Step 1: Understanding the Question:


We are asked to find the derivative \(\frac{dy}{dx}\) from an implicit equation.

The equation involves an inverse trigonometric function set equal to a constant.


Step 2: Key Formula or Approach:


The best approach is to simplify the equation first by removing the inverse trigonometric function.

Then, we will use implicit differentiation.


Step 3: Detailed Explanation:


The given equation is:
\[ \tan^{-1}(x^2 - y^2) = a \]

Apply the tangent function to both sides:
\[ \tan(\tan^{-1}(x^2 - y^2)) = \tan(a) \]
\[ x^2 - y^2 = \tan(a) \]

Since 'a' is a constant, tan(a) is also a constant. Let C = tan(a).
\[ x^2 - y^2 = C \]

Now, differentiate both sides with respect to x:
\[ \frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = \frac{d}{dx}(C) \]

Using the power rule and chain rule:
\[ 2x - 2y \frac{dy}{dx} = 0 \]

Rearrange to solve for \(\frac{dy}{dx}\):
\[ 2x = 2y \frac{dy}{dx} \]
\[ \frac{dy}{dx} = \frac{2x}{2y} = \frac{x}{y} \]


Step 4: Final Answer:


The derivative \(\frac{dy}{dx}\) is \(\frac{x}{y}\).
Quick Tip: When an implicit equation has an inverse trigonometric function equal to a constant, always eliminate the inverse function first.
This simplifies the equation significantly before you need to differentiate, avoiding a complicated chain rule.

Step 1: Understanding the Question:


We need to classify the given matrix A based on its properties.


Step 2: Detailed Explanation:


Let's check the definition of each option for the matrix \(A = \begin{bmatrix} 0 & 0 & -5
0 & 3 & 0
4.3 & 0 & 0 \end{bmatrix}\).

- (A) Skew-symmetric matrix: Requires \(A' = -A\) and zero diagonal elements. The diagonal elements here are 0, 3, 0. Since not all are zero, it cannot be skew-symmetric. (Also \(A' \neq -A\)).

- (B) Scalar matrix: A diagonal matrix where all diagonal elements are equal. Matrix A is not a diagonal matrix because it has non-zero off-diagonal elements (-5 and 4.3). So, it cannot be a scalar matrix.

- (C) Diagonal matrix: Requires all off-diagonal elements to be zero. Here, \(a_{13} = -5\) and \(a_{31} = 4.3\) are non-zero. So, it is not a diagonal matrix.

- (D) Square matrix: A matrix with an equal number of rows and columns. Matrix A has 3 rows and 3 columns (order 3 \(\times\) 3). Therefore, it is a square matrix.


Step 3: Final Answer:


The only classification that fits is a square matrix.

*(Note: There appears to be a typo in the matrix element '4.3' as it is unusual for standard exam questions. However, it does not affect the classification.)*
Quick Tip: Always start with the most general classification and move to more specific ones.
Every diagonal, scalar, and identity matrix is a square matrix, but not every square matrix fits the other categories.
Check the definitions carefully.

Step 1: Understanding the Question:


We must identify the function corresponding to the given graph by analyzing its key characteristics.


Step 2: Detailed Explanation:


Let's examine the features of the graph:

- Domain: The graph exists for x \(\leq\) -1 and x \(\geq\) 1. The domain is (\(-\infty, -1]\) \(\cup\) \([1, \infty)\).

- Range: The function's values are in the interval [0, \(\pi\)], but the value \(\pi/2\) is excluded. The range is \([0, \pi/2) \cup (\pi/2, \pi]\). The graph has a horizontal asymptote at y = \(\pi/2\).

- Key Points: The graph passes through (1, 0) and (-1, \(\pi\)).

- Behavior: The function is increasing on both parts of its domain.


Now, let's compare these features with the inverse trigonometric functions:

- (A) y = sec\(^{-1}\) x: The domain is (\(-\infty, -1]\) \(\cup\) \([1, \infty)\). The principal value range is \([0, \pi] - \{\pi/2\}\). Also, sec\(^{-1}\)(1) = 0 and sec\(^{-1}\)(-1) = \(\pi\). This perfectly matches all features of the graph.

- (C) y = cosec\(^{-1}\) x: The domain is the same, but the principal value range is \([-\pi/2, 0) \cup (0, \pi/2]\). This does not match the range of the graph.

- (B) y = sec x and (D) y = cosec x are periodic functions with vertical asymptotes and a range of (\(-\infty, -1]\) \(\cup\) \([1, \infty)\), which is completely different.


Step 3: Final Answer:


The graph shown is that of y = sec\(^{-1}\) x.
Quick Tip: Memorizing the graphs of the six inverse trigonometric functions is essential.
Pay special attention to the domain and range, which are swapped from the original trig functions (with restricted domains).
sec\(^{-1}\)x and cos\(^{-1}\)x share a similar range structure (related to [0, \(\pi\)]).
csc\(^{-1}\)x and sin\(^{-1}\)x share a similar range structure (related to [-\(\pi/2\), \(\pi/2\)]).

Step 1: Understanding the Question:


We need to evaluate the given expression by finding the principal values of the inverse trigonometric functions.


Step 2: Key Formula or Approach:


We use the following properties and standard values:

- Range of sec\(^{-1}\)(x) is \([0, \pi] - \{\pi/2\}\).

- Property: sec\(^{-1}\)(-x) = \(\pi\) - sec\(^{-1}\)(x).

- Standard value: sec(\(\pi/4\)) = \(\sqrt{2}\).

- Standard value: tan(\(\pi/6\)) = \(1/\sqrt{3}\).


Step 3: Detailed Explanation:


First, evaluate the term sec\(^{-1}\)(-\(\sqrt{2}\)):
\[ sec^{-1}(-\sqrt{2}) = \pi - sec^{-1}(\sqrt{2}) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \]


Next, evaluate the term tan\(^{-1}\)(\(1/\sqrt{3}\)):
\[ \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]


Now, perform the subtraction:
\[ Expression = \frac{3\pi}{4} - \frac{\pi}{6} \]

Find a common denominator, which is 12:
\[ = \frac{3\pi \times 3}{12} - \frac{\pi \times 2}{12} = \frac{9\pi - 2\pi}{12} = \frac{7\pi}{12} \]


Step 4: Final Answer:


The value of the expression is \(\frac{7\pi}{12}\).
Quick Tip: Remember the rules for handling negative arguments in inverse functions.
For \(\cos^{-1}, \sec^{-1}, \cot^{-1}\), the rule is \(\pi - f^{-1}(x)\).
For \(\sin^{-1}, \csc^{-1}, \tan^{-1}\), the rule is \(-f^{-1}(x)\).
This is a very common topic for exam questions.

Step 1: Understanding the Question:


Given the order of matrix A (n \(\times\) m), we need to determine the order of matrix B such that the products AB' and B'A are both defined.


Step 2: Key Formula or Approach:


For a matrix product XY to be defined, the number of columns in X must equal the number of rows in Y.

If matrix B has order p \(\times\) q, its transpose B' has order q \(\times\) p.


Step 3: Detailed Explanation:


Let the order of matrix B be p \(\times\) q. Then the order of B' is q \(\times\) p.

- For AB' to be defined: The product is (A)\(_{n \times m}\) (B')\(_{q \times p}\). The inner dimensions must match: \(m = q\).

- For B'A to be defined: The product is (B')\(_{q \times p}\) (A)\(_{n \times m}\). The inner dimensions must match: \(p = n\).


Step 4: Final Answer:


We found that p = n and q = m. Since the order of B is p \(\times\) q, the order of B is n \(\times\) m.
Quick Tip: Quickly solve this by setting up the dimension chains:
- For AB': (n \(\times\) \textbf{m}) (\textbf{q} \(\times\) p) \(\implies\) \textbf{m = q}.
- For B'A: (q \(\times\) \textbf{p}) (\textbf{n} \(\times\) m) \(\implies\) \textbf{p = n}.
The order of B (p \(\times\) q) is therefore n \(\times\) m.

Step 1: Understanding the Question:


We need to determine the property of the resultant matrix when two skew-symmetric matrices of the same order are added.


Step 2: Key Formula or Approach:


A matrix 'M' is skew-symmetric if \(M' = -M\). The property for the transpose of a sum is \((A + B)' = A' + B'\).


Step 3: Detailed Explanation:


Let A and B be two skew-symmetric matrices. This means \(A' = -A\) and \(B' = -B\).

Let C = A + B. We find the transpose of C:
\[ C' = (A + B)' = A' + B' \]

Substitute the properties of A and B:
\[ C' = (-A) + (-B) = -(A + B) = -C \]

Since \(C' = -C\), the sum C is also a skew-symmetric matrix.


Step 4: Final Answer:


The sum is always a skew-symmetric matrix.
Quick Tip: This is a standard closure property in matrix algebra.
- Symmetric + Symmetric = Symmetric
- Skew-symmetric + Skew-symmetric = Skew-symmetric
Remember these rules to answer such theoretical questions instantly.

Step 1: Understanding the Question:


We are asked to find the value of a differential expression involving a given function y and its derivatives.


Step 2: Detailed Explanation:


Find the first derivative, y\(_1\):
\[ y_1 = \frac{dy}{dx} = -a\sin(\log x) \cdot \frac{1}{x} + b\cos(\log x) \cdot \frac{1}{x} \]

Multiply by x to simplify for the next differentiation:
\[ xy_1 = -a \sin(\log x) + b \cos(\log x) \]

Differentiate this equation again using the product rule on the left:
\[ (1)y_1 + x(y_2) = -a\cos(\log x) \cdot \frac{1}{x} - b\sin(\log x) \cdot \frac{1}{x} \]
\[ xy_2 + y_1 = -\frac{1}{x} (a \cos(\log x) + b \sin(\log x)) \]

Multiply the entire equation by x:
\[ x^2y_2 + xy_1 = -(a \cos(\log x) + b \sin(\log x)) \]


Step 3: Final Answer:


The expression in the parenthesis is the original function y.

Therefore, \(x^2y_2 + xy_1 = -y\).
Quick Tip: This type of problem involves forming a Cauchy-Euler differential equation from its solution.
The key technique is to differentiate once, multiply by x to clear the fraction from the chain rule, and then differentiate again. This structure will naturally lead to the \(x^2y_2 + xy_1\) form.

Step 1: Understanding the Question:


For f(x) to be continuous at x = 0, the limit of the function as x approaches 0 must equal the function's value at x = 0. We need to find \(k = \lim_{x \to 0} f(x)\).


Step 2: Key Formula or Approach:


We can use L'Hôpital's Rule for the 0/0 indeterminate form or the standard limit \(\lim_{u \to 0} \frac{\log(1+u)}{u} = 1\).


Step 3: Detailed Explanation:


Using L'Hôpital's Rule:
\[ k = \lim_{x \to 0} \frac{\frac{d}{dx}(\log(1+ax) + \log(1-bx))}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{\frac{a}{1+ax} + \frac{-b}{1-bx}}{1} \]

Substitute x = 0:
\[ k = \frac{a}{1+0} - \frac{b}{1-0} = a - b \]


Step 4: Final Answer:


The value of k is a - b.
Quick Tip: L'Hôpital's Rule is a powerful tool for evaluating limits of indeterminate forms like 0/0 or \(\infty/\infty\).
It often simplifies the problem much faster than algebraic manipulation or using standard limit formulas.

Step 1: Understanding the Question:


A critical point of a function occurs where its first derivative is either zero or undefined.

We need to find the derivative of f(x) = x\(^x\) and set it to zero.


