CBSE Class 12 Mathematics Set 2- (65/3/2) Question Paper 2026 with Solution PDF

Concept:
The problem requires finding the first derivative of a given trigonometric relation with respect to \( x \). We can approach this problem either by using the standard formula for the derivative of the inverse trigonometric function \( \tan^{-1} x \) or by converting the inverse trigonometric expression into an explicit trigonometric form and employing implicit differentiation along with fundamental trigonometric identities.

Step 1: Expressing the equation explicitly in terms of \(x\).

The given equation is: \[ \tan^{-1} x = y \]
By taking the tangent of both sides, we can rewrite the inverse function as a direct trigonometric equation: \[ x = \tan y \]

Step 2: Differentiating implicitly with respect to \(x\).

Now, we differentiate both sides of the equation \( x = \tan y \) with respect to \( x \). Applying the chain rule on the right-hand side, we get: \[ \frac{d}{dx}(x) = \frac{d}{dx}(\tan y) \] \[ 1 = \sec^2 y \cdot \frac{dy}{dx} \]

Step 3: Solving for \( \frac{dy}{dx} \) and matching the options.

To isolate \( \frac{dy}{dx} \), we divide both sides by \( \sec^2 y \): \[ \frac{dy}{dx} = \frac{1}{\sec^2 y} \]
Using the reciprocal trigonometric identity where \( \frac{1}{\sec \theta} = \cos \theta \), we can simplify the expression: \[ \frac{dy}{dx} = \cos^2 y \]
This precisely matches option (D). Quick Tip: When evaluating derivatives involving inverse trigonometric expressions, if the standard answer \( \frac{1}{1+x^2} \) is not present in the options, look to convert the variable \( x \) back into terms of \( y \) using identities like \( \sec^2 y = 1 + \tan^2 y = 1 + x^2 \). Thus, \( \frac{1}{1+x^2} = \frac{1}{\sec^2 y} = \cos^2 y \).

Concept:
The problem asks for the rate of change of the volume \(V\) of a sphere with respect to its diameter \(D\). Thus, we need to find \( \frac{dV}{dD} \) and evaluate it when the radius is \(5\) cm.

The volume of a sphere is: \[ V=\frac{4}{3}\pi r^3 \]
and the diameter \(D\) is related to the radius \(r\) by: \[ D=2r. \]

Step 1: Express the volume in terms of the diameter.

Since \[ D=2r, \]
we have \[ r=\frac{D}{2}. \]
Substituting this into the volume formula: \[ V=\frac{4}{3}\pi\left(\frac{D}{2}\right)^3 \] \[ =\frac{4}{3}\pi\frac{D^3}{8} \] \[ V=\frac{\pi D^3}{6}. \]

Step 2: Differentiate the volume with respect to the diameter.

Differentiating with respect to \(D\): \[ \frac{dV}{dD} = \frac{\pi}{6}\frac{d}{dD}(D^3) \] \[ \frac{dV}{dD} = \frac{\pi}{6}(3D^2) \] \[ \frac{dV}{dD} = \frac{\pi D^2}{2}. \]

Step 3: Use the given radius.

The radius is given as \[ r=5 cm. \]
Therefore, the diameter is: \[ D=2r=2(5)=10 cm. \]

Substituting \(D=10\) into the expression for the rate of change: \[ \frac{dV}{dD} = \frac{\pi(10)^2}{2} \] \[ = \frac{100\pi}{2} \] \[ \frac{dV}{dD}=50\pi. \]

Hence, the required rate of change is: \[ \boxed{50\pi cm^3/cm}. \]

Therefore, the correct option is \(\boxed{(C)}\). Quick Tip: When the rate of change of volume with respect to diameter is required, differentiate \(V\) with respect to \(D\), not with respect to \(r\). Since \(D=2r\), a convenient result is \( \frac{dV}{dD}=2\pi r^2 \). For \(r=5\) cm, this gives \(50\pi cm^3/cm\).

Concept:
To solve the indefinite integral \( I = \int \frac{dx}{\sqrt{e^{-2x} - 1}} \), we need to simplify the integrand using algebraic manipulation of exponents and then apply a suitable substitution that transforms it into a standard integral form, specifically matching the standard form \( \int \frac{du}{\sqrt{1-u^2}} = \sin^{-1}u + C \).

Step 1: Simplify the expression inside the square root.

We can write \( e^{-2x} \) as \( \frac{1}{e^{2x}} \). Let us rewrite the integral: \[ I = \int \frac{dx}{\sqrt{\frac{1}{e^{2x}} - 1}} = \int \frac{dx}{\sqrt{\frac{1 - e^{2x}}{e^{2x}}}} \]
Taking \( e^{2x} \) out of the square root in the denominator gives \( e^x \): \[ I = \int \frac{dx}{\frac{\sqrt{1 - e^{2x}}}{e^x}} = \int \frac{e^x dx}{\sqrt{1 - (e^x)^2}} \]

Step 2: Using the substitution method.

