Let \(y = \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)\).


The range of the principal value branch for \(\cot^{-1}(x)\) is \((0, \pi)\).


This implies that \(\cot(y) = -\frac{1}{\sqrt{3}}\).


We know that the reference angle for which \(\cot(\theta) = \frac{1}{\sqrt{3}}\) is \(\theta = \frac{\pi}{3}\).


Since \(\cot(y)\) is negative, the angle \(y\) must be in the second quadrant to fall within the principal value range \((0, \pi)\).


The corresponding angle in the second quadrant is given by the identity \(\cot(\pi - \theta) = -\cot(\theta)\).


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


Since \(\frac{2\pi}{3}\) lies in the interval \((0, \pi)\), it is the principal value.
Quick Tip: The principal value branch of \(\cot^{-1}(x)\) is \((0, \pi)\). For a negative input, the angle will always be in the second quadrant \((\frac{\pi}{2}, \pi)\). Remember the identity \(\cot^{-1}(-x) = \pi - \cot^{-1}(x)\) for \(x > 0\).

Given that A is a 3 \(\times\) 3 diagonal matrix.


The diagonal elements are given as \(a_{11} = 1\), \(a_{22} = 5\), and \(a_{33} = -2\).


In a diagonal matrix, all non-diagonal elements are zero.


So, the matrix A can be written as:

\(A = \begin{pmatrix} 1 & 0 & 0
0 & 5 & 0
0 & 0 & -2 \end{pmatrix}\)


The determinant of a diagonal matrix is the product of its main diagonal elements.


Therefore, \(|A| = a_{11} \times a_{22} \times a_{33}\).

\(|A| = 1 \times 5 \times (-2) = -10\).
Quick Tip: The determinant of any diagonal or triangular (upper or lower) matrix is simply the product of the elements on its main diagonal. This is a significant time-saver in calculations.

We are given the relation \(A = kB\), where A and B are square matrices of order n, and k is a scalar.


We need to find the relationship between their determinants, \(|A|\) and \(|B|\).


Taking the determinant on both sides of the equation gives: \(|A| = |kB|\).


A key property of determinants states that for a square matrix M of order n and a scalar c, the determinant of their product is \(|cM| = c^n|M|\).


This is because the scalar c multiplies every element in the matrix M, and since M has n rows, the scalar can be factored out from each of the n rows.


Applying this property to \(|kB|\), we have:

\(|kB| = k^n|B|\).


Therefore, by substitution, we get \(|A| = k^n|B|\).
Quick Tip: Be careful not to confuse the property \(|kA| = k^n|A|\) with scalar multiplication of the determinant itself, which is just \(k|A|\). A scalar multiplies the entire matrix, affecting each of the 'n' rows, hence the power 'n'.

For a function \(f(x)\) to be continuous at a point \(x=c\), we must have \(\lim_{x \to c} f(x) = f(c)\).


For this problem, at \(x=0\), we must have \(\lim_{x \to 0} f(x) = f(0)\).


From the definition of the function, we are given \(f(0) = 1\).


Now, we evaluate the limit:

\(\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin^2(ax)}{x^2}\).


We can rewrite the expression as \(\lim_{x \to 0} \left(\frac{\sin(ax)}{x}\right)^2\).


To use the standard trigonometric limit \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\), we manipulate the expression:

\(\lim_{x \to 0} \left(\frac{\sin(ax)}{ax} \cdot a\right)^2 = \left(\lim_{x \to 0} \frac{\sin(ax)}{ax}\right)^2 \cdot a^2\).


As \(x \to 0\), \(ax \to 0\). The limit becomes \((1)^2 \cdot a^2 = a^2\).


For continuity, we must equate the limit with the function's value:

\(a^2 = f(0) = 1\).


Taking the square root on both sides gives \(a = \pm 1\).
Quick Tip: When dealing with continuity and limits involving trigonometric functions, always try to reduce the expression to the fundamental forms like \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) by algebraic manipulation.

The problem requires us to solve for \(x\) by equating two \(2 \times 2\) determinants.


The determinant of a \(2 \times 2\) matrix \(\begin{vmatrix} a & b
c & d \end{vmatrix}\) is calculated as \(ad - bc\).


First, we calculate the determinant on the left-hand side (LHS):


LHS = \((2x)(x) - (5)(12) = 2x^2 - 60\).


Next, we calculate the determinant on the right-hand side (RHS):


RHS = \((6)(3) - (-5)(4) = 18 - (-20) = 18 + 20 = 38\).


Now, we set the LHS equal to the RHS as per the given equation:

\(2x^2 - 60 = 38\).


Add 60 to both sides of the equation:

\(2x^2 = 38 + 60 = 98\).


Divide both sides by 2:

\(x^2 = \frac{98}{2} = 49\).


Take the square root of both sides to solve for \(x\):

\(x = \pm\sqrt{49} = \pm 7\).
Quick Tip: A common mistake when solving equations involving determinants is to equate the corresponding elements of the matrices. Always remember to first calculate the scalar value of each determinant and then equate those values.

We use the addition theorem of probability: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).


Substituting the given values:

\(0.9 = P(A) + P(B) - 0.4\).


Rearranging the equation to find the sum of probabilities of A and B:

\(P(A) + P(B) = 0.9 + 0.4 = 1.3\).


We need to find the value of \(P(\bar{A}) + P(\bar{B})\).


Using the complement rule, \(P(\bar{A}) = 1 - P(A)\) and \(P(\bar{B}) = 1 - P(B)\).


Therefore, \(P(\bar{A}) + P(\bar{B}) = (1 - P(A)) + (1 - P(B))\).

\(= 2 - (P(A) + P(B))\).


Substituting the value of \(P(A) + P(B)\) we found earlier:

\(= 2 - 1.3 = 0.7\).
Quick Tip: Remember the relationship for complements: \(P(\bar{A}) + P(\bar{B}) = 2 - (P(A) + P(B))\). This allows you to quickly solve for the sum of complements if you know the sum of the original probabilities.

The given matrix A is a scalar matrix. It can be written as the product of a scalar and the identity matrix.

\(A = 5 \begin{pmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{pmatrix} = 5I\).


We need to find \(A^3\).

\(A^3 = (5I)^3 = 5^3 \cdot I^3\).


We know that for the identity matrix \(I\), any integer power of \(I\) is \(I\) itself. So, \(I^3 = I\).


Therefore, \(A^3 = 125 \cdot I\).

\(A^3 = 125 \begin{pmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 125 & 0 & 0
0 & 125 & 0
0 & 0 & 125 \end{pmatrix}\).


Alternatively, for a diagonal matrix, raising it to a power \(n\) is equivalent to raising each diagonal element to the power \(n\).

\(A^3 = \begin{pmatrix} 5^3 & 0 & 0
0 & 5^3 & 0
0 & 0 & 5^3 \end{pmatrix} = \begin{pmatrix} 125 & 0 & 0
0 & 125 & 0
0 & 0 & 125 \end{pmatrix}\).
Quick Tip: For any scalar matrix \(A = kI\) of order \(m\), its \(n^{th}\) power is given by \(A^n = k^n I\). This is a very useful shortcut for powers of scalar and diagonal matrices.

We will examine each statement, which are properties of the transpose of a matrix.


(A) \((A')' = A\): The transpose of the transpose of a matrix is the original matrix itself. This statement is correct.


(B) \((kA)' = kA'\): The transpose of a matrix multiplied by a scalar is equal to the scalar multiplied by the transpose of the matrix. This statement is correct.


(C) \((A' + B')' = A + B\): Using the property that \((X+Y)' = X' + Y'\), we have \((A' + B')' = (A')' + (B')'\). Using the property from (A), this becomes \(A+B\). This statement is correct.


(D) \((AB)' = A'B'\): This statement is incorrect. The correct property for the transpose of a product of matrices is the reversal law, which states \((AB)' = B'A'\). The order of the matrices is reversed.


Therefore, the statement that is not correct is (D).
Quick Tip: The most commonly confused property of matrix transpose is the product rule. Always remember the "reversal law" or "socks and shoes rule": \((AB)' = B'A'\). You reverse the order, just like you take off your shoes before your socks.

First, we define the function \(y = x|x|\) piecewise.

\(y = f(x) = \begin{cases} x(x) = x^2, & if x \ge 0
x(-x) = -x^2, & if x < 0 \end{cases}\).


The required area is given by the definite integral \(\int_{-2}^{2} |f(x)| \,dx\).


We split the integral at \(x=0\) according to the piecewise definition.


Area = \(\int_{-2}^{0} |-x^2| \,dx + \int_{0}^{2} |x^2| \,dx\).


Since \(x^2 \ge 0\) for all real \(x\), we have \(|-x^2| = x^2\) and \(|x^2| = x^2\).


Area = \(\int_{-2}^{0} x^2 \,dx + \int_{0}^{2} x^2 \,dx\).


Now, we evaluate the integrals:

\(\int x^2 \,dx = \frac{x^3}{3}\).


Area = \(\left[\frac{x^3}{3}\right]_{-2}^{0} + \left[\frac{x^3}{3}\right]_{0}^{2}\).


Area = \(\left(\frac{0^3}{3} - \frac{(-2)^3}{3}\right) + \left(\frac{2^3}{3} - \frac{0^3}{3}\right)\).


Area = \(\left(0 - \frac{-8}{3}\right) + \left(\frac{8}{3} - 0\right) = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}\) square units.
Quick Tip: When integrating functions involving absolute values, always split the integral at the points where the expression inside the absolute value changes sign. For \(y = x|x|\), the critical point is \(x=0\).

