The given matrix is A = \(\begin{pmatrix} 5 & 0 & 0
0 & 5 & 0
0 & 0 & 5 \end{pmatrix}\).


This is a scalar matrix, which can be written as A = 5I, where I is the identity matrix of order 3.


We need to find A³.


A³ = (5I)³.


Using the property of scalar multiplication with matrices, (kM)ⁿ = kⁿMⁿ.


A³ = 5³I³ = 125I.


Since I³ = I (the identity matrix raised to any power is the identity matrix itself).


So, A³ = 125I = 125 \(\begin{pmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{pmatrix}\).


A³ = \(\begin{pmatrix} 125 & 0 & 0
0 & 125 & 0
0 & 0 & 125 \end{pmatrix}\).


Thus, option (B) is correct.
Quick Tip: For any diagonal matrix D with diagonal elements \(d_1, d_2, ..., d_n\), the matrix D raised to the power k will be a diagonal matrix with diagonal elements \(d_1^k, d_2^k, ..., d_n^k\). In this case, the elements are \(5^3, 5^3, 5^3\).

We know the addition rule for probability: P(A \(\cup\) B) = P(A) + P(B) - P(A \(\cap\) B).


Substituting the given values, we have:

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


From this, we can find the sum P(A) + P(B).


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


We need to find 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)).


P(\(\bar{A}\)) + P(\(\bar{B}\)) = 2 - (P(A) + P(B)).


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


P(\(\bar{A}\)) + P(\(\bar{B}\)) = 2 - 1.3 = 0.7.


Hence, the correct option is (D).
Quick Tip: This problem combines two fundamental probability rules: the addition rule and the complement rule. A useful identity to remember is P(\(\bar{A}\)) + P(\(\bar{B}\)) = 2 - (P(A) + P(B)), which can be derived as shown in the solution.

Let's analyze the dimensions (order) of the matrices A and B.


The order of matrix A is 2 rows \(\times\) 3 columns, i.e., 2 \(\times\) 3.


The order of matrix B is 3 rows \(\times\) 2 columns, i.e., 3 \(\times\) 2.


For the product AB to be defined, the number of columns of A must equal the number of rows of B.


Number of columns of A (3) = Number of rows of B (3).


So, AB is defined. The resulting matrix will have the order 2 \(\times\) 2.


For the product BA to be defined, the number of columns of B must equal the number of rows of A.


Number of columns of B (2) = Number of rows of A (2).


So, BA is also defined. The resulting matrix will have the order 3 \(\times\) 3.


Since both AB and BA are defined, option (C) is the correct statement.
Quick Tip: To check if a matrix product M\(_ {m \times n}\)N\(_ {p \times q}\) is defined, check if the inner dimensions match (n = p). The resulting matrix will have the outer dimensions (m \(\times\) q).

We are given an equation involving two determinants. First, we calculate the value of each determinant.


The determinant of a 2x2 matrix \(\begin{pmatrix} a & b
c & d \end{pmatrix}\) is given by \(ad - bc\).


Evaluating the left-hand side (LHS) determinant:

\(\begin{vmatrix} 2x & 5
12 & x \end{vmatrix} = (2x)(x) - (5)(12) = 2x^2 - 60\).


Evaluating the right-hand side (RHS) determinant:

\(\begin{vmatrix} 6 & -5
4 & 3 \end{vmatrix} = (6)(3) - (-5)(4) = 18 - (-20) = 18 + 20 = 38\).


Now, we set the two results equal to each other as per the given equation:

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


Solving for \(x\):

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

\(2x^2 = 98\).

\(x^2 = 49\).

\(x = \pm\sqrt{49} = \pm7\).


Therefore, the correct option is (C).
Quick Tip: When two determinants are equal, it does not mean their corresponding elements are equal. You must first evaluate each determinant to get a scalar value, and then solve the resulting algebraic equation.

For the function f(x) to be continuous at x = 0, the following condition must be met:

\(\lim_{x \to 0} f(x) = f(0)\).


We are given that \(f(0) = 1\).


Now we must evaluate the limit of f(x) as x approaches 0.

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


This can be rewritten as \(\lim_{x \to 0} \left(\frac{\sin(ax)}{x}\right)^2\).


To use the standard limit formula \(\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1\), we multiply and divide by 'a' inside the parenthesis.

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

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


Since \(\lim_{x \to 0} \frac{\sin(ax)}{ax} = 1\), the expression becomes:

\((a \cdot 1)^2 = a^2\).


For continuity, the limit must equal f(0).

\(a^2 = 1\).


Taking the square root, we get \(a = \pm1\).


Thus, option (C) is the correct answer.
Quick Tip: A key limit to memorize for calculus is \(\lim_{x \to 0} \frac{\sin(kx)}{x} = k\). Using this, we can quickly see that \(\lim_{x \to 0} (\frac{\sin(ax)}{x})^2 = (\lim_{x \to 0} \frac{\sin(ax)}{x})^2 = (a)^2 = a^2\). This saves calculation steps.

The matrix A is a 3 \(\times\) 3 diagonal matrix.


A diagonal matrix has non-zero elements only on its main diagonal. All other elements are zero.


Based on the given values, the matrix A is:


A = \(\begin{pmatrix} a_{11} & 0 & 0
0 & a_{22} & 0
0 & 0 & a_{33} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0
0 & 5 & 0
0 & 0 & -2 \end{pmatrix}\).


The determinant of a diagonal (or triangular) matrix is the product of its main diagonal elements.

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

\(|A| = 1 \times 5 \times (-2)\).

\(|A| = -10\).


Therefore, the correct option is (B).
Quick Tip: The determinant of any diagonal matrix is simply the product of its diagonal entries. This is a very useful shortcut that avoids the full cofactor expansion calculation.

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


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


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


First, we find the acute angle \(\alpha\) for which \(\cot(\alpha) = \frac{1}{\sqrt{3}}\).


We know that \(\cot(\frac{\pi}{3}) = \frac{1}{\sqrt{3}}\). So, \(\alpha = \frac{\pi}{3}\).


Since \(\cot(y)\) is negative, and the principal value range is \((0, \pi)\), y must lie in the second quadrant.


The formula for an angle in the second quadrant is \(\pi - \alpha\).


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

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


The value \(\frac{2\pi}{3}\) is in the range \((0, \pi)\), so it is the correct principal value.


Hence, option (D) is the correct answer.
Quick Tip: For inverse trigonometric functions, remember the principal value ranges. For \(\cot^{-1}(x)\), it's \((0, \pi)\). For a negative argument, the angle will be in Quadrant II, calculated as \(\pi - \alpha\), where \(\alpha\) is the reference angle in Quadrant I.

A matrix is called singular if its determinant is equal to zero.


Let the given matrix be A = \(\begin{pmatrix} 4+x & x-1
-2 & 3 \end{pmatrix}\).


For A to be singular, \(|A| = 0\).


We calculate the determinant:

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


Setting the determinant to zero:

\(3(4+x) + 2(x-1) = 0\).

\(12 + 3x + 2x - 2 = 0\).


Combine like terms:

\(5x + 10 = 0\).

\(5x = -10\).

\(x = \frac{-10}{5} = -2\).


Therefore, the value of x is -2, which corresponds to option (C).
Quick Tip: The terms 'singular matrix', 'non-invertible matrix', and 'matrix with determinant zero' all mean the same thing. If a question uses any of these phrases, immediately set the determinant of the matrix to zero.

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


Step 1: Check for continuity at x = 2.


For continuity, the left-hand limit (LHL), right-hand limit (RHL), and the function value f(2) must be equal.


Left-Hand Limit (LHL): \(\lim_{x \to 2^-} [x]\). For x slightly less than 2 (e.g., 1.999), \([x] = 1\). So, LHL = 1.


