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


Therefore, for the function \(\sin^{-1}(2x^2 - 3)\) to be defined, the expression inside the inverse sine must satisfy:

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


This inequality can be split into two separate inequalities:


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

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

\(2 \le 2x^2\)

\(1 \le x^2\)


This implies \(x \ge 1\) or \(x \le -1\). So, \(x \in (-\infty, -1] \cup [1, \infty)\).


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

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

\(2x^2 \le 4\)

\(x^2 \le 2\)


This implies \(-\sqrt{2} \le x \le \sqrt{2}\). So, \(x \in [-\sqrt{2}, \sqrt{2}]\).


The domain of the function is the intersection of the solutions from Part 1 and Part 2.


We need to find the intersection of \((-\infty, -1] \cup [1, \infty)\) and \([-\sqrt{2}, \sqrt{2}]\).


The common values are in the intervals \([-\sqrt{2}, -1]\) and \([1, \sqrt{2}]\).


Thus, the domain is \([-\sqrt{2}, -1] \cup [1, \sqrt{2}]\).
Quick Tip: To find the domain of an inverse trigonometric function like \(\sin^{-1}(f(x))\) or \(\cos^{-1}(f(x))\), always set the argument \(f(x)\) to be within the range \([-1, 1]\) and solve the resulting inequality.

Let the given matrix be A.

\(A = \begin{pmatrix} 0 & 1 & -2
-1 & 0 & -7
2 & 7 & 0 \end{pmatrix}\)


A matrix is skew-symmetric if its transpose is equal to its negative, i.e., \(A^T = -A\).


First, let's find the transpose of A, denoted by \(A^T\).

\(A^T = \begin{pmatrix} 0 & -1 & 2
1 & 0 & 7
-2 & -7 & 0 \end{pmatrix}\)


Next, let's find the negative of A, denoted by \(-A\).

\(-A = -1 \times \begin{pmatrix} 0 & 1 & -2
-1 & 0 & -7
2 & 7 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 2
1 & 0 & 7
-2 & -7 & 0 \end{pmatrix}\)


Comparing \(A^T\) and \(-A\), we see that they are identical.


Since \(A^T = -A\), the matrix A is a skew-symmetric matrix.


Additionally, we can observe that all the diagonal elements are zero and \(a_{ij} = -a_{ji}\) for all i, j, which are properties of a skew-symmetric matrix.
Quick Tip: A quick check for a skew-symmetric matrix has two parts: 1. All elements on the main diagonal must be zero. 2. The elements are mirrored across the diagonal with opposite signs (i.e., \(a_{ij} = -a_{ji}\)). If both conditions are met, the matrix is skew-symmetric.

For the function \(f(x)\) to be continuous on the interval \((0, 2)\), it must be continuous at the point where the definition of the function changes, which is \(x=1\).


The condition for continuity at a point \(x=c\) is that the Left-Hand Limit (LHL) equals the Right-Hand Limit (RHL) and equals the function's value at that point.

\(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)\)


First, we calculate the Left-Hand Limit (LHL) at \(x=1\). For \(x \to 1^-\), we use the definition \(f(x) = 3x - 2\).


LHL = \(\lim_{x \to 1^-} (3x - 2) = 3(1) - 2 = 1\).


Next, we calculate the Right-Hand Limit (RHL) at \(x=1\). For \(x \to 1^+\), we use the definition \(f(x) = 2x^2 + ax\).


RHL = \(\lim_{x \to 1^+} (2x^2 + ax) = 2(1)^2 + a(1) = 2 + a\).


For the function to be continuous at \(x=1\), we must have LHL = RHL.

\(1 = 2 + a\)


Solving for \(a\), we get:

\(a = 1 - 2 = -1\).


Thus, the value of \(a\) is -1.
Quick Tip: When dealing with continuity of piecewise functions, focus on the points where the function's definition changes. Equate the left-hand and right-hand limits at these points to find the values of any unknown constants.

The given function is \(y = \log_{2x}(\sqrt{2x})\).


We can simplify this expression using the properties of logarithms before differentiating.


First, rewrite the square root as a power of \(1/2\).

\(y = \log_{2x}((2x)^{1/2})\)


Using the logarithm property \(\log_b(a^c) = c \cdot \log_b(a)\), we can bring the exponent to the front.

\(y = \frac{1}{2} \log_{2x}(2x)\)


Now, using the logarithm property \(\log_b(b) = 1\), where the base and the argument are the same.


Here, the base is \(2x\) and the argument is \(2x\), so \(\log_{2x}(2x) = 1\).


Substituting this back into the expression for \(y\):

\(y = \frac{1}{2} \times 1 = \frac{1}{2}\)


The function simplifies to a constant value, \(y = 1/2\).


The derivative of a constant with respect to \(x\) is always zero.

\(\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2}\right) = 0\).
Quick Tip: Before applying complex differentiation rules like the quotient rule or chain rule, always check if the function can be simplified using algebraic or logarithmic properties. This can often reduce a complicated problem to a very simple one.

The function is \(f: N \to W\), where N = {1, 2, 3, ... is the set of natural numbers (domain) and W = {0, 1, 2, ... is the set of whole numbers (codomain).


Checking for Injectivity (One-to-one):


A function is injective if different inputs always produce different outputs.


Let's test a few inputs.


For \(n=1\) (odd), \(f(1) = 0\).


For \(n=3\) (odd), \(f(3) = 0\).


Here, we have two different inputs, \(n=1\) and \(n=3\), but they produce the same output, \(f(1) = f(3) = 0\).


Since different inputs lead to the same output, the function is not injective.


Checking for Surjectivity (Onto):


A function is surjective if every element in the codomain (W) has at least one pre-image in the domain (N).


Let's take an arbitrary element \(w \in W\). We need to see if there exists an \(n \in N\) such that \(f(n) = w\).


Case 1: If \(w = 0\).


We can choose any odd natural number for \(n\), for example \(n=1\). Then \(f(1) = 0\). So, \(0\) has a pre-image.


Case 2: If \(w > 0\) (i.e., \(w \in \{1, 2, 3, ...\}\)).


We need to find an \(n\) such that \(f(n) = w\). Since \(w \ne 0\), \(n\) must be even.


So we set \(\frac{n}{2} = w\), which gives \(n = 2w\).


For any positive whole number \(w\), \(n = 2w\) will be an even natural number. So, it is a valid pre-image in the domain N.


For example, if \(w=5\), \(n=10\), and \(f(10) = 10/2 = 5\).


Since every element in the codomain W (both 0 and all positive integers) has a pre-image in N, the function is surjective.


Since the function is surjective but not injective, it is surjective only.
Quick Tip: To disprove injectivity, you only need to find one counterexample of two different inputs giving the same output. To prove surjectivity, you must show that for any arbitrary element 'y' in the codomain, you can always find an element 'x' in the domain such that f(x) = y.

Let the given point be A(-2, 3, 5).


Let P be the foot of the perpendicular from point A to the y-axis.


Any point P on the y-axis has its x-coordinate and z-coordinate equal to zero.


So, the coordinates of P can be written as (0, y, 0) for some value y.


The direction ratios of the line segment AP are given by the differences in the coordinates:


Direction ratios of AP = \((0 - (-2), y - 3, 0 - 5) = (2, y-3, -5)\).


The direction ratios of the y-axis are (0, 1, 0).


Since the line segment AP is perpendicular to the y-axis, the dot product of their direction ratios must be zero.

\(a_1a_2 + b_1b_2 + c_1c_2 = 0\)

\((2)(0) + (y-3)(1) + (-5)(0) = 0\)

\(0 + y - 3 + 0 = 0\)

\(y - 3 = 0\)

\(y = 3\).


Therefore, the coordinates of the foot of the perpendicular P are (0, 3, 0).
Quick Tip: A simple way to find the foot of the perpendicular from a point (x, y, z) onto one of the coordinate axes is to set the other two coordinates to zero. For the y-axis, the foot of the perpendicular from (x, y, z) is (0, y, 0).

We are given det(A) = 4 and det[(AB)\(^{-1}\)] = \(\frac{1}{20}\).


We use the property of determinants that for any invertible matrix M, det(M\(^{-1}\)) = \(\frac{1}{det(M)}\).


Applying this property to the matrix (AB), we get:


det[(AB)\(^{-1}\)] = \(\frac{1}{det(AB)}\).


Substituting the given value:

\(\frac{1}{20} = \frac{1}{det(AB)}\)


This implies that det(AB) = 20.


Another property of determinants is that for matrices A and B of the same order, det(AB) = det(A) \(\cdot\) det(B).


Using this property, we can write:


det(A) \(\cdot\) det(B) = 20.


We are given det(A) = 4. Substituting this value:


4 \(\cdot\) det(B) = 20.


det(B) = \(\frac{20}{4}\).


det(B) = 5.
Quick Tip: Remember two key properties of determinants for this type of problem: 1. \(\det(M^{-1}) = 1/\det(M)\) 2. \(\det(MN) = \det(M) \det(N)\) Combining these gives \(\det((MN)^{-1}) = 1/(\det(M)\det(N))\).

The given function is \(f(x) = x^3 + 5x + 1\), with domain and codomain as R (the set of real numbers).


To check for one-one (injective):


We can check the derivative of the function, \(f'(x)\).

\(f'(x) = \frac{d}{dx}(x^3 + 5x + 1) = 3x^2 + 5\).


For any real number \(x\), \(x^2 \ge 0\).


Therefore, \(3x^2 \ge 0\).