The function is the greatest integer function, \(f(x) = [x]\). We need to check its continuity and differentiability at \(x=2\).


Step 1: Check for continuity at \(x=2\).


A function is continuous at \(x=c\) if \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)\).


Left-Hand Limit (LHL): \(\lim_{x \to 2^-} [x] = 1\). (For \(x\) slightly less than 2, like 1.999, the greatest integer is 1).


Right-Hand Limit (RHL): \(\lim_{x \to 2^+} [x] = 2\). (For \(x\) slightly greater than 2, like 2.001, the greatest integer is 2).


Function Value: \(f(2) = [2] = 2\).


Since the LHL \(\neq\) RHL, the limit does not exist, and the function is not continuous at \(x=2\).


Step 2: Check for differentiability at \(x=2\).


A fundamental theorem in calculus states that if a function is differentiable at a point, it must be continuous at that point.


Since we have already established that \(f(x) = [x]\) is not continuous at \(x=2\), it cannot be differentiable at \(x=2\).


Therefore, the function is neither continuous nor differentiable at \(x=2\).
Quick Tip: The greatest integer function \(f(x) = [x]\) is discontinuous at every integer value. A direct consequence of this is that it is also not differentiable at any integer value.

We start by using the double-angle identity for cosine: \(\cos 2A = 2\cos^2 A - 1\).


Applying this to the numerator, we get:


Numerator = \((\cos 2x - \cos 2\theta) = (2\cos^2 x - 1) - (2\cos^2 \theta - 1)\).

\(= 2\cos^2 x - 1 - 2\cos^2 \theta + 1 = 2(\cos^2 x - \cos^2 \theta)\).


Now, we use the difference of squares formula, \(a^2 - b^2 = (a-b)(a+b)\).

\(2(\cos^2 x - \cos^2 \theta) = 2(\cos x - \cos \theta)(\cos x + \cos \theta)\).


The integral becomes: \(\int \frac{2(\cos x - \cos \theta)(\cos x + \cos \theta)}{\cos x - \cos \theta} dx\).


Canceling the term \((\cos x - \cos \theta)\), the integral simplifies to:

\(\int 2(\cos x + \cos \theta) dx\).


We integrate term by term. Note that \(\cos \theta\) is a constant with respect to \(x\).

\(= 2 \left( \int \cos x \,dx + \int \cos \theta \,dx \right)\).

\(= 2 (\sin x + x \cos \theta) + C\).
Quick Tip: When integrating expressions with trigonometric functions of different angles (like \(x\) and \(\theta\)), remember that the term involving \(\theta\) is treated as a constant if the integration is with respect to \(x\).

This integral can be solved using the method of substitution.


Let \(u = 5x^2 + 1\).


Then, the differential is \(du = 10x \,dx\).


This implies that \(2x \,dx = \frac{du}{5}\).


Now, we need to change the limits of integration from \(x\) to \(u\).


When \(x = 0\), \(u = 5(0)^2 + 1 = 1\).


When \(x = 1\), \(u = 5(1)^2 + 1 = 6\).


Substituting \(u\) and the new limits into the integral:

\(\int_1^6 \frac{1}{u} \cdot \frac{du}{5} = \frac{1}{5} \int_1^6 \frac{1}{u} \,du\).

\(= \frac{1}{5} [\log|u|]_1^6\).

\(= \frac{1}{5} (\log 6 - \log 1)\).


Since \(\log 1 = 0\), the result is \(\frac{1}{5} \log 6\).
Quick Tip: For integrals of the form \(\int \frac{f'(x)}{f(x)} dx\), the result is \(\log|f(x)| + C\). Here, the derivative of \(5x^2+1\) is \(10x\). The numerator is \(2x\), so you can adjust the constant to get the answer quickly: \(\frac{1}{5} \int \frac{10x}{5x^2+1} dx = \frac{1}{5}\log(5x^2+1)\).

The slope of the curve is given by its first derivative, \(\frac{dy}{dx}\).


Let \(m(x) = \frac{dy}{dx} = \frac{d}{dx}(-x^3 + 3x^2 + 8x - 20)\).

\(m(x) = -3x^2 + 6x + 8\).


To find the value of \(x\) where the slope \(m(x)\) is maximum, we need to find the derivative of \(m(x)\) and set it to zero.

\(m'(x) = \frac{d}{dx}(-3x^2 + 6x + 8) = -6x + 6\).


Set \(m'(x) = 0\) to find the critical point:

\(-6x + 6 = 0 \implies 6x = 6 \implies x = 1\).


To confirm this is a maximum, we check the second derivative of the slope, \(m''(x)\).

\(m''(x) = -6\), which is negative. This confirms that the slope is maximum at \(x=1\).


Now, find the corresponding \(y\)-coordinate by substituting \(x=1\) into the original equation for the curve.

\(y = -(1)^3 + 3(1)^2 + 8(1) - 20 = -1 + 3 + 8 - 20 = 10 - 20 = -10\).


The point where the slope is maximum is \((1, -10)\).
Quick Tip: To find the maximum or minimum of a quantity (like slope), find the expression for that quantity, then use the first and second derivative tests on that expression.

The given differential equation is not in a standard linear form. We need to rearrange it.


Start with \(\frac{dx}{dy} = -\frac{1+\sin x}{x+y \cos x}\).


It's easier to work with \(\frac{dy}{dx}\). Inverting both sides gives:

\(\frac{dy}{dx} = -\frac{x+y \cos x}{1+\sin x}\).

\(\frac{dy}{dx} = -\frac{x}{1+\sin x} - \frac{y \cos x}{1+\sin x}\).


Rearranging to the standard linear form \(\frac{dy}{dx} + P(x)y = Q(x)\):

\(\frac{dy}{dx} + \left(\frac{\cos x}{1+\sin x}\right)y = -\frac{x}{1+\sin x}\).


Here, \(P(x) = \frac{\cos x}{1+\sin x}\).


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

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


To solve the integral, let \(u = 1 + \sin x\). Then \(du = \cos x \,dx\).


The integral becomes \(\int \frac{1}{u} du = \ln|u| = \ln|1+\sin x|\).


So, the integrating factor is \(I.F. = e^{\ln|1+\sin x|}\).


Using the property \(e^{\ln a} = a\), we get:

\(I.F. = 1 + \sin x\).
Quick Tip: If a differential equation doesn't immediately look linear in the form \(\frac{dy}{dx} + P(x)y = Q(x)\), try rearranging it or checking if it's linear in the form \(\frac{dx}{dy} + P(y)x = Q(y)\).

First, we analyze the constraints to determine the feasible region.


The constraints \(x \ge 0\) and \(y \ge 0\) restrict the region to the first quadrant.


For the constraint \(x + 2y \le 10\), we consider the line \(x + 2y = 10\).

The intercepts are \((10, 0)\) (Point D) and \((0, 5)\) (Point A).

Testing the origin \((0,0)\): \(0 + 2(0) \le 10\), which is \(0 \le 10\) (True). So, the region is on the origin side of this line.


For the constraint \(3x + y \le 15\), we consider the line \(3x + y = 15\).

The intercepts are \((5, 0)\) (Point E) and \((0, 15)\) (Point B).

Testing the origin \((0,0)\): \(3(0) + 0 \le 15\), which is \(0 \le 15\) (True). So, the region is on the origin side of this line.


The feasible region is the intersection of these shaded areas in the first quadrant.


The vertices of this region are:

- The origin, O(0,0).

- Point A(0,5), the y-intercept of \(x+2y=10\).

- Point E(5,0), the x-intercept of \(3x+y=15\).

- The intersection of the two lines, Point C(4,3).


The feasible region is the polygon formed by connecting these vertices: O-A-C-E.


Therefore, the correct feasible region is AOEC.
Quick Tip: To determine the feasible region for a set of linear inequalities, graph the boundary lines and use a test point (like the origin (0,0), if it's not on a line) to determine which side of each line to shade. The feasible region is the common shaded area.

The given differential equation is \(\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2} = \frac{d^2y}{dx^2}\).


To determine the degree, we must first clear any fractional powers of the derivatives.


We can do this by squaring both sides of the equation.

\(\left(\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}\right)^2 = \left(\frac{d^2y}{dx^2}\right)^2\).


This simplifies to \(\left[1 + \left(\frac{dy}{dx}\right)^2\right]^3 = \left(\frac{d^2y}{dx^2}\right)^2\).


Now the equation is a polynomial in its derivatives.


The order of a differential equation is the order of the highest derivative present.

The highest derivative is \(\frac{d^2y}{dx^2}\), so the order is 2.


The degree of a differential equation is the highest power of the highest-order derivative after the equation has been cleared of radicals and fractions.

The highest derivative is \(\frac{d^2y}{dx^2}\) and its power is 2. So, the degree is 2.


The sum of the order and degree is \(2 + 2 = 4\).
Quick Tip: Before determining the degree of a differential equation, always ensure it is expressed as a polynomial in its derivatives. This often involves eliminating radicals (like square roots or fractional powers) and denominators containing derivatives.

We start by expanding the given dot product:

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


Since the dot product is commutative (\(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\)), the middle terms cancel out.

\(\vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 512\).


Using the property \(\vec{v} \cdot \vec{v} = |\vec{v}|^2\), the equation becomes:

\(|\vec{a}|^2 - |\vec{b}|^2 = 512\).


We are also given the relation \(|\vec{a}| = 3|\vec{b}|\).


Substitute this relation into the equation:

\((3|\vec{b}|)^2 - |\vec{b}|^2 = 512\).

\(9|\vec{b}|^2 - |\vec{b}|^2 = 512\).

\(8|\vec{b}|^2 = 512\).