Right-Hand Limit (RHL): \(\lim_{x \to 2^+} [x]\). For x slightly greater than 2 (e.g., 2.001), \([x] = 2\). So, RHL = 2.


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


Since LHL \(\neq\) RHL (1 \(\neq\) 2), 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.


Conversely, if a function is not continuous at a point, it cannot be differentiable at that point.


Since we have already established that f(x) is not continuous at x = 2, it follows that f(x) is also not differentiable at x = 2.


Therefore, f is neither continuous nor differentiable at x = 2. This matches option (B).
Quick Tip: The greatest integer function f(x) = [x] is discontinuous (and therefore non-differentiable) at every integer point. It exhibits a 'jump' discontinuity at these points.

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


Let the slope be represented by a function S(x).


Given curve: \(y = -x^3 + 3x^2 + 8x - 20\).

\(S(x) = \frac{dy}{dx} = \frac{d}{dx}(-x^3 + 3x^2 + 8x - 20) = -3x^2 + 6x + 8\).


To find the value of x where the slope S(x) is maximum, we need to find the critical points of S(x). We do this by finding its derivative, S'(x), and setting it to zero.

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


Set \(S'(x) = 0\): \(-6x + 6 = 0 \implies 6x = 6 \implies x = 1\).


To verify that this corresponds to a maximum, we use the second derivative test on S(x).

\(S''(x) = \frac{d}{dx}(-6x + 6) = -6\).


Since \(S''(1) = -6 < 0\), the slope S(x) is indeed maximum at x = 1.


The question asks for the point on the original curve where the slope is maximum. We substitute x = 1 back into the equation for y.

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

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


The required point is (1, -10). This corresponds to option (A).
Quick Tip: To find the maximum or minimum of a rate of change (like slope), you need to differentiate the rate function itself and set that derivative to zero. This is an application of finding the maxima/minima of a function.

We start by simplifying the integrand \(\sqrt{1+\sin x}\).


Using the identities \(1 = \sin^2(\frac{x}{2}) + \cos^2(\frac{x}{2})\) and \(\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{2})\).

\(1 + \sin x = \sin^2(\frac{x}{2}) + \cos^2(\frac{x}{2}) + 2\sin(\frac{x}{2})\cos(\frac{x}{2})\).


This is the expansion of a perfect square: \((\sin(\frac{x}{2}) + \cos(\frac{x}{2}))^2\).


So, \(\sqrt{1+\sin x} = \sqrt{(\sin(\frac{x}{2}) + \cos(\frac{x}{2}))^2} = |\sin(\frac{x}{2}) + \cos(\frac{x}{2})|\).


Assuming the expression is positive (for the principal value of the square root), the integral becomes:

\(\int (\sin(\frac{x}{2}) + \cos(\frac{x}{2})) \,dx\).


Integrating term by term:

\(\int \sin(\frac{x}{2})\,dx + \int \cos(\frac{x}{2})\,dx\).

\(= \frac{-\cos(\frac{x}{2})}{1/2} + \frac{\sin(\frac{x}{2})}{1/2} + C\).

\(= -2\cos(\frac{x}{2}) + 2\sin(\frac{x}{2}) + C\).

\(= 2(\sin(\frac{x}{2}) - \cos(\frac{x}{2})) + C\).


This matches option (B).
Quick Tip: Expressions like \(\sqrt{1 \pm \sin x}\) and \(\sqrt{1 \pm \cos x}\) are common in integration. Always use trigonometric identities to convert them into perfect squares, which simplifies the integrand by removing the square root.

We will solve this definite integral using the method of substitution.


Let \(u = \sin x\).


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


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


When \(x = 0\), \(u = \sin(0) = 0\).


When \(x = \pi/2\), \(u = \sin(\pi/2) = 1\).


The integral transforms to:

\(\int_0^1 e^u \,du\).


The integral of \(e^u\) is simply \(e^u\).


Now, we evaluate the integral at the new limits:

\([e^u]_0^1 = e^1 - e^0\).

\(= e - 1\).


Therefore, the correct option is (C).
Quick Tip: When using substitution for definite integrals, it's often easier to change the limits of integration to the new variable. This avoids the need to substitute the original variable back into the result before evaluating.

First, let's define the function \(y = x|x|\) as a piecewise function.

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


The area is given by the integral of the absolute value of the function, but since the curve is below the x-axis for \(x<0\) and above for \(x>0\), we calculate the area as follows:


Area = \(\int_{-2}^{2} |x|x|\,dx\). We can split the integral based on the definition of the function.


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


Since \(x^2\) is always non-negative, this simplifies to:


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


Notice that the function \(y=x^2\) is an even function, and the integral is symmetric about the origin. Thus, we can write:


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


Now, we compute the integral:


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


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


Area = \(2 \left(\frac{8}{3}\right) = \frac{16}{3}\).


Thus, the correct option is (B).
Quick Tip: For calculating area, always be mindful of the parts of the curve that are below the x-axis. The area is \(\int_a^b |f(x)|\,dx\). If a function is symmetric, use the property of definite integrals to simplify the calculation, for example, \(\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx\) if f(x) is even.

First, we need to rearrange the given differential equation into the standard linear form \(\frac{dy}{dx} + P(x)y = Q(x)\).


The given equation is \((\frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}) \frac{dx}{dy} = 1\).


Taking the reciprocal, we get \(\frac{dy}{dx} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}\).


Rearranging the terms to match the standard form:

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


This is a linear differential equation with \(P(x) = \frac{1}{\sqrt{x}}\) and \(Q(x) = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}\).


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


I.F. = \(e^{\int \frac{1}{\sqrt{x}}dx} = e^{\int x^{-1/2}dx}\).


Integrating the exponent: \(\int x^{-1/2}dx = \frac{x^{1/2}}{1/2} = 2x^{1/2} = 2\sqrt{x}\).


So, the integrating factor is \(e^{2\sqrt{x}}\).


This corresponds to option (C).
Quick Tip: The first step in solving a linear differential equation is to correctly identify P(x) and Q(x) by arranging the equation in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\). The integrating factor formula is \(e^{\int P(x)dx}\).

First, let's identify the order of the differential equation.


The order is defined as the order of the highest derivative present in the equation.


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


Next, let's determine the degree. The degree is the highest power of the highest order derivative after the equation has been cleared of radicals and fractions in its derivatives.


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


To remove the fractional power (3/2), we square both sides of the equation.

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

\((1 + (\frac{dy}{dx})^2)^3 = (\frac{d^2y}{dx^2})^2\).


Now the equation is a polynomial in the derivatives. The highest order derivative is \(\frac{d^2y}{dx^2}\), and its power is 2.


So, the degree is 2.


The question asks for the sum of the order and degree.


Sum = Order + Degree = 2 + 2 = 4.


Thus, the correct option is (D).
Quick Tip: To find the degree of a differential equation, you must first ensure the equation is a polynomial in terms of the derivatives. This means eliminating all fractional powers and radicals involving \(y', y'', \dots\).

The feasible region is the set of all points (x, y) that satisfy all the given constraints simultaneously.


The constraints are:

1. \(x + 2y \le 10\)

2. \(3x + y \le 15\)

3. \(x \ge 0\)

4. \(y \ge 0\)


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


Let's find the corner points of the region bounded by these inequalities.


Point O: Intersection of \(x=0\) and \(y=0\), which is (0, 0).


Point A: Intersection of \(x=0\) and \(x + 2y = 10\). This gives \(0 + 2y = 10 \implies y=5\). So A is (0, 5).