This means \(f'(x) = 3x^2 + 5 \ge 5\).


Since \(f'(x) > 0\) for all \(x \in R\), the function \(f(x)\) is strictly increasing on its entire domain.


A strictly monotonic function (either strictly increasing or strictly decreasing) is always one-one.


Thus, \(f(x)\) is one-one.


To check for onto (surjective):


The function \(f(x)\) is a polynomial of odd degree (degree 3).


The range of any odd-degree polynomial function is the set of all real numbers, i.e., \((-\infty, \infty)\) or R.


The codomain of the function is given as R.


Since the range of \(f(x)\) is R, which is equal to its codomain, the function is onto.


As the function is both one-one and onto, it is a bijection on R.
Quick Tip: For polynomial functions on R: 1. To check if it's one-one, find the derivative. If the derivative is always positive or always negative, the function is strictly monotonic and therefore one-one. 2. If the polynomial has an odd degree, its range is all real numbers (R), making it onto if the codomain is R.

For a function \(f(x)\) to be decreasing for all real values of \(x\), its derivative \(f'(x)\) must be less than or equal to zero for all \(x \in R\).

\(f'(x) \le 0\) for all \(x \in R\).


First, we find the derivative of the given function:

\(f(x) = \sin x - \cos x - \lambda x + C\)

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


Now, we apply the condition for a decreasing function:

\(\cos x + \sin x - \lambda \le 0\) for all \(x \in R\).


This can be rewritten as:

\(\cos x + \sin x \le \lambda\) for all \(x \in R\).


This inequality must hold for all values of \(x\). Therefore, \(\lambda\) must be greater than or equal to the maximum value of the expression \(\cos x + \sin x\).


We know that the maximum value of an expression of the form \(a \sin x + b \cos x\) is \(\sqrt{a^2 + b^2}\).


For \(\cos x + \sin x\), we have \(a=1\) and \(b=1\).


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


Therefore, the condition becomes:

\(\lambda \ge \sqrt{2}\).
Quick Tip: Remember the range of the function \(a \sin x + b \cos x\) is \([-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]\). To ensure an inequality like \(a \sin x + b \cos x \le k\) holds for all \(x\), \(k\) must be greater than or equal to the maximum value, \(\sqrt{a^2+b^2}\).

We are given two conditions: AB = A and BA = B. We need to find an expression for \(A^2 + B^2\).


Let's first simplify \(A^2\).

\(A^2 = A \cdot A\)


Using the first given condition, we can substitute A with AB:

\(A^2 = (AB)A\)


Using the associative property of matrix multiplication, we get:

\(A^2 = A(BA)\)


Now, using the second given condition, we can substitute BA with B:

\(A^2 = AB\)


Finally, using the first condition again (AB = A), we get:

\(A^2 = A\). (This means A is an idempotent matrix).


Now, let's simplify \(B^2\).

\(B^2 = B \cdot B\)


Using the second given condition, we can substitute B with BA:

\(B^2 = (BA)B\)


Using the associative property, we get:

\(B^2 = B(AB)\)


Now, using the first given condition, we can substitute AB with A:

\(B^2 = BA\)


Finally, using the second condition again (BA = B), we get:

\(B^2 = B\). (This means B is also an idempotent matrix).


Now we can find the required expression:

\(A^2 + B^2 = A + B\).
Quick Tip: When dealing with abstract matrix algebra problems, the key is to strategically substitute the given relations into the expression you want to simplify. Associativity, i.e., (XY)Z = X(YZ), is often used to regroup terms for the next substitution.

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


Here, \(y = \sqrt{x}\), \(a=0\), and \(b=4\).


So, the required area is \(A = \int_{0}^{4} \sqrt{x} \,dx\).


We can write \(\sqrt{x}\) as \(x^{1/2}\).

\(A = \int_{0}^{4} x^{1/2} \,dx\).


Using the power rule for integration, \(\int x^n \,dx = \frac{x^{n+1}}{n+1}\):

\(A = \left[ \frac{x^{1/2 + 1}}{1/2 + 1} \right]_{0}^{4} = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{4} = \left[ \frac{2}{3}x^{3/2} \right]_{0}^{4}\).


Now, we evaluate the integral at the upper and lower limits:

\(A = \frac{2}{3} (4^{3/2} - 0^{3/2})\).

\(A = \frac{2}{3} ( (4^{1/2})^3 - 0 ) = \frac{2}{3} (2^3)\).

\(A = \frac{2}{3} (8) = \frac{16}{3}\).


Thus, the area is \(\frac{16}{3}\) square units.
Quick Tip: To find the area under a curve \(y=f(x)\) above the x-axis from \(x=a\) to \(x=b\), always set up the definite integral \(A = \int_{a}^{b} f(x) \,dx\). Remember that \(x^{3/2}\) is the same as \((\sqrt{x})^3\).

Let the integral be \(I = \int_{0}^{1} \frac{dx}{e^x + e^{-x}}\).


First, rewrite the integrand by multiplying the numerator and denominator by \(e^x\).

\(I = \int_{0}^{1} \frac{e^x \,dx}{e^x(e^x + e^{-x})} = \int_{0}^{1} \frac{e^x \,dx}{(e^x)^2 + e^x e^{-x}} = \int_{0}^{1} \frac{e^x \,dx}{(e^x)^2 + 1}\).


Now, use the substitution method. Let \(t = e^x\).


Then, \(dt = e^x \,dx\).


We also need to change the limits of integration.


When \(x=0\), \(t = e^0 = 1\).


When \(x=1\), \(t = e^1 = e\).


The integral becomes:

\(I = \int_{1}^{e} \frac{dt}{t^2 + 1}\).


The integral of \(\frac{1}{t^2+1}\) is \(\tan^{-1}(t)\).

\(I = [\tan^{-1}(t)]_{1}^{e}\).


Evaluating at the limits:

\(I = \tan^{-1}(e) - \tan^{-1}(1)\).


We know that \(\tan^{-1}(1) = \frac{\pi}{4}\).


So, \(I = \tan^{-1}(e) - \frac{\pi}{4}\).
Quick Tip: Integrals involving \(e^x\) and \(e^{-x}\) can often be simplified by multiplying the numerator and denominator by \(e^x\). This frequently leads to a form that is solvable by substituting \(t = e^x\). Remember the standard integral \(\int \frac{dx}{x^2+1} = \tan^{-1}(x)\).

The objective function is Z = ax + by, where a > 0 and b > 0.


According to the fundamental theorem of linear programming, the optimal value (maximum or minimum) of the objective function must occur at one of the corner points of the feasible region.


The problem states that the maximum value of Z is obtained at two different corner points: (0, 2) and (3, 0).


This means that the value of Z at (0, 2) is equal to the value of Z at (3, 0), and this value is the maximum.


Let's calculate the value of Z at these two points.


At point (0, 2):

\(Z_1 = a(0) + b(2) = 2b\).


At point (3, 0):

\(Z_2 = a(3) + b(0) = 3a\).


Since the maximum value occurs at both points, we must have \(Z_1 = Z_2\).

\(2b = 3a\).


This gives the relation between a and b as \(3a = 2b\).
Quick Tip: If the optimal value (maximum or minimum) of an LPP occurs at two adjacent corner points, then it also occurs at every point on the line segment connecting these two points. For this to happen, the values of the objective function at these two points must be equal.

We need to evaluate the integral \(I = \int e^{-3 \log x} \,dx\).


First, simplify the integrand using the properties of logarithms.


Using the property \(n \log m = \log m^n\), we have:

\(-3 \log x = \log x^{-3}\).


So, the integrand becomes \(e^{\log x^{-3}}\).


Using the property \(e^{\log y} = y\), we have:

\(e^{\log x^{-3}} = x^{-3}\).


Now the integral is greatly simplified:

\(I = \int x^{-3} \,dx\).


Using the power rule for integration, \(\int x^n \,dx = \frac{x^{n+1}}{n+1} + C\):

\(I = \frac{x^{-3+1}}{-3+1} + C = \frac{x^{-2}}{-2} + C\).

\(I = -\frac{1}{2x^2} + C\).


The problem states that the integral is equal to \(f(x) + C\).


By comparing the results, we find that \(f(x) = -\frac{1}{2x^2}\).
Quick Tip: Always simplify the integrand before attempting to integrate. The combination of exponential and logarithmic functions often simplifies using the properties \(a \log b = \log b^a\) and \(e^{\log c} = c\).

A function is continuous at a point \(x=c\) if the Left-Hand Limit (LHL), the Right-Hand Limit (RHL), and the function's value at that point are all equal.


That is, \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)\).


The function \(f(x)\) is a piecewise function, and the definition changes at \(x=1\). This is the only point where a discontinuity might occur. Let's check the continuity at \(x=1\).


First, calculate the Left-Hand Limit (LHL) at \(x=1\). For \(x \to 1^-\), we use \(f(x) = x\).


LHL = \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1\).


Next, calculate the Right-Hand Limit (RHL) at \(x=1\). For \(x \to 1^+\), we use \(f(x) = 5\).


RHL = \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5 = 5\).


Since the Left-Hand Limit (1) is not equal to the Right-Hand Limit (5), the limit of \(f(x)\) as \(x\) approaches 1 does not exist.


Therefore, the function \(f(x)\) is not continuous at \(x=1\).
Quick Tip: For piecewise-defined functions, the points to check for discontinuity are the 'break-points' where the function's rule changes. Compare the Left-Hand Limit and the Right-Hand Limit at these points. If they are not equal, the function is discontinuous there.

