2dy - (2xy + y^2)dx = 0\), the term \(x^2\) has degree 2, \(2xy\) has degree 2 (1+1), and \(y^2\) has degree 2. Since all terms have the same degree, the equation is homogeneous.

We need to find the principal values of the given inverse trigonometric functions.


First, let's evaluate \(\tan^{-1}(\sqrt{3})\).


Let \(x = \tan^{-1}(\sqrt{3})\). Then \(\tan(x) = \sqrt{3}\).


The principal value range for \(\tan^{-1}\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\).


The value of \(x\) in this range for which \(\tan(x) = \sqrt{3}\) is \(x = \frac{\pi}{3}\).


Next, let's evaluate \(\sec^{-1}(-2)\).


Let \(y = \sec^{-1}(-2)\). Then \(\sec(y) = -2\), which means \(\cos(y) = -\frac{1}{2}\).


The principal value range for \(\sec^{-1}\) is \([0, \pi] - \{\frac{\pi}{2}\}\).


In this range, \(\cos(y)\) is negative in the second quadrant.


The reference angle for \(\cos(\theta) = \frac{1}{2}\) is \(\frac{\pi}{3}\).


So, \(y = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\).


Now, we compute the final expression:

\(\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}\).
Quick Tip: It is crucial to remember the principal value branches for all six inverse trigonometric functions. For \(\sec^{-1}(x)\), the range is \([0, \pi] - \{\frac{\pi}{2}\}\). For negative inputs, the angle will be in the second quadrant.

To check for differentiability at a point, we use the first principle of derivatives.


The derivative of a function \(f(x)\) at a point \(x=c\) is given by the limit:

\(f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\), if the limit exists.


Here, \(f(x) = (x-1)^{1/3}\) and the point is \(c=1\).


First, we find \(f(1)\):

\(f(1) = (1-1)^{1/3} = 0^{1/3} = 0\).


Next, we find \(f(1+h)\):

\(f(1+h) = ((1+h)-1)^{1/3} = h^{1/3}\).


Now, we substitute these into the limit definition:

\(f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{h^{1/3} - 0}{h}\).

\(f'(1) = \lim_{h \to 0} \frac{h^{1/3}}{h^1} = \lim_{h \to 0} h^{1/3 - 1} = \lim_{h \to 0} h^{-2/3}\).

\(f'(1) = \lim_{h \to 0} \frac{1}{h^{2/3}}\).


As \(h\) approaches 0, the denominator \(h^{2/3}\) approaches 0, and the expression \(\frac{1}{h^{2/3}}\) approaches infinity.


Since the limit is not a finite number, the limit does not exist.


Therefore, the function \(f(x) = (x-1)^{1/3}\) is not differentiable at \(x=1\).
Quick Tip: A function fails to be differentiable at a point if its graph has a sharp corner, a cusp, or a vertical tangent. The graph of \(y = (x-1)^{1/3}\) has a vertical tangent line at \(x=1\).

We are given the function \(y = \log(x + \sqrt{x^2 + a^2})\).


We will use the chain rule to differentiate. Let \(u = x + \sqrt{x^2 + a^2}\).


Then \(y = \log(u)\), and \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).


First, \(\frac{dy}{du} = \frac{1}{u} = \frac{1}{x + \sqrt{x^2 + a^2}}\).


Next, we find \(\frac{du}{dx}\):

\(\frac{du}{dx} = \frac{d}{dx} (x + \sqrt{x^2 + a^2}) = \frac{d}{dx}(x) + \frac{d}{dx}(x^2 + a^2)^{1/2}\).

\(= 1 + \frac{1}{2}(x^2 + a^2)^{-1/2} \cdot \frac{d}{dx}(x^2 + a^2)\).

\(= 1 + \frac{1}{2\sqrt{x^2 + a^2}} \cdot (2x)\).

\(= 1 + \frac{x}{\sqrt{x^2 + a^2}}\).


Taking a common denominator:

\(\frac{du}{dx} = \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}}\).


Now, we multiply the parts to get \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \frac{x + \sqrt{x^2 + a^2}}{\sqrt{x^2 + a^2}}\).


The term \((x + \sqrt{x^2 + a^2})\) cancels out from the numerator and denominator.

\(\frac{dy}{dx} = \frac{1}{\sqrt{x^2 + a^2}}\).
Quick Tip: The derivative of \(\log(x + \sqrt{x^2 + a^2})\) is a standard result. Recognizing this pattern is useful as \(\frac{1}{\sqrt{x^2+a^2}}\) is the integrand for this logarithmic function (or for \(\sinh^{-1}(x/a)\)).

We are given the equation of the curve: \(y = 7x - x^3\).


The rate at which \(x\) increases is given as \(\frac{dx}{dt} = 2\) units/sec.


First, we find the slope of the curve, which we denote by \(m\). The slope is the first derivative of \(y\) with respect to \(x\).

\(m = \frac{dy}{dx} = \frac{d}{dx}(7x - x^3) = 7 - 3x^2\).


We need to find the rate at which the slope is changing, which is \(\frac{dm}{dt}\).


To find \(\frac{dm}{dt}\), we differentiate the expression for \(m\) with respect to time \(t\), using the chain rule.

\(\frac{dm}{dt} = \frac{d}{dt}(7 - 3x^2) = 0 - 3(2x) \frac{dx}{dt}\).

\(\frac{dm}{dt} = -6x \frac{dx}{dt}\).


Now we substitute the given values: \(x=5\) and \(\frac{dx}{dt} = 2\).

\(\frac{dm}{dt} = -6(5)(2)\).

\(\frac{dm}{dt} = -60\).


The negative sign indicates that the slope is decreasing.


Thus, the slope of the curve is changing at a rate of -60 units per second, or it is decreasing at 60 units per second.
Quick Tip: This is a related rates problem. The key steps are: 1. Identify the given rates and the rate to be found. 2. Write an equation relating the variables (in this case, the slope \(m\) and the variable \(x\)). 3. Differentiate this equation with respect to time \(t\). 4. Substitute the known values to solve for the unknown rate.

We are given magnitudes of the sum and difference of two vectors, and the magnitude of one of the vectors.


We use the parallelogram law identity for vectors, which states:

\(|\vec{a} + \vec{b}|^2 + |\vec{a} - \vec{b}|^2 = 2(|\vec{a}|^2 + |\vec{b}|^2)\).


Now, we substitute the given values into this identity.


Given: \(|\vec{a} + \vec{b}| = 60\), \(|\vec{a} - \vec{b}| = 40\), and \(|\vec{b}| = 46\).

\((60)^2 + (40)^2 = 2(|\vec{a}|^2 + (46)^2)\).