\(|\vec{b}|^2 = \frac{512}{8} = 64\).


Since magnitude must be non-negative, \(|\vec{b}| = \sqrt{64} = 8\).


Now, we find the magnitude of \(\vec{a}\):

\(|\vec{a}| = 3|\vec{b}| = 3(8) = 24\).


So, the respective values are \(|\vec{a}| = 24\) and \(|\vec{b}| = 8\).
Quick Tip: Remember the vector algebra identity \((\vec{a} - \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2\). This is analogous to the difference of squares formula \((x-y)(x+y) = x^2 - y^2\) and is very useful.

Let the position vector of point A be \(\vec{OA}\) and the position vector of point B be \(\vec{OB}\).


The vector \(\vec{a}\) is given as the position vector with its tip at (2, -3). So, \(\vec{a} = 2\hat{i} - 3\hat{j}\).


The coordinates of point A are (-4, 5), so its position vector is \(\vec{OA} = -4\hat{i} + 5\hat{j}\).


The vector from A to B is defined as \(\vec{AB} = \vec{OB} - \vec{OA}\).


We are given that \(\vec{AB} = \vec{a}\).


Therefore, \(\vec{OB} - \vec{OA} = \vec{a}\).


To find the position vector of B, we rearrange the equation: \(\vec{OB} = \vec{OA} + \vec{a}\).


Substitute the known vectors:

\(\vec{OB} = (-4\hat{i} + 5\hat{j}) + (2\hat{i} - 3\hat{j})\).


Combine the respective components:

\(\vec{OB} = (-4+2)\hat{i} + (5-3)\hat{j}\).

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


The coordinates of point B are the components of its position vector \(\vec{OB}\), which are (-2, 2).
Quick Tip: Always remember the "tip-minus-tail" rule for vectors: the vector from point A to point B, \(\vec{AB}\), is found by subtracting the position vector of the tail (A) from the position vector of the tip (B). \(\vec{AB} = \vec{OB} - \vec{OA}\).

First, let's verify Assertion (A).

The constraints are \(2x + y \ge 8\), \(x + 2y \ge 10\), and \(x, y \ge 0\).


For \(2x + y \ge 8\), testing the origin (0,0) gives \(0 \ge 8\), which is false. The region is away from the origin.


For \(x + 2y \ge 10\), testing the origin (0,0) gives \(0 \ge 10\), which is false. The region is away from the origin.

The shaded region in the graph correctly represents the common area satisfying these conditions in the first quadrant. So, Assertion (A) is true.


Next, let's verify Reason (R).

The corner points of the unbounded feasible region are A(0,8), B(2,4), and C(10,0).


Let's evaluate the objective function Z at these points:

Z(A) = \(50(0) + 70(8) = 560\).

Z(B) = \(50(2) + 70(4) = 100 + 280 = 380\).

Z(C) = \(50(10) + 70(0) = 500\).


The minimum value at a corner point is 380 at B(2,4).

For an unbounded region, this is the minimum value if the open half-plane defined by \(Z < 380\) has no points in common with the feasible region.


The inequality is \(50x + 70y < 380\). This region lies on the origin side of the line \(50x + 70y = 380\).

Since the feasible region is entirely on the other side of this line (away from the origin), there are no common points.

Thus, the minimum value is indeed 380. Reason (R) is a true statement.


Finally, let's connect Assertion and Reason.

Assertion (A) states that the region is correctly identified. Reason (R) correctly applies the theorem for finding the minimum of an objective function in an unbounded feasible region. The procedure described in (R) is the justification for concluding that a minimum exists for the LPP defined in (A). Therefore, R is the correct explanation for the solution of the LPP described in A.
Quick Tip: When dealing with an unbounded feasible region in an LPP, finding the minimum/maximum value at the corner points is not enough. You must also check the open half-plane condition (\(Z < Z_{min}\) for minimization, \(Z > Z_{max}\) for maximization) to ensure it does not overlap with the feasible region.

Let's analyze Assertion (A).

The function is \(f(x) = x^2\) with domain A = [-1, 1] and codomain A = [-1, 1].

A function is onto (surjective) if its range is equal to its codomain.


Let's find the range of \(f(x) = x^2\) for \(x \in [-1, 1]\).

The square of any real number is non-negative. The minimum value of \(x^2\) is at \(x=0\), which is 0. The maximum value occurs at \(x=1\) and \(x=-1\), which is 1.


So, the range of \(f\) is the interval [0, 1].

The codomain is given as A = [-1, 1].

Since Range [0, 1] \(\neq\) Codomain [-1, 1], the function is not onto. Thus, Assertion (A) is true.


Now, let's analyze Reason (R).

Reason (R) states that for an element \(y = -1\) in the codomain A, there is no corresponding \(x\) in the domain A such that \(f(x) = y\).


Let's check: \(f(x) = -1 \implies x^2 = -1\).

The solutions for \(x\) are \(x = \pm \sqrt{-1} = \pm i\). These are complex numbers, not real numbers, so they do not belong to the domain A = [-1, 1].


This shows that there is at least one element in the codomain (-1, in this case) that has no pre-image in the domain. This is the definition of a function not being onto.

Therefore, Reason (R) is true.


Finally, we connect them. Reason (R) gives a specific counterexample that proves Assertion (A) is correct. It directly explains why the function is not onto.
Quick Tip: To prove a function is not onto, you only need to find one element in the codomain that is not in the range (i.e., has no pre-image in the domain).

The domain of the function \(\sin^{-1}(u)\) is the interval \([-1, 1]\).


This means the argument of the function, \(u = x^2 - 3\), must satisfy the inequality:

\(-1 \le x^2 - 3 \le 1\).


We can split this into two separate inequalities.


First inequality:

\(x^2 - 3 \ge -1\)

\(x^2 \ge 2\).


This implies \(x \le -\sqrt{2}\) or \(x \ge \sqrt{2}\).


Second inequality:

\(x^2 - 3 \le 1\)

\(x^2 \le 4\).


This implies \(-2 \le x \le 2\).


The domain is the intersection of the solutions of both inequalities.


We need to find the values of \(x\) that satisfy both (\(x \le -\sqrt{2}\) or \(x \ge \sqrt{2}\)) and (\(-2 \le x \le 2\)).


The intersection is the set of numbers that are between -2 and \(-\sqrt{2}\) (inclusive) or between \(\sqrt{2}\) and 2 (inclusive).


Therefore, the domain is \([-2, -\sqrt{2}] \cup [\sqrt{2}, 2]\).
Quick Tip: To find the domain of a composite function like \(f(g(x))\), you must find the values of \(x\) such that \(g(x)\) is in the domain of \(f\). For inverse trigonometric functions, knowing their standard domains is crucial.

Let \(r_i\) be the inner radius and \(r_o\) be the outer radius of the hollow sphere.


The volume \(V\) of the hollow sphere is given by the formula:

\(V = \frac{4}{3}\pi (r_o^3 - r_i^3)\).


We are given that the volume is constant, so its derivative with respect to time \(t\) is zero, i.e., \(\frac{dV}{dt} = 0\).


We are also given that the inner radius increases at 2 cm/s, so \(\frac{dr_i}{dt} = 2\) cm/s.


Differentiating the volume formula with respect to time \(t\):

\(\frac{dV}{dt} = \frac{4}{3}\pi \left(3r_o^2 \frac{dr_o}{dt} - 3r_i^2 \frac{dr_i}{dt}\right)\).


Since \(\frac{dV}{dt} = 0\), we have:

\(0 = \frac{4}{3}\pi \left(3r_o^2 \frac{dr_o}{dt} - 3r_i^2 \frac{dr_i}{dt}\right)\).

\(r_o^2 \frac{dr_o}{dt} - r_i^2 \frac{dr_i}{dt} = 0\).

\(r_o^2 \frac{dr_o}{dt} = r_i^2 \frac{dr_i}{dt}\).


We need to find \(\frac{dr_o}{dt}\) at the instant when \(r_i = 2\) cm and \(r_o = 4\) cm.


Substitute the known values into the equation:

\((4)^2 \frac{dr_o}{dt} = (2)^2 (2)\).

\(16 \frac{dr_o}{dt} = 4 \times 2 = 8\).

\(\frac{dr_o}{dt} = \frac{8}{16} = \frac{1}{2}\) cm/s.


The rate of increase of the outer radius is \(\frac{1}{2}\) cm/s.
Quick Tip: In related rates problems, the first step is always to identify an equation that connects the variables. Then, differentiate that equation implicitly with respect to time (\(t\)) and substitute the known values to solve for the unknown rate.

The two points that trisect the line segment AB will divide it in the ratios 1:2 and 2:1.


Let P and Q be the points of trisection. P divides AB in the ratio 1:2, and Q divides AB in the ratio 2:1.


We use the section formula for a point dividing a line segment with endpoints \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) in the ratio \(m:n\):

\(\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)\).


For point P, \(m=1, n=2\), A(4, 1, -2) is \((x_1, y_1, z_1)\) and B(6, 2, -3) is \((x_2, y_2, z_2)\).

\(P_x = \frac{1(6) + 2(4)}{1+2} = \frac{6+8}{3} = \frac{14}{3}\).

\(P_y = \frac{1(2) + 2(1)}{1+2} = \frac{2+2}{3} = \frac{4}{3}\).

\(P_z = \frac{1(-3) + 2(-2)}{1+2} = \frac{-3-4}{3} = -\frac{7}{3}\).


So, the coordinates of the first point P are \(\left(\frac{14}{3}, \frac{4}{3}, -\frac{7}{3}\right)\).


For point Q, \(m=2, n=1\).