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


This equation can be solved using the method of separation of variables.


Rearrange the terms to group all y-terms with dy and all x-terms with dx.

\(y \,dy = -x \,dx\).


Now, integrate both sides of the equation.

\(\int y \,dy = \int -x \,dx\).

\(\frac{y^2}{2} = -\frac{x^2}{2} + C_1\), where \(C_1\) is the constant of integration.


To simplify, multiply the entire equation by 2.

\(y^2 = -x^2 + 2C_1\).


Rearrange the terms to the standard form of a conic section.

\(x^2 + y^2 = 2C_1\).


Let \(C = 2C_1\). Since \(C_1\) is an arbitrary constant, C is also an arbitrary constant. For a circle to exist, we must have \(C > 0\).


The equation \(x^2 + y^2 = C\) (where \(C>0\)) represents a family of circles centered at the origin (0, 0) with radius \(\sqrt{C}\).
Quick Tip: Recognize the standard forms of conic sections. The equation \(x^2 + y^2 = r^2\) represents a circle. The equation \(y^2 = 4ax\) represents a parabola. The equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) represents an ellipse. The equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) represents a hyperbola.

Let the vertex A be the origin.


Then the position vector of B is \(\vec{AB} = \hat{j} + \hat{k}\).


And the position vector of C is \(\vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k}\).


The median through A bisects the side BC. Let D be the midpoint of BC.


The vector representing the median is \(\vec{AD}\).


The position vector of the midpoint D is given by the midpoint formula: \(\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}\).

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


Combine the corresponding components:

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

\(\vec{AD} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k}\).


The length of the median is the magnitude of the vector \(\vec{AD}\).


Length = \(|\vec{AD}| = \sqrt{(\frac{3}{2})^2 + (0)^2 + (\frac{5}{2})^2}\).


Length = \(\sqrt{\frac{9}{4} + \frac{25}{4}} = \sqrt{\frac{9+25}{4}} = \sqrt{\frac{34}{4}}\).


Length = \(\frac{\sqrt{34}}{\sqrt{4}} = \frac{\sqrt{34}}{2}\) units.
Quick Tip: In a triangle ABC, the vector representing the median from vertex A to the side BC is given by \(\frac{\vec{AB} + \vec{AC}}{2}\). This is a direct application of the vector midpoint formula.

To determine if the function is increasing or decreasing, we need to analyze its first derivative, \(f'(x)\).


The function is \(f(x) = 2x + \cos x\).


Differentiating with respect to x:

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


A function is increasing if \(f'(x) > 0\), and decreasing if \(f'(x) < 0\).


We know that the range of the sine function is \(-1 \le \sin x \le 1\) for all real numbers \(x\).


Let's find the range of \(f'(x) = 2 - \sin x\).


The minimum value of \(f'(x)\) occurs when \(\sin x\) is at its maximum value of 1.


Minimum \(f'(x) = 2 - 1 = 1\).


The maximum value of \(f'(x)\) occurs when \(\sin x\) is at its minimum value of -1.


Maximum \(f'(x) = 2 - (-1) = 3\).


So, for all real values of \(x\), we have \(1 \le f'(x) \le 3\).


Since \(f'(x)\) is always positive (specifically, \(f'(x) \ge 1\)), the function \(f(x)\) is strictly increasing for all \(x \in R\).


Therefore, it does not have any maxima or minima.
Quick Tip: To analyze the monotonic behavior of a function, find its first derivative. If the derivative is always positive, the function is always increasing. If the derivative is always negative, the function is always decreasing. Knowing the range of trigonometric functions like sin(x) and cos(x) is crucial for this analysis.

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


Reason (R) states: "Two events are independent if the occurrence of one does not effect the occurrence of the other." This is the conceptual definition of independent events. So, Reason (R) is a true statement.


Next, let's analyze the Assertion (A).


Assertion (A) states: "If A and B are two events such that P(A \(\cap\) B) = 0, then A and B are independent events."


The mathematical condition for two events A and B to be independent is P(A \(\cap\) B) = P(A) \(\cdot\) P(B).


The condition P(A \(\cap\) B) = 0 means that events A and B are mutually exclusive (they cannot happen at the same time).


If A and B are mutually exclusive, for them to be independent, we must have P(A) \(\cdot\) P(B) = 0.


This only holds true if P(A) = 0 or P(B) = 0 (or both).


However, if we consider two non-impossible, mutually exclusive events (e.g., getting a 1 and getting a 2 on a single die roll), then P(A) > 0 and P(B) > 0.


In this case, P(A \(\cap\) B) = 0, but P(A) \(\cdot\) P(B) > 0.


Since \(0 \ne P(A) \cdot P(B)\), the events are dependent. In fact, they are strongly dependent, as the occurrence of one guarantees the non-occurrence of the other.


Therefore, the Assertion (A) is false.


Since Assertion (A) is false and Reason (R) is true, the correct option is (D).
Quick Tip: Do not confuse 'mutually exclusive' with 'independent'. - Mutually Exclusive: P(A \(\cap\) B) = 0. The events cannot happen together. - Independent: P(A \(\cap\) B) = P(A)P(B). The occurrence of one doesn't affect the probability of the other. Two non-impossible events cannot be both mutually exclusive and independent.

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


Reason (R) states: "A feasible region is defined as the region that satisfies all the constraints." This is the correct definition of a feasible region in the context of LPP. So, Reason (R) is true.


Next, let's analyze the Assertion (A).


Assertion (A) states: "In a Linear Programming Problem, if the feasible region is empty, then the Linear Programming Problem has no solution."


An optimal solution (maximum or minimum) to an LPP must be a point within the feasible region.


If the feasible region is empty, it means there are no points (x, y) that satisfy all the given constraints simultaneously.


Since there are no valid points from which to choose a solution, the LPP cannot have a solution.


Therefore, Assertion (A) is true.


Finally, let's check if (R) explains (A).


The assertion (A) concludes there is no solution because the feasible region is empty. The reason (R) defines what a feasible region is (the set of all points satisfying the constraints). The logic flows directly: A solution must satisfy all constraints. The set of all such points is the feasible region (as per R). If this set is empty, no such point exists, and thus no solution exists (as per A).


Therefore, Reason (R) provides the correct explanation for Assertion (A).
Quick Tip: In LPP, three outcomes are possible regarding the solution: 1. A unique or multiple optimal solutions exist (if the feasible region is bounded and non-empty, or in some cases, unbounded). 2. An unbounded solution exists (the objective function can be increased/decreased indefinitely within an unbounded feasible region). 3. No solution exists (the feasible region is empty, also known as an infeasible LPP).

We are given the matrix A = \(\begin{pmatrix} 2 & -2
-2 & 2 \end{pmatrix}\) and the equation A\(^2\) = kA.


First, let's compute A\(^2\).


A\(^2\) = A \(\cdot\) A = \(\begin{pmatrix} 2 & -2
-2 & 2 \end{pmatrix} \begin{pmatrix} 2 & -2
-2 & 2 \end{pmatrix}\).


A\(^2\) = \(\begin{pmatrix} (2)(2) + (-2)(-2) & (2)(-2) + (-2)(2)
(-2)(2) + (2)(-2) & (-2)(-2) + (2)(2) \end{pmatrix}\).


A\(^2\) = \(\begin{pmatrix} 4 + 4 & -4 - 4
-4 - 4 & 4 + 4 \end{pmatrix} = \begin{pmatrix} 8 & -8
-8 & 8 \end{pmatrix}\).


Now, let's compute kA.


kA = k \(\begin{pmatrix} 2 & -2
-2 & 2 \end{pmatrix} = \begin{pmatrix} 2k & -2k
-2k & 2k \end{pmatrix}\).


We are given A\(^2\) = kA.

\(\begin{pmatrix} 8 & -8
-8 & 8 \end{pmatrix} = \begin{pmatrix} 2k & -2k
-2k & 2k \end{pmatrix}\).


By comparing the corresponding elements of the matrices, we get:

\(8 = 2k \implies k = 4\).

\(-8 = -2k \implies k = 4\).


So, the value of k is 4.
Quick Tip: When performing matrix multiplication, remember the "row-by-column" rule. To find the element in the i-th row and j-th column of the product, take the dot product of the i-th row of the first matrix and the j-th column of the second matrix.

Let the given expression be \(y = \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\).


To simplify this expression, we use a trigonometric substitution. Let \(x = \tan \theta\).


This implies that \(\theta = \tan^{-1} x\).


Now substitute \(x = \tan \theta\) into the expression.

\(y = \sin^{-1}\left(\frac{\tan \theta}{\sqrt{1+\tan^2 \theta}}\right)\).


Using the trigonometric identity \(1 + \tan^2 \theta = \sec^2 \theta\).

\(y = \sin^{-1}\left(\frac{\tan \theta}{\sqrt{\sec^2 \theta}}\right) = \sin^{-1}\left(\frac{\tan \theta}{\sec \theta}\right)\).


Now express \(\tan \theta\) and \(\sec \theta\) in terms of \(\sin \theta\) and \(\cos \theta\).

\(y = \sin^{-1}\left(\frac{\sin \theta / \cos \theta}{1 / \cos \theta}\right) = \sin^{-1}\left(\frac{\sin \theta}{\cos \theta} \cdot \cos \theta\right)\).

\(y = \sin^{-1}(\sin \theta)\).

\(y = \theta\).


Since we initially set \(\theta = \tan^{-1} x\), we can substitute it back.

\(y = \tan^{-1} x\).