Point E: Intersection of \(y=0\) and \(3x + y = 15\). This gives \(3x + 0 = 15 \implies x=5\). So E is (5, 0).


Point C: Intersection of \(x + 2y = 10\) and \(3x + y = 15\). From the second equation, \(y=15-3x\). Substituting into the first: \(x + 2(15-3x) = 10 \implies x + 30 - 6x = 10 \implies -5x = -20 \implies x=4\). Then \(y = 15 - 3(4) = 3\). So C is (4, 3).


The feasible region is the polygon formed by connecting these vertices in order: O(0,0), A(0,5), C(4,3), E(5,0).


This corresponds to the region labeled AOEC in the given graph (assuming O is the origin).


Therefore, the correct option is (B).
Quick Tip: The feasible region in an LPP is the intersection of the half-planes defined by all constraints. To identify it, plot the boundary lines and use a test point (like (0,0)) to determine which side of each line satisfies the inequality.

Let O be the origin (0, 0).


The position vector \(\vec{a}\) with its tip at (2, -3) can be written as \(\vec{a} = 2\hat{i} - 3\hat{j}\).


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


Let the coordinates of point B be (x, y). Its position vector is \(\vec{OB} = x\hat{i} + y\hat{j}\).


The vector \(\vec{AB}\) is defined as the position vector of B minus the position vector of A.

\(\vec{AB} = \vec{OB} - \vec{OA}\).


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


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


Rearranging to solve for \(\vec{OB}\):

\(\vec{OB} = \vec{a} + \vec{OA}\).


Substituting the vector components:

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

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

\(x\hat{i} + y\hat{j} = -2\hat{i} + 2\hat{j}\).


By comparing the components, we get \(x = -2\) and \(y = 2\).


Therefore, the coordinates of B are (-2, 2). This matches option (C).
Quick Tip: Remember the fundamental vector relationship: \(\vec{AB} = \vec{OB} - \vec{OA}\), where O is the origin. This can be rearranged to find the position vector of the endpoint B: \(\vec{OB} = \vec{OA} + \vec{AB}\).

We are given two equations:

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

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


First, let's expand the dot product in the first equation. This is similar to the algebraic identity \((x-y)(x+y) = x^2 - y^2\).

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


Since \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\) (dot product is commutative) and \(\vec{v} \cdot \vec{v} = |\vec{v}|^2\), the equation simplifies to:

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


Now, we substitute the second given equation, \(|\vec{a}| = 3|\vec{b}|\), into this result.

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

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

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

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


Taking the square root (magnitude is always non-negative), we get \(|\vec{b}| = 8\).


Finally, we find the magnitude of \(\vec{a}\) using the second equation:

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


The respective values are \(|\vec{a}|=24\) and \(|\vec{b}|=8\). This is option (C).
Quick Tip: The vector identity \((\vec{a} - \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2\) is a direct application of the distributive property of the dot product and is very useful for solving problems of this type quickly.

Step 1: Analyze Assertion (A).


The assertion makes two claims: (1) The shaded region is the feasible region for the given constraints. (2) The minimum value of Z is 380 at B(2,4).


Let's check the corner points of the feasible region defined by \(2x+y \ge 8, x+2y \ge 10, x, y \ge 0\).


The vertices are A(0,8), B(2,4) and C(10,0). The shading is above these lines in the first quadrant, which is correct for 'greater than or equal to' inequalities. So the first claim is true.


Let's check the value of Z at these corner points:


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


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


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


The minimum value among the corner points is 380. Since the region is unbounded, we must verify this is the true minimum. The second claim is also true. Thus, Assertion (A) is true.


Step 2: Analyze Reason (R).


Reason (R) states that the region (open half-plane) defined by \(50x + 70y < 380\) has no points in common with the feasible region.


This is the standard test to confirm a minimum value in an unbounded LPP. If the open half-plane corresponding to \(Z < Z_{min}\) has no intersection with the feasible region, then \(Z_{min}\) is the true minimum.


The boundary line \(50x+70y=380\) passes through the point B(2,4). The feasible region lies on and above this line. The region \(50x+70y < 380\) lies strictly below this line. Therefore, there are no common points. Reason (R) is true.


Step 3: Relate Assertion (A) and Reason (R).


Reason (R) provides the mathematical justification required to confirm that the value found at a corner point (380) is indeed the minimum for an unbounded region. It directly explains why the second part of Assertion (A) is correct.


Therefore, both A and R are true, and R is the correct explanation of A.
Quick Tip: For an unbounded feasible region in a minimization LPP, find the minimum value at the corner points, let's call it \(Z_{min}\). To confirm it's the true minimum, you must check that the open half-plane \(Z < Z_{min}\) has no points in common with the feasible region.

Step 1: Analyze Assertion (A).


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


The domain is A = [-1, 1].


The codomain is also A = [-1, 1].


Let's find the range of the function f(x) = x² for the given domain.


For any x in [-1, 1], the value of x² will always be non-negative.


The minimum value of x² is 0 (when x = 0).


The maximum value of x² is 1 (when x = -1 or x = 1).


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


Since the range [0, 1] is a proper subset of the codomain [-1, 1], the range is not equal to the codomain. Therefore, the function f is not onto. Assertion (A) is true.


Step 2: Analyze Reason (R).


Reason (R) provides an example to show why the function is not onto. It considers the value y = -1, which is an element of the codomain A.


For f to be onto, there must exist an x in the domain A such that f(x) = y, i.e., x² = -1.


The solutions to x² = -1 are x = \(\pm\sqrt{-1} = \pm i\).


These solutions are complex numbers, not real numbers. They do not belong to the domain A = [-1, 1].


This shows that there is no pre-image for y = -1 in the domain. This is a valid proof that the function is not onto. Thus, Reason (R) is true.


Step 3: Relate Assertion (A) and Reason (R).


Reason (R) provides a specific counterexample (y=-1) that demonstrates that the range is not equal to the codomain, thereby proving that the function is not onto. It is a direct and correct explanation for Assertion (A).


Therefore, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Quick Tip: To check if a function is onto, compare its range with its given codomain. If the range is smaller than the codomain, the function is not onto. Finding even one element in the codomain that has no pre-image in the domain is sufficient proof.

The domain of the inverse cosine function, \(\cos^{-1}(u)\), is defined for \(-1 \le u \le 1\).


In this problem, the argument is \(u = x^2 - 4\).


Therefore, we must solve the inequality:

\(-1 \le x^2 - 4 \le 1\).


We can split this into two separate inequalities:


1) \(x^2 - 4 \ge -1\)

\(x^2 \ge 3\)


This implies \(x \le -\sqrt{3}\) or \(x \ge \sqrt{3}\). In interval notation, this is \((-\infty, -\sqrt{3}] \cup [\sqrt{3}, \infty)\).


2) \(x^2 - 4 \le 1\)

\(x^2 \le 5\)


This implies \(-\sqrt{5} \le x \le \sqrt{5}\). In interval notation, this is \([-\sqrt{5}, \sqrt{5}]\).


The domain of f(x) is the intersection of the solutions of both inequalities.


We need to find the intersection of \((-\infty, -\sqrt{3}] \cup [\sqrt{3}, \infty)\) and \([-\sqrt{5}, \sqrt{5}]\).


Visualizing on a number line, the common region is from \(-\sqrt{5}\) to \(-\sqrt{3}\) and from \(\sqrt{3}\) to \(\sqrt{5}\).


Thus, the domain of the function is \([-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]\).
Quick Tip: To find the domain of a composite function like \(g(h(x))\), first find the domain of the outer function \(g(u)\), and then find the values of \(x\) for which \(h(x)\) lies within that domain.

Let S be the surface area, V be the volume, and r be the radius of the spherical balloon.