\(3600 + 1600 = 2(|\vec{a}|^2 + 2116)\).

\(5200 = 2(|\vec{a}|^2 + 2116)\).


Divide both sides by 2:

\(2600 = |\vec{a}|^2 + 2116\).


Now, solve for \(|\vec{a}|^2\):

\(|\vec{a}|^2 = 2600 - 2116\).

\(|\vec{a}|^2 = 484\).


Take the square root to find the magnitude \(|\vec{a}|\). Since magnitude is a non-negative quantity:

\(|\vec{a}| = \sqrt{484} = 22\).
Quick Tip: The parallelogram law, \(|\vec{a} + \vec{b}|^2 + |\vec{a} - \vec{b}|^2 = 2(|\vec{a}|^2 + |\vec{b}|^2)\), provides a direct and quick way to solve problems involving the magnitudes of vector sums and differences.

Let the given points be A = (K, -11, 2), B = (0, -2, 2), and C = (2, 4, 2).


For three points to be collinear, the vectors formed by any two pairs of these points must be parallel.


Let's find the vectors \(\vec{AB}\) and \(\vec{BC}\).

\(\vec{AB} = (Position vector of B) - (Position vector of A)\).

\(\vec{AB} = (0 - K)\hat{i} + (-2 - (-11))\hat{j} + (2 - 2)\hat{k}\).

\(\vec{AB} = -K\hat{i} + 9\hat{j} + 0\hat{k}\).

\(\vec{BC} = (Position vector of C) - (Position vector of B)\).

\(\vec{BC} = (2 - 0)\hat{i} + (4 - (-2))\hat{j} + (2 - 2)\hat{k}\).

\(\vec{BC} = 2\hat{i} + 6\hat{j} + 0\hat{k}\).


If the points A, B, and C are collinear, then the vectors \(\vec{AB}\) and \(\vec{BC}\) are parallel.


This means that their corresponding components must be in proportion.

\(\frac{-K}{2} = \frac{9}{6}\).


We can simplify the ratio on the right side:

\(\frac{-K}{2} = \frac{3}{2}\).


Now, solve for K:

\(-K = 3\).

\(K = -3\).
Quick Tip: Three points A, B, and C are collinear if the vector \(\vec{AB}\) is a scalar multiple of \(\vec{BC}\), i.e., \(\vec{AB} = \lambda \vec{BC}\). A simpler method is to equate the ratios of their corresponding direction components.

To find the angle between two lines, we first need their direction ratios. We get these by writing the equations in standard symmetric form: \(\frac{x}{a} = \frac{y}{b} = \frac{z}{c}\).


For the first line: \(2x = 3y = -z\).


To convert to standard form, we divide by the LCM of the coefficients' magnitudes, which is LCM(2, 3, 1) = 6.

\(\frac{2x}{6} = \frac{3y}{6} = \frac{-z}{6} \implies \frac{x}{3} = \frac{y}{2} = \frac{z}{-6}\).


So, the direction ratios of the first line are \((a_1, b_1, c_1) = (3, 2, -6)\).


For the second line: \(6x = -y = -4z\).


To convert to standard form, we divide by the LCM of the coefficients' magnitudes, which is LCM(6, 1, 4) = 12.

\(\frac{6x}{12} = \frac{-y}{12} = \frac{-4z}{12} \implies \frac{x}{2} = \frac{y}{-12} = \frac{z}{-3}\).


So, the direction ratios of the second line are \((a_2, b_2, c_2) = (2, -12, -3)\).


The angle \(\theta\) between the two lines is given by the formula:

\(\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\).


Let's first calculate the dot product of the direction vectors: \(a_1a_2 + b_1b_2 + c_1c_2\).

\(= (3)(2) + (2)(-12) + (-6)(-3)\).

\(= 6 - 24 + 18\).

\(= 0\).


Since the dot product of the direction ratios is zero, the lines are perpendicular to each other.


Therefore, the angle between them is \(\theta = 90^\circ\) or \(\frac{\pi}{2}\) radians.
Quick Tip: Before finding direction ratios, always make sure the line equations are in standard form where the coefficients of \(x, y, z\) in the numerators are 1. If the dot product of the direction vectors (\(a_1a_2 + b_1b_2 + c_1c_2\)) is zero, you can immediately conclude the lines are perpendicular without calculating the magnitudes.

The given function is \(f(x) = 3x^4 - 4x^3 - 12x^2 + 5\).


To find the intervals of increasing and decreasing, we first need to find the derivative of the function.

\(f'(x) = \frac{d}{dx}(3x^4 - 4x^3 - 12x^2 + 5) = 12x^3 - 12x^2 - 24x\).


Next, we find the critical points by setting the derivative equal to zero, \(f'(x) = 0\).

\(12x^3 - 12x^2 - 24x = 0\).


Factor out \(12x\):

\(12x(x^2 - x - 2) = 0\).


Factor the quadratic expression:

\(12x(x - 2)(x + 1) = 0\).


The critical points are \(x = 0\), \(x = 2\), and \(x = -1\).


These points divide the number line into four intervals: \((-\infty, -1)\), \((-1, 0)\), \((0, 2)\), and \((2, \infty)\).


We now test the sign of \(f'(x)\) in each interval.


Interval \((-\infty, -1)\): Let's pick \(x=-2\). \(f'(-2) = 12(-2)(-2-2)(-2+1) = (-)(-)(-) = negative\). So, \(f(x)\) is decreasing.


Interval \((-1, 0)\): Let's pick \(x=-0.5\). \(f'(-0.5) = 12(-0.5)(-0.5-2)(-0.5+1) = (-)(-)(+) = positive\). So, \(f(x)\) is increasing.


Interval \((0, 2)\): Let's pick \(x=1\). \(f'(1) = 12(1)(1-2)(1+1) = (+)(-)(+) = negative\). So, \(f(x)\) is decreasing.


Interval \((2, \infty)\): Let's pick \(x=3\). \(f'(3) = 12(3)(3-2)(3+1) = (+)(+)(+) = positive\). So, \(f(x)\) is increasing.


(a) The function is strictly increasing where \(f'(x) > 0\), which is in the intervals \((-1, 0)\) and \((2, \infty)\).


(b) The function is strictly decreasing where \(f'(x) < 0\), which is in the intervals \((-\infty, -1)\) and \((0, 2)\).
Quick Tip: To determine where a function is increasing or decreasing, first find its derivative \(f'(x)\). Then, find the critical points where \(f'(x)=0\) or is undefined. These points partition the domain into intervals. Test a point from each interval in \(f'(x)\) to check its sign: positive means increasing, negative means decreasing.