\(Q_x = \frac{2(6) + 1(4)}{2+1} = \frac{12+4}{3} = \frac{16}{3}\).

\(Q_y = \frac{2(2) + 1(1)}{2+1} = \frac{4+1}{3} = \frac{5}{3}\).

\(Q_z = \frac{2(-3) + 1(-2)}{2+1} = \frac{-6-2}{3} = -\frac{8}{3}\).


So, the coordinates of the second point Q are \(\left(\frac{16}{3}, \frac{5}{3}, -\frac{8}{3}\right)\).


The two points of trisection are \(\left(\frac{14}{3}, \frac{4}{3}, -\frac{7}{3}\right)\) and \(\left(\frac{16}{3}, \frac{5}{3}, -\frac{8}{3}\right)\).
Quick Tip: Trisection means dividing into three equal parts. This requires two points, which divide the segment in the ratios 1:2 and 2:1 respectively.

Let \(y = \frac{\sin x}{\sqrt{\cos x}}\). We use the quotient rule: \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}\).


Here, \(u = \sin x\) and \(v = \sqrt{\cos x} = (\cos x)^{1/2}\).


First, find the derivatives of \(u\) and \(v\):

\(u' = \frac{d}{dx}(\sin x) = \cos x\).

\(v' = \frac{d}{dx}((\cos x)^{1/2}) = \frac{1}{2}(\cos x)^{-1/2} \cdot (-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}}\).


Now apply the quotient rule:

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

\(\frac{dy}{dx} = \frac{\cos^{3/2} x + \frac{\sin^2 x}{2\sqrt{\cos x}}}{\cos x}\).


To simplify the numerator, find a common denominator:

\(\frac{dy}{dx} = \frac{\frac{2(\cos^{3/2} x)(\sqrt{\cos x}) + \sin^2 x}{2\sqrt{\cos x}}}{\cos x} = \frac{2\cos^2 x + \sin^2 x}{2\sqrt{\cos x} \cdot \cos x}\).

\(\frac{dy}{dx} = \frac{2\cos^2 x + \sin^2 x}{2\cos^{3/2} x}\).


Using the identity \(\sin^2 x = 1 - \cos^2 x\):

\(\frac{dy}{dx} = \frac{2\cos^2 x + (1 - \cos^2 x)}{2\cos^{3/2} x} = \frac{1 + \cos^2 x}{2\cos^{3/2} x}\).
Quick Tip: When using the quotient rule with complex terms, calculate the derivatives of the numerator (\(u'\)) and the denominator (\(v'\)) separately first to avoid errors in the main formula.

Given the function \(y = 5 \cos x - 3 \sin x\).


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

\(\frac{dy}{dx} = \frac{d}{dx}(5 \cos x - 3 \sin x) = 5(-\sin x) - 3(\cos x) = -5 \sin x - 3 \cos x\).


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

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(-5 \sin x - 3 \cos x) = -5(\cos x) - 3(-\sin x) = -5 \cos x + 3 \sin x\).


Now, we need to show that \(\frac{d^2y}{dx^2} + y = 0\).


Substitute the expressions for \(\frac{d^2y}{dx^2}\) and \(y\) into the left-hand side (LHS):


LHS = \((-5 \cos x + 3 \sin x) + (5 \cos x - 3 \sin x)\).


Combine like terms:


LHS = \((-5 \cos x + 5 \cos x) + (3 \sin x - 3 \sin x) = 0 + 0 = 0\).


Since LHS = 0, which is the right-hand side (RHS), the relation is proved.
Quick Tip: Functions of the form \(y = A \cos(kx) + B \sin(kx)\) are solutions to the second-order linear homogeneous differential equation \(\frac{d^2y}{dx^2} + k^2y = 0\). This is a fundamental concept in differential equations.

Let the given vectors be \(\vec{a} = 3\hat{i} - 2\hat{j} + \hat{k}\) and \(\vec{b} = 4\hat{i} + 3\hat{j} - 2\hat{k}\).


A vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is given by their cross product, \(\vec{c} = \vec{a} \times \vec{b}\).

\(\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -2 & 1
4 & 3 & -2 \end{vmatrix}\).

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

\(\vec{c} = \hat{i}(4 - 3) - \hat{j}(-6 - 4) + \hat{k}(9 + 8)\).

\(\vec{c} = 1\hat{i} - (-10)\hat{j} + 17\hat{k} = \hat{i} + 10\hat{j} + 17\hat{k}\).


This vector \(\vec{c}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\). Now we need a vector with magnitude 5.


First, find the magnitude of \(\vec{c}\):

\(|\vec{c}| = \sqrt{1^2 + 10^2 + 17^2} = \sqrt{1 + 100 + 289} = \sqrt{390}\).


The unit vector in the direction of \(\vec{c}\) is \(\hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{\hat{i} + 10\hat{j} + 17\hat{k}}{\sqrt{390}}\).


The required vector is a vector with magnitude 5 in the direction of \(\hat{c}\).


Required vector = \(5 \hat{c} = 5 \left(\frac{\hat{i} + 10\hat{j} + 17\hat{k}}{\sqrt{390}}\right) = \frac{5}{\sqrt{390}}(\hat{i} + 10\hat{j} + 17\hat{k})\).
Quick Tip: The process to find a vector with a specific magnitude 'M' in a certain direction is: 1. Find any vector in that direction. 2. Normalize it to get a unit vector. 3. Multiply the unit vector by the desired magnitude M.

We are given two conditions:


1) \(\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}\)


2) \(\vec{a} \times \vec{b} = \vec{a} \times \vec{c}\)


From the first condition, we can write:

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


Using the distributive property of the dot product:

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


This implies that either \(\vec{a} = \vec{0}\), or \(\vec{b} - \vec{c} = \vec{0}\), or \(\vec{a}\) is perpendicular to \((\vec{b} - \vec{c})\).


From the second condition, we can write:

\(\vec{a} \times \vec{b} - \vec{a} \times \vec{c} = \vec{0}\).


Using the distributive property of the cross product:

\(\vec{a} \times (\vec{b} - \vec{c}) = \vec{0}\).


This implies that either \(\vec{a} = \vec{0}\), or \(\vec{b} - \vec{c} = \vec{0}\), or \(\vec{a}\) is parallel to \((\vec{b} - \vec{c})\).


We are given that \(\vec{a} \neq \vec{0}\).


So we have two possibilities for the vector \((\vec{b} - \vec{c})\):


- \((\vec{b} - \vec{c}) = \vec{0}\), which means \(\vec{b} = \vec{c}\).


- Or, \(\vec{a}\) is simultaneously perpendicular and parallel to the vector \((\vec{b} - \vec{c})\).


A non-zero vector (\(\vec{a}\)) cannot be both parallel and perpendicular to another non-zero vector.


Therefore, the only way for \(\vec{a}\) to be both parallel and perpendicular to \((\vec{b} - \vec{c})\) is if the vector \((\vec{b} - \vec{c})\) is the zero vector.


So, in either case, we must have \((\vec{b} - \vec{c}) = \vec{0}\).


This implies \(\vec{b} = \vec{c}\). Hence proved.
Quick Tip: The conditions \(\vec{a} \cdot \vec{x} = 0\) and \(\vec{a} \times \vec{x} = \vec{0}\) for a non-zero vector \(\vec{a}\) together imply that \(\vec{x} = \vec{0}\). This is a powerful result in vector algebra.

The given function is \(f(x) = \sin 3x - \cos 3x\).


For a function to be strictly increasing, its first derivative must be greater than zero, i.e., \(f'(x) > 0\).


First, we find the derivative of \(f(x)\):

\(f'(x) = \frac{d}{dx}(\sin 3x - \cos 3x) = 3\cos 3x - (-3\sin 3x) = 3(\cos 3x + \sin 3x)\).


Now, we set \(f'(x) > 0\):

\(3(\cos 3x + \sin 3x) > 0 \implies \sin 3x + \cos 3x > 0\).


To solve this inequality, we can write the expression in the form \(R\sin(3x+\alpha)\).

\(f'(x) = 3\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin 3x + \frac{1}{\sqrt{2}}\cos 3x\right) = 3\sqrt{2}\sin(3x + \frac{\pi}{4})\).


We need \(3\sqrt{2}\sin(3x + \frac{\pi}{4}) > 0\), which means \(\sin(3x + \frac{\pi}{4}) > 0\).


The sine function is positive when its angle is in the interval \((2k\pi, (2k+1)\pi)\) for any integer \(k\).


The domain for \(x\) is \(0 < x < \frac{\pi}{2}\).


This implies the domain for the angle \(3x + \frac{\pi}{4}\) is:

\(0 < 3x < \frac{3\pi}{2} \implies \frac{\pi}{4} < 3x + \frac{\pi}{4} < \frac{3\pi}{2} + \frac{\pi}{4} = \frac{7\pi}{4}\).


We need to find where the sine is positive within the angle range \((\frac{\pi}{4}, \frac{7\pi}{4})\).


This occurs when the angle is between \(\frac{\pi}{4}\) and \(\pi\).


So, we must have \(\frac{\pi}{4} < 3x + \frac{\pi}{4} < \pi\).


Subtracting \(\frac{\pi}{4}\) from all parts of the inequality:

\(0 < 3x < \pi - \frac{\pi}{4} = \frac{3\pi}{4}\).


Dividing by 3:

\(0 < x < \frac{\pi}{4}\).


Thus, the function is strictly increasing on the interval \((0, \frac{\pi}{4})\).
Quick Tip: To solve inequalities like \(\sin \theta + \cos \theta > 0\), it is often easiest to convert the expression to the form \(R\sin(\theta + \alpha)\) or \(R\cos(\theta - \alpha)\). This simplifies finding the intervals where the function is positive or negative.