Therefore, the simplified form is \(\tan^{-1} x\).
Quick Tip: Recognize standard substitution forms for inverse trigonometric functions. For expressions involving \(\sqrt{1+x^2}\), the substitution \(x = \tan \theta\) or \(x = \cot \theta\) is usually effective. For \(\sqrt{1-x^2}\), use \(x=\sin\theta\). For \(\sqrt{x^2-1}\), use \(x=\sec\theta\).

For the function \(f(x) = \sin^{-1}\sqrt{x-1}\) to be defined, two conditions must be met.


Condition 1: The expression inside the square root must be non-negative.

\(x - 1 \ge 0 \implies x \ge 1\).


This gives us the interval \([1, \infty)\).


Condition 2: The argument of the inverse sine function, \(\sqrt{x-1}\), must be in the interval \([-1, 1]\).

\(-1 \le \sqrt{x-1} \le 1\).


Since the square root of a real number is always non-negative, this inequality can be simplified to:

\(0 \le \sqrt{x-1} \le 1\).


Squaring all parts of the inequality gives:

\(0^2 \le (\sqrt{x-1})^2 \le 1^2\).

\(0 \le x - 1 \le 1\).


Adding 1 to all parts of the inequality:

\(0+1 \le x-1+1 \le 1+1\).

\(1 \le x \le 2\).


This gives us the interval \([1, 2]\).


The domain of the function is the intersection of the results from both conditions.


The intersection of \([1, \infty)\) and \([1, 2]\) is \([1, 2]\).


Thus, the domain of \(\sin^{-1}\sqrt{x-1}\) is \([1, 2]\).
Quick Tip: When finding the domain of a composite function, you must satisfy the domain requirements of both the outer function and the inner function(s). Work from the inside out.

The given curve is the equation of an ellipse centered at the origin, with semi-major axis \(a=3\) and semi-minor axis \(b=2\).


The region bounded by the ellipse and the x-axis corresponds to the area of the upper half of the ellipse, where \(y \ge 0\).


First, we solve the equation for y in terms of x.

\(\frac{y^2}{4} = 1 - \frac{x^2}{9} = \frac{9-x^2}{9}\).

\(y^2 = \frac{4}{9}(9-x^2)\).


For the upper half of the ellipse, we take the positive square root:

\(y = \sqrt{\frac{4}{9}(9-x^2)} = \frac{2}{3}\sqrt{9-x^2}\).


The ellipse intersects the x-axis at \(x = -3\) and \(x = 3\). So, the limits of integration are from -3 to 3.


Area \(A = \int_{-3}^{3} y \,dx = \int_{-3}^{3} \frac{2}{3}\sqrt{9-x^2} \,dx\).


The integral \(\int \sqrt{a^2-x^2} \,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)\).


Here, \(a=3\).

\(A = \frac{2}{3} \left[ \frac{x}{2}\sqrt{9-x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) \right]_{-3}^{3}\).


Evaluating at the upper limit (\(x=3\)):

\(\frac{2}{3} \left[ \frac{3}{2}\sqrt{9-9} + \frac{9}{2}\sin^{-1}\left(\frac{3}{3}\right) \right] = \frac{2}{3} \left[ 0 + \frac{9}{2}\sin^{-1}(1) \right] = \frac{2}{3} \left[ \frac{9}{2} \cdot \frac{\pi}{2} \right] = \frac{3\pi}{2}\).


Evaluating at the lower limit (\(x=-3\)):

\(\frac{2}{3} \left[ \frac{-3}{2}\sqrt{9-9} + \frac{9}{2}\sin^{-1}\left(\frac{-3}{3}\right) \right] = \frac{2}{3} \left[ 0 + \frac{9}{2}\sin^{-1}(-1) \right] = \frac{2}{3} \left[ \frac{9}{2} \cdot \left(-\frac{\pi}{2}\right) \right] = -\frac{3\pi}{2}\).


Area = (Value at upper limit) - (Value at lower limit) = \(\frac{3\pi}{2} - \left(-\frac{3\pi}{2}\right) = \frac{3\pi}{2} + \frac{3\pi}{2} = 3\pi\).


The area is \(3\pi\) square units.
Quick Tip: The area of a full ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is \(\pi ab\). The area bounded by the ellipse and the x-axis is half of this, i.e., \(\frac{1}{2}\pi ab\). For this problem, \(a=3, b=2\), so the area is \(\frac{1}{2}\pi(3)(2) = 3\pi\). This can be used to quickly verify your integration result.

For a function to be increasing on an interval, its first derivative must be greater than or equal to zero on that interval.


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


First, find the derivative, \(f'(x)\).

\(f'(x) = 4x - a\).


The condition for the function to be increasing on \([2, 4]\) is \(f'(x) \ge 0\) for all \(x \in [2, 4]\).

\(4x - a \ge 0\) for all \(x \in [2, 4]\).


This can be rewritten as \(a \le 4x\) for all \(x \in [2, 4]\).


For this inequality to hold for the entire interval, 'a' must be less than or equal to the minimum value that \(4x\) can take on the interval \([2, 4]\).


The function \(g(x) = 4x\) is an increasing function, so its minimum value on \([2, 4]\) occurs at the left endpoint, \(x=2\).


Minimum value of \(4x\) on \([2, 4]\) is \(4(2) = 8\).


Therefore, the condition on 'a' is \(a \le 8\).


The question asks for the "least value of 'a'". The condition \(a \le 8\) provides an upper bound for 'a', but not a lower bound. Any value of 'a' from \(-\infty\) up to 8 will satisfy the condition. This indicates a likely typo in the question, which should probably ask for the "greatest value of 'a'". If it were the greatest value, the answer would be 8.
Quick Tip: To ensure a function is increasing on an interval, its derivative \(f'(x)\) must be \(\ge 0\) throughout that interval. This often translates to finding an upper or lower bound for a parameter by analyzing the minimum or maximum value of the derivative on the interval.

To show that the function \(f(x)\) is increasing, we need to show that its derivative, \(f'(x)\), is greater than or equal to zero for all \(x \ge 1\).


The function is \(f(x) = x + \frac{1}{x} = x + x^{-1}\).


Differentiating with respect to x:

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


Now we need to analyze the sign of \(f'(x)\) for the given domain \(x \ge 1\).


We can rewrite \(f'(x)\) as a single fraction:

\(f'(x) = \frac{x^2 - 1}{x^2}\).


Consider the domain \(x \ge 1\).


The denominator, \(x^2\), is always positive since \(x \ge 1\).


For the numerator, if \(x \ge 1\), then \(x^2 \ge 1\), which implies \(x^2 - 1 \ge 0\).


Since the numerator is non-negative (\(\ge 0\)) and the denominator is positive (\(>0\)), their ratio \(f'(x)\) must be non-negative.


So, \(f'(x) \ge 0\) for all \(x \ge 1\).


Therefore, the function \(f(x) = x + \frac{1}{x}\) is an increasing function on the interval \([1, \infty)\).
Quick Tip: When analyzing the sign of a rational expression (a fraction), determine the sign of the numerator and the denominator separately over the given interval. Then combine them to find the overall sign of the expression.

Let V be the volume of water in the cylinder, h be the height of the water, and r be the radius of the cylinder.


We are given the following information:


Rate of change of volume, \(\frac{dV}{dt} = -5\) cm\(^3\)/s (negative because the volume is decreasing).


Radius of the cylinder, \(r = 15\) cm (this is a constant).


We need to find the rate of change of height, \(\frac{dh}{dt}\).


The formula for the volume of a cylinder is \(V = \pi r^2 h\).


Since the radius \(r\) is constant, we can substitute its value:

\(V = \pi (15)^2 h = 225\pi h\).


Now, we differentiate this equation with respect to time, t, to relate the rates.

\(\frac{d}{dt}(V) = \frac{d}{dt}(225\pi h)\).

\(\frac{dV}{dt} = 225\pi \frac{dh}{dt}\).


Substitute the given value of \(\frac{dV}{dt}\):

\(-5 = 225\pi \frac{dh}{dt}\).


Now, solve for \(\frac{dh}{dt}\).

\(\frac{dh}{dt} = \frac{-5}{225\pi} = -\frac{1}{45\pi}\).


The negative sign indicates that the height is decreasing.


The rate at which the height of the water is decreasing is \(\frac{1}{45\pi}\) cm/s.


(Note: The given height of 2 metres is extra information not needed for the calculation, as the rate of decrease in height is constant for a cylinder).
Quick Tip: In related rates problems involving geometric shapes, first write down the formula connecting the variables (e.g., volume, height, radius). Then, differentiate the formula with respect to time, treating variables that change over time as functions of t. Finally, substitute the given rates and values to solve for the unknown rate. Always check for consistent units.

Let the integral be \(I = \int \frac{\sqrt{x}}{1 + x^{3/2}} dx\).


We can solve this integral using the substitution method.


Let \(t = 1 + x^{3/2}\).


Differentiating both sides with respect to x, we get:

\(\frac{dt}{dx} = \frac{d}{dx}(1 + x^{3/2}) = \frac{3}{2}x^{(3/2 - 1)} = \frac{3}{2}x^{1/2} = \frac{3}{2}\sqrt{x}\).


Rearranging the terms, we have \(\sqrt{x} \,dx = \frac{2}{3} \,dt\).


Now, substitute \(t\) and \(\frac{2}{3} \,dt\) back into the integral.

\(I = \int \frac{1}{t} \left(\frac{2}{3} \,dt\right) = \frac{2}{3} \int \frac{1}{t} \,dt\).