We use partial fraction decomposition to break down the integrand.


Let \(\frac{x^2 - x + 1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+1}\).


Multiplying both sides by \((x-1)(x^2+1)\) gives:

\(x^2 - x + 1 = A(x^2+1) + (Bx+C)(x-1)\).


To find A, let \(x=1\):

\(1^2 - 1 + 1 = A(1^2+1) + 0 \implies 1 = 2A \implies A = \frac{1}{2}\).


To find B and C, we can expand and equate coefficients.

\(x^2 - x + 1 = Ax^2 + A + Bx^2 - Bx + Cx - C\).

\(x^2 - x + 1 = (A+B)x^2 + (-B+C)x + (A-C)\).


Equating coefficients of \(x^2\): \(1 = A+B \implies 1 = \frac{1}{2} + B \implies B = \frac{1}{2}\).


Equating constant terms: \(1 = A-C \implies 1 = \frac{1}{2} - C \implies C = -\frac{1}{2}\).


So the integral becomes:

\(\int \left( \frac{1/2}{x-1} + \frac{(1/2)x - 1/2}{x^2+1} \right) dx\).

\(= \frac{1}{2} \int \frac{1}{x-1} dx + \frac{1}{2} \int \frac{x}{x^2+1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx\).

\(= \frac{1}{2} \ln|x-1| + \frac{1}{4} \int \frac{2x}{x^2+1} dx - \frac{1}{2} \tan^{-1}(x)\).

\(= \frac{1}{2} \ln|x-1| + \frac{1}{4} \ln(x^2+1) - \frac{1}{2} \tan^{-1}(x) + C\).
Quick Tip: When using partial fractions, if you have a linear factor \((x-a)\), the numerator is a constant \(A\). If you have an irreducible quadratic factor \((x^2+b)\), the numerator is a linear term \(Bx+C\).

First, simplify the integrand within the given interval \([1, 4]\).


For any \(x\) in \([1, 4]\), \(x\) is positive, so \(|x| = x\).


Next, we need to handle the term \(|3-x|\). The definition of absolute value depends on the sign of the inner expression.


The expression \(3-x\) changes sign at \(x=3\).


Case 1: For \(x \le 3\), \(3-x \ge 0\), so \(|3-x| = 3-x\).


Case 2: For \(x > 3\), \(3-x < 0\), so \(|3-x| = -(3-x) = x-3\).


We split the integral at \(x=3\):

\(I = \int_{1}^{4} (x + |3-x|) dx = \int_{1}^{3} (x + |3-x|) dx + \int_{3}^{4} (x + |3-x|) dx\).


Now apply the correct definition for \(|3-x|\) in each part.

\(I = \int_{1}^{3} (x + (3-x)) dx + \int_{3}^{4} (x + (x-3)) dx\).


Simplify the integrands:

\(I = \int_{1}^{3} 3 dx + \int_{3}^{4} (2x-3) dx\).


Evaluate the first integral:

\(\int_{1}^{3} 3 dx = [3x]_{1}^{3} = 3(3) - 3(1) = 9 - 3 = 6\).


Evaluate the second integral:

\(\int_{3}^{4} (2x-3) dx = [x^2 - 3x]_{3}^{4} = (4^2 - 3(4)) - (3^2 - 3(3)) = (16-12) - (9-9) = 4 - 0 = 4\).


Add the results:

\(I = 6 + 4 = 10\).
Quick Tip: When integrating an absolute value function \(|f(x)|\), find the points where \(f(x)=0\). Use these points to split the interval of integration. Then, in each sub-interval, replace \(|f(x)|\) with either \(f(x)\) or \(-f(x)\), depending on its sign.

The given differential equation is \(\frac{dy}{dx} = 1 + x^2 + y^2 + x^2y^2\).


We can factor the right-hand side by grouping terms:

\(\frac{dy}{dx} = (1+x^2) + y^2(1+x^2) = (1+x^2)(1+y^2)\).


This is a separable differential equation. We can separate the variables \(x\) and \(y\).

\(\frac{dy}{1+y^2} = (1+x^2)dx\).


Now, integrate both sides:

\(\int \frac{1}{1+y^2} dy = \int (1+x^2) dx\).

\(\tan^{-1}(y) = x + \frac{x^3}{3} + C\).


This is the general solution. To find the particular solution, we use the initial condition \(y=1\) when \(x=0\).


Substitute these values into the general solution:

\(\tan^{-1}(1) = 0 + \frac{0^3}{3} + C\).

\(\frac{\pi}{4} = C\).


Substitute the value of C back into the general solution to get the particular solution:

\(\tan^{-1}(y) = x + \frac{x^3}{3} + \frac{\pi}{4}\).
Quick Tip: The first step in solving a first-order differential equation is to identify its type. Look for possibilities like variable separable, homogeneous, linear, or exact. Factoring the expressions is often a key to identifying the type.

The given equation is \(2xy \frac{dy}{dx} = x^2 + 3y^2\).


Rearrange it into the form \(\frac{dy}{dx} = f(x,y)\):

\(\frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}\).


This is a homogeneous differential equation because each term (\(x^2, y^2, xy\)) has a degree of 2.


We use the substitution \(y = vx\), which implies \(\frac{dy}{dx} = v + x \frac{dv}{dx}\).


Substitute these into the equation:

\(v + x \frac{dv}{dx} = \frac{x^2 + 3(vx)^2}{2x(vx)} = \frac{x^2(1+3v^2)}{2x^2v} = \frac{1+3v^2}{2v}\).


Now, separate the variables \(v\) and \(x\):

\(x \frac{dv}{dx} = \frac{1+3v^2}{2v} - v = \frac{1+3v^2 - 2v^2}{2v} = \frac{1+v^2}{2v}\).

\(\frac{2v}{1+v^2} dv = \frac{1}{x} dx\).


Integrate both sides:

\(\int \frac{2v}{1+v^2} dv = \int \frac{1}{x} dx\).


The left side is of the form \(\int \frac{f'(v)}{f(v)} dv = \ln|f(v)|\).

\(\ln|1+v^2| = \ln|x| + C\).


Let the constant of integration be \(C = \ln|k|\).

\(\ln(1+v^2) = \ln|kx|\). (Since \(1+v^2\) is always positive, we can drop the absolute value)

\(1+v^2 = kx\).


Substitute back \(v = \frac{y}{x}\):

\(1 + \left(\frac{y}{x}\right)^2 = kx\).

\(1 + \frac{y^2}{x^2} = kx\).

\(\frac{x^2+y^2}{x^2} = kx\).