The formulas for surface area and volume are:

\(S = 4\pi r^2\) and \(V = \frac{4}{3}\pi r^3\).


We are given the rate of change of surface area, \(\frac{dS}{dt} = 5\) mm²/s.


We need to find the rate of change of volume, \(\frac{dV}{dt}\), when \(r = 8\) mm.


First, differentiate the surface area formula with respect to time t:

\(\frac{dS}{dt} = \frac{d}{dt}(4\pi r^2) = 8\pi r \frac{dr}{dt}\).


Substitute the given values to find \(\frac{dr}{dt}\) when \(r=8\):

\(5 = 8\pi (8) \frac{dr}{dt} \implies 5 = 64\pi \frac{dr}{dt}\).

\(\frac{dr}{dt} = \frac{5}{64\pi}\) mm/s.


Next, differentiate the volume formula with respect to time t:

\(\frac{dV}{dt} = \frac{d}{dt}(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}\).


Now substitute \(r=8\) and the value of \(\frac{dr}{dt}\) we just found:

\(\frac{dV}{dt} = 4\pi (8)^2 \left(\frac{5}{64\pi}\right)\).

\(\frac{dV}{dt} = 4\pi (64) \left(\frac{5}{64\pi}\right)\).


The terms \(64\pi\) cancel out.

\(\frac{dV}{dt} = 4 \times 5 = 20\).


The volume of the balloon is increasing at a rate of 20 mm³/s.
Quick Tip: In related rates problems, identify the given rate and the rate you need to find. Write down the geometric formula that connects the variables. Differentiate the formula with respect to time (using the chain rule) and then substitute the known values.

Let \(y = \frac{\sin x}{\sqrt{\cos x}}\). We use the quotient rule for differentiation: \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\).


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:

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

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


To simplify the numerator, find a common denominator of \(2\sqrt{\cos x}\):

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

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


Rewrite \(2\cos^2 x\) as \(\cos^2 x + \cos^2 x\):

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


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

\(\frac{dy}{dx} = \frac{\cos^2 x + 1}{2(\cos x)^{3/2}}\).
Quick Tip: When differentiating complex fractions, especially those involving radicals and trigonometric functions, be systematic. Apply the main rule (like quotient rule) first, find the derivatives of the parts separately (using chain rule if needed), and then substitute and simplify carefully.

We are given the function \(y = 5 \cos x - 3 \sin x\).


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

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


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

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

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


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


Substitute the expressions for \(\frac{d^2y}{dx^2}\) and \(y\) into the equation:


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


Combine like terms:


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


LHS = \(0 + 0 = 0\).


Since LHS = RHS (0 = 0), the relation is proven.
Quick Tip: Functions of the form \(y = A \cos(kx) + B \sin(kx)\) are solutions to the second-order linear homogeneous differential equation \(y'' + k^2y = 0\). Recognizing this pattern can help you anticipate the result.

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

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


Expanding the determinant:

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

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

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


Now, we find the magnitude of this vector \(\vec{c}\):

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


To get a vector of magnitude 5, we first find the unit vector in the direction of \(\vec{c}\), which is \(\hat{c} = \frac{\vec{c}}{|\vec{c}|}\).

\(\hat{c} = \frac{\hat{i} + 10\hat{j} + 17\hat{k}}{\sqrt{390}}\).


The required vector is 5 times this unit vector:


Required Vector = \(5\hat{c} = \frac{5(\hat{i} + 10\hat{j} + 17\hat{k})}{\sqrt{390}}\).
Quick Tip: To find a vector of a specific magnitude 'M' in the direction of a given vector '\(\vec{v}\)', the formula is \(M \cdot \frac{\vec{v}}{|\vec{v}|}\). The cross product \(\vec{a} \times \vec{b}\) is always the first step to find a vector perpendicular to both \(\vec{a}\) and \(\vec{b}\).

We are given two conditions:

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

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


Let's manipulate the first equation:

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


Using the distributive property of the dot product:

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


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


Now let's manipulate the second equation:

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


Using the distributive property of the cross product:

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


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


We have two conclusions for the vector \((\vec{b} - \vec{c})\): it is either the zero vector, or it is simultaneously perpendicular and parallel to the non-zero vector \(\vec{a}\).


A non-zero vector cannot be both parallel and perpendicular to another non-zero vector.


Therefore, the only possibility that satisfies both conditions is that the vector \((\vec{b} - \vec{c})\) must be the zero vector.

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


This leads to \(\vec{b} = \vec{c}\). Hence proved.
Quick Tip: This is a standard vector identity proof. The key is to rearrange the given equations to isolate the term \((\vec{b} - \vec{c})\) and then interpret the geometric meaning of the dot product being zero (perpendicular) and the cross product being zero (parallel).

Let the two points that trisect the line segment AB be P and Q.


Point P divides the segment AB in the ratio 1:2.


Point Q divides the segment AB in the ratio 2:1.


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

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


Given points are A(4, 1, -2) and B(6, 2, -3). So, \((x_1, y_1, z_1) = (4, 1, -2)\) and \((x_2, y_2, z_2) = (6, 2, -3)\).


For point P (ratio m:n = 1:2):

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

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

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


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


For point Q (ratio m:n = 2:1):

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

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

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


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


The two points are \((\frac{14}{3}, \frac{4}{3}, -\frac{7}{3})\) and \((\frac{16}{3}, \frac{5}{3}, -\frac{8}{3})\).
Quick Tip: Trisection points divide a line segment into three equal parts. There are always two such points. The first divides the segment in the ratio 1:2, and the second divides it in the ratio 2:1.

For a function f(x) to be decreasing, its first derivative must be less than or equal to zero, i.e., \(f'(x) \le 0\).


First, we find the derivative of the given function:

\(f'(x) = \frac{d}{dx}(\sqrt{3} \sin x - \cos x - 2ax + 6)\).

\(f'(x) = \sqrt{3} \cos x - (-\sin x) - 2a = \sqrt{3} \cos x + \sin x - 2a\).


For f(x) to be decreasing for all \(x \in \mathbb{R}\), we must have:

\(\sqrt{3} \cos x + \sin x - 2a \le 0\) for all \(x \in \mathbb{R}\).


This can be rewritten as \(\sqrt{3} \cos x + \sin x \le 2a\).


To find the condition on 'a', we need to find the maximum value of the expression \(\sqrt{3} \cos x + \sin x\).


For an expression of the form \(A \cos x + B \sin x\), the maximum value is \(\sqrt{A^2 + B^2}\).


Here, \(A = \sqrt{3}\) and \(B = 1\).


Maximum value = \(\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2\).


The inequality becomes: (Maximum value of LHS) \(\le 2a\).

\(2 \le 2a\).


Dividing by 2, we get \(1 \le a\), or \(a \ge 1\).
Quick Tip: To find the range of an expression like \(A \cos x + B \sin x\), convert it to the form \(R \cos(x - \alpha)\) or \(R \sin(x + \alpha)\), where \(R = \sqrt{A^2 + B^2}\). The minimum and maximum values will be \(-R\) and \(R\) respectively.

This integral can be solved using substitution.


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


The integral transforms into:

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


We use partial fraction decomposition. Let \(\frac{1}{(u+3)(u-5)} = \frac{A}{u+3} + \frac{B}{u-5}\).

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


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


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


So, the integral becomes:

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

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

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


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: When an integral contains \(x^2\) and its derivative \(2x\), a substitution like \(u = x^2\) is very effective. This often simplifies the problem to a standard rational function integral that can be solved with partial fractions.

The integrand contains absolute value functions, so we need to split the integral at the points where the expressions inside the absolute value change sign. The critical points are \(x=2\) and \(x=4\).