Step 2: Key Formula or Approach:


To differentiate a function of the form \(g(x)^{h(x)}\), we must use logarithmic differentiation.

Let y = x\(^x\). Take the natural logarithm of both sides.


Step 3: Detailed Explanation:


Let y = x\(^x\). The domain of this function requires x > 0.

Take the natural log:
\[ \ln y = \ln(x^x) = x \ln x \]

Differentiate both sides implicitly with respect to x, using the product rule on the right:
\[ \frac{1}{y} \frac{dy}{dx} = (1)(\ln x) + x\left(\frac{1}{x}\right) = \ln x + 1 \]

Solve for the derivative:
\[ \frac{dy}{dx} = y (\ln x + 1) \]

Substitute y = x\(^x\) back in:
\[ f'(x) = x^x (\ln x + 1) \]

To find the critical point, set f'(x) = 0:
\[ x^x (\ln x + 1) = 0 \]

Since x \(>\) 0, the term x\(^x\) is always positive. Therefore, we only need to solve:
\[ \ln x + 1 = 0 \implies \ln x = -1 \]

Convert to exponential form:
\[ x = e^{-1} = \frac{1}{e} \]


Step 4: Final Answer:


The critical point occurs at x = e\(^{-1}\).
Quick Tip: Logarithmic differentiation is the standard method for functions where the variable appears in both the base and the exponent.
The process is: Let y = f(x), take ln of both sides, differentiate implicitly, and solve for y'.
Remember that the domain of x\(^x\) is x > 0.

Step 1: Understanding the Question:


We need to solve a first-order differential equation. The structure allows for the separation of variables.


Step 2: Key Formula or Approach:


First, isolate \(\frac{dy}{dx}\). Then, rearrange the equation so all y-terms are on one side with dy and all x-terms are on the other with dx. Finally, integrate both sides.


Step 3: Detailed Explanation:


Start with the equation: \( \log\left(\frac{dy}{dx}\right) = 3x + 4y \).

Exponentiate both sides:
\[ \frac{dy}{dx} = e^{3x + 4y} = e^{3x} \cdot e^{4y} \]

Separate the variables:
\[ e^{-4y} dy = e^{3x} dx \]

Integrate both sides:
\[ \int e^{-4y} dy = \int e^{3x} dx \implies \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C_1 \]

Multiply by -12 to clear fractions and match the options' form:
\[ 3e^{-4y} = -4e^{3x} - 12C_1 \]

Rearrange to one side:
\[ 3e^{-4y} + 4e^{3x} + 12C_1 = 0 \]

This matches the form of option (D).


Step 4: Final Answer:


The solution is \(3e^{-4y} + 4e^{3x} + 12C = 0\).
Quick Tip: A differential equation is separable if it can be written as \(f(y) dy = g(x) dx\). This is often the simplest type of DE to solve.
Don't be concerned by the form of the constant (C, 12C, etc.), as it's an arbitrary constant. Focus on matching the variable terms.

Step 1: Understanding the Question:


We must find the minimum and maximum of Z = x + 2y by evaluating it at the given corner points of the feasible region and then verify the given statements.


Step 2: Detailed Explanation:


Evaluate Z at each vertex:


- At P(\(\frac{3}{13}, \frac{24}{13}\)): \( Z = \frac{3}{13} + 2(\frac{24}{13}) = \frac{51}{13} \approx 3.92 \)


- At Q(\(\frac{3}{2}, \frac{15}{4}\)): \( Z = \frac{3}{2} + 2(\frac{15}{4}) = \frac{3}{2} + \frac{15}{2} = 9 \) (Maximum)


- At R(\(\frac{7}{2}, \frac{3}{4}\)): \( Z = \frac{7}{2} + 2(\frac{3}{4}) = \frac{7}{2} + \frac{3}{2} = 5 \)


- At S(\(\frac{18}{7}, \frac{2}{7}\)): \( Z = \frac{18}{7} + 2(\frac{2}{7}) = \frac{22}{7} \approx 3.14 \) (Minimum)


Now, check the statements:


(A) Z is minimum at S. This is true.

(B) Z is maximum at R. This is false.

(C) Value at P \(>\) Value at Q (\(3.92 > 9\)). This is false.

(D) Value at Q \(<\) Value at R (\(9 < 5\)). This is false.


Step 3: Final Answer:


The only correct statement is (A).
Quick Tip: In an LPP, the optimal (maximum or minimum) value of the objective function, if it exists, will always occur at one of the corner points (vertices) of the feasible region.
The process is to simply test all the given vertices.

Step 1: Understand Order and Degree


- Order: The order of the highest derivative present in the equation.

- Degree: The highest power (positive integer) of the highest-order derivative after the equation has been expressed as a polynomial in its derivatives (i.e., cleared of radicals and fractions involving derivatives).


Step 2: Find the Order


The derivatives present in the equation are \(\frac{d^2y}{dx^2}\) (second order) and \(\frac{dy}{dx}\) (first order).

The highest order among these is 2.

Therefore, the order is 2.


Step 3: Find the Degree


The equation is already a polynomial in its derivatives. To find the degree, we need to identify the highest power of the highest-order derivative, which is \(\frac{d^2y}{dx^2}\).

Let's expand the left side of the equation using the formula \((a-b)^2 = a^2 - 2ab + b^2\), where \(a = \left( \frac{d^2y}{dx^2} \right)^2\) and \(b = 1\).
\[ \left( \left( \frac{d^2y}{dx^2} \right)^2 \right)^2 - 2\left( \frac{d^2y}{dx^2} \right)^2(1) + (1)^2 = \frac{dy}{dx} \]
\[ \left( \frac{d^2y}{dx^2} \right)^4 - 2\left( \frac{d^2y}{dx^2} \right)^2 + 1 = \frac{dy}{dx} \]

In this expanded form, we can see that the highest-order derivative, \(\frac{d^2y}{dx^2}\), appears with powers of 4 and 2.

The highest power is 4.

Therefore, the degree is 4.


Step 4: Final Answer and Conclusion


The order of the differential equation is 2, and the degree is 4.

Since this result (2, 4) does not match any of the given options (A: 2,2, B: 2, not defined, C: 1,2, D: 1, not defined), there is an error in the question paper's provided options.
Quick Tip: To find the degree, you must first clear any parentheses by expanding the expression.
The degree is the power of the highest derivative after the equation is fully expanded into a polynomial form.
Be careful not to mistake an intermediate power for the final degree.

Step 1: Understanding the Question:


We are given the derivative of a function and an initial condition. We need to find the original function by integrating and solving for the constant of integration.


Step 2: Integrate f'(x)


First, expand and rewrite f'(x): \(f'(x) = 3x^2 + 6x - 4x^{-3} + 5\).

Integrate term by term:
\[ f(x) = \int (3x^2 + 6x - 4x^{-3} + 5) dx \]
\[ f(x) = x^3 + 3x^2 - 4\frac{x^{-2}}{-2} + 5x + C = x^3 + 3x^2 + \frac{2}{x^2} + 5x + C \]


Step 3: Use the Initial Condition to Find C


We are given f(1) = 0.
\[ (1)^3 + 3(1)^2 + \frac{2}{1^2} + 5(1) + C = 0 \]
\[ 1 + 3 + 2 + 5 + C = 0 \implies 11 + C = 0 \implies C = -11 \]


Step 4: Final Answer:


The function is \(f(x) = x^3 + 3x^2 + \frac{2}{x^2} + 5x - 11\).
Quick Tip: This is a standard initial value problem. The procedure is always the same:
1. Integrate the derivative to get the general solution with a constant C.
2. Substitute the given point (x, y) into the general solution to solve for C.
3. Write the final particular solution with the value of C.

Step 1: Understanding the Question:


We need to analyze the constraints to determine if a feasible region exists for this LPP.


Step 2: Detailed Explanation:


Analyze the constraints mathematically, disregarding the given graph which may be misleading.

- Constraint 1: \(x + y \leq 4\).

- Constraint 2: \(3x + 3y \geq 18\), which simplifies to \(x + y \geq 6\).

We are looking for points (x,y) that satisfy both \(x+y \leq 4\) AND \(x+y \geq 6\).

It is logically impossible for the sum of two numbers to be both less than or equal to 4 and greater than or equal to 6 simultaneously.


Step 3: Final Answer:


The constraints are contradictory, meaning there is no point (x, y) that satisfies all conditions. The feasible region is an empty set.

Therefore, the solution to the LPP does not exist.
Quick Tip: Always perform a quick mathematical check on the constraints in an LPP, especially if a graph is provided.
Contradictory constraints (e.g., \(f(x,y) \leq k_1\) and \(f(x,y) \geq k_2\) where \(k_1 < k_2\)) are a common exam trick that leads to "no feasible solution".

Step 1: Understanding the Region


The curve y\(^2\) = x is a parabola opening to the right, symmetric about the x-axis.

The region is bounded by this curve and the vertical lines x=0 and x=1.

The area consists of two symmetric parts, one above the x-axis (\(y=\sqrt{x}\)) and one below (\(y=-\sqrt{x}\)).


Step 2: Setting up the Integral


We can find the area of the top half and double it.
\[ Area = 2 \times \int_{0}^{1} y_{upper} \, dx = 2 \int_{0}^{1} \sqrt{x} \, dx = 2 \int_{0}^{1} x^{1/2} \, dx \]


Step 3: Evaluating the Integral

\[ Area = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} = 2 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{4}{3} [1^{3/2} - 0^{3/2}] = \frac{4}{3} \]


Step 4: Final Answer:


The area of the region is \(\frac{4}{3}\) square units.
Quick Tip: Using symmetry is a powerful tool to simplify area calculations.
For curves symmetric about the x-axis, integrate the top half and double it.
For curves symmetric about the y-axis, integrate the right half and double it.

Step 1: Identify the Integral Form


The integral has the structure \(\int e^x g(x) dx\), which suggests the special form \(\int e^x [f(x) + f'(x)] dx = e^x f(x) + C\).


Step 2: Manipulate the Algebraic Term


We rewrite the rational function to fit the required form:
\[ \frac{x+5}{(x+6)^2} = \frac{(x+6) - 1}{(x+6)^2} = \frac{x+6}{(x+6)^2} - \frac{1}{(x+6)^2} = \frac{1}{x+6} - \frac{1}{(x+6)^2} \]

If we let \(f(x) = \frac{1}{x+6}\), then its derivative is \(f'(x) = -\frac{1}{(x+6)^2}\).

The expression is perfectly in the form \(f(x) + f'(x)\).


Step 3: Apply the Formula

\[ \int e^x \left[ \frac{1}{x+6} + \left(-\frac{1}{(x+6)^2}\right) \right] dx = e^x \cdot f(x) + C = e^x \cdot \frac{1}{x+6} + C \]


Step 4: Final Answer:


The integral evaluates to \(\frac{e^x}{x+6} + C\).
Quick Tip: Anytime you encounter an integral with \(e^x\) multiplied by another function, your first thought should be to check for the \(e^x[f(x) + f'(x)]\) pattern.
This is a very common and time-saving integration trick in exams.

Step 1: Understanding the Question


We need the range of \(|\lambda \vec{a}|\), which is the magnitude of a scaled vector.


Step 2: Key Formula or Approach


Use the property \(|\lambda \vec{a}| = |\lambda| |\vec{a}|\), where \(|\lambda|\) is the absolute value of the scalar.


Step 3: Detailed Explanation


Given \(|\vec{a}| = 5\) and \(-2 \leq \lambda \leq 1\).

First, find the range of \(|\lambda|\). In the interval [-2, 1], the smallest \(|\lambda|\) can be is 0 (when \(\lambda=0\)) and the largest is |-2|=2. So, \(0 \leq |\lambda| \leq 2\).