We are given the vector equation \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\).


To find the angle between \(\vec{a}\) and \(\vec{b}\), we isolate these two vectors on one side.

\(\vec{a} + \vec{b} = -\vec{c}\).


Now, we take the magnitude squared (or dot product with itself) of both sides.

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


Using the property \(|\vec{u}+\vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2\vec{u}\cdot\vec{v}\), we expand the left side.

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


Let \(\theta\) be the angle between vectors \(\vec{a}\) and \(\vec{b}\). Then \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\).

\(|\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\theta = |\vec{c}|^2\).


Substitute the given magnitudes: \(|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7\).

\((3)^2 + (5)^2 + 2(3)(5)\cos\theta = (7)^2\).

\(9 + 25 + 30\cos\theta = 49\).

\(34 + 30\cos\theta = 49\).

\(30\cos\theta = 49 - 34 = 15\).

\(\cos\theta = \frac{15}{30} = \frac{1}{2}\).


Therefore, the angle \(\theta = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}\) or \(60^\circ\).
Quick Tip: When given a vector sum equal to zero (e.g., \(\vec{a}+\vec{b}+\vec{c}=\vec{0}\)), you can find the angle between any two vectors by isolating them on one side, moving the third to the other side, and then squaring the magnitudes of both sides.

We need to prove the given identity. Let's start by considering the square of the left-hand side (LHS).


LHS\(^2 = \left(\frac{1}{2}|\vec{a} - \vec{b}|\right)^2 = \frac{1}{4}|\vec{a} - \vec{b}|^2\).


We know that \(|\vec{u}|^2 = \vec{u} \cdot \vec{u}\). So, \(|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})\).


Expanding the dot product:

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

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


We are given that \(\vec{a}\) and \(\vec{b}\) are unit vectors, so \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\).


Also, \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = (1)(1)\cos\theta = \cos\theta\).


Substituting these values:

\(|\vec{a} - \vec{b}|^2 = 1^2 - 2\cos\theta + 1^2 = 2 - 2\cos\theta = 2(1 - \cos\theta)\).


Using the trigonometric half-angle identity \(1 - \cos\theta = 2\sin^2\frac{\theta}{2}\):

\(|\vec{a} - \vec{b}|^2 = 2(2\sin^2\frac{\theta}{2}) = 4\sin^2\frac{\theta}{2}\).


Now, substitute this back into the expression for LHS\(^2\):


LHS\(^2 = \frac{1}{4}(4\sin^2\frac{\theta}{2}) = \sin^2\frac{\theta}{2}\).


Taking the square root of both sides:


LHS = \(\sqrt{\sin^2\frac{\theta}{2}} = \sin\frac{\theta}{2}\) (assuming \(\sin\frac{\theta}{2} \ge 0\), which is standard).


Hence, \(\frac{1}{2}|\vec{a} - \vec{b}| = \sin\frac{\theta}{2}\). Proved.
Quick Tip: The identities \(|\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}\) and \(|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b}\) are fundamental in vector algebra and frequently used in proofs involving vector magnitudes.

The given equation is \(x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x\).


This is a homogeneous differential equation. To solve it, we first isolate \(\frac{dy}{dx}\).

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


Let \(y = vx\). Then \(\frac{dy}{dx} = v + x\frac{dv}{dx}\).


Substituting this into the equation:

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


Subtracting \(v\) from both sides gives:

\(x\frac{dv}{dx} = \frac{1}{\cos(v)}\).


Now, we separate the variables:

\(\cos(v) \,dv = \frac{1}{x} \,dx\).


Integrate both sides of the equation:

\(\int \cos(v) \,dv = \int \frac{1}{x} \,dx\).

\(\sin(v) = \ln|x| + C\), where C is the constant of integration.


Finally, substitute back \(v = \frac{y}{x}\):

\(\sin\left(\frac{y}{x}\right) = \ln|x| + C\).


This is the general solution to the differential equation.
Quick Tip: To identify a homogeneous differential equation, check if it can be written in the form \(\frac{dy}{dx} = F\left(\frac{y}{x}\right)\). If so, the substitution \(y=vx\) will always successfully separate the variables.

Let C be the event that the student buys a colouring book, and B be the event that she buys a box of colours.


We are given the following probabilities:

\(P(C) = 0.7\)

\(P(B) = 0.2\)

\(P(C|B) = 0.3\) (probability of buying a colouring book given she bought a box of colours).


(i) Find the probability of buying both, which is \(P(C \cap B)\).


Using the formula for conditional probability, \(P(C|B) = \frac{P(C \cap B)}{P(B)}\).


We can rearrange this to solve for \(P(C \cap B)\):

\(P(C \cap B) = P(C|B) \times P(B)\).

\(P(C \cap B) = 0.3 \times 0.2 = 0.06\).


The probability that she buys both is 0.06.


(ii) Find the probability of buying a box of colours given she bought a colouring book, which is \(P(B|C)\).


Using the conditional probability formula again:

\(P(B|C) = \frac{P(B \cap C)}{P(C)}\).


Since \(P(B \cap C) = P(C \cap B)\), we can use the value from part (i).

\(P(B|C) = \frac{0.06}{0.7} = \frac{6}{70} = \frac{3}{35}\).


The probability that she buys a box of colours given she bought a colouring book is \(\frac{3}{35}\).
Quick Tip: Master the conditional probability formula: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\). It's the key to solving most problems involving "given that" scenarios and forms the basis of Bayes' Theorem.

Total fruits in the box = 6 apples + 4 oranges = 10 fruits.


The probability of drawing an orange (success) is \(p = P(O) = \frac{4}{10} = \frac{2}{5}\).


The probability of drawing an apple (failure) is \(q = P(A) = \frac{6}{10} = \frac{3}{5}\).


Since the fruit is replaced after each draw, the trials are independent. This is a binomial distribution with \(n=3\) trials.


Let \(X\) be the random variable representing the number of oranges drawn. \(X\) can take values 0, 1, 2, 3.


The probability mass function is \(P(X=k) = \binom{3}{k} p^k q^{3-k} = \binom{3}{k} \left(\frac{2}{5}\right)^k \left(\frac{3}{5}\right)^{3-k}\).


(i) The probability distribution of X:

\(P(X=0) = \binom{3}{0} \left(\frac{2}{5}\right)^0 \left(\frac{3}{5}\right)^3 = 1 \cdot 1 \cdot \frac{27}{125} = \frac{27}{125}\).

\(P(X=1) = \binom{3}{1} \left(\frac{2}{5}\right)^1 \left(\frac{3}{5}\right)^2 = 3 \cdot \frac{2}{5} \cdot \frac{9}{25} = \frac{54}{125}\).

\(P(X=2) = \binom{3}{2} \left(\frac{2}{5}\right)^2 \left(\frac{3}{5}\right)^1 = 3 \cdot \frac{4}{25} \cdot \frac{3}{5} = \frac{36}{125}\).

\(P(X=3) = \binom{3}{3} \left(\frac{2}{5}\right)^3 \left(\frac{3}{5}\right)^0 = 1 \cdot \frac{8}{125} \cdot 1 = \frac{8}{125}\).


The distribution can be represented in a table:

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

\hline
P(X) & \(\frac{27}{125}\) & \(\frac{54}{125}\) & \(\frac{36}{125}\) & \(\frac{8}{125}\)

\hline
\end{tabular


(ii) The expectation of the random variable, \(E(X)\):


For a binomial distribution, the expectation (mean) is given by \(E(X) = np\).

\(E(X) = 3 \times \frac{2}{5} = \frac{6}{5} = 1.2\).


Alternatively, using the distribution table:

\(E(X) = \sum x_i P(X=x_i) = 0\left(\frac{27}{125}\right) + 1\left(\frac{54}{125}\right) + 2\left(\frac{36}{125}\right) + 3\left(\frac{8}{125}\right)\).

\(E(X) = \frac{0 + 54 + 72 + 24}{125} = \frac{150}{125} = \frac{6}{5}\).
Quick Tip: Recognizing a scenario as a binomial distribution (fixed number of independent trials, constant probability of success) allows you to use the shortcuts for mean (\(E(X) = np\)) and variance (\(Var(X) = npq\)), which is much faster than calculating from the full probability distribution table.

This integral is well-suited for a substitution.


Let \(u = x^2\). Then, the differential is \(du = 2x \,dx\).


Substituting \(u\) and \(du\) into the integral, we get:

\(\int \frac{du}{(u+3)(u-5)}\).


Now we use the method of partial fractions to decompose the integrand.

\(\frac{1}{(u+3)(u-5)} = \frac{A}{u+3} + \frac{B}{u-5}\).


Multiplying by \((u+3)(u-5)\), we get \(1 = A(u-5) + B(u+3)\).


To find B, let \(u=5\): \(1 = A(0) + B(5+3) \implies 1 = 8B \implies B = \frac{1}{8}\).


To find A, let \(u=-3\): \(1 = A(-3-5) + B(0) \implies 1 = -8A \implies A = -\frac{1}{8}\).


The integral becomes:

\(\int \left(\frac{-1/8}{u+3} + \frac{1/8}{u-5}\right) du = \frac{1}{8} \int \left(\frac{1}{u-5} - \frac{1}{u+3}\right) du\).


Integrating with respect to \(u\):

\(= \frac{1}{8} (\ln|u-5| - \ln|u+3|) + C\).


Using the logarithm property \(\ln a - \ln b = \ln(a/b)\):

\(= \frac{1}{8} \ln\left|\frac{u-5}{u+3}\right| + C\).