The interval of integration is \([1, 4]\). We split this interval at \(x=2\).


The integral becomes \(\int_1^2 (|x-2| + |x-4|) \,dx + \int_2^4 (|x-2| + |x-4|) \,dx\).


Case 1: 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 integral is \(\int_1^2 ((2-x) + (4-x)) \,dx = \int_1^2 (6-2x) \,dx\).

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


Case 2: 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 integral is \(\int_2^4 ((x-2) + (4-x)) \,dx = \int_2^4 2 \,dx\).

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


The total value of the integral is the sum of the results from both cases.


Total value = \(3 + 4 = 7\).
Quick Tip: When integrating absolute value functions, always split the integral at the "critical points" where the argument of the absolute value function equals zero. This allows you to rewrite the function as a piecewise function without absolute value bars.

Rearrange the given differential equation:

\(x \,dy = -[x \sin^2(\frac{y}{x}) - y] \,dx\).

\(\frac{dy}{dx} = \frac{-[x \sin^2(\frac{y}{x}) - y]}{x} = -\sin^2(\frac{y}{x}) + \frac{y}{x}\).


This is a homogeneous differential equation. Let \(y = vx\).


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


Substitute into the equation:

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

\(x\frac{dv}{dx} = -\sin^2(v)\).


Separate the variables:

\(\frac{dv}{\sin^2(v)} = -\frac{dx}{x}\).

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


Integrate both sides:

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

\(-\cot(v) = -\ln|x| + C\).

\(\cot(v) = \ln|x| - C\). Let \(C_1 = -C\).

\(\cot(v) = \ln|x| + C_1\).


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

\(\cot(\frac{y}{x}) = \ln|x| + C_1\).


Now apply the given condition: \(y=\frac{\pi}{4}\) when \(x=1\).

\(\cot(\frac{\pi/4}{1}) = \ln|1| + C_1\).

\(\cot(\frac{\pi}{4}) = 0 + C_1\).

\(1 = C_1\).


Therefore, the particular solution is \(\cot(\frac{y}{x}) = \ln|x| + 1\).
Quick Tip: Recognize a homogeneous differential equation if it can be written in the form \(\frac{dy}{dx} = F(\frac{y}{x})\). The standard substitution is always \(y = vx\), which transforms the equation into a separable one in terms of \(v\) and \(x\).

The problem is to maximize Z = 5x + 10y subject to the given constraints.


First, we identify the corner points of the feasible region defined by the inequalities.


The vertices are the intersection points of the boundary lines:


1. Intersection of \(x+y=60\) and \(x-2y=0\): Substituting \(x=2y\) gives \(2y+y=60 \implies 3y=60 \implies y=20, x=40\). Point A(40, 20).


2. Intersection of \(x+2y=120\) and \(x-2y=0\): Substituting \(x=2y\) gives \(2y+2y=120 \implies 4y=120 \implies y=30, x=60\). Point B(60, 30).


3. Intersection of \(x+2y=120\) and \(y=0\) (x-axis): \(x=120\). Point C(120, 0).


4. Intersection of \(x+y=60\) and \(y=0\) (x-axis): \(x=60\). Point D(60, 0).


All four points satisfy all the given constraints and form the vertices of the feasible region.


Now, we evaluate the objective function Z at each corner point:


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


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


The maximum value of Z is 600.


This maximum value occurs at two adjacent corner points, B(60, 30) and C(120, 0).


When the optimal value occurs at two adjacent vertices, it also occurs at every point on the line segment connecting them.


The points giving the maximum value are (60, 30) and (120, 0).
Quick Tip: The optimal solution (maximum or minimum) of an LPP always occurs at one of the corner points (vertices) of the feasible region. If the optimal value is found at two adjacent vertices, then all points on the line segment joining these two vertices are also optimal solutions.

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 rearrange the equation to isolate these vectors:

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


Now, we take the magnitude squared of both sides. This is equivalent to taking the dot product of each side with itself.

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


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:

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


We know that \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\), where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).

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


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 \(\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}\) or 60°.
Quick Tip: When given a relationship like \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\) and magnitudes, a good strategy to find the angle between two vectors is to isolate them on one side of the equation and then square the magnitudes of both sides.

We need to prove the identity. Let's start by considering the square of the magnitude \(|\vec{a} - \vec{b}|\).

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


Expanding the dot product:

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


Using the properties \(\vec{u} \cdot \vec{u} = |\vec{u}|^2\) and \(\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}\):

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


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 into the expression:

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


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

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


Taking the square root of both sides:

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


(We take the positive root as magnitude is non-negative and for \(0 \le \theta \le \pi\), \(\sin(\theta/2) \ge 0\)).


Finally, dividing both sides by 2:

\(\frac{1}{2}|\vec{a} - \vec{b}| = \sin\frac{\theta}{2}\). Hence proved.
Quick Tip: The expressions \(|\vec{a} + \vec{b}|^2\) and \(|\vec{a} - \vec{b}|^2\) are frequently used in vector proofs. They relate vector magnitudes to the dot product and can be simplified using trigonometric identities, especially half-angle formulas.

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) We need to find the probability that she buys both, which is P(C \(\cap\) B).


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


Rearranging the formula: \(P(C \cap B) = P(C|B) \times P(B)\).

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


The probability of buying both is 0.06.


(ii) We need to find the probability that she buys a box of colours given she bought a colouring book, which is P(B|C).


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


Since \(P(B \cap C) = P(C \cap B)\), we have:


P(B|C) = \(\frac{0.06}{0.7}\).


P(B|C) = \(\frac{6/100}{7/10} = \frac{6}{100} \times \frac{10}{7} = \frac{6}{70} = \frac{3}{35}\).


The probability of buying a box of colours given she bought a colouring book is \(\frac{3}{35}\).
Quick Tip: The core formula for conditional probability is \(P(A|B) = P(A \cap B) / P(B)\). This can be rearranged to find the intersection probability: \(P(A \cap B) = P(A|B) \times P(B)\). This is often called the multiplication rule of probability.

This is a sequence of Bernoulli trials since the fruit is replaced after each draw.


Total fruits = 6 apples + 4 oranges = 10.


Probability of drawing an orange (success), \(p = \frac{4}{10} = \frac{2}{5}\).


Probability of drawing an apple (failure), \(q = 1 - p = \frac{6}{10} = \frac{3}{5}\).


The number of trials is \(n=3\). Let X be the random variable for the number of oranges drawn. X can take values 0, 1, 2, 3.


The probability mass function is given by the binomial distribution: \(P(X=k) = \binom{n}{k} p^k q^{n-k}\).


(i) Probability distribution:


P(X=0) = \(\binom{3}{0} (\frac{2}{5})^0 (\frac{3}{5})^{3-0} = 1 \times 1 \times \frac{27}{125} = \frac{27}{125}\).


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


P(X=2) = \(\binom{3}{2} (\frac{2}{5})^2 (\frac{3}{5})^{3-2} = 3 \times \frac{4}{25} \times \frac{3}{5} = \frac{36}{125}\).


P(X=3) = \(\binom{3}{3} (\frac{2}{5})^3 (\frac{3}{5})^{3-3} = 1 \times \frac{8}{125} \times 1 = \frac{8}{125}\).


The distribution is: X=0, P(X)=27/125; X=1, P(X)=54/125; X=2, P(X)=36/125; X=3, P(X)=8/125.


(ii) Expectation of X:


The expectation (mean) of a binomial distribution is given by the formula \(E(X) = np\).