Now find the range of \(|\lambda \vec{a}| = |\lambda| \cdot 5\):
\[ 0 \cdot 5 \leq 5|\lambda| \leq 2 \cdot 5 \]
\[ 0 \leq |\lambda \vec{a}| \leq 10 \]


Step 4: Final Answer:


The correct range is [0, 10]. The provided options are incorrect. Option (D) [-10, 5] would be the range of \(\lambda|\vec{a}|\), which is a different quantity.
Quick Tip: Be extremely careful with absolute value signs. The magnitude of a vector is always non-negative.
The absolute value of a scalar is also always non-negative.
The question asks for \(|\lambda \vec{a}|\), not \(\lambda |\vec{a}|\), which is a crucial distinction.

Step 1: Understanding the Question


The meeting is cancelled if at least one member does not turn up.

The easiest way to calculate the probability of "at least one" event is to use the complement rule:

P(at least one) = 1 - P(none).

In this context, P(meeting cancelled) = 1 - P(meeting is not cancelled).

The meeting is not cancelled only if all three members are present.


Step 2: Find the Probabilities of Each Member Being Present


Let A, B, C be the events that the respective members are present.

Let A', B', C' be the events that they are not present.

We are given:

- P(A') = 0.10 \(\implies\) P(A) = 1 - 0.10 = 0.90

- P(B') = 0.20 \(\implies\) P(B) = 1 - 0.20 = 0.80

- P(C') = 0.05 \(\implies\) P(C) = 1 - 0.05 = 0.95


Step 3: Calculate the Probability that the Meeting is Held


The meeting is held if A AND B AND C are all present. Since their attendance is independent, we multiply their probabilities:

P(Meeting Held) = P(A \(\cap\) B \(\cap\) C) = P(A) \(\times\) P(B) \(\times\) P(C)
\[ = 0.90 \times 0.80 \times 0.95 = 0.72 \times 0.95 = 0.684 \]


Step 4: Calculate the Probability that the Meeting is Cancelled


P(Meeting Cancelled) = 1 - P(Meeting Held)
\[ = 1 - 0.684 = 0.316 \]


Step 5: Final Answer:


The probability of the meeting being cancelled is 0.316.
Quick Tip: For problems asking for the probability of "at least one" of several events, using the complement is almost always the fastest method.
1. Identify the complement event (in this case, "all members are present").
2. Calculate the probability of the complement event.
3. Subtract this from 1.

Step 1: Analyze the Reason


The Reason (R) states three fundamental facts: the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\), and the geometric definitions of the magnitude of the cross product and the dot product. All three statements are correct.


Step 2: Connect the Reason to the Assertion


The statements in the Reason are the building blocks for Lagrange's identity. Let's derive it:
\(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = (|\vec{a}||\vec{b}|\sin\theta)^2 + (|\vec{a}||\vec{b}|\cos\theta)^2\)
\(= |\vec{a}|^2|\vec{b}|^2(\sin^2\theta + \cos^2\theta) = |\vec{a}|^2|\vec{b}|^2(1) = |\vec{a}|^2|\vec{b}|^2\).

So, the Reason directly explains the identity needed to evaluate the Assertion.


Step 3: Verify the Assertion


Using the identity, the given equation becomes \(|\vec{a}|^2|\vec{b}|^2 = 256\).

Substitute \(|\vec{b}| = 8\), so \(|\vec{b}|^2 = 64\):
\(|\vec{a}|^2 \cdot 64 = 256 \implies |\vec{a}|^2 = 4 \implies |\vec{a}| = 2\).

The Assertion (A) is true.


Step 4: Final Answer:


Both A and R are true, and R is the correct explanation of A.
Quick Tip: Lagrange's identity, \(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2\), is a key formula in vector algebra.
For Assertion-Reason questions, always verify three things: Is A true? Is R true? Does R correctly and directly explain A?

Step 1: Analyze the Reason


The Reason (R) states the rule for finding the domain of the sum of two functions: it's the intersection of their individual domains. This is a correct mathematical definition. So, Reason (R) is true.


Step 2: Verify the Assertion using the Reason


Let's find the domains of f(x) and g(x):

- Domain of f(x) = e\(^x\) is all real numbers, R or (\(-\infty, \infty\)).

- Domain of g(x) = log x is all positive real numbers, (0, \(\infty\)).

According to the rule in R, the domain of their sum is the intersection:

Dom(f + g) = R \(\cap\) (0, \(\infty\)) = (0, \(\infty\)).

The Assertion (A) claims the domain is R, which is incorrect. So, Assertion (A) is false.


Step 3: Final Answer:


Assertion (A) is false, but Reason (R) is true.
Quick Tip: When combining functions, the domain of the resulting function is always the most restrictive of the original functions' domains.
Functions with inherent domain restrictions (like logarithms, square roots, and rational functions) are the ones to watch out for.

Step 1: Understanding the Question:


We are given the position vectors for points A (\(\vec{a}\)) and B (\(\vec{b}\)).

Point C lies on the line passing through B and A, extended beyond A.

The condition is given by the vector equation \(\vec{BC} = 3\vec{BA}\). We need to find the position vector of C, let's call it \(\vec{c}\).


Step 2: Key Formula or Approach:


We express the vectors in terms of the position vectors of their endpoints:
\(\vec{BC} = \vec{c} - \vec{b}\)
\(\vec{BA} = \vec{a} - \vec{b}\)

We will substitute these into the given equation and solve for \(\vec{c}\).


Step 3: Detailed Explanation:


The given vector relation is:
\[ \vec{BC} = 3\vec{BA} \]

Substitute the expressions from Step 2:
\[ \vec{c} - \vec{b} = 3(\vec{a} - \vec{b}) \]

Distribute the scalar 3 on the right side:
\[ \vec{c} - \vec{b} = 3\vec{a} - 3\vec{b} \]

To solve for \(\vec{c}\), add \(\vec{b}\) to both sides of the equation:
\[ \vec{c} = 3\vec{a} - 3\vec{b} + \vec{b} \]
\[ \vec{c} = 3\vec{a} - 2\vec{b} \]


Step 4: Final Answer:


The position vector of point C is \(3\vec{a} - 2\vec{b}\).
Quick Tip: Remember the formula for a vector between two points: the vector from point P to point Q is given by (position vector of Q) - (position vector of P).
Drawing a simple diagram can also help visualize the relative positions of A, B, and C to confirm your answer. C is an external point dividing the line segment.

Step 1: Understanding the Question:


A vector is inclined at equal angles to the coordinate axes, which gives information about its direction cosines.

We are also given its magnitude. We need to find the vector itself.


Step 2: Key Formula or Approach:


Let the equal angles be \(\alpha\). The direction cosines are l = cos(\(\alpha\)), m = cos(\(\alpha\)), n = cos(\(\alpha\)). So, l = m = n.

The fundamental identity for direction cosines is \(l^2 + m^2 + n^2 = 1\).

A vector \(\vec{r}\) can be expressed as \(\vec{r} = |\vec{r}| (l\hat{i} + m\hat{j} + n\hat{k})\).


Step 3: Detailed Explanation:


Since l = m = n, the identity becomes:
\[ l^2 + l^2 + l^2 = 1 \implies 3l^2 = 1 \implies l^2 = \frac{1}{3} \implies l = \pm \frac{1}{\sqrt{3}} \]

So, the direction cosines are \(l=m=n = \pm \frac{1}{\sqrt{3}}\).

We are given the magnitude \(|\vec{r}| = 5\sqrt{3}\).

Now we can write the vector \(\vec{r}\):
\[ \vec{r} = |\vec{r}| (l\hat{i} + m\hat{j} + n\hat{k}) \]
\[ \vec{r} = 5\sqrt{3} \left( \pm \frac{1}{\sqrt{3}}\hat{i} \pm \frac{1}{\sqrt{3}}\hat{j} \pm \frac{1}{\sqrt{3}}\hat{k} \right) \]

The sign must be the same for all components.
\[ \vec{r} = \pm \frac{5\sqrt{3}}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) = \pm 5 (\hat{i} + \hat{j} + \hat{k}) \]


Step 4: Final Answer:


The vector \(\vec{r}\) is \(5\hat{i} + 5\hat{j} + 5\hat{k}\) or \(-5\hat{i} - 5\hat{j} - 5\hat{k}\).
Quick Tip: A vector equally inclined to the coordinate axes is always parallel to the vector \(\hat{i} + \hat{j} + \hat{k}\) or its negative.
Its unit vector is always \(\pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\). This can be used as a shortcut.

Step 1: Understanding the Question:


We need to find the set of all valid input values (x-values) for which the function f(x) is defined.


Step 2: Key Formula or Approach:


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

This means the argument 'u' must satisfy the inequality: \(-1 \leq u \leq 1\).

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


Step 3: Detailed Explanation:


We apply the domain constraint to the argument of our function:
\[ -1 \leq -x^2 \leq 1 \]

This is a compound inequality, which we can split into two parts:

1) \(-1 \leq -x^2 \implies x^2 \leq 1 \implies |x| \leq 1 \implies -1 \leq x \leq 1\).

2) \(-x^2 \leq 1 \implies x^2 \geq -1\). This is true for all real numbers x, since a square is always non-negative.

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

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


Step 4: Final Answer:


The domain of the function f(x) = sin\(^{-1}\)(– x\(^2\)) is [-1, 1].
Quick Tip: To find the domain of a composite function, start with the outer function.
Identify the domain requirement for the outer function (here, sin\(^{-1}\)) and apply it as an inequality to the inner function (here, -x\(^2\)).
Then, solve the resulting inequality for x.

Step 1: Understanding the Question:


A function is increasing where its first derivative is positive (f'(x) > 0).

We need to find the derivative of f(x) and then solve the inequality f'(x) > 0.


Step 2: Find the Derivative


The function is \(f(x) = x + x^{-1}\).
\[ f'(x) = \frac{d}{dx}(x + x^{-1}) = 1 - 1x^{-2} = 1 - \frac{1}{x^2} \]


Step 3: Solve the Inequality f'(x) > 0

\[ 1 - \frac{1}{x^2} > 0 \]
\[ 1 > \frac{1}{x^2} \]
\[ x^2 > 1 \]

Taking the square root of both sides, we get:
\[ |x| > 1 \]

This inequality holds true when \(x > 1\) or \(x < -1\).


Step 4: Final Answer:


The function is increasing on the intervals (-\(\infty\), -1) and (1, \(\infty\)).
Quick Tip: To find intervals of increasing/decreasing behavior:
1. Find the first derivative, f'(x).
2. Find the critical points by solving f'(x) = 0.
3. Test the sign of f'(x) in the intervals between the critical points.
f'(x) > 0 implies increasing.
f'(x) < 0 implies decreasing.

Step 1: Understanding the Question:


We are asked to find the derivative of one function with respect to another. This is a form of parametric differentiation.


Step 2: Key Formula or Approach:


Let \(u = \sqrt{e^{\sqrt{2x}}}\) and \(v = e^{\sqrt{2x}}\). We need to find \(\frac{du}{dv}\).

The simplest approach is to first express u directly in terms of v.