Finally, substitute back \(u=x^2\):

\(= \frac{1}{8} \ln\left|\frac{x^2-5}{x^2+3}\right| + C\).
Quick Tip: Look for substitutions before resorting to more complex methods. If the numerator is the derivative (or a multiple of the derivative) of a part of the denominator, a substitution is usually the most efficient approach. Here, \(2x\) is the derivative of \(x^2\).

The integrand contains absolute value functions. We need to split the integral at the points where the arguments of the absolute value functions become zero.


The critical points are \(x=2\) and \(x=4\). Since the interval of integration is \([1, 4]\), we must split the integral at \(x=2\).


The integral can be written as:

\(\int_1^2 (|x-2| + |x-4|) dx + \int_2^4 (|x-2| + |x-4|) dx\).


Now we evaluate the absolute values in each interval:


For the interval \([1, 2]\):

\(x-2 \le 0\), so \(|x-2| = -(x-2) = 2-x\).

\(x-4 < 0\), so \(|x-4| = -(x-4) = 4-x\).


The first integral is \(\int_1^2 ((2-x) + (4-x)) dx = \int_1^2 (6-2x) dx\).


For the interval \([2, 4]\):

\(x-2 \ge 0\), so \(|x-2| = x-2\).

\(x-4 \le 0\), so \(|x-4| = -(x-4) = 4-x\).


The second integral is \(\int_2^4 ((x-2) + (4-x)) dx = \int_2^4 (2) dx\).


Now we evaluate both integrals:

\(\int_1^2 (6-2x) dx = [6x - x^2]_1^2 = (6(2) - 2^2) - (6(1) - 1^2) = (12-4) - (6-1) = 8 - 5 = 3\).

\(\int_2^4 (2) dx = [2x]_2^4 = 2(4) - 2(2) = 8 - 4 = 4\).


The total value is the sum of the results from the two integrals:


Total value = \(3 + 4 = 7\).
Quick Tip: When integrating an absolute value function \(|f(x)|\), always split the integral at the roots of \(f(x)=0\) that fall within the integration limits. Then, replace \(|f(x)|\) with \(f(x)\) where it's positive and \(-f(x)\) where it's negative.

First, we identify the feasible region by graphing the constraints. The corner points of this region determine the potential locations for the maximum value.


The constraints are:

1. \(x + 2y \le 120\)

2. \(x + y \ge 60\)

3. \(x - 2y \ge 0 \implies x \ge 2y\)

4. \(x \ge 0, y \ge 0\)


Let's find the corner points (vertices) of the feasible region by finding the intersection of the boundary lines.


Vertex A: Intersection of \(x+y=60\) and \(x-2y=0\).

Substitute \(x=2y\) into the first equation: \(2y+y=60 \implies 3y=60 \implies y=20\). Then \(x=2(20)=40\). So, A = (40, 20).


Vertex B: Intersection of \(x+2y=120\) and \(x-2y=0\).

Substitute \(x=2y\) into the first equation: \(2y+2y=120 \implies 4y=120 \implies y=30\). Then \(x=2(30)=60\). So, B = (60, 30).


Vertex C: Intersection of \(x+y=60\) and the x-axis (\(y=0\)).
\(x+0=60 \implies x=60\). So, C = (60, 0).


Vertex D: Intersection of \(x+2y=120\) and the x-axis (\(y=0\)).
\(x+0=120 \implies x=120\). So, D = (120, 0).


The vertices of the feasible region are A(40, 20), B(60, 30), D(120, 0), and C(60, 0).


Now, evaluate the objective function \(Z = 5x + 10y\) at each vertex:


At A(40, 20): \(Z = 5(40) + 10(20) = 200 + 200 = 400\).


At B(60, 30): \(Z = 5(60) + 10(30) = 300 + 300 = 600\).


At D(120, 0): \(Z = 5(120) + 10(0) = 600 + 0 = 600\).


At C(60, 0): \(Z = 5(60) + 10(0) = 300 + 0 = 300\).


The maximum value of Z is 600.


This maximum value occurs at two adjacent vertices, B(60, 30) and D(120, 0).


When the maximum (or minimum) value of an LPP occurs at two adjacent corner points, it also occurs at every point on the line segment connecting them.


The line segment connecting B and D is part of the line \(x+2y=120\).


Therefore, the maximum value of Z=600 occurs at all points on the line segment joining (60, 30) and (120, 0).
Quick Tip: If the optimal value of the objective function in an LPP is achieved at more than one corner point, then it is achieved at every point on the line segment connecting these corner points. This results in infinitely many optimal solutions.

First, we define the function \(y = 1 + |x + 1|\) as a piecewise function.


The expression inside the absolute value, \(x+1\), is zero at \(x=-1\).


Case 1: When \(x \ge -1\), \(|x+1| = x+1\).

So, \(y = 1 + (x+1) = x+2\).


Case 2: When \(x < -1\), \(|x+1| = -(x+1)\).

So, \(y = 1 - (x+1) = -x\).


The function is \(y = \begin{cases} -x, & if x < -1
x+2, & if x \ge -1 \end{cases}\).


The area of the region is given by the definite integral of the function from \(x=-2\) to \(x=2\).


We must split the integral at the critical point \(x=-1\).


Area = \(\int_{-2}^{2} y \,dx = \int_{-2}^{-1} (-x) \,dx + \int_{-1}^{2} (x+2) \,dx\).


Now, we evaluate the first integral:

\(\int_{-2}^{-1} (-x) \,dx = \left[-\frac{x^2}{2}\right]_{-2}^{-1} = \left(-\frac{(-1)^2}{2}\right) - \left(-\frac{(-2)^2}{2}\right) = \left(-\frac{1}{2}\right) - \left(-\frac{4}{2}\right) = -\frac{1}{2} + 2 = \frac{3}{2}\).


Next, we evaluate the second integral:

\(\int_{-1}^{2} (x+2) \,dx = \left[\frac{x^2}{2} + 2x\right]_{-1}^{2} = \left(\frac{2^2}{2} + 2(2)\right) - \left(\frac{(-1)^2}{2} + 2(-1)\right)\).

\(= (2+4) - \left(\frac{1}{2} - 2\right) = 6 - \left(-\frac{3}{2}\right) = 6 + \frac{3}{2} = \frac{15}{2}\).


The total area is the sum of the two parts:


Total Area = \(\frac{3}{2} + \frac{15}{2} = \frac{18}{2} = 9\) square units.
Quick Tip: To find the area for a curve involving an absolute value, always split the integral at the points where the expression inside the absolute value changes sign. Sketching the graph helps visualize the regions you are integrating.

Let the speeds of the three runners be \(x, y,\) and \(z\) km/h.


From the problem statement, we can form the following system of linear equations:

1. \(x + y + z = 6\)

2. \(x + 2z = 7 \implies x + 0y + 2z = 7\)

3. \(3x + y + z = 12\)


This system can be written in the matrix form \(AX = B\), where:
\(A = \begin{pmatrix} 1 & 1 & 1
1 & 0 & 2
3 & 1 & 1 \end{pmatrix}, X = \begin{pmatrix} x
y
z \end{pmatrix}, B = \begin{pmatrix} 6
7
12 \end{pmatrix}\).


The solution is given by \(X = A^{-1}B\). First, we find the determinant of A.
\(|A| = 1(0-2) - 1(1-6) + 1(1-0) = -2 + 5 + 1 = 4\).


Since \(|A| \neq 0\), the inverse exists. Now we find the adjoint of A. The matrix of cofactors is:
\(C = \begin{pmatrix} (0-2) & -(1-6) & (1-0)
-(1-1) & (1-3) & -(1-3)
(2-0) & -(2-1) & (0-1) \end{pmatrix} = \begin{pmatrix} -2 & 5 & 1
0 & -2 & 2
2 & -1 & -1 \end{pmatrix}\).


The adjoint of A is the transpose of the cofactor matrix:

adj(A) = \(C^T = \begin{pmatrix} -2 & 0 & 2
5 & -2 & -1
1 & 2 & -1 \end{pmatrix}\).


The inverse of A is \(A^{-1} = \frac{1}{|A|}adj(A) = \frac{1}{4} \begin{pmatrix} -2 & 0 & 2
5 & -2 & -1
1 & 2 & -1 \end{pmatrix}\).


Now we solve for X:
\(X = A^{-1}B = \frac{1}{4} \begin{pmatrix} -2 & 0 & 2
5 & -2 & -1
1 & 2 & -1 \end{pmatrix} \begin{pmatrix} 6
7
12 \end{pmatrix}\).

\(X = \frac{1}{4} \begin{pmatrix} (-2)(6) + (0)(7) + (2)(12)
(5)(6) + (-2)(7) + (-1)(12)
(1)(6) + (2)(7) + (-1)(12) \end{pmatrix} = \frac{1}{4} \begin{pmatrix} -12+24
30-14-12
6+14-12 \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 12
4
8 \end{pmatrix}\).

\(X = \begin{pmatrix} 3
1
2 \end{pmatrix}\).


Therefore, the speeds are \(x=3\) km/h, \(y=1\) km/h, and \(z=2\) km/h.
Quick Tip: The matrix method for solving linear equations involves three main steps: 1. Set up the equation \(AX=B\). 2. Calculate the inverse matrix \(A^{-1}\) using the formula \(A^{-1} = \frac{1}{|A|}adj(A)\). 3. Find the solution using \(X=A^{-1}B\).

This is a problem of differentiating one function with respect to another. Let the functions be:
\(u = a^{t+\frac{1}{t}}\) and \(v = \left(t+\frac{1}{t}\right)^a\).