\(x^2 + y^2 = kx^3\).
Quick Tip: To solve a homogeneous differential equation, rearrange it to the form \(\frac{dy}{dx} = F(\frac{y}{x})\), then use the substitution \(y=vx\). This will always transform it into a variable-separable equation in terms of \(v\) and \(x\).

First, we compute the two vectors for which we need a perpendicular vector.


Let \(\vec{p} = \vec{a} - \vec{b}\).

\(\vec{p} = (\hat{i} + 2\hat{j} + \hat{k}) - (2\hat{i} + \hat{j}) = (1-2)\hat{i} + (2-1)\hat{j} + (1-0)\hat{k} = -\hat{i} + \hat{j} + \hat{k}\).


Let \(\vec{q} = \vec{c} - \vec{b}\).

\(\vec{q} = (3\hat{i} - 4\hat{j} - 5\hat{k}) - (2\hat{i} + \hat{j}) = (3-2)\hat{i} + (-4-1)\hat{j} + (-5-0)\hat{k} = \hat{i} - 5\hat{j} - 5\hat{k}\).


A vector perpendicular to both \(\vec{p}\) and \(\vec{q}\) is given by their cross product, \(\vec{r} = \vec{p} \times \vec{q}\).

\(\vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 1 & 1
1 & -5 & -5 \end{vmatrix}\).

\(\vec{r} = \hat{i}(1(-5) - 1(-5)) - \hat{j}((-1)(-5) - 1(1)) + \hat{k}((-1)(-5) - 1(1))\).

\(\vec{r} = \hat{i}(-5+5) - \hat{j}(5-1) + \hat{k}(5-1)\).

\(\vec{r} = 0\hat{i} - 4\hat{j} + 4\hat{k}\).


To find the unit vector, we divide \(\vec{r}\) by its magnitude, \(|\vec{r}|\).

\(|\vec{r}| = \sqrt{0^2 + (-4)^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}\).


The unit vector is \(\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{-4\hat{j} + 4\hat{k}}{4\sqrt{2}}\).

\(\hat{r} = \frac{-\hat{j} + \hat{k}}{\sqrt{2}} = -\frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}\).
Quick Tip: The cross product \(\vec{a} \times \vec{b}\) of two vectors yields a third vector that is orthogonal (perpendicular) to the plane containing \(\vec{a}\) and \(\vec{b}\). To get a unit vector, simply divide the resulting cross product vector by its own magnitude.

The objective function is \(Z = ax + by\).


The corner points of the feasible region are O(0,0), A(5,0), B(3,4), and C(0,5).


According to the Corner Point Theorem, the maximum value of Z must occur at one of these corner points.


We are given that the maximum of Z occurs at two points, B(3,4) and C(0,5).


This implies that the value of Z at point B is equal to the value of Z at point C, and this value is the maximum.


Value of Z at B(3,4):

\(Z_B = a(3) + b(4) = 3a + 4b\).


Value of Z at C(0,5):

\(Z_C = a(0) + b(5) = 5b\).


Set these two values equal to each other:

\(3a + 4b = 5b\).


Subtract \(4b\) from both sides:

\(3a = b\).


This is the condition on a and b.


We should also verify that this value is indeed greater than the value of Z at the other corner points.


Value of Z at O(0,0) is \(a(0)+b(0)=0\).


Value of Z at A(5,0) is \(a(5)+b(0)=5a\).


The maximum value is \(5b = 5(3a) = 15a\). Since \(a>0\), we have \(15a > 5a\) and \(15a>0\). The condition is verified.
Quick Tip: If the optimal value (maximum or minimum) of an objective function in an LPP occurs at two adjacent corner points, then it also occurs at every point on the line segment connecting these two points. This happens when the objective function line is parallel to the boundary constraint line connecting those two points.

Let X be the random variable representing the number of doublets obtained in three throws of a pair of dice.


The possible values for X are 0, 1, 2, and 3.


First, let's find the probability of getting a doublet in a single throw of a pair of dice.


Total possible outcomes = \(6 \times 6 = 36\).


Favorable outcomes (doublets) are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). There are 6 favorable outcomes.


Let \(p\) be the probability of success (getting a doublet), so \(p = \frac{6}{36} = \frac{1}{6}\).


Let \(q\) be the probability of failure (not getting a doublet), so \(q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}\).


This scenario follows a binomial distribution with \(n=3\) trials and probability of success \(p=\frac{1}{6}\).


The probability mass function is \(P(X=k) = \binom{n}{k} p^k q^{n-k} = \binom{3}{k} (\frac{1}{6})^k (\frac{5}{6})^{3-k}\).


Now we calculate the probability for each value of X.

\(P(X=0) = \binom{3}{0} (\frac{1}{6})^0 (\frac{5}{6})^3 = 1 \cdot 1 \cdot \frac{125}{216} = \frac{125}{216}\).

\(P(X=1) = \binom{3}{1} (\frac{1}{6})^1 (\frac{5}{6})^2 = 3 \cdot \frac{1}{6} \cdot \frac{25}{36} = \frac{75}{216}\).

\(P(X=2) = \binom{3}{2} (\frac{1}{6})^2 (\frac{5}{6})^1 = 3 \cdot \frac{1}{36} \cdot \frac{5}{6} = \frac{15}{216}\).

\(P(X=3) = \binom{3}{3} (\frac{1}{6})^3 (\frac{5}{6})^0 = 1 \cdot \frac{1}{216} \cdot 1 = \frac{1}{216}\).


The probability distribution is:

\begin{tabular{|c|c|c|c|c|
\hline
X & 0 & 1 & 2 & 3

\hline
P(X) & \(\frac{125}{216}\) & \(\frac{75}{216}\) & \(\frac{15}{216}\) & \(\frac{1}{216}\)

\hline
\end{tabular
Quick Tip: Recognize scenarios that fit a binomial distribution: a fixed number of independent trials, only two outcomes (success/failure) for each trial, and a constant probability of success. The formula \(P(X=k) = \binom{n}{k} p^k q^{n-k}\) is fundamental.

We are given that E and F are independent events.

\(P(E) = p\) and \(P(F) = 2p\).


The probability that exactly one of the events occurs is given by:

\(P(exactly one of E, F) = P(E and not F) + P(F and not E)\).

\(P(E \cap F') + P(E' \cap F)\).


Since the events are independent, we have:

\(= P(E)P(F') + P(E')P(F)\).


We also need the probabilities of the complementary events:

\(P(E') = 1 - P(E) = 1 - p\).

\(P(F') = 1 - P(F) = 1 - 2p\).


Substitute all these into the equation:

\(p(1 - 2p) + (1 - p)(2p) = \frac{5}{9}\).


Expand the terms:

\((p - 2p^2) + (2p - 2p^2) = \frac{5}{9}\).