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


Alternatively, using the distribution table:

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

\(E(X) = \frac{0 + 54 + 72 + 24}{125} = \frac{150}{125} = \frac{6}{5}\).
Quick Tip: Recognize scenarios with a fixed number of independent trials, constant probability of success, and only two outcomes as a binomial distribution. This allows you to use the shortcuts \(P(X=k) = \binom{n}{k}p^k q^{n-k}\) and \(E(X)=np\).

The given curves are \(y = x^2\), \(y=9\), and \(x=0\) (the y-axis).


The region is bounded in the first quadrant.


We need to find the area of the region enclosed by these curves. It's easier to integrate with respect to y.


First, we express x in terms of y from the equation \(y = x^2\). This gives \(x = \sqrt{y}\) (since we are in the first quadrant, x is positive).


The limits of integration for y are from 0 to 9.


The formula for the area is \(A = \int_{c}^{d} x \,dy\).

\(A = \int_{0}^{9} \sqrt{y} \,dy = \int_{0}^{9} y^{1/2} \,dy\).


Now, we perform the integration:

\(A = \left[ \frac{y^{3/2}}{3/2} \right]_{0}^{9} = \left[ \frac{2}{3} y^{3/2} \right]_{0}^{9}\).


Evaluate the integral at the limits:

\(A = \frac{2}{3} (9^{3/2} - 0^{3/2})\).

\(A = \frac{2}{3} ((\sqrt{9})^3) = \frac{2}{3} (3^3) = \frac{2}{3} (27)\).

\(A = 2 \times 9 = 18\).


The area of the region is 18 square units.


(A rough sketch showing the parabola \(y=x^2\) and the line \(y=9\), with the area shaded in the first quadrant).
Quick Tip: When finding the area bounded by curves, decide whether to integrate with respect to x or y. If the region is defined by functions of y (like x = g(y)) and horizontal lines, integrating with respect to y is usually simpler.

Let the number of chairs, tables, and beds be x, y, and z respectively.


From the given information, we can form a system of linear equations:


1. Total furniture: \(x + y + z = 45\).


2. Beds exceed chairs by 8: \(z = x + 8 \implies -x + z = 8 \implies -x + 0y + z = 8\).


3. Chairs and beds are twice the tables: \(x + z = 2y \implies x - 2y + z = 0\).


This system can be written in matrix form AX = B:

\(A = \begin{pmatrix} 1 & 1 & 1
-1 & 0 & 1
1 & -2 & 1 \end{pmatrix}\), \(X = \begin{pmatrix} x
y
z \end{pmatrix}\), \(B = \begin{pmatrix} 45
8
0 \end{pmatrix}\).


The solution is given by X = \(A⁻¹B\). First, we find the determinant of A.

\(|A| = 1(0 - (-2)) - 1(-1 - 1) + 1(2 - 0) = 2 - (-2) + 2 = 6\). Since \(|A| \neq 0\), the inverse exists.


Next, we find the adjugate of A. The matrix of cofactors is:

\(C = \begin{pmatrix} (0+2) & -(-1-1) & (2-0)
-(1+2) & (1-1) & -(-2-1)
(1-0) & -(1+1) & (0+1) \end{pmatrix} = \begin{pmatrix} 2 & 2 & 2
-3 & 0 & 3
1 & -2 & 1 \end{pmatrix}\).


adj(A) = \(C^T = \begin{pmatrix} 2 & -3 & 1
2 & 0 & -2
2 & 3 & 1 \end{pmatrix}\).

\(A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{6} \begin{pmatrix} 2 & -3 & 1
2 & 0 & -2
2 & 3 & 1 \end{pmatrix}\).


Now, we find X:

\(X = \frac{1}{6} \begin{pmatrix} 2 & -3 & 1
2 & 0 & -2
2 & 3 & 1 \end{pmatrix} \begin{pmatrix} 45
8
0 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 2(45) - 3(8) + 1(0)
2(45) + 0(8) - 2(0)
2(45) + 3(8) + 1(0) \end{pmatrix}\).

\(X = \frac{1}{6} \begin{pmatrix} 90 - 24
90
90 + 24 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 66
90
114 \end{pmatrix} = \begin{pmatrix} 11
15
19 \end{pmatrix}\).


Therefore, \(x=11\) (chairs), \(y=15\) (tables), and \(z=19\) (beds).
Quick Tip: To solve a system of linear equations using the matrix method (AX=B), follow these steps: 1. Form the matrices A, X, and B. 2. Calculate the determinant |A|. 3. Find the adjugate of A. 4. Calculate the inverse \(A⁻¹ = (1/|A|)adj(A)\). 5. Find the solution X = A⁻¹B.

This problem asks for the derivative of one function with respect to another. We can use parametric differentiation.


Let \(y = a^{t + \frac{1}{t}}\) and let \(z = t + \frac{1}{t}\).


We need to find \(\frac{dy}{dz}\).


The formula for parametric differentiation is \(\frac{dy}{dz} = \frac{dy/dt}{dz/dt}\).


First, find \(\frac{dy}{dt}\):

\(\frac{dy}{dt} = \frac{d}{dt} \left( a^{t + \frac{1}{t}} \right)\).


Using the chain rule and the formula \(\frac{d}{dx}(a^u) = a^u \ln(a) \frac{du}{dx}\):

\(\frac{dy}{dt} = a^{t + \frac{1}{t}} \ln(a) \cdot \frac{d}{dt} \left( t + \frac{1}{t} \right)\).

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


Next, find \(\frac{dz}{dt}\):

\(\frac{dz}{dt} = \frac{d}{dt} \left( t + \frac{1}{t} \right) = 1 - \frac{1}{t^2}\).


Now, we can find \(\frac{dy}{dz}\):

\(\frac{dy}{dz} = \frac{a^{t + \frac{1}{t}} \ln(a) \left( 1 - \frac{1}{t^2} \right)}{1 - \frac{1}{t^2}}\).


Assuming \(1 - \frac{1}{t^2} \neq 0\) (i.e., \(t \neq \pm 1\)), the terms cancel out.

\(\frac{dy}{dz} = a^{t + \frac{1}{t}} \ln a\).


Alternatively, using direct substitution: Let \(u = t + \frac{1}{t}\). The problem is to find \(\frac{d(a^u)}{du}\), which is simply \(a^u \ln a\). Substituting back gives \(a^{t + \frac{1}{t}} \ln a\).
Quick Tip: When asked to differentiate a function \(f(x)\) with respect to another function \(g(x)\), you can use the substitution method. Let \(u = g(x)\), rewrite \(f(x)\) in terms of \(u\), and then find \(\frac{df}{du}\).

This is an implicit differentiation problem involving terms of the form \(f(x)^{g(x)}\). We use logarithmic differentiation for each term.


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


Differentiating with respect to x gives \(\frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0\).


For \(u = y^x\): \(\ln u = x \ln y\). Differentiating implicitly: \(\frac{1}{u}\frac{du}{dx} = 1 \cdot \ln y + x \cdot \frac{1}{y}\frac{dy}{dx}\).

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


For \(v = x^y\): \(\ln v = y \ln x\). Differentiating implicitly: \(\frac{1}{v}\frac{dv}{dx} = \frac{dy}{dx} \ln x + y \cdot \frac{1}{x}\).

\(\frac{dv}{dx} = v \left( \frac{dy}{dx} \ln x + \frac{y}{x} \right) = x^y \left( \frac{dy}{dx} \ln x + \frac{y}{x} \right)\).


For \(w = x^x\): \(\ln w = x \ln x\). Differentiating implicitly: \(\frac{1}{w}\frac{dw}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1\).

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


Now substitute these back into the main differentiated equation:

\(y^x \left( \ln y + \frac{x}{y}\frac{dy}{dx} \right) + x^y \left( \frac{dy}{dx} \ln x + \frac{y}{x} \right) + x^x(1 + \ln x) = 0\).