Step 3: Detailed Explanation:


By observing the expressions for u and v, we can see that:
\[ u = \sqrt{v} = v^{1/2} \]

Now, we can differentiate u directly with respect to v:
\[ \frac{du}{dv} = \frac{d}{dv}(v^{1/2}) = \frac{1}{2}v^{-1/2} = \frac{1}{2\sqrt{v}} \]

Substitute the original expression for v back into the result:
\[ \frac{du}{dv} = \frac{1}{2\sqrt{e^{\sqrt{2x}}}} \]


Step 4: Final Answer:


The derivative of \(\sqrt{e^{\sqrt{2x}}}\) with respect to \(e^{\sqrt{2x}}\) is \( \frac{1}{2\sqrt{e^{\sqrt{2x}}}} \).
Quick Tip: When asked to differentiate f(x) with respect to g(x), always check for a simple algebraic relationship between f and g first.
If you can write f as a function of g (or vice versa), the differentiation becomes much simpler than using the chain rule \(\frac{df/dx}{dg/dx}\).

Step 1: Understanding the Question:


We need to find \(\frac{dy}{dx}\) for an implicit equation where variables appear in both the base and the exponent. This requires logarithmic differentiation.


Step 2: Key Formula or Approach:


Take the natural logarithm (ln) of both sides to bring the exponents down, then differentiate implicitly with respect to x using the product rule.


Step 3: Detailed Explanation:


Start with the equation:
\[ x^y = y^x \]

Take the natural logarithm of both sides:
\[ \ln(x^y) = \ln(y^x) \]
\[ y \ln x = x \ln y \]

Now, differentiate both sides with respect to x, using the product rule:
\[ \frac{d}{dx}(y \ln x) = \frac{d}{dx}(x \ln y) \]
\[ \left(\frac{dy}{dx} \cdot \ln x + y \cdot \frac{1}{x}\right) = \left(1 \cdot \ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx}\right) \]

Group all terms with \(\frac{dy}{dx}\) on one side:
\[ \frac{dy}{dx} \ln x - \frac{x}{y} \frac{dy}{dx} = \ln y - \frac{y}{x} \]

Factor out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \left(\ln x - \frac{x}{y}\right) = \ln y - \frac{y}{x} \]
\[ \frac{dy}{dx} \left(\frac{y \ln x - x}{y}\right) = \frac{x \ln y - y}{x} \]

Isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{y(x \ln y - y)}{x(y \ln x - x)} \]


Step 4: Final Answer:


The derivative is \(\frac{dy}{dx} = \frac{y(x \ln y - y)}{x(y \ln x - x)}\).
Quick Tip: Logarithmic differentiation is essential for functions of the form \(f(x)^{g(x)}\).
The process is always:
1. Take ln of both sides.
2. Use log properties to simplify.
3. Differentiate implicitly, remembering the product rule and chain rule.
4. Isolate \(\frac{dy}{dx}\).

Step 1: Understand the Condition for Perpendicular Lines


Two lines are perpendicular if and only if the dot product of their direction vectors (or direction ratios) is equal to zero.

Our first task is to find the correct direction ratios for each line by converting their equations to the standard form \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\).


Step 2: Standardize the Line Equations


For line \(l_1\):
\[ \frac{1-x}{-3} = \frac{-(x-1)}{-3} = \frac{x-1}{3} \]
\[ \frac{3y-2}{2\lambda} = \frac{3(y-2/3)}{2\lambda} = \frac{y-2/3}{2\lambda/3} \]
\[ \frac{z-3}{3} is already in standard form. \]

So, the direction ratios for \(l_1\) are \(\langle a_1, b_1, c_1 \rangle = \langle 3, \frac{2\lambda}{3}, 3 \rangle\).


For line \(l_2\):
\[ \frac{x-1}{3\lambda} is already in standard form. \]
\[ \frac{1-y}{1} = \frac{-(y-1)}{1} = \frac{y-1}{-1} \]
\[ \frac{2z-5}{3} = \frac{2(z-5/2)}{3} = \frac{z-5/2}{3/2} \]

So, the direction ratios for \(l_2\) are \(\langle a_2, b_2, c_2 \rangle = \langle 3\lambda, -1, \frac{3}{2} \rangle\).


Step 3: Apply the Perpendicularity Condition


The condition for perpendicular lines is \(a_1a_2 + b_1b_2 + c_1c_2 = 0\).

Substituting the direction ratios we found:
\[ (3)(3\lambda) + \left(\frac{2\lambda}{3}\right)(-1) + (3)\left(\frac{3}{2}\right) = 0 \]


Step 4: Solve for \(\lambda\)

\[ 9\lambda - \frac{2\lambda}{3} + \frac{9}{2} = 0 \]

To simplify, let's move the constant term to the right side:
\[ 9\lambda - \frac{2\lambda}{3} = -\frac{9}{2} \]

Combine the terms with \(\lambda\) using a common denominator:
\[ \frac{27\lambda - 2\lambda}{3} = -\frac{9}{2} \]
\[ \frac{25\lambda}{3} = -\frac{9}{2} \]

Cross-multiply to solve for \(\lambda\):
\[ 2 \times 25\lambda = 3 \times (-9) \]
\[ 50\lambda = -27 \]
\[ \lambda = -\frac{27}{50} \]
Quick Tip: The most critical step in problems involving direction ratios is to ensure the line equations are in standard form.
The coefficients of x, y, and z in the numerators \textbf{must be +1}.
For example, \(\frac{1-x}{a}\) must be converted to \(\frac{x-1}{-a}\) before identifying the direction ratio.
Failing to do this is the most common source of error.

Step 1: Understanding the Question:


We are given a matrix equation that results in the zero matrix (O). We need to solve for the unknown value x.

We will perform the matrix multiplications step-by-step.


Step 2: Perform the First Multiplication


Let's multiply the first two matrices. A (1\(\times\)3) matrix multiplied by a (3\(\times\)3) matrix will result in a (1\(\times\)3) matrix.
\[ [1 \quad x \quad 1] \begin{bmatrix} 1 & 3 & 2
2 & 5 & 1
15 & 3 & 2 \end{bmatrix} \]
\[ = [(1)(1) + (x)(2) + (1)(15) \quad (1)(3) + (x)(5) + (1)(3) \quad (1)(2) + (x)(1) + (1)(2)] \]
\[ = [1 + 2x + 15 \quad 3 + 5x + 3 \quad 2 + x + 2] \]
\[ = [2x + 16 \quad 5x + 6 \quad x + 4] \]


Step 3: Perform the Second Multiplication


Now, multiply this result by the third matrix. A (1\(\times\)3) matrix multiplied by a (3\(\times\)1) matrix will result in a (1\(\times\)1) matrix.
\[ [2x + 16 \quad 5x + 6 \quad x + 4] \begin{bmatrix} 1
2
x \end{bmatrix} = [0] \]
\[ (2x + 16)(1) + (5x + 6)(2) + (x + 4)(x) = 0 \]


Step 4: Solve the Resulting Equation


Expand and simplify the equation:
\[ 2x + 16 + 10x + 12 + x^2 + 4x = 0 \]

Combine like terms to form a standard quadratic equation:
\[ x^2 + (2x + 10x + 4x) + (16 + 12) = 0 \]
\[ x^2 + 16x + 28 = 0 \]

Factor the quadratic equation:
\[ (x + 2)(x + 14) = 0 \]

The solutions are \(x = -2\) and \(x = -14\).


Step 5: Final Answer:


The possible values for x are -2 and -14.
Quick Tip: When multiplying a chain of matrices, work from left to right.
Keep careful track of the dimensions of the resulting matrix at each step.
The final step in this problem leads to a standard algebraic equation (in this case, quadratic) to solve.

Step 1: Understanding the Problem


We need to find the shortest distance from a given point P to a given line in 3D space.


Step 2: Key Formula or Approach


The distance (d) of a point P with position vector \(\vec{p}\) from a line \(\vec{r} = \vec{a} + \lambda\vec{b}\) is given by the formula:
\[ d = \frac{|(\vec{p} - \vec{a}) \times \vec{b}|}{|\vec{b}|} \]

From the given line equation, we identify a point A on the line and the line's direction vector \(\vec{b}\).


Step 3: Detailed Explanation


The given point is P(2, 4, -1), so its position vector is \(\vec{p} = 2\hat{i} + 4\hat{j} - \hat{k}\).

The line equation \(\frac{x-(-5)}{1} = \frac{y-(-3)}{4} = \frac{z-6}{-9}\) shows that a point on the line is A(-5, -3, 6), so \(\vec{a} = -5\hat{i} - 3\hat{j} + 6\hat{k}\).

The direction vector of the line is \(\vec{b} = 1\hat{i} + 4\hat{j} - 9\hat{k}\).


First, calculate the vector from A to P, which is \((\vec{p} - \vec{a})\):
\[ \vec{p} - \vec{a} = (2 - (-5))\hat{i} + (4 - (-3))\hat{j} + (-1 - 6)\hat{k} = 7\hat{i} + 7\hat{j} - 7\hat{k} \]

Next, calculate the cross product \((\vec{p} - \vec{a}) \times \vec{b}\):
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
7 & 7 & -7
1 & 4 & -9 \end{vmatrix} = \hat{i}(-63 - (-28)) - \hat{j}(-63 - (-7)) + \hat{k}(28 - 7) \]
\[ = -35\hat{i} + 56\hat{j} + 21\hat{k} \]

Now, find the magnitude of this cross product:
\[ |(\vec{p} - \vec{a}) \times \vec{b}| = \sqrt{(-35)^2 + 56^2 + 21^2} = \sqrt{1225 + 3136 + 441} = \sqrt{4802} \]

Next, find the magnitude of the direction vector \(\vec{b}\):
\[ |\vec{b}| = \sqrt{1^2 + 4^2 + (-9)^2} = \sqrt{1 + 16 + 81} = \sqrt{98} \]

Finally, calculate the distance:
\[ d = \frac{\sqrt{4802}}{\sqrt{98}} = \sqrt{\frac{4802}{98}} = \sqrt{49} = 7 \]


Step 4: Final Answer:


The distance of the point P from the line is 7 units.
Quick Tip: The vector formula \(d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}\) is generally the fastest method for this type of problem.
The numerator represents the area of the parallelogram formed by vectors \(\vec{AP}\) and \(\vec{b}\). Dividing by the base \(|\vec{b}|\) gives the height, which is the perpendicular distance.

Step 1: Understanding the Problem


We need to find the equation of a line that passes through a given point A and has a direction parallel to the vector \(\vec{BC}\).


Step 2: Key Formula or Approach


The vector equation of a line is \(\vec{r} = \vec{a} + \lambda\vec{d}\), where \(\vec{a}\) is the position vector of a point on the line and \(\vec{d}\) is its direction vector.


Step 3: Detailed Explanation


The line passes through point A, so its position vector is given:
\[ \vec{a} = 3\hat{i} - \hat{j} - 2\hat{k} \]

The line is parallel to BC, so its direction vector \(\vec{d}\) is \(\vec{BC}\):
\[ \vec{d} = \vec{BC} = (position vector of C) - (position vector of B) \]
\[ \vec{d} = (\hat{i} + 5\hat{j} + 3\hat{k}) - (\hat{i} + 2\hat{j} - \hat{k}) = 0\hat{i} + 3\hat{j} + 4\hat{k} \]

Vector Equation:
\[ \vec{r} = (3\hat{i} - \hat{j} - 2\hat{k}) + \lambda(3\hat{j} + 4\hat{k}) \]

Cartesian Equation:

The point is A(3, -1, -2) and the direction ratios are \(<0, 3, 4>\).

The equation is \(\frac{x-3}{0} = \frac{y-(-1)}{3} = \frac{z-(-2)}{4}\).

When a direction ratio is 0, we write it separately:
\[ x-3 = 0 \implies x = 3 \]

So the equations are \(x = 3, \frac{y+1}{3} = \frac{z+2}{4}\).