\(3p - 4p^2 = \frac{5}{9}\).


Multiply by 9 to clear the fraction:

\(27p - 36p^2 = 5\).


Rearrange into a standard quadratic equation form:

\(36p^2 - 27p + 5 = 0\).


We can solve this by factoring or using the quadratic formula. Let's factor it.

\((12p - 5)(3p - 1) = 36p^2 - 12p - 15p + 5 = 36p^2 - 27p + 5 = 0\).


The solutions are:

\(12p - 5 = 0 \implies p = \frac{5}{12}\).


or

\(3p - 1 = 0 \implies p = \frac{1}{3}\).


Both values are valid as they lead to probabilities between 0 and 1 for both P(E) and P(F).
Quick Tip: For independent events E and F, the probability of "exactly one" occurring can be calculated as \(P(E) + P(F) - 2P(E)P(F)\). This is an alternative formula derived from \(P(E \cup F) - P(E \cap F)\).

First, we find the inverse of the matrix \(A = \begin{bmatrix} 2 & -3 & 5
3 & 2 & -4
1 & 1 & -2 \end{bmatrix}\).


The determinant of A is:

\(|A| = 2(2(-2) - (-4)(1)) - (-3)(3(-2) - (-4)(1)) + 5(3(1) - 2(1))\).

\(|A| = 2(-4+4) + 3(-6+4) + 5(3-2) = 2(0) + 3(-2) + 5(1) = -6 + 5 = -1\).


Since \(|A| \neq 0\), the inverse exists. Now we find the adjoint of A. The cofactors are:

\(C_{11} = (-4 - (-4)) = 0\).

\(C_{12} = -(-6 - (-4)) = 2\).

\(C_{13} = (3 - 2) = 1\).

\(C_{21} = -(6 - 5) = -1\).

\(C_{22} = (-4 - 5) = -9\).

\(C_{23} = -(2 - (-3)) = -5\).

\(C_{31} = (12 - 10) = 2\).

\(C_{32} = -(-8 - 15) = 23\).

\(C_{33} = (4 - (-9)) = 13\).


The cofactor matrix is \(C = \begin{bmatrix} 0 & 2 & 1
-1 & -9 & -5
2 & 23 & 13 \end{bmatrix}\).


The adjoint of A is the transpose of C: \(adj(A) = C^T = \begin{bmatrix} 0 & -1 & 2
2 & -9 & 23
1 & -5 & 13 \end{bmatrix}\).


The inverse is \(A^{-1} = \frac{1}{|A|}adj(A) = \frac{1}{-1} \begin{bmatrix} 0 & -1 & 2
2 & -9 & 23
1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2
-2 & 9 & -23
-1 & 5 & -13 \end{bmatrix}\).


Now, we solve the system of equations. The system can be written in matrix form as \(AX=B\), where:

\(X = \begin{bmatrix} x
y
z \end{bmatrix}\) and \(B = \begin{bmatrix} 11
-5
-3 \end{bmatrix}\).


The solution is given by \(X = A^{-1}B\).

\(X = \begin{bmatrix} 0 & 1 & -2
-2 & 9 & -23
-1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11
-5
-3 \end{bmatrix}\).

\(X = \begin{bmatrix} 0(11) + 1(-5) + (-2)(-3)
-2(11) + 9(-5) + (-23)(-3)
-1(11) + 5(-5) + (-13)(-3) \end{bmatrix} = \begin{bmatrix} 0 - 5 + 6
-22 - 45 + 69
-11 - 25 + 39 \end{bmatrix} = \begin{bmatrix} 1
2
3 \end{bmatrix}\).


Therefore, \(x=1, y=2, z=3\).
Quick Tip: When asked to find the inverse and then solve a system of equations, always check if the matrix of coefficients of the system is the same as the matrix for which you found the inverse. This saves significant time.

Let \(y = x^{\sin x} + (\sin x)^x\).


Let \(u = x^{\sin x}\) and \(v = (\sin x)^x\). Then \(y = u + v\), and \(\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}\).


First, find the derivative of \(u = x^{\sin x}\) using logarithmic differentiation.

\(\ln u = \ln(x^{\sin x}) = \sin x \ln x\).


Differentiating with respect to \(x\):

\(\frac{1}{u}\frac{du}{dx} = (\cos x)(\ln x) + (\sin x)\left(\frac{1}{x}\right)\).

\(\frac{du}{dx} = u \left( \cos x \ln x + \frac{\sin x}{x} \right) = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right)\).


Next, find the derivative of \(v = (\sin x)^x\) using logarithmic differentiation.

\(\ln v = \ln((\sin x)^x) = x \ln(\sin x)\).


Differentiating with respect to \(x\):

\(\frac{1}{v}\frac{dv}{dx} = (1)(\ln(\sin x)) + (x)\left(\frac{1}{\sin x} \cdot \cos x\right)\).

\(\frac{1}{v}\frac{dv}{dx} = \ln(\sin x) + x \cot x\).

\(\frac{dv}{dx} = v (\ln(\sin x) + x \cot x) = (\sin x)^x (\ln(\sin x) + x \cot x)\).


Finally, combine the two derivatives:

\(\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right) + (\sin x)^x (\ln(\sin x) + x \cot x)\).
Quick Tip: For functions of the form \(f(x)^{g(x)}\), logarithmic differentiation is the required method. Never differentiate term by term as in \(x^n\). When there is a sum of such functions, differentiate each part separately.

We are given the function \(y = x + \tan x\).


First, we find the first derivative, \(\frac{dy}{dx}\).

\(\frac{dy}{dx} = \frac{d}{dx}(x + \tan x) = 1 + \sec^2 x\).


Next, we find the second derivative, \(\frac{d^2y}{dx^2}\).

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(1 + \sec^2 x) = 0 + 2(\sec x) \cdot \frac{d}{dx}(\sec x)\).

\(\frac{d^2y}{dx^2} = 2 \sec x (\sec x \tan x) = 2 \sec^2 x \tan x\).


Now, we substitute these results into the left-hand side (LHS) of the equation we need to prove:


LHS = \(\cos^2 x \frac{d^2y}{dx^2} - 2y + 2x\).


Substitute the expression for the second derivative:


LHS = \(\cos^2 x (2 \sec^2 x \tan x) - 2y + 2x\).


Since \(\sec^2 x = \frac{1}{\cos^2 x}\), these terms cancel out:


LHS = \(\cos^2 x \left(2 \frac{1}{\cos^2 x} \tan x\right) - 2y + 2x = 2 \tan x - 2y + 2x\).