We need to find \(\frac{du}{dv}\), which can be calculated as \(\frac{du}{dv} = \frac{du/dt}{dv/dt}\).


First, we find \(\frac{du}{dt}\):

Using the chain rule for exponential functions (\(\frac{d}{dx}c^x = c^x \ln c\)), we have:
\(\frac{du}{dt} = a^{t+\frac{1}{t}} \cdot \ln(a) \cdot \frac{d}{dt}\left(t+\frac{1}{t}\right)\).
\(\frac{d}{dt}\left(t+t^{-1}\right) = 1 - t^{-2} = 1 - \frac{1}{t^2}\).

So, \(\frac{du}{dt} = a^{t+\frac{1}{t}} \cdot \ln(a) \cdot \left(1 - \frac{1}{t^2}\right)\).


Next, we find \(\frac{dv}{dt}\):

Using the power rule and chain rule (\(\frac{d}{dx}x^n = nx^{n-1}\)), we have:
\(\frac{dv}{dt} = a \left(t+\frac{1}{t}\right)^{a-1} \cdot \frac{d}{dt}\left(t+\frac{1}{t}\right)\).
\(\frac{dv}{dt} = a \left(t+\frac{1}{t}\right)^{a-1} \cdot \left(1 - \frac{1}{t^2}\right)\).


Finally, we calculate \(\frac{du}{dv}\):
\(\frac{du}{dv} = \frac{a^{t+\frac{1}{t}} \cdot \ln(a) \cdot \left(1 - \frac{1}{t^2}\right)}{a \left(t+\frac{1}{t}\right)^{a-1} \cdot \left(1 - \frac{1}{t^2}\right)}\).


The term \(\left(1 - \frac{1}{t^2}\right)\) cancels out from the numerator and denominator.
\(\frac{du}{dv} = \frac{a^{t+\frac{1}{t}} \ln(a)}{a \left(t+\frac{1}{t}\right)^{a-1}}\).
Quick Tip: When asked to differentiate \(f(x)\) with respect to \(g(x)\), use parametric differentiation. Let \(u=f(x)\) and \(v=g(x)\), and find \(\frac{du}{dv} = \frac{du/dx}{dv/dx}\).

The given equation involves variables in both the base and the exponent, so we must use logarithmic differentiation.

Let \(u = y^x\), \(v = x^y\), and \(w = x^x\). The equation is \(u+v+w = a^b\).

Differentiating with respect to \(x\), we get \(\frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0\) (since \(a^b\) is a constant).


1. Differentiate \(u = y^x\):
\(\ln u = x \ln y\).
\(\frac{1}{u}\frac{du}{dx} = 1 \cdot \ln y + x \cdot \frac{1}{y} \frac{dy}{dx}\).
\(\frac{du}{dx} = y^x \left(\ln y + \frac{x}{y}\frac{dy}{dx}\right)\).


2. Differentiate \(v = x^y\):
\(\ln v = y \ln x\).
\(\frac{1}{v}\frac{dv}{dx} = \frac{dy}{dx} \ln x + y \cdot \frac{1}{x}\).
\(\frac{dv}{dx} = x^y \left(\ln x \frac{dy}{dx} + \frac{y}{x}\right)\).


3. Differentiate \(w = x^x\):
\(\ln w = x \ln x\).
\(\frac{1}{w}\frac{dw}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1\).
\(\frac{dw}{dx} = x^x(1 + \ln x)\).


Now substitute these back into the main differentiated equation:
\(y^x \left(\ln y + \frac{x}{y}\frac{dy}{dx}\right) + x^y \left(\ln x \frac{dy}{dx} + \frac{y}{x}\right) + x^x(1 + \ln x) = 0\).


Expand and group terms with \(\frac{dy}{dx}\):
\(y^x \ln y + y^x \frac{x}{y}\frac{dy}{dx} + x^y \ln x \frac{dy}{dx} + x^y \frac{y}{x} + x^x(1+\ln x) = 0\).
\(\frac{dy}{dx} \left(x y^{x-1} + x^y \ln x\right) = -[y^x \ln y + y x^{y-1} + x^x(1+\ln x)]\).


Finally, solve for \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = -\frac{y^x \ln y + y x^{y-1} + x^x(1+\ln x)}{x y^{x-1} + x^y \ln x}\).
Quick Tip: For equations of the form \(f(x,y) + g(x,y) + ... = C\), where the terms are complex functions like \(x^y\), differentiate each term implicitly one by one, then collect all the \(\frac{dy}{dx}\) terms to solve.

First, let's write the equation of the line in standard form.
\(\frac{x+2}{5} = \frac{y+1}{2} = \frac{-(z-4)}{-3} \implies \frac{x+2}{5} = \frac{y+1}{2} = \frac{z-4}{3}\).


Let the line be L. The direction vector of the line L is \(\vec{d} = \langle 5, 2, 3 \rangle\).

Let P be the given point (1, 1, 4). Let Q be the foot of the perpendicular from P to the line L.


Any point Q on the line L can be represented by a parameter \(\lambda\):
\(\frac{x+2}{5} = \frac{y+1}{2} = \frac{z-4}{3} = \lambda\).
\(x = 5\lambda - 2\), \(y = 2\lambda - 1\), \(z = 3\lambda + 4\).

So, the coordinates of Q are \((5\lambda - 2, 2\lambda - 1, 3\lambda + 4)\).


The vector \(\vec{PQ}\) is given by the difference in coordinates of Q and P.
\(\vec{PQ} = \langle (5\lambda - 2 - 1), (2\lambda - 1 - 1), (3\lambda + 4 - 4) \rangle\).
\(\vec{PQ} = \langle 5\lambda - 3, 2\lambda - 2, 3\lambda \rangle\).


Since Q is the foot of the perpendicular, the vector \(\vec{PQ}\) must be perpendicular to the direction vector of the line, \(\vec{d}\).

Their dot product must be zero: \(\vec{PQ} \cdot \vec{d} = 0\).
\((5\lambda - 3)(5) + (2\lambda - 2)(2) + (3\lambda)(3) = 0\).
\(25\lambda - 15 + 4\lambda - 4 + 9\lambda = 0\).
\(38\lambda - 19 = 0\).
\(38\lambda = 19 \implies \lambda = \frac{19}{38} = \frac{1}{2}\).


Now, substitute \(\lambda = \frac{1}{2}\) into the coordinates of Q to find the foot of the perpendicular.
\(x = 5\left(\frac{1}{2}\right) - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2}\).
\(y = 2\left(\frac{1}{2}\right) - 1 = 1 - 1 = 0\).
\(z = 3\left(\frac{1}{2}\right) + 4 = \frac{3}{2} + \frac{8}{2} = \frac{11}{2}\).


The foot of the perpendicular is the point \(\left(\frac{1}{2}, 0, \frac{11}{2}\right)\).
Quick Tip: To find the foot of a perpendicular from a point P to a line, find a general point Q on the line in terms of a parameter \(\lambda\). Then, form the vector \(\vec{PQ}\) and use the condition \(\vec{PQ} \cdot \vec{d} = 0\), where \(\vec{d}\) is the direction vector of the line, to solve for \(\lambda\).

Let the given line be L and the given point be P(-1, -1, 2).

Any point Q on the line L can be represented by a parameter \(\lambda\):
\(\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-4}{3} = \lambda\).

The coordinates of Q are \((3\lambda + 1, 2\lambda - 1, 3\lambda + 4)\).


The distance between points P and Q is given by the distance formula:
\(PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2}\).


We are given that this distance is \(2\sqrt{2}\). So, \(PQ^2 = (2\sqrt{2})^2 = 8\).
\(PQ^2 = ((3\lambda+1) - (-1))^2 + ((2\lambda-1) - (-1))^2 + ((3\lambda+4) - 2)^2 = 8\).
\((3\lambda + 2)^2 + (2\lambda)^2 + (3\lambda + 2)^2 = 8\).
\(2(3\lambda + 2)^2 + 4\lambda^2 = 8\).
\(2(9\lambda^2 + 12\lambda + 4) + 4\lambda^2 = 8\).
\(18\lambda^2 + 24\lambda + 8 + 4\lambda^2 = 8\).
\(22\lambda^2 + 24\lambda = 0\).
\(2\lambda(11\lambda + 12) = 0\).


This gives two possible values for \(\lambda\): \(\lambda=0\) or \(\lambda = -\frac{12}{11}\).


We find the coordinates of the point(s) Q by substituting these values of \(\lambda\).


Case 1: \(\lambda = 0\).
\(x = 3(0) + 1 = 1\).
\(y = 2(0) - 1 = -1\).
\(z = 3(0) + 4 = 4\).

The first point is \((1, -1, 4)\).


Case 2: \(\lambda = -\frac{12}{11}\).
\(x = 3(-\frac{12}{11}) + 1 = -\frac{36}{11} + \frac{11}{11} = -\frac{25}{11}\).
\(y = 2(-\frac{12}{11}) - 1 = -\frac{24}{11} - \frac{11}{11} = -\frac{35}{11}\).
\(z = 3(-\frac{12}{11}) + 4 = -\frac{36}{11} + \frac{44}{11} = \frac{8}{11}\).

The second point is \(\left(-\frac{25}{11}, -\frac{35}{11}, \frac{8}{11}\right)\).


There are two such points on the line.
Quick Tip: To find a point on a line at a specific distance from another point, represent the point on the line parametrically, use the distance formula to set up an equation, and solve for the parameter. This will often result in a quadratic equation, yielding two possible points.

(i)

A bijective function must be both one-one (injective) and onto (surjective).