Step 4: Final Answer:


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

Cartesian Equations: \(x = 3, \frac{y+1}{3} = \frac{z+2}{4}\).
Quick Tip: When a direction ratio in a Cartesian equation is zero (e.g., for the x-component), it signifies that the line is parallel to the yz-plane.
This means all points on the line share the same x-coordinate, which is expressed as \(x = x_1\).

Step 1: Identify the Problem and Boundary Lines


The problem is to minimize Z = x + 4y subject to the given constraints.

First, we plot the boundary lines:

- L1: \(2x + y = 1000\). Intercepts are (500, 0) and (0, 1000).

- L2: \(x + 2y = 800\). Intercepts are (800, 0) and (0, 400).

The feasible region is the unbounded area on or above both lines in the first quadrant.


Step 2: Find the Corner Points of the Feasible Region


The vertices of the feasible region are:

- Point A (y-intercept): The highest y-intercept on the boundary is from L1. A = (0, 1000).

- Point B (Intersection): We solve the system \(2x + y = 1000\) and \(x + 2y = 800\).

This yields the point B = (400, 200).

- Point C (x-intercept): The rightmost x-intercept on the boundary is from L2. C = (800, 0).


Step 3: Evaluate the Objective Function at Corner Points


We evaluate Z = x + 4y at each vertex:

- At A(0, 1000): \(Z = 0 + 4(1000) = 4000\)

- At B(400, 200): \(Z = 400 + 4(200) = 1200\)

- At C(800, 0): \(Z = 800 + 4(0) = 800\)

The minimum value among the vertices is M = 800.


Step 4: Check for Unbounded Region


Since the region is unbounded, we must test if a value smaller than 800 is possible by checking the open half-plane \(x + 4y < 800\).

For any point (x, y) in the feasible region, we know \(x + 2y \geq 800\). Since \(y \geq 0\), it follows that \(2y \geq 0\).

Therefore, \(Z = x + 4y = (x + 2y) + 2y \geq 800 + 2y \geq 800\).

This confirms that no value of Z can be less than 800. The minimum is valid.


Step 5: Final Answer:


The minimum value of Z is 800.

\begin{figure[h!]
\centering
\begin{tikzpicture[
scale=0.9,
x=0.007cm, y=0.007cm, % Scale coordinates to fit page
every node/.style={font=\small
]

% Draw axes
\draw[->, thick] (0,0) -- (1150,0) node[right] {\(x\);
\draw[->, thick] (0,0) -- (0,1150) node[above] {\(y\);

% Draw ticks and labels for axes
\foreach \x in {200, 400, 600, 800, 1000 {
\draw (\x, 15) -- (\x, -15) node[below] {\x;

\foreach \y in {200, 400, 600, 800, 1000 {
\draw (15, \y) -- (-15, \y) node[left] {\y;

\node[below left] at (0,0) {0;

% Define coordinates for corner points
\coordinate (A) at (0, 1000);
\coordinate (B) at (400, 200);
\coordinate (C) at (800, 0);

% Shade the feasible region (unbounded)
\fill[cyan!20, opacity=0.6] (A) -- (B) -- (C) -- (1150,0) -- (1150,1150) -- (0,1150) -- cycle;

% Draw the boundary lines of the constraints
% Line 1: 2x + y = 1000 (intercepts at (500,0) and (0,1000))
\draw[blue, thick] (0, 1000) -- (500, 0)
node[pos=0.7, above, sloped, rotate=0, black, xshift=-15pt] {\(2x+y \geq 1000\);

% Line 2: x + 2y = 800 (intercepts at (800,0) and (0,400))
\draw[red, thick] (0, 400) -- (800, 0)
node[pos=0.3, below, sloped, rotate=0, black, xshift=0pt] {\(x+2y \geq 800\);

% Highlight the boundary of the feasible region
\draw[black, ultra thick] (A) -- (B) -- (C);

% Mark and label the corner points
\filldraw[black] (A) circle (3pt) node[above right=2pt] {A(0, 1000);
\filldraw[black] (B) circle (3pt) node[below right=5pt] {B(400, 200);
\filldraw[black] (C) circle (3pt) node[above right=2pt] {C(800, 0);

% Label the feasible region
\node[text=black, align=center] at (750, 600) {Feasible Region;

\end{tikzpicture
\end{figure Quick Tip: For minimization problems with an unbounded feasible region, the minimum value M found at a vertex is only valid if the open half-plane \(Z < M\) has no points in common with the feasible region.
Always perform this check.

Step 1: Domain and Range


The relation R consists of pairs (x, y) such that y = 41 - 2x, with x, y \(\in\) N (natural numbers \(\geq 1\)).

For y to be a natural number, \(41 - 2x \geq 1 \implies 40 \geq 2x \implies 20 \geq x\).

Since x must also be a natural number, the domain is Domain = {1, 2, 3, ..., 20}.

The corresponding y values range from y(1)=39 down to y(20)=1.

The range is the set of odd numbers from 1 to 39. Range = {1, 3, 5, ..., 39}.


Step 2: Checking Properties


- Reflexive: For (a, a) \(\in\) R, we need 2a + a = 41, or 3a = 41. This gives a = 41/3, which is not a natural number. Thus, the relation is not reflexive.


- Symmetric: If (a, b) \(\in\) R, then (b, a) \(\in\) R.

Let's take a counterexample. (1, 39) is in R because 2(1) + 39 = 41.

But for (39, 1) to be in R, we need 2(39) + 1 = 41, which is false (79 \(\neq\) 41). Thus, the relation is not symmetric.


- Transitive: If (a, b) \(\in\) R and (b, c) \(\in\) R, then (a, c) \(\in\) R.

Let (a, b) \(\in\) R \(\implies\) b = 41 - 2a. Let (b, c) \(\in\) R \(\implies\) c = 41 - 2b.

Substitute b: \(c = 41 - 2(41 - 2a) = 4a - 41\).

For (a, c) to be in R, we need 2a + c = 41. Let's check: \(2a + (4a - 41) = 6a - 41\). This is not equal to 41 in general. Thus, the relation is not transitive.


Step 3: Final Answer:


An equivalence relation must be reflexive, symmetric, AND transitive. Since this relation is none of these, it is not an equivalence relation.
Quick Tip: To disprove a property (reflexive, symmetric, or transitive), you only need to find one single counterexample.
To prove it, you must show it holds true for all general cases.

Step 1: Understanding the Question


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


Step 2: Proving One-one (Injective)


We consider three cases for inputs n\(_1\) and n\(_2\).

- Case 1 (Both even): If f(n\(_1\)) = f(n\(_2\)), then n\(_1\) - 1 = n\(_2\) - 1 \(\implies\) n\(_1\) = n\(_2\).

- Case 2 (Both odd): If f(n\(_1\)) = f(n\(_2\)), then n\(_1\) + 1 = n\(_2\) + 1 \(\implies\) n\(_1\) = n\(_2\).

- Case 3 (One even, one odd): Let n\(_1\) be even and n\(_2\) be odd. Then f(n\(_1\)) = n\(_1\) - 1 (which is odd) and f(n\(_2\)) = n\(_2\) + 1 (which is even). An odd number can never equal an even number, so f(n\(_1\)) \(\neq\) f(n\(_2\)).

In all cases, different inputs give different outputs. Thus, f is one-one.


Step 3: Proving Onto (Surjective)


We must show that for any natural number y in the codomain, there exists a pre-image n in the domain.

- Case 1 (y is odd): We are looking for an input n such that f(n) = y. Since y is odd, the input must have been even (as f(even) = even-1 = odd). So, we solve \(n-1 = y \implies n = y+1\). If y is an odd natural number, n = y+1 is an even natural number. So a pre-image exists.

- Case 2 (y is even): We are looking for an input n such that f(n) = y. Since y is even, the input must have been odd (as f(odd) = odd+1 = even). So, we solve \(n+1 = y \implies n = y-1\). If y is an even natural number (\(\geq 2\)), n = y-1 is an odd natural number. So a pre-image exists.

Every natural number in the codomain has a pre-image. Thus, f is onto.


Step 4: Final Answer:


Since the function is both one-one and onto, it is a bijection.
Quick Tip: For piecewise functions defined by properties like even/odd, always structure your proofs for injectivity and surjectivity into cases.
For surjectivity (onto), work backwards: pick an arbitrary 'y' from the codomain and find the 'x' from the domain that maps to it.

Step 1: Understanding the Question:


The expression inside the sin\(^{-1}\) function matches the triple angle formula for sine, suggesting a trigonometric substitution.


Step 2: Key Formula or Approach:


We use the identity: \(\sin(3\theta) = 3\sin\theta - 4\sin^3\theta\).

Let \(x = \sin\theta\), which means \(\theta = \sin^{-1}x\).

We must check that the given domain for x keeps \(3\theta\) within the principal value branch of sin\(^{-1}\), which is [-\(\pi/2\), \(\pi/2\)].


Step 3: Detailed Explanation:


Given \(-\frac{1}{2} < x < \frac{1}{2}\), we substitute \(x = \sin\theta\):
\[ -\frac{1}{2} < \sin\theta < \frac{1}{2} \implies -\frac{\pi}{6} < \theta < \frac{\pi}{6} \]

Multiplying by 3 gives:
\[ -\frac{\pi}{2} < 3\theta < \frac{\pi}{2} \]

This range is valid for the simplification.

Substitute into the function:
\[ y = \sin^{-1}(3\sin\theta - 4\sin^3\theta) = \sin^{-1}(\sin(3\theta)) \]

Since \(3\theta\) is in the principal branch, we have \(y = 3\theta\).

Substitute back \(\theta = \sin^{-1}x\):
\[ y = 3\sin^{-1}x \]

Now, differentiate:
\[ \frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\sin^{-1}x) = \frac{3}{\sqrt{1-x^2}} \]


Step 4: Final Answer:


The derivative is \(\frac{dy}{dx} = \frac{3}{\sqrt{1-x^2}}\).
Quick Tip: Recognizing trigonometric identities is essential for simplifying the differentiation of inverse trig functions.
The expression \(3x - 4x^3\) is a strong indicator for the substitution \(x = \sin\theta\).
Always verify the domain to ensure the simplification \( \sin^{-1}(\sin u) = u \) is valid.

Step 1: Understanding the Question:


The expression inside the cos\(^{-1}\) function matches the double angle formula for cosine, suggesting a trigonometric substitution.


Step 2: Key Formula or Approach:


We use the identity: \(\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}\).

Let \(x = \tan\theta\), which means \(\theta = \tan^{-1}x\).

We check the domain of x to ensure \(2\theta\) is in the principal value branch of cos\(^{-1}\), which is [0, \(\pi\)].


Step 3: Detailed Explanation:


Given \(0 < x < 1\), we substitute \(x = \tan\theta\):
\[ 0 < \tan\theta < 1 \implies 0 < \theta < \frac{\pi}{4} \]

Multiplying by 2 gives:
\[ 0 < 2\theta < \frac{\pi}{2} \]

This range is valid for the simplification.

Substitute into the function:
\[ y = \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) = \cos^{-1}(\cos(2\theta)) \]

Since \(2\theta\) is in the principal branch, we have \(y = 2\theta\).

Substitute back \(\theta = \tan^{-1}x\):
\[ y = 2\tan^{-1}x \]

Now, differentiate:
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}x) = \frac{2}{1+x^2} \]


Step 4: Final Answer:


The derivative is \(\frac{dy}{dx} = \frac{2}{1+x^2}\).
Quick Tip: The form \(\frac{1-x^2}{1+x^2}\) is a classic indicator for the substitution \(x = \tan\theta\).
Knowing your double and triple angle formulas for sin, cos, and tan is key to solving these problems quickly.