Now, substitute the original expression for \(y\), which is \(y = x + \tan x\):


LHS = \(2 \tan x - 2(x + \tan x) + 2x\).


LHS = \(2 \tan x - 2x - 2 \tan x + 2x\).


LHS = \(0\).


Since LHS = 0, which is the right-hand side, the identity is proven.
Quick Tip: In 'prove that' differentiation problems, after finding the required derivatives, substitute them into the given expression. Then, use the original function to substitute for 'y' if it appears in the expression to simplify it to the final result.

The area enclosed by the curve \(x=y^2\) (or \(y=\sqrt{x}\) for the upper half) and the line \(x=4\) is symmetric about the x-axis.


The total area \(A_{total}\) is from \(x=0\) to \(x=4\).

\(A_{total} = \int_{0}^{4} \sqrt{x} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^4 = \frac{2}{3}[4^{3/2} - 0^{3/2}] = \frac{2}{3}(8) = \frac{16}{3}\). (Considering only the area above the x-axis)


The line \(x=a\) divides this area into two equal parts. This means the area from \(x=0\) to \(x=a\) is half of the total area.


Area from \(x=0\) to \(x=a\) is \(A_{0 \to a} = \int_{0}^{a} \sqrt{x} \, dx\).


We are given that \(A_{0 \to a} = \frac{1}{2} A_{total}\).


So, \(\int_{0}^{a} \sqrt{x} \, dx = \frac{1}{2} \left( \frac{16}{3} \right) = \frac{8}{3}\).


Let's evaluate the integral:

\(\left[ \frac{2}{3}x^{3/2} \right]_0^a = \frac{8}{3}\).

\(\frac{2}{3}(a^{3/2} - 0) = \frac{8}{3}\).

\(\frac{2}{3}a^{3/2} = \frac{8}{3}\).


Multiply both sides by \(\frac{3}{2}\):

\(a^{3/2} = 4\).


To find \(a\), raise both sides to the power of \(\frac{2}{3}\):

\(a = (4)^{2/3}\).
Quick Tip: When a problem involves symmetric regions, you can simplify calculations by finding the area of one half (e.g., above the x-axis) and then using that for the required conditions, as the ratios will be the same.

The given lines are in the form \(\vec{r} = \vec{a}_1 + \lambda \vec{b}_1\) and \(\vec{r} = \vec{a}_2 + \mu \vec{b}_2\).


Here, \(\vec{a}_1 = 4\hat{i} - \hat{j} + 2\hat{k}\), \(\vec{b}_1 = \hat{i} + 2\hat{j} - 3\hat{k}\).


And \(\vec{a}_2 = 2\hat{i} + \hat{j} - \hat{k}\), \(\vec{b}_2 = 3\hat{i} + 2\hat{j} - 4\hat{k}\).


The shortest distance \(d\) between two skew lines is given by the formula:

\(d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|\).


First, calculate \(\vec{a}_2 - \vec{a}_1\):

\(\vec{a}_2 - \vec{a}_1 = (2-4)\hat{i} + (1-(-1))\hat{j} + (-1-2)\hat{k} = -2\hat{i} + 2\hat{j} - 3\hat{k}\).


Next, calculate the cross product \(\vec{b}_1 \times \vec{b}_2\):

\(\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & -3
3 & 2 & -4 \end{vmatrix} = \hat{i}(-8 - (-6)) - \hat{j}(-4 - (-9)) + \hat{k}(2 - 6)\).

\(= -2\hat{i} - 5\hat{j} - 4\hat{k}\).


Now, find the magnitude of the cross product:

\(|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-2)^2 + (-5)^2 + (-4)^2} = \sqrt{4 + 25 + 16} = \sqrt{45} = 3\sqrt{5}\).


Next, find the dot product in the numerator:

\((\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-2)(-2) + (2)(-5) + (-3)(-4) = 4 - 10 + 12 = 6\).


Finally, calculate the distance:

\(d = \left| \frac{6}{3\sqrt{5}} \right| = \frac{2}{\sqrt{5}}\) units.
Quick Tip: The numerator of the shortest distance formula, \((\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)\), is a scalar triple product. It can be computed efficiently as a 3x3 determinant with rows formed by the components of \((\vec{a}_2 - \vec{a}_1)\), \(\vec{b}_1\), and \(\vec{b}_2\).

The position vector of point P is \(\vec{p} = 5\hat{i} + 4\hat{j} + 2\hat{k}\).


The equation of the line is \(\vec{r} = \vec{a} + \lambda\vec{b}\), with \(\vec{a} = -\hat{i} + 3\hat{j} + \hat{k}\) and direction vector \(\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}\).


Let M be the foot of the perpendicular from P to the line. The position vector of M, \(\vec{m}\), lies on the line for some value of \(\lambda\).

\(\vec{m} = (-1+2\lambda)\hat{i} + (3+3\lambda)\hat{j} + (1-\lambda)\hat{k}\).


The vector \(\vec{PM}\) is perpendicular to the line's direction vector \(\vec{b}\). Thus, their dot product is zero: \(\vec{PM} \cdot \vec{b} = 0\).


First, find the vector \(\vec{PM} = \vec{m} - \vec{p}\).

\(\vec{PM} = [(-1+2\lambda)-5]\hat{i} + [(3+3\lambda)-4]\hat{j} + [(1-\lambda)-2]\hat{k}\).

\(\vec{PM} = (2\lambda-6)\hat{i} + (3\lambda-1)\hat{j} + (-\lambda-1)\hat{k}\).


Now, set the dot product to zero:

\(((2\lambda-6)\hat{i} + (3\lambda-1)\hat{j} + (-\lambda-1)\hat{k}) \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0\).

\(2(2\lambda-6) + 3(3\lambda-1) - 1(-\lambda-1) = 0\).

\(4\lambda - 12 + 9\lambda - 3 + \lambda + 1 = 0\).

\(14\lambda - 14 = 0 \implies \lambda = 1\).


Substitute \(\lambda=1\) into the position vector of M to find its coordinates.

\(\vec{m} = (-1+2(1))\hat{i} + (3+3(1))\hat{j} + (1-1)\hat{k} = \hat{i} + 6\hat{j} + 0\hat{k}\).


The coordinates of the foot of the perpendicular M are (1, 6, 0).


Now, find the length of the perpendicular, which is the magnitude of \(\vec{PM}\) when \(\lambda=1\).

\(\vec{PM} = (2(1)-6)\hat{i} + (3(1)-1)\hat{j} + (-1-1)\hat{k} = -4\hat{i} + 2\hat{j} - 2\hat{k}\).