Expand and group the terms containing \(\frac{dy}{dx}\):

\(y^x \ln y + y^x \frac{x}{y}\frac{dy}{dx} + x^y \frac{dy}{dx} \ln x + x^y \frac{y}{x} + x^x(1 + \ln x) = 0\).

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


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 implicit functions containing variable powers like \(f(x)^{g(x)}\), do not differentiate directly. Differentiate each term separately using logarithmic differentiation, then substitute back and solve for \(\frac{dy}{dx}\).

First, we write the equation of the line in standard form.

\(\frac{x+2}{5} = \frac{y+1}{2} = \frac{-(z-4)}{-3} \implies \frac{x-(-2)}{5} = \frac{y-(-1)}{2} = \frac{z-4}{3}\).


The direction ratios of the line are \((5, 2, 3)\). Let this be vector \(\vec{d}\).


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


Any point Q on the line can be represented by a parameter \(\lambda\):

\(\frac{x+2}{5} = \frac{y+1}{2} = \frac{z-4}{3} = \lambda\).


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


The direction ratios of the line segment PQ are given by the coordinates of Q minus the coordinates of P:

\(\vec{PQ} = (5\lambda - 2 - 1, 2\lambda - 1 - 1, 3\lambda + 4 - 4) = (5\lambda - 3, 2\lambda - 2, 3\lambda)\).


Since PQ is perpendicular to the given line, the dot product of their direction vectors must be zero: \(\vec{PQ} \cdot \vec{d} = 0\).

\(5(5\lambda - 3) + 2(2\lambda - 2) + 3(3\lambda) = 0\).

\(25\lambda - 15 + 4\lambda - 4 + 9\lambda = 0\).

\(38\lambda - 19 = 0\).

\(\lambda = \frac{19}{38} = \frac{1}{2}\).


Now substitute this value of \(\lambda\) back into the coordinates of Q to find the foot of the perpendicular.

\(x = 5(\frac{1}{2}) - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2}\).

\(y = 2(\frac{1}{2}) - 1 = 1 - 1 = 0\).

\(z = 3(\frac{1}{2}) + 4 = \frac{3}{2} + \frac{8}{2} = \frac{11}{2}\).


The coordinates of the foot of the perpendicular are \((\frac{1}{2}, 0, \frac{11}{2})\).
Quick Tip: To find the foot of a perpendicular from a point to a line, 1. Write a general point on the line using a parameter \(\lambda\). 2. Find the direction ratios of the line segment connecting the given point and the general point. 3. Use the dot product condition for perpendicularity 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 as \(2\sqrt{2}\).


Using the distance formula, \(PQ^2 = (2\sqrt{2})^2 = 8\).

\(PQ^2 = (x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2\).

\(8 = ((3\lambda + 1) - (-1))^2 + ((2\lambda - 1) - (-1))^2 + ((3\lambda + 4) - 2)^2\).

\(8 = (3\lambda + 2)^2 + (2\lambda)^2 + (3\lambda + 2)^2\).

\(8 = 2(3\lambda + 2)^2 + 4\lambda^2\).

\(8 = 2(9\lambda^2 + 12\lambda + 4) + 4\lambda^2\).

\(8 = 18\lambda^2 + 24\lambda + 8 + 4\lambda^2\).

\(0 = 22\lambda^2 + 24\lambda\).

\(0 = 2\lambda(11\lambda + 12)\).


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


For \(\lambda = 0\):


The point is \((3(0)+1, 2(0)-1, 3(0)+4) = (1, -1, 4)\).


For \(\lambda = -\frac{12}{11}\):

\(x = 3(-\frac{12}{11}) + 1 = -\frac{36}{11} + \frac{11}{11} = -\frac{25}{11}\).

\(y = 2(-\frac{12}{11}) - 1 = -\frac{24}{11} - \frac{11}{11} = -\frac{35}{11}\).

\(z = 3(-\frac{12}{11}) + 4 = -\frac{36}{11} + \frac{44}{11} = \frac{8}{11}\).


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


There are two such points on the line.
Quick Tip: To find a point on a line at a given distance from another point, 1. Write a general point on the line using a parameter \(\lambda\). 2. Use the distance formula between the given point and the general point. 3. Set this distance equal to the given value and solve the resulting quadratic equation for \(\lambda\).

The length and breadth of the square base are given as x m, and the height is y m.


The volume (V) of the cuboidal box is given by \(V = length \times breadth \times height\).

\(V = x \cdot x \cdot y = x^2 y\).


Since the volume V is constant, we can express y in terms of x and V:

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


The total surface area (S) of the closed box is the sum of the area of the top and bottom faces and the area of the four sides.

\(S = (Area of base) + (Area of top) + 4 \times (Area of one side)\).

\(S = (x^2) + (x^2) + 4(xy) = 2x^2 + 4xy\).


To express S in terms of x and V only, we substitute the expression for y:

\(S = 2x^2 + 4x \left(\frac{V}{x^2}\right)\).

\(S = 2x^2 + \frac{4V}{x}\).
Quick Tip: In optimization problems, the first step is to establish two equations: a primary equation for the quantity to be optimized (e.g., surface area S) and a secondary (constraint) equation that relates the variables (e.g., volume V). Use the constraint to express the primary equation in terms of a single variable.

From the previous part, the surface area S is given by the function:

\(S(x) = 2x^2 + \frac{4V}{x}\).


To differentiate with respect to x, we can write the expression as:

\(S(x) = 2x^2 + 4Vx^{-1}\).


Now, we apply the power rule for differentiation, \(\frac{d}{dx}(x^n) = nx^{n-1}\).

\(\frac{dS}{dx} = \frac{d}{dx}(2x^2) + \frac{d}{dx}(4Vx^{-1})\).

\(\frac{dS}{dx} = 2(2x) + 4V(-1x^{-2})\).

\(\frac{dS}{dx} = 4x - 4Vx^{-2}\).


This can be written as:

\(\frac{dS}{dx} = 4x - \frac{4V}{x^2}\).
Quick Tip: When differentiating expressions with variables in the denominator, like \(\frac{k}{x^n}\), it's often easier to rewrite them using negative exponents, i.e., \(kx^{-n}\), before applying the power rule.

To find the minimum surface area, we set the first derivative \(\frac{dS}{dx}\) to zero.


From the previous part, \(\frac{dS}{dx} = 4x - \frac{4V}{x^2}\).


Setting \(\frac{dS}{dx} = 0\):

\(4x - \frac{4V}{x^2} = 0\).

\(4x = \frac{4V}{x^2}\).

\(4x^3 = 4V \implies x^3 = V\).


We also know from the problem setup that the volume is \(V = x^2 y\).


Substituting this expression for V into our result:

\(x^3 = x^2 y\).


Since x is a dimension of the box, \(x \neq 0\), so we can divide both sides by \(x^2\).

\(x = y\).


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

\(\frac{d^2S}{dx^2} = \frac{d}{dx}(4x - 4Vx^{-2}) = 4 - 4V(-2x^{-3}) = 4 + \frac{8V}{x^3}\).


Since volume V and dimension x must be positive, \(\frac{d^2S}{dx^2} > 0\), which confirms a local minimum.


The relation for minimum surface area is \(x=y\), meaning the box must be a cube.
Quick Tip: To find the maximum or minimum of a function, find its critical points by setting the first derivative to zero. Then use the second derivative test: if \(f''(c) > 0\), it's a minimum at \(x=c\); if \(f''(c) < 0\), it's a maximum.

We are given the volume V as a function of x, with S being a constant surface area.

\(V(x) = \frac{1}{4}(Sx - 2x^3)\).


To find the value of x that maximizes V, we first find the derivative \(\frac{dV}{dx}\).