Step 1: Understand the Two-Stage Process and Define Events


This is a problem of total probability because the final outcome (drawing a white ball from Bag II) depends on the outcome of a previous event (which color ball was transferred from Bag I).

Let's define the events:

- \(W_1\): The event that a white ball is transferred from Bag I to Bag II.

- \(B_1\): The event that a black ball is transferred from Bag I to Bag II.

- \(W_2\): The event that the final ball drawn from Bag II is white.

Our goal is to find the total probability of \(W_2\).


Step 2: Calculate Initial Probabilities from Bag I


Bag I contains 4 white + 5 black = 9 total balls.

The probability of transferring a white ball is:
\[ P(W_1) = \frac{4}{9} \]

The probability of transferring a black ball is:
\[ P(B_1) = \frac{5}{9} \]


Step 3: Analyze the Two Possible Cases (Paths)


Case 1: A white ball is transferred.

The probability of this path starting is \(P(W_1) = 4/9\).

After transferring 1 white ball, Bag II contains (6+1) white and 7 black balls, for a new total of 7 W + 7 B = 14 balls.

The probability of drawing a white ball from Bag II, given that a white ball was transferred, is:
\[ P(W_2 | W_1) = \frac{7}{14} = \frac{1}{2} \]


Case 2: A black ball is transferred.

The probability of this path starting is \(P(B_1) = 5/9\).

After transferring 1 black ball, Bag II contains 6 white and (7+1) black balls, for a new total of 6 W + 8 B = 14 balls.

The probability of drawing a white ball from Bag II, given that a black ball was transferred, is:
\[ P(W_2 | B_1) = \frac{6}{14} = \frac{3}{7} \]


Step 4: Apply the Law of Total Probability


The total probability of drawing a white ball from Bag II is the sum of the probabilities of the two mutually exclusive paths.
\[ P(W_2) = P(W_1) \cdot P(W_2 | W_1) + P(B_1) \cdot P(W_2 | B_1) \]
\[ P(W_2) = \left(\frac{4}{9}\right) \cdot \left(\frac{1}{2}\right) + \left(\frac{5}{9}\right) \cdot \left(\frac{3}{7}\right) \]
\[ P(W_2) = \frac{4}{18} + \frac{15}{63} \]

Simplify the fractions:
\[ P(W_2) = \frac{2}{9} + \frac{5}{21} \]

Find a common denominator (63):
\[ P(W_2) = \frac{2 \times 7}{63} + \frac{5 \times 3}{63} = \frac{14}{63} + \frac{15}{63} = \frac{29}{63} \]


Step 5: Final Answer:


The probability that the ball drawn from Bag II is white is \(\frac{29}{63}\).
Quick Tip: This is a classic Law of Total Probability problem. A tree diagram is an excellent way to visualize it:
- The first set of branches represents the first event (drawing from Bag I).
- The second set of branches represents the second event (drawing from Bag II), with probabilities conditional on the first branch.
- To find the total probability of a final outcome, multiply the probabilities along each path leading to it, and then add the results of all such paths.

Step 1: Identifying the Type of Equation


Rearrange the equation to find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{x^2y}{x^3 + y^3} \]

This is a homogeneous differential equation because each term has the same degree (3).


Step 2: Substitution


Use the substitution \(y = vx\). This implies \(\frac{dy}{dx} = v + x\frac{dv}{dx}\).

Substitute into the equation:
\[ v + x\frac{dv}{dx} = \frac{x^2(vx)}{x^3 + (vx)^3} = \frac{vx^3}{x^3(1+v^3)} = \frac{v}{1+v^3} \]

Isolate the terms for separation:
\[ x\frac{dv}{dx} = \frac{v}{1+v^3} - v = \frac{v - v(1+v^3)}{1+v^3} = \frac{-v^4}{1+v^3} \]


Step 3: Separation and Integration


Separate the variables:
\[ \frac{1+v^3}{v^4} dv = -\frac{1}{x} dx \implies \left( v^{-4} + \frac{1}{v} \right) dv = -\frac{1}{x} dx \]

Integrate both sides:
\[ \int (v^{-4} + \frac{1}{v}) dv = \int -\frac{1}{x} dx \]
\[ \frac{v^{-3}}{-3} + \ln|v| = -\ln|x| + C \]


Step 4: Back-Substitution


Substitute back \(v = \frac{y}{x}\):
\[ -\frac{1}{3(y/x)^3} + \ln\left|\frac{y}{x}\right| = -\ln|x| + C \]
\[ -\frac{x^3}{3y^3} + \ln|y| - \ln|x| = -\ln|x| + C \]
\[ -\frac{x^3}{3y^3} + \ln|y| = C \]


Step 5: Final Answer:


The general solution is \(-\frac{x^3}{3y^3} + \ln|y| = C\).
Quick Tip: To check if a DE is homogeneous, replace x with \(\lambda x\) and y with \(\lambda y\). If the \(\lambda\)s cancel out, it is homogeneous.
The substitution \(y=vx\) will always transform it into a separable equation.

Step 1: Identifying the Type of Equation


Rearrange the equation into the standard linear form, \(\frac{dy}{dx} + P(x)y = Q(x)\).
\[ \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4x^2}{1+x^2} \]

This is a linear DE with \(P(x) = \frac{2x}{1+x^2}\) and \(Q(x) = \frac{4x^2}{1+x^2}\).


Step 2: Finding the Integrating Factor (I.F.)


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

The integral is \(\ln(1+x^2)\). So, I.F. = \(e^{\ln(1+x^2)} = 1+x^2\).


Step 3: Finding the General Solution


The solution is \(y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx\).
\[ y(1+x^2) = \int \frac{4x^2}{1+x^2} \cdot (1+x^2) dx = \int 4x^2 dx \]
\[ y(1+x^2) = \frac{4x^3}{3} + C \]


Step 4: Applying the Initial Condition


Given y(0) = 0, substitute x = 0, y = 0:
\[ 0(1+0) = \frac{4(0)}{3} + C \implies C = 0 \]

The particular solution is \(y(1+x^2) = \frac{4x^3}{3}\).


Step 5: Final Answer:


The particular solution is \(y = \frac{4x^3}{3(1+x^2)}\).
Quick Tip: The process for solving first-order linear DEs is algorithmic:
1. Put in standard form \(\frac{dy}{dx} + P(x)y = Q(x)\).
2. Find Integrating Factor \(I.F. = e^{\int P(x)dx}\).
3. Solution is \(y \cdot I.F. = \int Q(x) \cdot I.F. dx\).
4. Use the initial condition to find C.

Step 1: Understand the Region and Setup


We need to find the area of the central vertical strip of the ellipse defined by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), bounded by the vertical lines \(x = -a/2\) and \(x = a/2\).

The ellipse is symmetric about both the x-axis and the y-axis. The total area of the strip is twice the area of the portion above the x-axis.

First, we express y in terms of x for the upper half of the ellipse:
\[ \frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} \implies y^2 = \frac{b^2}{a^2}(a^2 - x^2) \implies y = \frac{b}{a}\sqrt{a^2 - x^2} \]


Step 2: Set up the Definite Integral for the Area


The total area A is twice the integral of the upper-half function from \(x = -a/2\) to \(x = a/2\).
\[ A = 2 \int_{-a/2}^{a/2} y \, dx = 2 \int_{-a/2}^{a/2} \frac{b}{a}\sqrt{a^2 - x^2} \, dx \]

Since the integrand \(\sqrt{a^2 - x^2}\) is an even function (its graph is symmetric about the y-axis) and the integration interval is symmetric about the origin, we can simplify the integral:
\[ A = 2 \cdot \left( 2 \int_{0}^{a/2} \frac{b}{a}\sqrt{a^2 - x^2} \, dx \right) = \frac{4b}{a} \int_{0}^{a/2} \sqrt{a^2 - x^2} \, dx \]


Step 3: Evaluate the Integral


We use the standard integration formula:
\[ \int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \]

Applying this to our definite integral:
\[ A = \frac{4b}{a} \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right]_{0}^{a/2} \]

Now, we evaluate the expression at the upper and lower limits.

- At the upper limit x = a/2:

\( \frac{a/2}{2}\sqrt{a^2 - (a/2)^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{a/2}{a}\right) \)

\( = \frac{a}{4}\sqrt{a^2 - \frac{a^2}{4}} + \frac{a^2}{2}\sin^{-1}\left(\frac{1}{2}\right) \)

\( = \frac{a}{4}\sqrt{\frac{3a^2}{4}} + \frac{a^2}{2}\left(\frac{\pi}{6}\right) = \frac{a}{4} \cdot \frac{a\sqrt{3}}{2} + \frac{a^2\pi}{12} = \frac{a^2\sqrt{3}}{8} + \frac{a^2\pi}{12} \)

- At the lower limit x = 0:

\( \frac{0}{2}\sqrt{a^2 - 0} + \frac{a^2}{2}\sin^{-1}(0) = 0 + 0 = 0 \)

The value of the definite integral is \(\left(\frac{a^2\sqrt{3}}{8} + \frac{a^2\pi}{12}\right) - 0\).


Step 4: Calculate the Total Area


Multiply the result of the integral by the constant factor \(\frac{4b}{a}\):
\[ A = \frac{4b}{a} \left( \frac{a^2\sqrt{3}}{8} + \frac{a^2\pi}{12} \right) \]
\[ A = \frac{4ba^2\sqrt{3}}{8a} + \frac{4ba^2\pi}{12a} \]
\[ A = \frac{ab\sqrt{3}}{2} + \frac{ab\pi}{3} \]

Factoring out 'ab' gives the final expression:
\[ A = ab \left(\frac{\sqrt{3}}{2} + \frac{\pi}{3}\right) \]


Step 5: Final Answer:


The area of the specified region of the ellipse is \(ab\left(\frac{\pi}{3} + \frac{\sqrt{3}}{2}\right)\) square units.
Quick Tip: The area of an ellipse is a standard application of definite integrals.
The formula \(\int \sqrt{a^2 - x^2} dx\) is one of the most important ones to memorize for this section.
Using the property of even functions (\(\int_{-k}^{k} f(x)dx = 2\int_{0}^{k} f(x)dx\)) is highly recommended as it simplifies the arithmetic by making the lower limit of integration zero.

Step 1: Understanding the Problem


This is an integral of a rational function where the degree of the numerator (2) is less than the degree of the denominator (3). We use partial fraction decomposition.


Step 2: Partial Fraction Decomposition


The denominator has a repeated linear factor \((x-1)^2\) and a distinct linear factor \((x+3)\). The decomposition is:
\[ \frac{x^2 + 1}{(x-1)^2(x+3)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+3} \]

Multiplying by the denominator gives:
\[ x^2 + 1 = A(x-1)(x+3) + B(x+3) + C(x-1)^2 \]

- Set x = 1: \( 2 = 4B \implies B = \frac{1}{2} \)

- Set x = -3: \( 10 = 16C \implies C = \frac{5}{8} \)

- Equate coefficients of \(x^2\): \( 1 = A + C \implies A = 1 - \frac{5}{8} = \frac{3}{8} \)


Step 3: Integration


Substitute the constants and integrate:
\[ \int \left( \frac{3/8}{x-1} + \frac{1/2}{(x-1)^2} + \frac{5/8}{x+3} \right) dx \]
\[ = \frac{3}{8} \int \frac{1}{x-1} dx + \frac{1}{2} \int (x-1)^{-2} dx + \frac{5}{8} \int \frac{1}{x+3} dx \]
\[ = \frac{3}{8} \ln|x-1| + \frac{1}{2} \frac{(x-1)^{-1}}{-1} + \frac{5}{8} \ln|x+3| + K \]
\[ = \frac{3}{8} \ln|x-1| - \frac{1}{2(x-1)} + \frac{5}{8} \ln|x+3| + K \]


Step 4: Final Answer:


The integral is \(\frac{3}{8} \ln|x-1| + \frac{5}{8} \ln|x+3| - \frac{1}{2(x-1)} + K\).
Quick Tip: For partial fractions with repeated linear factors like \((x-a)^n\), the decomposition must include terms for all powers from 1 to n.
e.g., \(\frac{A}{x-a} + \frac{B}{(x-a)^2} + \dots + \frac{Z}{(x-a)^n}\).