The integral of \(\frac{1}{t}\) is \(\ln|t|\).

\(I = \frac{2}{3} \ln|t| + C\).


Now, substitute back the value of \(t = 1 + x^{3/2}\).


Since \(x \ge 0\), the term \(1 + x^{3/2}\) is always positive, so we can drop the absolute value bars.

\(I = \frac{2}{3} \ln(1 + x^{3/2}) + C\).
Quick Tip: When the integrand is a fraction, check if the numerator is related to the derivative of the denominator (or part of the denominator). In this case, the derivative of \(x^{3/2}\) is a multiple of \(\sqrt{x}\), which makes substitution a perfect strategy.

First, we find the point of intersection of the two lines.


Let the first line be \(L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = \lambda\).


Any point on \(L_1\) can be written as \(P_1(2\lambda+1, 3\lambda+2, 4\lambda+3)\).


Let the second line be \(L_2: \frac{x-4}{5} = \frac{y-1}{2} = \frac{z}{1} = \mu\).


Any point on \(L_2\) can be written as \(P_2(5\mu+4, 2\mu+1, \mu)\).


At the point of intersection, the coordinates must be equal.

\(2\lambda+1 = 5\mu+4 \implies 2\lambda - 5\mu = 3\) ---(1)

\(3\lambda+2 = 2\mu+1 \implies 3\lambda - 2\mu = -1\) ---(2)

\(4\lambda+3 = \mu\) ---(3)


Substitute equation (3) into equation (2):

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

\(3\lambda - 8\lambda - 6 = -1\).

\(-5\lambda = 5 \implies \lambda = -1\).


Now, find the point of intersection by substituting \(\lambda = -1\) into the coordinates for \(P_1\).

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

\(y = 3(-1) + 2 = -1\).

\(z = 4(-1) + 3 = -1\).


The point of intersection is P(–1, –1, –1).


Now, we find the distance between P(–1, –1, –1) and the given point Q(–1, –5, –10).


Using the distance formula, \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\).

\(d = \sqrt{(-1 - (-1))^2 + (-1 - (-5))^2 + (-1 - (-10))^2}\).

\(d = \sqrt{(0)^2 + (4)^2 + (9)^2} = \sqrt{0 + 16 + 81} = \sqrt{97}\).


The required distance is \(\sqrt{97}\) units.
Quick Tip: To find the intersection of two lines in 3D, write the general coordinates of a point on each line using different parameters (like \(\lambda\) and \(\mu\)). Then, equate the corresponding x, y, and z coordinates to get a system of three linear equations. Solve any two to find the parameters and check if they satisfy the third.

To prove that f is a bijection, we must prove that it is both injective (one-one) and surjective (onto).


Injectivity (One-one):


Let \(x_1, x_2 \in R^+\) (the domain) such that \(f(x_1) = f(x_2)\).

\(\log_a x_1 = \log_a x_2\).


By definition of the logarithm function, if the logs of two numbers to the same base are equal, the numbers must be equal.


Therefore, \(x_1 = x_2\).


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


Surjectivity (Onto):


Let \(y\) be an arbitrary element in the codomain, \(R\). We need to find a pre-image \(x\) in the domain \(R^+\) such that \(f(x)=y\).

\(y = \log_a x\).


By converting the logarithmic form to exponential form, we get:

\(x = a^y\).


Since \(a>0\), the value of \(a^y\) is always a positive real number for any \(y \in R\).


Thus, for any \(y \in R\), there exists an \(x = a^y \in R^+\) which is its pre-image.


Therefore, the function \(f\) is surjective.


Since \(f\) is both injective and surjective, it is a bijection.
Quick Tip: For proofs of bijections: 1. Injective: Start with \(f(x_1) = f(x_2)\) and show algebraically that it must lead to \(x_1 = x_2\). 2. Surjective: Start with an arbitrary \(y\) in the codomain, set \(y = f(x)\), and solve for \(x\). Then, show that this value of \(x\) is always valid (i.e., it exists within the domain).

(i) Elements of R:


We test each element \(x \in A\) to find a corresponding \(y \in B\) such that \(x+y=6\).


If \(x = 1\), then \(1 + y = 6 \implies y = 5\). Since \(5 \in B\), the pair (1, 5) is in R.


If \(x = 2\), then \(2 + y = 6 \implies y = 4\). Since \(4 \in B\), the pair (2, 4) is in R.


If \(x = 3\), then \(3 + y = 6 \implies y = 3\). Since \(3 \notin B\), there is no pair starting with 3.


Therefore, the elements of R are R = {(1, 5), (2, 4).


(ii) Is R a function?


A relation R from set A to set B is a function if every element of set A has exactly one image in set B.


In this case, the element \(3 \in A\) does not have any image in B under the relation R.


Since not all elements of A are mapped to an element in B, the relation R is not a function from A to B.


(iii) Domain and Range of R:


The domain of a relation is the set of all first elements of the ordered pairs.


Domain of R = {1, 2.


The range of a relation is the set of all second elements of the ordered pairs.


Range of R = {4, 5.
Quick Tip: For a relation from a set A to a set B to be a function, two conditions must be met: 1. Every element in A must be used as a first element in a pair. 2. No element in A can be the first element in more than one pair. If even one element of A is left out, it's not a function.

(i) Determine the value of p.


For a probability distribution, the sum of all probabilities must be equal to 1.

\(\sum P(X_i) = P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1\).

\(p + \frac{p}{3} + \frac{p}{6} + \frac{p}{12} = 1\).


To add the fractions, we find a common denominator, which is 12.

\(\frac{12p}{12} + \frac{4p}{12} + \frac{2p}{12} + \frac{p}{12} = 1\).

\(\frac{12p + 4p + 2p + p}{12} = 1 \implies \frac{19p}{12} = 1\).

\(p = \frac{12}{19}\).


(ii) Calculate P(X \(\ge\) 1).


This can be calculated as \(1 - P(X < 1)\), which is \(1 - P(X=0)\).

\(P(X \ge 1) = 1 - P(X=0) = 1 - p\).

\(P(X \ge 1) = 1 - \frac{12}{19} = \frac{19-12}{19} = \frac{7}{19}\).


(iii) Calculate expectation of X, E(X).


The expectation is given by the formula \(E(X) = \sum X_i P(X_i)\).

\(E(X) = (0 \cdot p) + (1 \cdot \frac{p}{3}) + (2 \cdot \frac{p}{6}) + (3 \cdot \frac{p}{12})\).

\(E(X) = 0 + \frac{p}{3} + \frac{2p}{6} + \frac{3p}{12} = \frac{p}{3} + \frac{p}{3} + \frac{p}{4}\).

\(E(X) = \frac{2p}{3} + \frac{p}{4} = \frac{8p + 3p}{12} = \frac{11p}{12}\).


Substitute the value \(p = \frac{12}{19}\).

\(E(X) = \frac{11}{12} \left(\frac{12}{19}\right) = \frac{11}{19}\).
Quick Tip: Key formulas for probability distributions: 1. Sum of probabilities: \(\sum P(X_i) = 1\). 2. Expectation (Mean): \(E(X) = \mu = \sum X_i P(X_i)\). 3. Variance: \(Var(X) = \sigma^2 = E(X^2) - [E(X)]^2\), where \(E(X^2) = \sum X_i^2 P(X_i)\).

Let A be the event that a person uses public transport, and B be the event that a person uses a bicycle.


We are given the probabilities:


P(A) = 50% = 0.50


P(B) = 35% = 0.35


P(A \(\cap\) B) = 20% = 0.20 (This is the probability of using both)


(i) The person uses only public transport.


This corresponds to the event A but not B, which is \(P(A \cap B')\).


The formula is \(P(A only) = P(A) - P(A \cap B)\).

\(P(A only) = 0.50 - 0.20 = 0.30\).


(ii) The person uses a bicycle, given that they also use the public transport.


This is a conditional probability, \(P(B|A)\).


The formula is \(P(B|A) = \frac{P(A \cap B)}{P(A)}\).

\(P(B|A) = \frac{0.20}{0.50} = \frac{2}{5} = 0.4\).


(iii) The person uses neither public transport nor a bicycle.


This corresponds to the event not A and not B, which is \(P(A' \cap B')\).


By De Morgan's law, \(P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)\).


First, we calculate \(P(A \cup B)\) using the addition rule of probability.

\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).

\(P(A \cup B) = 0.50 + 0.35 - 0.20 = 0.65\).


Now, \(P(neither) = 1 - P(A \cup B) = 1 - 0.65 = 0.35\).
Quick Tip: Remember the key probability formulas: - \(P(A only) = P(A) - P(A \cap B)\) - Conditional Probability: \(P(B|A) = P(A \cap B) / P(A)\) - Union of events: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) - Complement (Neither): \(P(A' \cap B') = 1 - P(A \cup B)\)

For the function \(f(x)\) to be continuous at \(x = -1\), the limit of the function as \(x\) approaches -1 must be equal to the value of the function at \(x = -1\).


That is, \(\lim_{x \to -1} f(x) = f(-1)\).


From the function definition, we have \(f(-1) = k\).


Now, we need to calculate the limit:

\(\lim_{x \to -1} f(x) = \lim_{x \to -1} \frac{x^2-2x-3}{x+1}\).


Substituting \(x=-1\) directly into the expression gives \(\frac{(-1)^2 - 2(-1) - 3}{-1+1} = \frac{1+2-3}{0} = \frac{0}{0}\), which is an indeterminate form.