Length = \(|\vec{PM}| = \sqrt{(-4)^2 + 2^2 + (-2)^2} = \sqrt{16+4+4} = \sqrt{24} = 2\sqrt{6}\) units.
Quick Tip: To find the foot of the perpendicular from a point to a line, represent a general point on the line using the parameter \(\lambda\). Form the vector from the given point to this general point. The dot product of this vector and the line's direction vector must be zero. Solve for \(\lambda\) to find the specific point.

Let the rectangular part have width (length) \(2x\) and height (breadth) \(2y\).


The semi-circular opening is surmounted on the top length, so its diameter is \(2x\).


The radius of the semi-circle is therefore \(r = x\).


The perimeter of the window consists of the two vertical sides of the rectangle, the bottom side of the rectangle, and the arc of the semi-circle.


Length of the bottom side = \(2x\).


Length of the two vertical sides = \(2y + 2y = 4y\).


Length of the semi-circular arc = \(\frac{1}{2} \times (circumference of a full circle) = \frac{1}{2} \times 2\pi r = \pi r = \pi x\).


The total perimeter is the sum of these lengths:


Perimeter \(P = 2x + 4y + \pi x\).


We are given that the perimeter is 10 m.


Therefore, the relation between x and y is \(10 = 2x + 4y + \pi x\).
Quick Tip: When calculating the perimeter of a composite shape, be careful to only include the outer boundary. The internal line segment where the rectangle and semi-circle meet is not part of the perimeter.

The total area (A) is the sum of the area of the rectangular part and the area of the semi-circular part.


Area of rectangle = (width) \(\times\) (height) = \((2x)(2y) = 4xy\).


Area of semi-circle = \(\frac{1}{2} \pi r^2 = \frac{1}{2} \pi x^2\).


Total Area \(A = 4xy + \frac{1}{2} \pi x^2\).


From part (i), we have the relation from the perimeter: \(10 = 2x + 4y + \pi x\).


We need to express the area in terms of \(x\) only, so we solve this relation for \(y\).

\(4y = 10 - 2x - \pi x\).

\(y = \frac{10 - 2x - \pi x}{4}\).


Now substitute this expression for \(y\) into the area formula.

\(A(x) = 4x \left( \frac{10 - 2x - \pi x}{4} \right) + \frac{1}{2} \pi x^2\).

\(A(x) = x(10 - 2x - \pi x) + \frac{1}{2} \pi x^2\).

\(A(x) = 10x - 2x^2 - \pi x^2 + \frac{1}{2} \pi x^2\).

\(A(x) = 10x - 2x^2 - \frac{1}{2} \pi x^2\).
Quick Tip: In optimization problems, you'll typically have two equations: one for the quantity to be optimized (like area) and one constraint (like perimeter). Use the constraint equation to express one variable in terms of the other, then substitute it into the optimization equation.

From part (ii), the area of the window as a function of \(x\) is:

\(A(x) = 10x - 2x^2 - \frac{\pi}{2}x^2\).


To find the value of \(x\) that maximizes the area, we find the first derivative of \(A(x)\) and set it to zero.

\(\frac{dA}{dx} = \frac{d}{dx} \left( 10x - 2x^2 - \frac{\pi}{2}x^2 \right) = 10 - 4x - \frac{\pi}{2}(2x) = 10 - 4x - \pi x\).


Set the derivative equal to zero:

\(10 - 4x - \pi x = 0\).

\(10 = 4x + \pi x = x(4+\pi)\).

\(x = \frac{10}{4+\pi}\).


To confirm this is a maximum, we check the second derivative:

\(\frac{d^2A}{dx^2} = -4 - \pi\). Since this is negative, the area is maximized at this value of \(x\).


Now we find the corresponding value of \(y\) using the relation from part (i).

\(y = \frac{10 - 2x - \pi x}{4} = \frac{10 - x(2+\pi)}{4}\).


Substitute the value of \(x\):

\(y = \frac{10 - \frac{10}{4+\pi}(2+\pi)}{4} = \frac{1}{4} \left[ 10 \left( 1 - \frac{2+\pi}{4+\pi} \right) \right]\).

\(y = \frac{10}{4} \left[ \frac{(4+\pi) - (2+\pi)}{4+\pi} \right] = \frac{5}{2} \left[ \frac{2}{4+\pi} \right] = \frac{5}{4+\pi}\).
Quick Tip: To find the maximum or minimum of a function, use the first derivative test: find critical points by setting \(f'(x)=0\). Then use the second derivative test: if \(f''(x) < 0\) at the critical point, it's a local maximum; if \(f''(x) > 0\), it's a local minimum.

In this version of the problem, the dimensions of the rectangle are length = \(x\) and breadth = \(y\).


The semi-circle is surmounted on the length \(x\), so its diameter is \(x\).


The radius of the semi-circle is \(r = \frac{x}{2}\).


The perimeter of the window consists of the bottom side, two vertical sides, and the semi-circular arc.


Perimeter \(P = x + 2y + (arc of semi-circle) = x + 2y + \pi r = x + 2y + \pi\left(\frac{x}{2}\right)\).


We are given that the perimeter is 10 m.

\(10 = x + 2y + \frac{\pi x}{2}\).


To express the area in terms of \(x\), we first solve this constraint equation for \(y\).

\(2y = 10 - x - \frac{\pi x}{2}\).

\(y = 5 - \frac{x}{2} - \frac{\pi x}{4}\).


Now, we write the formula for the total area, \(A\).


Area \(A = (Area of rectangle) + (Area of semi-circle) = xy + \frac{1}{2}\pi r^2\).

\(A = xy + \frac{1}{2}\pi\left(\frac{x}{2}\right)^2 = xy + \frac{\pi x^2}{8}\).


Substitute the expression for \(y\) into the area formula:

\(A(x) = x\left(5 - \frac{x}{2} - \frac{\pi x}{4}\right) + \frac{\pi x^2}{8}\).

\(A(x) = 5x - \frac{x^2}{2} - \frac{\pi x^2}{4} + \frac{\pi x^2}{8}\).


Combine the terms with \(x^2\):

\(A(x) = 5x - \frac{x^2}{2} - \frac{2\pi x^2}{8} + \frac{\pi x^2}{8}\).

\(A(x) = 5x - \frac{x^2}{2} - \frac{\pi x^2}{8}\).
Quick Tip: Always pay close attention to how variables are defined. In this problem, changing the rectangle's dimensions from (2x, 2y) in part (i) to (x, y) in this part fundamentally changes all the initial formulas for perimeter and area. You must re-derive everything from scratch.

Let A, B, and C be the events that a selected student is from category A, B, and C respectively.


Total students = 60. Number in A = 6, Number in B = 26, Number in C = \(60-6-26 = 28\).