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


Set the first derivative to zero to find critical points:

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

\(S - 6x^2 = 0\).

\(S = 6x^2 \implies x^2 = \frac{S}{6}\).


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


To confirm this corresponds to a maximum, we perform the second derivative test.

\(\frac{d^2V}{dx^2} = \frac{d}{dx} \left[ \frac{1}{4}(S - 6x^2) \right] = \frac{1}{4}(-12x) = -3x\).


Since x is a positive length, \(\frac{d^2V}{dx^2} = -3x\) will be negative.


A negative second derivative confirms that the volume V is maximum at \(x = \sqrt{\frac{S}{6}}\). Hence shown.
Quick Tip: This problem is the dual of the first part. In part (a), volume was fixed and surface area was minimized. Here, surface area is fixed and volume is maximized. For a cuboid with a square base, both conditions lead to the same result: the shape must be a cube.

A function is bijective if it is both one-to-one (injective) and onto (surjective).


(i) Analysis of the function f(x) = Roll Number of student x:


1. One-to-one (Injective):


By definition, every student in a class is assigned a different and unique roll number.


Therefore, if \(x_1\) and \(x_2\) are two different students (\(x_1 \neq x_2\)), then their roll numbers will also be different (\(f(x_1) \neq f(x_2)\)).


Hence, the function f is one-to-one.


2. Onto (Surjective):


A function is onto if its range is equal to its codomain.


The domain A is the set of 30 students.


The range of the function is the set of the 30 distinct roll numbers assigned to these students. This is a finite set with 30 elements.


The codomain of the function is given as N, the set of all natural numbers \(\{1, 2, 3, ... \}\), which is an infinite set.


Since the range (a set of 30 numbers) is not equal to the codomain (all natural numbers), the function is not onto.


Conclusion: Since the function f is not onto, it cannot be bijective.


(ii) Reason:


The function f is not bijective because it fails the condition for being surjective (onto). The reason is that the range of f, which is the set of 30 roll numbers, is a proper subset of the codomain N. For example, the natural number 100 (assuming it is not a roll number in the class) has no pre-image in the set A.
Quick Tip: A function f: A \(\rightarrow\) B is bijective if it is both injective (one-to-one) and surjective (onto). To prove a function is not bijective, you only need to show that one of these conditions fails.

Let the set of roll numbers be A = \(\{1, 2, 3, ..., 30\}\). The relation is R on A.


Listing the elements of R = \(\{(x, y) \in A \times A : y=3x\}\):


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

If x=2, y=6. (2, 6) \(\in\) R.

...

If x=10, y=30. (10, 30) \(\in\) R.

If x=11, y=33, but \(33 \notin A\). So we stop.

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


Checking the properties:


1. Reflexive: For R to be reflexive, \((x, x) \in R\) for all \(x \in A\). This means \(x=3x\), which is only true for \(x=0\). Since \(0 \notin A\), there is no \(x \in A\) for which \((x,x) \in R\). For example, \((1,1) \notin R\). Thus, R is not reflexive.


2. Symmetric: For R to be symmetric, if \((x, y) \in R\), then \((y, x) \in R\). Let's take \((1, 3) \in R\). For \((3, 1)\) to be in R, the condition must be \(1 = 3(3)\), which is false. Thus, R is not symmetric.


3. Transitive: For R to be transitive, if \((x, y) \in R\) and \((y, z) \in R\), then \((x, z) \in R\). Let's take \((x,y) = (1, 3) \in R\). Now we need a pair starting with 3, so \((y,z) = (3, 9) \in R\). We must check if \((x,z)=(1, 9) \in R\). The condition is \(9=3(1)\), which is true. Let's try another one: \((2, 6) \in R\) and \((6, 18) \in R\). We must check if \((2, 18) \in R\). The condition is \(18 = 3(2)\), which is \(18=6\). This is false. Therefore, R is not transitive.
Quick Tip: To disprove a property (reflexive, symmetric, or transitive), you only need to find a single counterexample. To prove it, you must show it holds for all general cases.

Let the set of roll numbers be A = \(\{1, 2, 3, ..., 30\}\). The relation R is on A.


Listing the elements of R = \(\{(x, y) \in A \times A : y=x^3\}\):


If x=1, y=1³=1. (1, 1) \(\in\) R.

If x=2, y=2³=8. (2, 8) \(\in\) R.

If x=3, y=3³=27. (3, 27) \(\in\) R.

If x=4, y=4³=64, but \(64 \notin A\). So we stop.

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


To determine if R is a function from A to A, we check two conditions:

1. Every element in the domain A must have an image.

2. Each element in the domain A must have exactly one image.


The domain is the set of all roll numbers, A = \(\{1, 2, ..., 30\}\).


The relation R only provides images for the elements {1, 2, 3. Elements such as 4, 5, ..., 30 from the domain A do not have any image in the set A under the relation R.


For example, for the element 4 \(\in\) A, there is no pair (4, y) in R.


Since not every element of the domain A has an image, the relation R is not a function from A to A.
Quick Tip: A relation R from a set A to a set B is a function only if the domain of R is equal to the entire set A and no element of A is mapped to more than one element in B.

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

Let G be the event that a chosen seed germinates.


Total number of seeds = 10 (brinjal) + 12 (cabbage) + 8 (radish) = 30.


The probabilities of choosing each type of seed are:
\(P(B) = \frac{10}{30} = \frac{1}{3}\).
\(P(C) = \frac{12}{30} = \frac{2}{5}\).
\(P(R) = \frac{8}{30} = \frac{4}{15}\).


The given conditional probabilities of germination are:
\(P(G|B) = 25% = 0.25\).
\(P(G|C) = 35% = 0.35\).
\(P(G|R) = 40% = 0.40\).


We need to find the total probability of germination, P(G).

Using the Law of Total Probability:
\(P(G) = P(B)P(G|B) + P(C)P(G|C) + P(R)P(G|R)\).

\(P(G) = \left(\frac{1}{3}\right)(0.25) + \left(\frac{2}{5}\right)(0.35) + \left(\frac{4}{15}\right)(0.40)\).

\(P(G) = \frac{0.25}{3} + \frac{0.70}{5} + \frac{1.60}{15}\).

\(P(G) = \frac{1}{12} + \frac{14}{100} + \frac{160}{1500} = \frac{1}{12} + \frac{7}{50} + \frac{8}{75}\).


To sum these fractions, we find a common denominator, which is 300.

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

\(P(G) = \frac{25 + 42 + 32}{300} = \frac{99}{300} = \frac{33}{100} = 0.33\).
Quick Tip: The Law of Total Probability is used to find the probability of an event that can occur through several disjoint paths. The formula is \(P(A) = \sum P(A|B_i)P(B_i)\), where the events \(B_i\) form a partition of the sample space.

We need to find the conditional probability P(C|G), which is the probability that the seed was a cabbage seed, given that it germinated.


We use Bayes' Theorem:

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


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

\(P(G|C) = 0.35 = \frac{35}{100} = \frac{7}{20}\).

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

\(P(G) = \frac{99}{300} = \frac{33}{100}\).


First, calculate the numerator, \(P(G|C)P(C)\):

\(P(G|C)P(C) = (0.35) \times \left(\frac{2}{5}\right) = 0.14 = \frac{14}{100}\).


Now, apply Bayes' formula:

\(P(C|G) = \frac{14/100}{33/100}\).

\(P(C|G) = \frac{14}{33}\).
Quick Tip: Bayes' Theorem is used to update the probability of a hypothesis based on new evidence. The formula \(P(A|B) = \frac{P(B|A)P(A)}{P(B)}\) calculates the "reverse" conditional probability.