We can simplify the expression by factoring the numerator.

\(x^2 - 2x - 3 = (x-3)(x+1)\).


So, the limit becomes:

\(\lim_{x \to -1} \frac{(x-3)(x+1)}{x+1}\).


For \(x \neq -1\), we can cancel the \((x+1)\) term.

\(\lim_{x \to -1} (x-3)\).


Now, substitute \(x = -1\) into the simplified expression:

\(-1 - 3 = -4\).


For continuity, we must have \(f(-1) = \lim_{x \to -1} f(x)\).


Therefore, \(k = -4\).
Quick Tip: When evaluating a limit for a rational function that results in the 0/0 indeterminate form, always try to factor the numerator and denominator to see if a common factor can be canceled. L'Hôpital's rule is another alternative if factoring is difficult.

To check for differentiability at a point, we must check if the Left-Hand Derivative (LHD) and the Right-Hand Derivative (RHD) exist and are equal.


First, let's write \(f(x)\) as a piecewise function.


The definition of \(|x|\) is \(|x| = \begin{cases} x, & x \ge 0
-x, & x < 0 \end{cases}\).


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


The value of the function at \(x=0\) is \(f(0) = 0|0| = 0\).


Now, we calculate the Right-Hand Derivative (RHD) at \(x=0\).

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


For \(h > 0\), we use \(f(h) = h^2\).

\(RHD = \lim_{h \to 0^+} \frac{h^2}{h} = \lim_{h \to 0^+} h = 0\).


Next, we calculate the Left-Hand Derivative (LHD) at \(x=0\).

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


For \(h < 0\), we use \(f(h) = -h^2\).

\(LHD = \lim_{h \to 0^-} \frac{-h^2}{h} = \lim_{h \to 0^-} (-h) = 0\).


Since LHD = RHD = 0, the derivative exists at \(x=0\).


Therefore, the function \(f(x) = x|x|\) is differentiable at \(x=0\), and its derivative is \(f'(0) = 0\).
Quick Tip: When checking differentiability of functions involving absolute values, it's almost always best to first rewrite the function as a piecewise-defined function without the absolute value bars. Then, apply the limit definition of the derivative from the left and the right at the point in question.

The feasible region is the shaded area OABCD. We need to find the inequalities that define this region.


First, we identify the non-negativity constraints. Since the region is in the first quadrant:

\(x \ge 0\)

\(y \ge 0\)


Next, we find the equations of the lines forming the boundaries AB, BC, and CD.


Line through A(0, 200) and B(50, 250):


Slope \(m = \frac{250-200}{50-0} = \frac{50}{50} = 1\). The y-intercept is 200.


Equation: \(y = x + 200 \implies y - x = 200\).


To find the inequality, we test a point in the feasible region, like (50, 50). \(50 - 50 = 0\). Since \(0 \le 200\), the inequality is \(y - x \le 200\).


Line through B(50, 250) and C(150, 150):


Slope \(m = \frac{150-250}{150-50} = \frac{-100}{100} = -1\).


Equation: \(y - 250 = -1(x - 50) \implies y - 250 = -x + 50 \implies x + y = 300\).


Test point (50, 50): \(50 + 50 = 100\). Since \(100 \le 300\), the inequality is \(x + y \le 300\).


Line through C(150, 150) and D(200, 0):


Slope \(m = \frac{0-150}{200-150} = \frac{-150}{50} = -3\).


Equation: \(y - 0 = -3(x - 200) \implies y = -3x + 600 \implies 3x + y = 600\).


Test point (50, 50): \(3(50) + 50 = 200\). Since \(200 \le 600\), the inequality is \(3x + y \le 600\).


The set of all constraints is:


1. \(y - x \le 200\) (or \(x-y \ge -200\))


2. \(x + y \le 300\)


3. \(3x + y \le 600\)


4. \(x \ge 0\)


5. \(y \ge 0\)
Quick Tip: To determine the inequality sign (\(\le\) or \(\ge\)) for a boundary line in an LPP graph, first find the equation of the line. Then, pick a test point that is clearly inside the feasible region (the origin (0,0) is easiest if applicable). Substitute the test point's coordinates into your equation. If the resulting statement is true, use that inequality; if false, use the opposite inequality.

The given equation is \(y = 4x - \frac{1}{2}x^2\), where y is the height and x is the number of days.


(i) Find the rate of growth of the plant with respect to sunlight.


The rate of growth is the derivative of the height y with respect to the number of days x, which is \(\frac{dy}{dx}\).

\(\frac{dy}{dx} = \frac{d}{dx} \left(4x - \frac{1}{2}x^2\right)\).

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


So, the rate of growth of the plant is \((4-x)\) cm/day.


(ii) In how many days will the plant attain its maximum height? What is the maximum height?


For the plant to attain its maximum height, the rate of growth must be zero.


Set \(\frac{dy}{dx} = 0\).

\(4 - x = 0 \implies x = 4\).


To confirm this is a maximum, we can check the second derivative, \(\frac{d^2y}{dx^2}\).

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


Since \(\frac{d^2y}{dx^2} < 0\), the height is maximum at \(x=4\).


So, the plant will attain its maximum height in 4 days.


To find the maximum height, substitute \(x=4\) back into the original equation for y.

\(y_{max} = 4(4) - \frac{1}{2}(4)^2 = 16 - \frac{1}{2}(16) = 16 - 8 = 8\).


The maximum height of the plant is 8 cm.
Quick Tip: To find the maximum or minimum value of a function, follow these steps: 1. Find the first derivative of the function. 2. Set the first derivative to zero and solve for the critical points. 3. Find the second derivative. If it's negative at a critical point, it's a local maximum. If it's positive, it's a local minimum. 4. Substitute the critical point back into the original function to find the maximum/minimum value.

Let the adjacent sides of the parallelogram be represented by the vectors \(\vec{p}\) and \(\vec{q}\).


Then the diagonals are given by \(\vec{a} = \vec{p} + \vec{q}\) and \(\vec{b} = \vec{p} - \vec{q}\).


The area of the parallelogram is given by \(|\vec{p} \times \vec{q}|\).


Let's compute the cross product of the diagonals:

\(\vec{a} \times \vec{b} = (\vec{p} + \vec{q}) \times (\vec{p} - \vec{q})\)

\(= \vec{p} \times \vec{p} - \vec{p} \times \vec{q} + \vec{q} \times \vec{p} - \vec{q} \times \vec{q}\).


Since the cross product of a vector with itself is the zero vector (\(\vec{p} \times \vec{p} = \vec{0}\)) and \(\vec{q} \times \vec{p} = -(\vec{p} \times \vec{q})\):

\(\vec{a} \times \vec{b} = \vec{0} - \vec{p} \times \vec{q} - \vec{p} \times \vec{q} - \vec{0} = -2(\vec{p} \times \vec{q})\).


Taking the magnitude of both sides:

\(|\vec{a} \times \vec{b}| = |-2(\vec{p} \times \vec{q})| = 2|\vec{p} \times \vec{q}|\).


Since Area = \(|\vec{p} \times \vec{q}|\), we have \(|\vec{a} \times \vec{b}| = 2 \times Area\).


Therefore, Area = \(\frac{1}{2}|\vec{a} \times \vec{b}|\). (Hence proved)


Now, let the given diagonals be \(\vec{a} = 2\hat{i} - \hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} + 3\hat{j} - \hat{k}\).

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

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

\(= \hat{i}(1 - 3) - \hat{j}(-2 - 1) + \hat{k}(6 + 1) = -2\hat{i} + 3\hat{j} + 7\hat{k}\).

\(|\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + (3)^2 + (7)^2} = \sqrt{4 + 9 + 49} = \sqrt{62}\).


Area = \(\frac{1}{2}|\vec{a} \times \vec{b}| = \frac{1}{2}\sqrt{62}\) square units.
Quick Tip: There are two important area formulas for a parallelogram using vectors: 1. If \(\vec{p}\) and \(\vec{q}\) are adjacent sides, Area = \(|\vec{p} \times \vec{q}|\). 2. If \(\vec{a}\) and \(\vec{b}\) are diagonals, Area = \(\frac{1}{2}|\vec{a} \times \vec{b}|\). Don't confuse the two formulas.

Let the required line be L. It passes through the point A(1, 2, –4), so its position vector is \(\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}\).


Let the two given lines be \(L_1\) and \(L_2\).


The direction vector of \(L_1\) is \(\vec{b_1} = 3\hat{i} - 16\hat{j} + 7\hat{k}\).


The direction vector of \(L_2\) is \(\vec{b_2} = 3\hat{i} + 8\hat{j} - 5\hat{k}\).


Since the required line L is perpendicular to both \(L_1\) and \(L_2\), its direction vector \(\vec{b}\) will be parallel to the cross product of \(\vec{b_1}\) and \(\vec{b_2}\).

\(\vec{b} = \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -16 & 7
3 & 8 & -5 \end{vmatrix}\).

\(= \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3))\).

\(= \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48)\).

\(\vec{b} = 24\hat{i} + 36\hat{j} + 72\hat{k}\).


We can simplify the direction vector by taking out the greatest common divisor, which is 12.


Let's use the simplified direction vector \(\vec{d} = \frac{1}{12}\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}\).


Vector Equation of the line:


The vector equation is \(\vec{r} = \vec{a} + \lambda\vec{d}\).

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


Cartesian Equation of the line:


The cartesian equation is \(\frac{x-x_1}{d_1} = \frac{y-y_1}{d_2} = \frac{z-z_1}{d_3}\).