Probabilities of selecting a student from each category:

\(P(A) = \frac{6}{60} = \frac{1}{10}\), \(P(B) = \frac{26}{60} = \frac{13}{30}\), \(P(C) = \frac{28}{60} = \frac{7}{15}\).


Let E be the event that a student is unable to get good marks.


We are given the conditional probabilities:

\(P(E|A) = 0.002\), \(P(E|B) = 0.02\), \(P(E|C) = 0.20\).


We need to find the total probability of event E, \(P(E)\). We use the Law of Total Probability:

\(P(E) = P(A)P(E|A) + P(B)P(E|B) + P(C)P(E|C)\).

\(P(E) = \left(\frac{6}{60}\right)(0.002) + \left(\frac{26}{60}\right)(0.02) + \left(\frac{28}{60}\right)(0.20)\).

\(P(E) = \frac{1}{60} [6(0.002) + 26(0.02) + 28(0.20)]\).

\(P(E) = \frac{1}{60} [0.012 + 0.52 + 5.6]\).

\(P(E) = \frac{1}{60} [6.132]\).

\(P(E) = 0.1022\).
Quick Tip: The Law of Total Probability is used to find the probability of an event when the sample space is partitioned into several mutually exclusive events. The formula is \(P(E) = \sum_{i=1}^{n} P(A_i)P(E|A_i)\).

We are given that the student was unable to get good marks, so event E has occurred.


We need to find the probability that this student is NOT of category A, which is \(P(A' | E)\).


It is easier to first find the probability that the student IS of category A, given E, which is \(P(A | E)\), and then use the complement rule.


Using Bayes' Theorem: \(P(A|E) = \frac{P(A)P(E|A)}{P(E)}\).


From the previous part, we have all the required values:

\(P(A)P(E|A) = \frac{6}{60} \times 0.002 = \frac{0.012}{60}\).

\(P(E) = \frac{6.132}{60}\).


So, \(P(A|E) = \frac{0.012/60}{6.132/60} = \frac{0.012}{6.132} = \frac{12}{6132}\).


Simplifying the fraction: \(\frac{12}{6132} = \frac{1}{511}\).


Now, we can find the probability that the student is NOT of category A:

\(P(A'|E) = 1 - P(A|E) = 1 - \frac{1}{511} = \frac{510}{511}\).
Quick Tip: Bayes' Theorem is used to find a conditional probability \(P(A|B)\) when you know the reverse conditional probability \(P(B|A)\) and the individual probabilities \(P(A)\) and \(P(B)\). The formula is \(P(A|B) = \frac{P(B|A)P(A)}{P(B)}\).

A function \(f\) is one-one (injective) if for every two distinct elements \(x_1, x_2\) in the domain, their images are distinct, i.e., \(f(x_1) \neq f(x_2)\).


Equivalently, if \(f(x_1) = f(x_2)\), then it must be that \(x_1 = x_2\).


The given function is \(f(x) = x^2\) with domain and codomain as the set of all real numbers, R.


Let's test this with a counterexample.


Choose two distinct real numbers, for instance, \(x_1 = -2\) and \(x_2 = 2\).


Calculate their images under \(f\):

\(f(x_1) = f(-2) = (-2)^2 = 4\).

\(f(x_2) = f(2) = (2)^2 = 4\).


Here, we have \(f(-2) = f(2)\) but \(-2 \neq 2\).


Since two different inputs produce the same output, the function is not one-one.
Quick Tip: To prove a function is not one-one, you only need to find a single counterexample of two different inputs that produce the same output.

A function \(f : A \rightarrow B\) is onto (surjective) if its range is equal to its codomain B.


This means that for every element \(y\) in the codomain B, there must exist at least one element \(x\) in the domain A such that \(f(x) = y\).


The given function is \(f(x) = x^2\) with domain R and codomain R.


The range of \(f(x) = x^2\) is the set of all possible output values. The square of any real number is always non-negative.


So, the range of \(f\) is \([0, \infty)\).


The codomain of \(f\) is given as R, the set of all real numbers.


Since the range \([0, \infty)\) is not equal to the codomain R, the function is not onto.


For example, let's take an element from the codomain that is not in the range, say \(y = -4\).


There is no real number \(x\) such that \(f(x) = x^2 = -4\).


Therefore, the function is not onto.
Quick Tip: To check if a function is onto, compare its range to its specified codomain. If the range is a proper subset of the codomain, the function is not onto.

The function is \(f(x) = x^2\), with both the domain and codomain being the set of natural numbers N = \{1, 2, 3, ...\.


One-one check:


Let \(x_1, x_2\) be two elements in the domain N such that \(f(x_1) = f(x_2)\).


This means \(x_1^2 = x_2^2\).


Taking the square root, we get \(x_1 = \pm x_2\).


However, since the domain is N, both \(x_1\) and \(x_2\) must be positive.


Therefore, we must have \(x_1 = x_2\).


Since \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), the function is one-one.


Onto check:


The range of the function is the set of all output values, which are the squares of natural numbers.


Range = \(\{f(1), f(2), f(3), ...\} = \{1^2, 2^2, 3^2, ...\} = \{1, 4, 9, 16, ...\}\).


The codomain is given as N = \{1, 2, 3, 4, ...\.


The range is a proper subset of the codomain. For example, the number 2 is in the codomain N, but there is no natural number \(x\) such that \(f(x) = x^2 = 2\).


Therefore, the function is not onto.
Quick Tip: The properties of a function (like being one-one or onto) heavily depend on its specified domain and codomain. The same formula \(f(x)=x^2\) can have different properties with different domains/codomains.

The function is \(f(x) = x^2\).


The domain is the set of natural numbers, N = \{1, 2, 3, ...\.


The codomain is the set of perfect squares, S = \{1, 4, 9, 16, ...\.


One-one check:


This is the same as in the previous part. Let \(x_1, x_2 \in N\) such that \(f(x_1) = f(x_2)\).


This implies \(x_1^2 = x_2^2\). Since \(x_1, x_2\) are positive, this means \(x_1 = x_2\).


Therefore, the function is one-one.


Onto check:


The range of the function is the set of all outputs when the inputs are natural numbers.


Range = \(\{f(1), f(2), f(3), ...\} = \{1^2, 2^2, 3^2, ...\} = \{1, 4, 9, 16, ...\}\).


The codomain is given as S = \{1, 4, 9, 16, ...\.


In this case, the range of the function is exactly equal to its codomain.


Therefore, the function is onto.


Since the function is both one-one and onto, it is a bijective function.
Quick Tip: A function becomes onto if its codomain is restricted to be exactly equal to its range. This problem illustrates how changing the codomain can change the surjective property of a function.