One-one (Injective): A function is one-one if different elements in the domain have different images. In a class, every student is assigned a unique roll number. If \(x_1\) and \(x_2\) are two different students, their roll numbers \(f(x_1)\) and \(f(x_2)\) will also be different. Therefore, the function \(f\) is one-one.


Onto (Surjective): A function is onto if its range is equal to its codomain. The domain A has 30 students, so the range of \(f\) is the set of 30 roll numbers (e.g., \{1, 2, ..., 30\). The codomain is N, the set of all natural numbers \{1, 2, 3, ...\. Since the range is a finite set of 30 elements and the codomain N is an infinite set, the range is not equal to the codomain. Therefore, the function \(f\) is not onto.


Since the function is not onto, it is not a bijective function.


(ii)

The function \(f\) is not bijective because it is not onto. The reason it is not onto is that the codomain N (all natural numbers) is an infinite set, while the range (the set of 30 roll numbers of the students) is a finite set. There exist natural numbers (like 31, 32, etc.) in the codomain that are not the roll number of any student in the class, meaning they have no pre-image in the domain A.


(iii) (a)

The set of roll numbers is S = \{1, 2, 3, ..., 30\. The relation is R = \{(x, y) : x, y \(\in\) S, y = 3x\.


Elements of R: We find pairs (x, y) that satisfy the condition.

If x=1, y=3. If x=2, y=6. ... If x=10, y=30. If x=11, y=33 (not in S).

R = \{(1, 3), (2, 6), (3, 9), (4, 12), (5, 15), (6, 18), (7, 21), (8, 24), (9, 27), (10, 30)\.


Reflexive: For R to be reflexive, (x, x) must be in R for all x \(\in\) S. For x=1, (1, 1) is not in R because 1 \(\neq\) 3(1). So, R is not reflexive.


Symmetric: For R to be symmetric, if (x, y) \(\in\) R, then (y, x) \(\in\) R. We have (1, 3) \(\in\) R. But (3, 1) is not in R because 1 \(\neq\) 3(3). So, R is not symmetric.


Transitive: For R to be transitive, if (x, y) \(\in\) R and (y, z) \(\in\) R, then (x, z) \(\in\) R. We have (1, 3) \(\in\) R and (3, 9) \(\in\) R. For transitivity, (1, 9) must be in R. This is true as 9 = 3(1). Let's check another case. We have (2, 6) \(\in\) R and (6, 18) \(\in\) R. For transitivity, (2, 18) must be in R. But this is not true as 18 \(\neq\) 3(2). So, R is not transitive.


(iii) (b)

The relation is R = \{(x, y) : x, y \(\in\) S, y = \(x^3\)\, where S = \{1, 2, ..., 30\.


Elements of R:

If x=1, y=\(1^3\)=1. So, (1, 1) \(\in\) R.

If x=2, y=\(2^3\)=8. So, (2, 8) \(\in\) R.

If x=3, y=\(3^3\)=27. So, (3, 27) \(\in\) R.

If x=4, y=\(4^3\)=64 (not in S).

R = \{(1, 1), (2, 8), (3, 27)\.


Is R a function?: For a relation from set S to set S to be a function, every element of S must have exactly one image in S.

In the relation R, the roll numbers 4, 5, ..., 30 from the set S do not have any image. For example, there is no pair in R starting with 4.

Since not all elements of the domain S = \{1, 2, ..., 30\ are mapped to an element in S, the relation R is not a function from S to S.
Quick Tip: A function from set A to set B is a rule that assigns each element of A to exactly one element of B. Remember the key conditions: it must be defined for every element in the domain A, and each of these elements must have only one image.

Let B, C, and R be the events of selecting a brinjal, cabbage, and radish seed, respectively.

Let G be the event that a selected seed germinates.

Total seeds = 10 (Brinjal) + 12 (Cabbage) + 8 (Radish) = 30 seeds.


Probabilities of selecting a seed type:

P(B) = \(\frac{10}{30} = \frac{1}{3}\).

P(C) = \(\frac{12}{30} = \frac{2}{5}\).

P(R) = \(\frac{8}{30} = \frac{4}{15}\).


Given conditional probabilities of germination:

P(G|B) = 25% = 0.25 = \(\frac{1}{4}\).

P(G|C) = 35% = 0.35 = \(\frac{7}{20}\).

P(G|R) = 40% = 0.40 = \(\frac{2}{5}\).


(i) Probability of a randomly chosen seed to germinate, P(G):

We use the Law of Total Probability:

\(P(G) = P(B)P(G|B) + P(C)P(G|C) + P(R)P(G|R)\).
\(P(G) = \left(\frac{1}{3}\right)\left(\frac{1}{4}\right) + \left(\frac{2}{5}\right)\left(\frac{7}{20}\right) + \left(\frac{4}{15}\right)\left(\frac{2}{5}\right)\).

\(P(G) = \frac{1}{12} + \frac{14}{100} + \frac{8}{75}\).
\(P(G) = \frac{1}{12} + \frac{7}{50} + \frac{8}{75}\).


The least common multiple of 12, 50, and 75 is 300.
\(P(G) = \frac{1 \times 25}{300} + \frac{7 \times 6}{300} + \frac{8 \times 4}{300}\).

\(P(G) = \frac{25 + 42 + 32}{300} = \frac{99}{300} = \frac{33}{100}\).

The probability of a randomly chosen seed germinating is 0.33.


(ii) Probability that it is a cabbage seed, given that it germinates, P(C|G):


We use Bayes' Theorem:
\(P(C|G) = \frac{P(C)P(G|C)}{P(G)}\).


We have the values for the numerator and denominator from part (i).

Numerator = \(P(C)P(G|C) = \left(\frac{2}{5}\right)\left(\frac{7}{20}\right) = \frac{14}{100} = \frac{7}{50}\).


Denominator = \(P(G) = \frac{33}{100}\).
\(P(C|G) = \frac{7/50}{33/100} = \frac{7}{50} \times \frac{100}{33} = \frac{7 \times 2}{33} = \frac{14}{33}\).

The required probability is \(\frac{14}{33}\).
Quick Tip: Bayes' Theorem problems often follow this structure: Part (i) asks for the total probability of an outcome (e.g., P(G)) using the Law of Total Probability, and Part (ii) asks for a reverse conditional probability (e.g., P(C|G)) which then uses the result from Part (i) as its denominator.

The box has a square base, so length = breadth = \(x\). Height = \(y\).

The volume is \(V = x^2y\) (constant).

The surface area is \(S = 2(x^2 + xy + yx) = 2x^2 + 4xy\).


(i) Express S in terms of x and V:

From the volume equation, we can express \(y\) in terms of \(x\) and \(V\): \(y = \frac{V}{x^2}\).

Substitute this into the surface area formula:
\(S(x) = 2x^2 + 4x\left(\frac{V}{x^2}\right)\).
\(S(x) = 2x^2 + \frac{4V}{x}\).


(ii) Find \(\frac{dS}{dx}\):

We differentiate the expression for \(S(x)\) from part (i) with respect to \(x\).
\(S(x) = 2x^2 + 4Vx^{-1}\).
\(\frac{dS}{dx} = \frac{d}{dx}(2x^2 + 4Vx^{-1}) = 4x - 4Vx^{-2}\).
\(\frac{dS}{dx} = 4x - \frac{4V}{x^2}\).


(iii) (a) Relation between x and y for minimum S:

To find the minimum surface area, we set the first derivative to zero:

\(\frac{dS}{dx} = 0 \implies 4x - \frac{4V}{x^2} = 0\).
\(4x = \frac{4V}{x^2}\).
\(x^3 = V\).


To check if this is a minimum, we find the second derivative:
\(\frac{d^2S}{dx^2} = 4 + \frac{8V}{x^3}\). Since \(x\) and \(V\) are positive, \(\frac{d^2S}{dx^2} > 0\), which confirms a minimum.


Now, substitute \(V = x^2y\) into the condition \(x^3 = V\):
\(x^3 = x^2y\).


Since \(x \neq 0\), we can divide by \(x^2\) to get \(x=y\).

The relation for minimum surface area is that the height is equal to the side of the square base, meaning the box is a cube.


(iii) (b) Show V is maximum for \(x=\sqrt{S/6}\) when S is constant:


We are given \(V(x) = \frac{1}{4}(Sx - 2x^3)\), where S is a constant.

To find the maximum volume, we differentiate V with respect to x and set the derivative to zero.

\(\frac{dV}{dx} = \frac{d}{dx}\left(\frac{1}{4}(Sx - 2x^3)\right) = \frac{1}{4}(S - 6x^2)\).

Set \(\frac{dV}{dx} = 0\):

\(\frac{1}{4}(S - 6x^2) = 0\).
\(S - 6x^2 = 0\).
\(6x^2 = S\).
\(x^2 = \frac{S}{6}\).


Since length \(x\) must be positive, \(x = \sqrt{\frac{S}{6}}\).

To confirm this is a maximum, we find the second derivative:
\(\frac{d^2V}{dx^2} = \frac{d}{dx}\left(\frac{1}{4}(S - 6x^2)\right) = \frac{1}{4}(-12x) = -3x\).


Since \(x > 0\), \(\frac{d^2V}{dx^2}\) is negative, which confirms that the volume is maximum at this value of \(x\). Hence proved.
Quick Tip: For optimization problems involving geometric shapes, the key steps are: 1. Write formulas for the quantity to be optimized (e.g., Area) and the given constraint (e.g., Volume). 2. Use the constraint to express the optimized quantity as a function of a single variable. 3. Find the derivative, set it to zero to find critical points, and use the second derivative test to confirm it's a maximum or minimum.