Step 1: Understanding the Problem


The 'x' in the numerator of a definite integral from 0 to 'a' is a strong hint to use the "King's property".


Step 2: Applying Properties of Definite Integrals


Let \( I = \int_{0}^{\pi/2} \frac{x}{\sin x + \cos x} dx \quad (1) \).

Using the property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx\):
\[ I = \int_{0}^{\pi/2} \frac{\frac{\pi}{2} - x}{\sin(\frac{\pi}{2}-x) + \cos(\frac{\pi}{2}-x)} dx = \int_{0}^{\pi/2} \frac{\frac{\pi}{2} - x}{\cos x + \sin x} dx \quad (2) \]

Adding equations (1) and (2):
\[ 2I = \int_{0}^{\pi/2} \frac{x + (\frac{\pi}{2} - x)}{\sin x + \cos x} dx = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{1}{\sin x + \cos x} dx \]


Step 3: Evaluating the New Integral


To evaluate \(\int \frac{1}{\sin x + \cos x} dx\), we write \(\sin x + \cos x = \sqrt{2}\sin(x + \frac{\pi}{4})\).
\[ 2I = \frac{\pi}{2\sqrt{2}} \int_{0}^{\pi/2} \csc(x + \frac{\pi}{4}) dx \]
\[ 2I = \frac{\pi}{2\sqrt{2}} [\ln|\csc(x + \frac{\pi}{4}) - \cot(x + \frac{\pi}{4})|]_{0}^{\pi/2} \]
\[ 2I = \frac{\pi}{2\sqrt{2}} \left[ \ln|\csc(\frac{3\pi}{4}) - \cot(\frac{3\pi}{4})| - \ln|\csc(\frac{\pi}{4}) - \cot(\frac{\pi}{4})| \right] \]
\[ 2I = \frac{\pi}{2\sqrt{2}} \left[ \ln|\sqrt{2} - (-1)| - \ln|\sqrt{2} - 1| \right] = \frac{\pi}{2\sqrt{2}} \ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) \]

Rationalizing the argument of the log: \( \frac{\sqrt{2}+1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} = (\sqrt{2}+1)^2 \).
\[ 2I = \frac{\pi}{2\sqrt{2}} \ln((\sqrt{2}+1)^2) = \frac{\pi}{2\sqrt{2}} \cdot 2\ln(\sqrt{2}+1) = \frac{\pi}{\sqrt{2}} \ln(\sqrt{2}+1) \]
\[ I = \frac{\pi}{2\sqrt{2}} \ln(\sqrt{2}+1) \]


Step 4: Final Answer:


The value of the integral is \(\frac{\pi}{2\sqrt{2}} \ln(\sqrt{2}+1)\).
Quick Tip: The King's property is one of the most useful tools for definite integrals.
When adding the original integral (I) and the transformed integral (I), the 'x' term often cancels, leaving a much simpler integral to solve.

Step 1: Understand the Goal


To show that two lines intersect, we need to demonstrate that they have a common point. We will do this by finding the equations of both lines and then solving the system of equations for their point of intersection. If a consistent solution exists, they intersect.


Step 2: Find the Vector Equations of the Two Lines


Line 1 (passing through A and B):

- A point on the line is A(0, -1, -1), so its position vector is \(\vec{a_1} = 0\hat{i} - 1\hat{j} - 1\hat{k}\).

- The direction vector of the line is \(\vec{d_1} = \vec{AB} = \vec{B} - \vec{A}\).

\(\vec{d_1} = (4-0)\hat{i} + (5-(-1))\hat{j} + (1-(-1))\hat{k} = 4\hat{i} + 6\hat{j} + 2\hat{k}\).

We can use a simpler, parallel vector by dividing by 2: \(\vec{d_1} = 2\hat{i} + 3\hat{j} + 1\hat{k}\).

- The vector equation for Line 1 is:

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


Line 2 (passing through C and D):

- A point on the line is C(3, 9, 4), so its position vector is \(\vec{a_2} = 3\hat{i} + 9\hat{j} + 4\hat{k}\).

- The direction vector of the line is \(\vec{d_2} = \vec{CD} = \vec{D} - \vec{C}\).

\(\vec{d_2} = (-4-3)\hat{i} + (4-9)\hat{j} + (4-4)\hat{k} = -7\hat{i} - 5\hat{j} + 0\hat{k}\).

- The vector equation for Line 2 is:

\(\vec{r} = (3\hat{i} + 9\hat{j} + 4\hat{k}) + \mu(-7\hat{i} - 5\hat{j})\).


Step 3: Set Up System of Equations for Intersection


If the lines intersect, there must be values for the parameters \(\lambda\) and \(\mu\) such that the position vectors are equal. We can write the parametric equations for each line and set them equal.

From Line 1: \(x = 2\lambda\), \(y = -1+3\lambda\), \(z = -1+\lambda\).

From Line 2: \(x = 3-7\mu\), \(y = 9-5\mu\), \(z = 4\).


Equating the corresponding components:

(i) \(2\lambda = 3-7\mu\)

(ii) \(-1+3\lambda = 9-5\mu\)

(iii) \(-1+\lambda = 4\)


Step 4: Solve the System and Verify Consistency


From the simplest equation, (iii), we can directly solve for \(\lambda\):
\[ -1+\lambda = 4 \implies \lambda = 5 \]

Now, substitute \(\lambda = 5\) into equation (i) to find \(\mu\):
\[ 2(5) = 3 - 7\mu \]
\[ 10 = 3 - 7\mu \implies 7 = -7\mu \implies \mu = -1 \]

We have found potential values for \(\lambda\) and \(\mu\). To confirm that the lines intersect, these values must also satisfy the remaining equation, (ii).

Check by substituting \(\lambda=5\) and \(\mu=-1\) into equation (ii):

LHS: \(-1 + 3(5) = -1 + 15 = 14\).

RHS: \(9 - 5(-1) = 9 + 5 = 14\).

Since LHS = RHS (14 = 14), the values are consistent.


Step 5: Final Answer and Conclusion


Because a consistent set of values for \(\lambda\) (\(\lambda=5\)) and \(\mu\) (\(\mu=-1\)) exists, the two lines have a common point and therefore intersect.

*(To find the point of intersection, we can substitute either parameter back into its line's equation. Using \(\lambda=5\) in Line 1 gives the point \((2(5), -1+3(5), -1+5) = (10, 14, 4)\).)*
Quick Tip: To show two lines in 3D space intersect:
1. Write the vector/parametric equations for both lines.
2. Equate the corresponding x, y, and z components to get a system of 3 equations with 2 variables (\(\lambda, \mu\)).
3. Solve any two of the equations to find values for \(\lambda\) and \(\mu\). A simple equation (like one with only one variable) is the best place to start.
4. Substitute these values into the third (unused) equation. If it holds true, the lines intersect. If it does not hold true, the lines are skew.

Step 1: Express y in terms of h and x.

The ladder, the wall, and the ground form a right-angled triangle.

- The length of the ladder is the hypotenuse, 'h'.

- The height on the wall is one leg, 'x'.

- The distance between the wall and the foot of the ladder is the other leg, 'y'.

By the Pythagorean theorem: \[ x^2 + y^2 = h^2 \]
Solving for y: \[ y^2 = h^2 - x^2 \] \[ y = \sqrt{h^2 - x^2} \quad (since distance y must be positive) \]

Step 2: Write an expression for the area (A).

The area of the right triangle is given by A = \(\frac{1}{2} \times base \times height\).

Here, the base is y and the height is x. \[ A = \frac{1}{2} yx \]
Substitute the expression for y from Step 1: \[ A = \frac{1}{2} x \sqrt{h^2 - x^2} \] Quick Tip: Drawing a simple diagram is the best way to start any geometry-based word problem. Clearly labeling the sides helps in correctly applying fundamental theorems like the Pythagorean theorem.

Step 1: Find the derivative of A with respect to x.

The area function is \(A(x) = \frac{1}{2} x \sqrt{h^2 - x^2}\). We use the product rule \((uv)' = u'v + uv'\) to differentiate.

Let \(u = x\) and \(v = \sqrt{h^2 - x^2}\). Then \(u' = 1\) and \(v' = \frac{1}{2\sqrt{h^2 - x^2}}(-2x) = \frac{-x}{\sqrt{h^2 - x^2}}\). \[ \frac{dA}{dx} = \frac{1}{2} \left[ (1) \sqrt{h^2 - x^2} + x \left( \frac{-x}{\sqrt{h^2 - x^2}} \right) \right] \] \[ \frac{dA}{dx} = \frac{1}{2} \left[ \sqrt{h^2 - x^2} - \frac{x^2}{\sqrt{h^2 - x^2}} \right] \]
Combine the terms by finding a common denominator: \[ \frac{dA}{dx} = \frac{1}{2} \left[ \frac{(h^2 - x^2) - x^2}{\sqrt{h^2 - x^2}} \right] = \frac{h^2 - 2x^2}{2\sqrt{h^2 - x^2}} \]

Step 2: Find the critical point.

Critical points occur where \(\frac{dA}{dx} = 0\) or is undefined. We set the numerator to zero: \[ h^2 - 2x^2 = 0 \] \[ 2x^2 = h^2 \] \[ x^2 = \frac{h^2}{2} \] \[ x = \frac{h}{\sqrt{2}} \quad (since height x must be positive) \]
This is the critical point.
Quick Tip: For optimization problems, critical points are the candidates for maxima or minima. They are found by setting the first derivative equal to zero. Remember to use the product and chain rules correctly.

Step 1: Use the Second Derivative Test.

We need to find the second derivative, \(\frac{d^2A}{dx^2}\), and evaluate its sign at the critical point \(x = \frac{h}{\sqrt{2}}\).

Our first derivative is \(\frac{dA}{dx} = \frac{h^2 - 2x^2}{2\sqrt{h^2 - x^2}}\).

Using the quotient rule \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\), where \(u = h^2 - 2x^2\) and \(v = 2\sqrt{h^2 - x^2}\).
\(u' = -4x\)
\(v' = 2 \cdot \frac{-x}{\sqrt{h^2 - x^2}} = \frac{-2x}{\sqrt{h^2 - x^2}}\)
\[ \frac{d^2A}{dx^2} = \frac{(-4x)(2\sqrt{h^2 - x^2}) - (h^2 - 2x^2)\left(\frac{-2x}{\sqrt{h^2 - x^2}}\right)}{(2\sqrt{h^2 - x^2})^2} \] \[ = \frac{-8x\sqrt{h^2 - x^2} + \frac{2x(h^2 - 2x^2)}{\sqrt{h^2 - x^2}}}{4(h^2 - x^2)} \]
Multiply numerator and denominator by \(\sqrt{h^2 - x^2}\): \[ = \frac{-8x(h^2 - x^2) + 2x(h^2 - 2x^2)}{4(h^2 - x^2)^{3/2}} = \frac{-8xh^2 + 8x^3 + 2xh^2 - 4x^3}{4(h^2 - x^2)^{3/2}} = \frac{4x^3 - 6xh^2}{4(h^2 - x^2)^{3/2}} = \frac{2x^3 - 3xh^2}{2(h^2 - x^2)^{3/2}} \]

Step 2: Evaluate the second derivative at the critical point.