Here \((x_1, y_1, z_1) = (1, 2, -4)\) and \((d_1, d_2, d_3) = (2, 3, 6)\).

\(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-(-4)}{6}\).

\(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}\).
Quick Tip: A line perpendicular to two other lines has a direction vector that is parallel to the cross product of the direction vectors of the two given lines. Remember to simplify the resulting direction vector by factoring out any common scalar multiple.

Let \(I = \int_{0}^{3/2} |x \cos(\pi x)| dx\). We need to analyze the sign of the integrand \(f(x) = x \cos(\pi x)\) in the interval \([0, 3/2]\).


In the interval \([0, 3/2]\), \(x \ge 0\). So the sign is determined by \(\cos(\pi x)\).

\(\cos(\theta)\) is positive when \(0 \le \theta < \pi/2\) and negative when \(\pi/2 < \theta \le 3\pi/2\).


Let \(\theta = \pi x\).

\(\cos(\pi x) \ge 0\) when \(0 \le \pi x \le \pi/2 \implies 0 \le x \le 1/2\).

\(\cos(\pi x) \le 0\) when \(\pi/2 \le \pi x \le 3\pi/2 \implies 1/2 \le x \le 3/2\).


So, we split the integral at \(x = 1/2\).

\(I = \int_{0}^{1/2} x \cos(\pi x) dx + \int_{1/2}^{3/2} -x \cos(\pi x) dx\).


We use integration by parts for \(\int x \cos(\pi x) dx\). Let \(u=x, dv=\cos(\pi x)dx\). Then \(du=dx, v=\frac{\sin(\pi x)}{\pi}\).

\(\int x \cos(\pi x) dx = x\frac{\sin(\pi x)}{\pi} - \int \frac{\sin(\pi x)}{\pi} dx = \frac{x\sin(\pi x)}{\pi} + \frac{\cos(\pi x)}{\pi^2}\).


Now, evaluate the definite integrals:

\(I_1 = \int_{0}^{1/2} x \cos(\pi x) dx = \left[ \frac{x\sin(\pi x)}{\pi} + \frac{\cos(\pi x)}{\pi^2} \right]_{0}^{1/2}\).

\(= \left(\frac{(1/2)\sin(\pi/2)}{\pi} + \frac{\cos(\pi/2)}{\pi^2}\right) - \left(0 + \frac{\cos(0)}{\pi^2}\right) = \left(\frac{1/2 \cdot 1}{\pi} + 0\right) - \left(\frac{1}{\pi^2}\right) = \frac{1}{2\pi} - \frac{1}{\pi^2}\).

\(I_2 = \int_{1/2}^{3/2} x \cos(\pi x) dx = \left[ \frac{x\sin(\pi x)}{\pi} + \frac{\cos(\pi x)}{\pi^2} \right]_{1/2}^{3/2}\).

\(= \left(\frac{(3/2)\sin(3\pi/2)}{\pi} + \frac{\cos(3\pi/2)}{\pi^2}\right) - \left(\frac{(1/2)\sin(\pi/2)}{\pi} + \frac{\cos(\pi/2)}{\pi^2}\right)\).

\(= \left(\frac{(3/2)(-1)}{\pi} + 0\right) - \left(\frac{1/2 \cdot 1}{\pi} + 0\right) = -\frac{3}{2\pi} - \frac{1}{2\pi} = -\frac{4}{2\pi} = -\frac{2}{\pi}\).


The total integral is \(I = I_1 - I_2\).

\(I = \left(\frac{1}{2\pi} - \frac{1}{\pi^2}\right) - \left(-\frac{2}{\pi}\right) = \frac{1}{2\pi} - \frac{1}{\pi^2} + \frac{2}{\pi} = \frac{5}{2\pi} - \frac{1}{\pi^2}\).
Quick Tip: When integrating a function with an absolute value, the first step is always to find the points where the expression inside the absolute value is zero. Use these points to split the integral into sub-intervals where the expression has a constant sign. Then remove the absolute value bars accordingly (positive or negative).

Let \(I = \int \frac{dx}{\sin x + \sin 2x}\).


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

\(I = \int \frac{dx}{\sin x + 2 \sin x \cos x} = \int \frac{dx}{\sin x(1 + 2\cos x)}\).


Multiply the numerator and denominator by \(\sin x\):

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


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

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


Let \(u = \cos x\), so \(du = -\sin x \,dx\). This gives \(\sin x \,dx = -du\).

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


Now, we use partial fractions. Let \(\frac{1}{(u-1)(u+1)(2u+1)} = \frac{A}{u-1} + \frac{B}{u+1} + \frac{C}{2u+1}\).

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


If \(u=1\): \(1 = A(2)(3) \implies A = 1/6\).


If \(u=-1\): \(1 = B(-2)(-1) \implies B = 1/2\).


If \(u=-1/2\): \(1 = C(-3/2)(1/2) \implies 1 = -3/4 C \implies C = -4/3\).

\(I = \int \left( \frac{1/6}{u-1} + \frac{1/2}{u+1} - \frac{4/3}{2u+1} \right) du\).

\(I = \frac{1}{6}\ln|u-1| + \frac{1}{2}\ln|u+1| - \frac{4}{3} \frac{\ln|2u+1|}{2} + C_{final}\).

\(I = \frac{1}{6}\ln|u-1| + \frac{1}{2}\ln|u+1| - \frac{2}{3}\ln|2u+1| + C_{final}\).


Substitute back \(u = \cos x\):

\(I = \frac{1}{6}\ln|\cos x - 1| + \frac{1}{2}\ln|\cos x + 1| - \frac{2}{3}\ln|2\cos x + 1| + C_{final}\).
Quick Tip: Integrals of trigonometric functions that don't fit standard forms can often be solved by using identities to rewrite the expression, followed by a substitution. If the substitution leads to a rational function (a polynomial fraction), the method of partial fractions is a powerful tool to break it down into simpler integrals.

Part 1: Proof


By definition of an inverse, a matrix M is the inverse of N if MN = NM = I.


We want to show that \(\frac{1}{k}A^{-1}\) is the inverse of \((kA)\). Let's multiply them.

\((kA) \left(\frac{1}{k}A^{-1}\right) = (k \cdot \frac{1}{k}) (A \cdot A^{-1})\). (Using associativity and commutativity of scalars)

\(= 1 \cdot I = I\).


Also, \(\left(\frac{1}{k}A^{-1}\right)(kA) = (\frac{1}{k} \cdot k) (A^{-1} \cdot A) = 1 \cdot I = I\).


Since their product is the identity matrix I, we have proved that \((kA)^{-1} = \frac{1}{k}A^{-1}\).


Part 2: Calculation


We need to calculate \((3A)^{-1}\), which from the proof is equal to \(\frac{1}{3}A^{-1}\).


First, we must find \(A^{-1}\) for the matrix \(A = \begin{pmatrix} 2 & -1 & 1
-1 & 2 & -1
1 & -1 & 2 \end{pmatrix}\).


The inverse is given by \(A^{-1} = \frac{1}{\det(A)}adj(A)\).


Calculate determinant of A:

\(\det(A) = 2(2 \cdot 2 - (-1)(-1)) - (-1)((-1)(2) - (-1)(1)) + 1((-1)(-1) - (2)(1))\).

\(= 2(4 - 1) + 1(-2 + 1) + 1(1 - 2) = 2(3) + 1(-1) + 1(-1) = 6 - 1 - 1 = 4\).


Since \(\det(A) = 4 \ne 0\), the inverse exists.


Calculate the adjugate of A (matrix of cofactors transposed):


The matrix of cofactors is:

\(C_{11} = (4-1)=3\), \(C_{12} = -(-2+1)=1\), \(C_{13} = (1-2)=-1\).

\(C_{21} = -(-2+1)=1\), \(C_{22} = (4-1)=3\), \(C_{23} = -(-2+1)=1\).

\(C_{31} = (1-2)=-1\), \(C_{32} = -(-2+1)=1\), \(C_{33} = (4-1)=3\).


Cofactor matrix \(C = \begin{pmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{pmatrix}\).

\(adj(A) = C^T = \begin{pmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{pmatrix}\). (The matrix is symmetric)

\(A^{-1} = \frac{1}{4} \begin{pmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{pmatrix}\).


Finally, calculate \((3A)^{-1} = \frac{1}{3}A^{-1}\).

\((3A)^{-1} = \frac{1}{3} \left( \frac{1}{4} \begin{pmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{pmatrix} \right) = \frac{1}{12} \begin{pmatrix} 3 & 1 & -1
1 & 3 & 1
-1 & 1 & 3 \end{pmatrix}\).
Quick Tip: The process for finding a matrix inverse is mechanical: 1. Calculate the determinant. If it's zero, the inverse doesn't exist. 2. Calculate the matrix of cofactors. Remember the checkerboard pattern of signs. 3. Transpose the cofactor matrix to get the adjugate. 4. The inverse is the adjugate divided by the determinant.

Let M be the event that a student has the misconception.


Let M' be the event that a student does not have the misconception.


Let B be the event that a student answers 'Bijoy'.


From the problem statement, we are given the following probabilities:


P(M) = 40% = 0.40.


P(M') = 1 - P(M) = 1 - 0.40 = 0.60.


The probability that a student with the misconception answers 'Bijoy' is P(B|M) = 80% = 0.80.


The probability that a student without the misconception does NOT answer 'Bijoy' is 90%.