At the critical point, \(x = \frac{h}{\sqrt{2}}\), we have \(x^2 = \frac{h^2}{2}\). The term \(h^2 - 2x^2\) in the numerator of \(\frac{d^2A}{dx^2}\) becomes \(h^2 - 2(\frac{h^2}{2}) = 0\).
Let's re-examine the expression for \(\frac{d^2A}{dx^2}\) before simplifying the numerator: \[ \frac{d^2A}{dx^2} = \frac{(-4x)(2\sqrt{h^2 - x^2}) - (h^2 - 2x^2)\left(\frac{-2x}{\sqrt{h^2 - x^2}}\right)}{4(h^2 - x^2)} \]
When \(x^2 = h^2/2\), the term \(h^2 - 2x^2 = 0\). So the second part of the numerator vanishes. \[ \frac{d^2A}{dx^2}\bigg|_{x=h/\sqrt{2}} = \frac{(-4(h/\sqrt{2}))(2\sqrt{h^2 - h^2/2}) - 0}{4(h^2 - h^2/2)} = \frac{(-4h/\sqrt{2})(2\sqrt{h^2/2})}{4(h^2/2)} = \frac{(-4h/\sqrt{2})(2h/\sqrt{2})}{2h^2}\]

\[= \frac{-8h^2/2}{2h^2} = \frac{-4h^2}{2h^2} = -2 \]


Since \(\frac{d^2A}{dx^2} = -2 < 0\) at the critical point, the area A is maximum at \(x = \frac{h}{\sqrt{2}}\).
Quick Tip: The First Derivative Test is often algebraically simpler than the Second Derivative Test. If the sign of the derivative changes from positive to negative at a critical point, it's a maximum. If it changes from negative to positive, it's a minimum.

Step 1: Set up the related rates problem.

We are given:
- Length of ladder, h = 5 m.
- The distance y is decreasing at 2 m/s, so \(\frac{dy}{dt} = -2\) m/s.
- We need to find \(\frac{dx}{dt}\) when y = 3 m.
The relationship between x and y is given by the Pythagorean theorem: \[ x^2 + y^2 = h^2 \implies x^2 + y^2 = 5^2 = 25 \]

Step 2: Differentiate with respect to time (t).

Differentiate the equation \(x^2 + y^2 = 25\) implicitly with respect to time t: \[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(25) \] \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]

Step 3: Find the value of x at the given instant.

We need to find x when y = 3 m. \[ x^2 + 3^2 = 25 \] \[ x^2 + 9 = 25 \] \[ x^2 = 16 \implies x = 4 m \quad (since height x must be positive) \]

Step 4: Solve for \(\frac{dx}{dt}\).

Substitute the known values (x=4, y=3, \(\frac{dy}{dt}=-2\)) into the differentiated equation: \[ 2(4) \frac{dx}{dt} + 2(3)(-2) = 0 \] \[ 8 \frac{dx}{dt} - 12 = 0 \] \[ 8 \frac{dx}{dt} = 12 \] \[ \frac{dx}{dt} = \frac{12}{8} = \frac{3}{2} = 1.5 m/s \]
Since the result is positive, the height x is increasing.


Final Answer:

The height on the wall (x) is increasing at a rate of 1.5 m/s.
Quick Tip: In related rates problems, the key is to find an equation connecting the variables and then differentiate it implicitly with respect to time, \(t\). Pay close attention to the signs of the rates (positive for increasing quantities, negative for decreasing).

Step 1: Define events and list probabilities.

Let A, B, and C be the events that the chosen smartphone is from company A, B, and C, respectively.

Let D be the event that the chosen smartphone is defective.

We are given the following probabilities from the inventory:
- P(A) = 25% = 0.25
- P(B) = 35% = 0.35
- P(C) = 40% = 0.40
We are also given the conditional probabilities of a phone being defective, given the company:
- P(D|A) = 5% = 0.05
- P(D|B) = 4% = 0.04
- P(D|C) = 2% = 0.02

Step 2: Apply the Law of Total Probability.

The probability of the phone being defective, P(D), is the sum of the probabilities of it being a defective phone from each company. \[ P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C) \] \[ P(D) = (0.25)(0.05) + (0.35)(0.04) + (0.40)(0.02) \] \[ P(D) = 0.0125 + 0.0140 + 0.0080 \] \[ P(D) = 0.0345 \]

Final Answer:

The probability that the smartphone was defective is 0.0345 or 3.45%.
Quick Tip: The Law of Total Probability is perfect for finding the overall probability of an event that can occur via several distinct paths. Structure the problem by listing the probability of each path and the conditional probability of the event along that path.

Step 1: Identify the required conditional probability.

We need to find the probability that the phone was made by company B, given that it is defective. This is the conditional probability P(B|D).


Step 2: Apply Bayes' Theorem.

Bayes' theorem states: \[ P(B|D) = \frac{P(B) P(D|B)}{P(D)} \]
We have all the necessary values from the problem statement and part (i).
- P(B) = 0.35
- P(D|B) = 0.04
- P(D) = 0.0345 (calculated in part i)

Step 3: Calculate the probability.
\[ P(B|D) = \frac{0.35 \times 0.04}{0.0345} \] \[ P(B|D) = \frac{0.0140}{0.0345} \]
To simplify the fraction, multiply the numerator and denominator by 10000: \[ P(B|D) = \frac{140}{345} \]
Divide both by 5: \[ P(B|D) = \frac{28}{69} \]

Final Answer:

The probability that the defective smartphone was manufactured by company B is \(\frac{28}{69}\).
Quick Tip: Bayes' Theorem helps "reverse" conditional probability. Use it when you know the outcome (the phone is defective) and want to find the probability of a specific cause (it came from company B). The denominator is almost always the total probability calculated in the first part of the question.

Step 1: Define the variables.

Let the price of one pen be Rs x.

Let the price of one notepad be Rs y.

Let the price of one eraser be Rs z.


Step 2: Form the linear equations based on the given information.

From Neha's purchase: \(4x + 3y + 2z = 60\)

From Rani's purchase: \(2x + 4y + 6z = 90\)

From Sam's purchase: \(6x + 2y + 3z = 70\)


Step 3: Express the system of equations in matrix form AX = B.

The system can be written as: \[ \begin{bmatrix} 4 & 3 & 2
2 & 4 & 6
6 & 2 & 3 \end{bmatrix} \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 60
90
70 \end{bmatrix} \]
Where: \[ A = \begin{bmatrix} 4 & 3 & 2
2 & 4 & 6
6 & 2 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x
y
z \end{bmatrix}, \quad B = \begin{bmatrix} 60
90
70 \end{bmatrix} \] Quick Tip: When converting a word problem to a matrix equation, ensure the variables are consistent in each equation. The coefficients of these variables form the rows of matrix A, the variables form matrix X, and the constants on the right side form matrix B.

Step 1: Write the matrix A. \[ A = \begin{bmatrix} 4 & 3 & 2
2 & 4 & 6
6 & 2 & 3 \end{bmatrix} \]

Step 2: Calculate the determinant \(|A|\).

We expand along the first row: \[ |A| = 4(4 \cdot 3 - 6 \cdot 2) - 3(2 \cdot 3 - 6 \cdot 6) + 2(2 \cdot 2 - 4 \cdot 6) \] \[ |A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24) \] \[ |A| = 4(0) - 3(-30) + 2(-20) \] \[ |A| = 0 + 90 - 40 = 50 \]

Step 3: Confirm if A\(^{-1}\) exists.

The inverse of a matrix A exists if and only if its determinant is non-zero (\(|A| \neq 0\)).

Since \(|A| = 50 \neq 0\), the matrix A is non-singular.

Therefore, it is possible to find A\(^{-1}\).
Quick Tip: A matrix is called "singular" if its determinant is zero, and "non-singular" otherwise. Only non-singular matrices are invertible. This is a fundamental concept in matrix algebra.

Step 1: Find the cofactor matrix of A.

The cofactors are:

\( C_{11} = (12-12) = 0 \)
\( C_{12} = -(6-36) = 30 \)
\( C_{13} = (4-24) = -20 \)
\( C_{21} = -(9-4) = -5 \)
\( C_{22} = (12-12) = 0 \)
\( C_{23} = -(8-18) = 10 \)
\( C_{31} = (18-8) = 10 \)
\( C_{32} = -(24-4) = -20 \)
\( C_{33} = (16-6) = 10 \)

The cofactor matrix is \(\begin{bmatrix} 0 & 30 & -20
-5 & 0 & 10
10 & -20 & 10 \end{bmatrix}\).


Step 2: Find the adjugate (adjoint) of A.

The adjugate of A is the transpose of the cofactor matrix. \[ adj(A) = \begin{bmatrix} 0 & -5 & 10
30 & 0 & -20
-20 & 10 & 10 \end{bmatrix} \]

Step 3: Find the inverse of A.

The formula for the inverse is \(A^{-1} = \frac{1}{|A|} adj(A)\). We know \(|A| = 50\). \[ A^{-1} = \frac{1}{50} \begin{bmatrix} 0 & -5 & 10
30 & 0 & -20
-20 & 10 & 10 \end{bmatrix} \]

Step 4: Write the formula to find X.

The solution to the system of equations AX = B is given by: \[ X = A^{-1}B \] Quick Tip: The most common error in finding a matrix inverse is confusing the cofactor matrix with the adjugate matrix. Remember: adjugate is the \textbf{transpose} of the cofactors. The formula \(X = A^{-1}B\) is crucial for solving linear systems.

Step 1: Calculate A\(^2\). \[ A^2 = A \cdot A = \begin{bmatrix} 4 & 3 & 2
2 & 4 & 6
6 & 2 & 3 \end{bmatrix} \begin{bmatrix} 4 & 3 & 2
2 & 4 & 6
6 & 2 & 3 \end{bmatrix} \] \[ A^2 = \begin{bmatrix} (16+6+12) & (12+12+4) & (8+18+6)
(8+8+36) & (6+16+12) & (4+24+18)
(24+4+18) & (18+8+6) & (12+12+9) \end{bmatrix} \] \[ A^2 = \begin{bmatrix} 34 & 28 & 32
52 & 34 & 46
46 & 32 & 33 \end{bmatrix} \]

Step 2: Calculate 8I.

I is the 3x3 identity matrix. \[ 8I = 8 \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 8 & 0 & 0
0 & 8 & 0
0 & 0 & 8 \end{bmatrix} \]

Step 3: Calculate A\(^2\) - 8I. \[ A^2 - 8I = \begin{bmatrix} 34 & 28 & 32
52 & 34 & 46
46 & 32 & 33 \end{bmatrix} - \begin{bmatrix} 8 & 0 & 0
0 & 8 & 0
0 & 0 & 8 \end{bmatrix} \] \[ A^2 - 8I = \begin{bmatrix} (34-8) & (28-0) & (32-0)
(52-0) & (34-8) & (46-0)
(46-0) & (32-0) & (33-8) \end{bmatrix} \] \[ A^2 - 8I = \begin{bmatrix} 26 & 28 & 32
52 & 26 & 46
46 & 32 & 25 \end{bmatrix} \] Quick Tip: Matrix multiplication (\(A^2\)) is done row-by-column. Scalar multiplication (8I) multiplies every element by the scalar. Matrix subtraction is performed element-wise. Keep the operations distinct to avoid errors.