Therefore, the probability that a student without the misconception answers 'Bijoy' is P(B|M') = 100% - 90% = 10% = 0.10.


The question asks for the probability of a student not having a misconception AND answering 'Bijoy', which is P(M' \(\cap\) B).


Using the multiplication rule of probability:


P(M' \(\cap\) B) = P(B|M') \(\times\) P(M').


P(M' \(\cap\) B) = 0.10 \(\times\) 0.60 = 0.06.
Quick Tip: The phrase "probability of A and B" corresponds to the intersection of events, P(A \(\cap\) B). This can often be calculated using the formula P(A \(\cap\) B) = P(A|B)P(B) or P(B|A)P(A).

Let M be the event that a student has the misconception, and B be the event that a student answers 'Bijoy'.


We have P(M) = 0.40, P(M') = 0.60, P(B|M) = 0.80, and P(B|M') = 0.10.


We want to find the total probability of a student answering 'Bijoy', which is P(B).


A student can answer 'Bijoy' in two mutually exclusive ways:


1. The student has the misconception and answers 'Bijoy' (event M \(\cap\) B).


2. The student does not have the misconception and answers 'Bijoy' (event M' \(\cap\) B).


Using the Law of Total Probability: P(B) = P(M \(\cap\) B) + P(M' \(\cap\) B).


We can calculate each term:


P(M \(\cap\) B) = P(B|M) \(\times\) P(M) = 0.80 \(\times\) 0.40 = 0.32.


P(M' \(\cap\) B) = P(B|M') \(\times\) P(M') = 0.10 \(\times\) 0.60 = 0.06.


Now, add these probabilities:


P(B) = 0.32 + 0.06 = 0.38.


The probability that a randomly selected student answers 'Bijoy' is 0.38 or 38%.
Quick Tip: The Law of Total Probability is used to find the probability of an event when it can occur via several mutually exclusive scenarios. 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.

Let M be the event that a student has the misconception, and B be the event that a student answers 'Bijoy'.


We need to find the probability that a student has the misconception GIVEN that they answered 'Bijoy'. This is the conditional probability P(M|B).


Using Bayes' Theorem, the formula is:

\(P(M|B) = \frac{P(B|M)P(M)}{P(B)}\).


From the previous parts, we have calculated the necessary components:


P(B|M)P(M) = P(M \(\cap\) B) = 0.32.


P(B) = 0.38.


Now, substitute these values into the formula:

\(P(M|B) = \frac{0.32}{0.38}\).

\(P(M|B) = \frac{32}{38} = \frac{16}{19}\).
Quick Tip: Bayes' Theorem is used to "reverse" conditional probability. If you know P(A|B), you can use it to find P(B|A). The general formula is \(P(B|A) = \frac{P(A|B)P(B)}{P(A)}\). Remember it as "Probability of cause, given the effect".

Two lines are parallel if and only if their direction vectors are proportional (i.e., one is a scalar multiple of the other).


The equation of the track for Line A is \(l_1: \frac{x-2}{3} = \frac{y+1}{-2} = \frac{z-3}{4}\).


The direction vector for Line A is \(\vec{b_1} = 3\hat{i} - 2\hat{j} + 4\hat{k}\).


The equation of the track for Line B is \(l_2: \frac{x-1}{2} = \frac{y-3}{1} = \frac{z+2}{-3}\).


The direction vector for Line B is \(\vec{b_2} = 2\hat{i} + 1\hat{j} - 3\hat{k}\).


To check for parallelism, we see if there exists a scalar \(k\) such that \(\vec{b_1} = k\vec{b_2}\).


Comparing the components:


For \(\hat{i}\): \(3 = k(2) \implies k = 3/2\).


For \(\hat{j}\): \(-2 = k(1) \implies k = -2\).


For \(\hat{k}\): \(4 = k(-3) \implies k = -4/3\).


Since we get different values of \(k\) for each component, the direction vectors are not proportional.


Therefore, the two metro tracks are not parallel.
Quick Tip: To quickly check if two vectors \(\vec{a} = \) and \(\vec{b} = \) are parallel, check if the ratios of their corresponding components are equal: \(\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}\).

The required line is parallel to Line A's track, \(l_1\).


This means the required line will have the same direction vector as \(l_1\).


The direction vector of \(l_1\) is \(\vec{b_1} = 3\hat{i} - 2\hat{j} + 4\hat{k}\). The direction ratios are (3, -2, 4).


The line passes through the given point P(1, –2, –3).


The equation of a line passing through a point \((x_1, y_1, z_1)\) with direction ratios \((a, b, c)\) is given by:

\(\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}\).


Substituting the point (1, -2, -3) and the direction ratios (3, -2, 4):

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

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


This is the required cartesian equation of the line.
Quick Tip: Parallel lines share the same direction vector. If a line is parallel to a known line and passes through a given point, you have all the information needed to write its equation in either vector or cartesian form.

Line A (\(l_1\)) passes through point \(\vec{a_1} = 2\hat{i} - \hat{j} + 3\hat{k}\) and has direction \(\vec{b_1} = 3\hat{i} - 2\hat{j} + 4\hat{k}\).


Line B (\(l_2\)) passes through point \(\vec{a_2} = \hat{i} + 3\hat{j} - 2\hat{k}\) and has direction \(\vec{b_2} = 2\hat{i} + \hat{j} - 3\hat{k}\).


The lines are skew (not parallel, as shown in part i). The shortest distance is given by the formula:

\(d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}\).


First, calculate \(\vec{a_2} - \vec{a_1}\):

\(\vec{a_2} - \vec{a_1} = (1-2)\hat{i} + (3-(-1))\hat{j} + (-2-3)\hat{k} = -\hat{i} + 4\hat{j} - 5\hat{k}\).


Next, calculate the cross product \(\vec{b_1} \times \vec{b_2}\):

\(\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -2 & 4
2 & 1 & -3 \end{vmatrix} = \hat{i}(6-4) - \hat{j}(-9-8) + \hat{k}(3-(-4)) = 2\hat{i} + 17\hat{j} + 7\hat{k}\).


Now, calculate the dot product for the numerator:

\((\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-1)(2) + (4)(17) + (-5)(7) = -2 + 68 - 35 = 31\).


Next, calculate the magnitude of the cross product for the denominator:

\(|\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + 17^2 + 7^2} = \sqrt{4 + 289 + 49} = \sqrt{342}\).


Finally, compute the distance:

\(d = \frac{|31|}{\sqrt{342}} = \frac{31}{\sqrt{342}}\) units.
Quick Tip: The shortest distance between two skew lines is the length of the projection of the vector connecting a point on each line onto the common normal vector. The common normal vector is found by the cross product of the direction vectors of the lines.

We are given the differential equation for Newton's Law of Cooling:

\(\frac{d(T(t))}{dt} = -k(T(t) - 25)\).


This can be written as \(\frac{dT}{dt} = -k(T-25)\).


We solve this using the method of separation of variables.

\(\frac{dT}{T-25} = -k \, dt\).


Integrating both sides:

\(\int \frac{dT}{T-25} = \int -k \, dt\).

\(\ln|T-25| = -kt + C\), where C is the constant of integration.


Since the processor is cooling from 85°C to room temperature 25°C, T will always be greater than 25, so we can drop the absolute value.

\(\ln(T-25) = -kt + C\).


To solve for T, we exponentiate both sides:

\(T-25 = e^{-kt+C} = e^C \cdot e^{-kt}\).


Let \(A = e^C\) be another constant.

\(T(t) = 25 + A e^{-kt}\).


Now, we use the initial condition T(0) = 85°C to find the value of A.

\(85 = 25 + A e^{-k(0)} = 25 + A e^0 = 25 + A\).

\(A = 85 - 25 = 60\).


Substituting the value of A back into the equation, we get the expression for the temperature:

\(T(t) = 25 + 60 e^{-kt}\).
Quick Tip: Differential equations of the form \(\frac{dy}{dt} = k(y-a)\) are common in growth and decay problems. The general solution is always of the form \(y(t) = a + C e^{kt}\), where C is a constant determined by the initial conditions.

From the previous part, the expression for the temperature at time t is:

\(T(t) = 25 + 60 e^{-kt}\).


We are given \(k = 0.03\) and we need to find the time \(t\) when the temperature \(T(t) = 40\)°C.


Substitute the known values into the equation:

\(40 = 25 + 60 e^{-0.03t}\).


First, solve for the exponential term.

\(40 - 25 = 60 e^{-0.03t}\).

\(15 = 60 e^{-0.03t}\).

\(\frac{15}{60} = e^{-0.03t}\).

\(\frac{1}{4} = e^{-0.03t}\).


To solve for \(t\), take the natural logarithm (log\(_e\)) of both sides.

\(\ln\left(\frac{1}{4}\right) = \ln\left(e^{-0.03t}\right)\).


Using the logarithm property \(\ln(a^b) = b\ln(a)\) and \(\ln(e)=1\):

\(\ln(4^{-1}) = -0.03t\).

\(-\ln(4) = -0.03t\).

\(t = \frac{\ln(4)}{0.03}\).


Now, substitute the given value \(\ln(4) = 1.3863\).

\(t = \frac{1.3863}{0.03} = \frac{138.63}{3}\).

\(t = 46.21\).


It will take 46.21 minutes for the processor's temperature to reach 40°C.
Quick Tip: To solve an exponential equation of the form \(a = b \cdot e^{cx}\), first isolate the exponential term (\(e^{cx} = a/b\)), then take the natural logarithm of both sides to bring the exponent down.