Let us substitute \( u = e^x \).
Differentiating both sides with respect to \( x \) yields: \[ du = e^x dx \]
Substituting these components into our integral expression gives: \[ I = \int \frac{du}{\sqrt{1 - u^2}} \]

Step 3: Integrating using standard formulas.

The integral is now in a standard form whose solution is well known: \[ \int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1}(u) + C \]
Substituting back the original value of \( u = e^x \): \[ I = \sin^{-1}(e^x) + C \]
This perfectly aligns with option (C). Quick Tip: Whenever an integral contains terms like \( e^{-x} \) or \( e^{-2x} \) inside a radical or denominator, multiplying the numerator and denominator by a suitable power of \( e^x \) often clears the negative exponents and reveals a straightforward substitution.

Concept:
This problem involves using the properties of definite integrals, specifically the behavior of even functions over a symmetric interval \( [-a, a] \). Recall that if a function \( f(x) \) satisfies \( f(-x) = f(x) \), it is called an even function, and its integral over a symmetric domain satisfies: \[ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx = 2 \int_{-a}^{0} f(x) \, dx \]

Step 1: Determine the parity (even/odd property) of the integrand.

Let our integrand function be defined as: \[ f(x) = 1 - |x| \]
Let us substitute \( -x \) in place of \( x \): \[ f(-x) = 1 - |-x| \]
Since the absolute value function removes negative signs, \( |-x| = |x| \). Therefore: \[ f(-x) = 1 - |x| = f(x) \]
Since \( f(-x) = f(x) \), the function \( f(x) \) is strictly an even function.

Step 2: Split and evaluate using interval properties.

For an even function, the total area under the curve from \( -1 \) to \( 1 \) is twice the area of either the positive half or the negative half: \[ \int_{-1}^{1} (1 - |x|) \, dx = 2 \int_{0}^{1} (1 - |x|) \, dx = 2 \int_{-1}^{0} (1 - |x|) \, dx \]

Step 3: Analyze the definition of absolute value in the intervals.

Let's check the behavior of \( |x| \) within the interval of integration for the options given:
- In the interval \( [-1, 0] \), \( x \le 0 \), which implies by definition that \( |x| = -x \).
Substituting this into our integral expression over \( [-1, 0] \): \[ 2 \int_{-1}^{0} (1 - (-x)) \, dx = 2 \int_{-1}^{0} (1 + x) \, dx \]
This precisely matches option (B). Quick Tip: For definite integrals with symmetric limits \( [-a, a] \), always evaluate if the function is even or odd. If even, remember that \( \int_{-a}^{a} f(x)dx = 2\int_{-a}^{0} f(x)dx \), and then replace \( |x| \) with \( -x \) since \( x \) is negative in that domain.

Concept:
The problem asks for the area of a region bounded by a circle \( x^2 + y^2 = 9 \) and a horizontal straight line \( y = 1 \). The shaded region lies above the line \( y = 1 \) inside the circle up to its highest point \( y = 3 \). We can set up the integration with respect to the y-axis to find the area directly.

Step 1: Identify the boundaries of the region.

The circle equation is given by: \[ x^2 + y^2 = 9 \]
This is a circle centered at the origin \( (0,0) \) with radius \( r = \sqrt{9} = 3 \).
The lower boundary of the shaded region is the line \( y = 1 \).
The upper limit of the shaded region along the y-axis is the peak of the circle, where \( x = 0 \), giving \( y = 3 \).
Thus, the limits of integration for \( y \) are from \( y = 1 \) to \( y = 3 \).

Step 2: Express \( x \) as a function of \( y \).

From the circle equation, solve for \( x \): \[ x^2 = 9 - y^2 \quad \Rightarrow \quad x = \pm \sqrt{9 - y^2} \]
The right half of the circle corresponds to \( x = +\sqrt{9 - y^2} \) and the left half corresponds to \( x = -\sqrt{9 - y^2} \).

Step 3: Set up the area integral.

By symmetry across the y-axis, the total area is twice the area contained in the first quadrant between \( y = 1 \) and \( y = 3 \): \[ Area = 2 \int_{1}^{3} x \, dy = 2 \int_{1}^{3} \sqrt{9 - y^2} \, dy \]
This expression directly matches option (B). Quick Tip: Integrating along the y-axis is often much cleaner when the region is bounded symmetrically by horizontal boundaries. The formula used is \(Area = \int_{c}^{d} (x_{right} - x_{left}) \, dy\).

Concept:
A first-order differential equation of the form \[ \frac{dy}{dx}=F(x,y) \]
is called a homogeneous differential equation if \(F(x,y)\) is a homogeneous function of degree zero. Equivalently, \(F(x,y)\) must be expressible as a function of the ratio \(y/x\) (or \(x/y\)).

A homogeneous function of degree zero satisfies: \[ F(\lambda x,\lambda y)=F(x,y). \]

We test each of the four given functions.

Step 1: Test function (i).

Given: \[ F(x,y)=3x+2y. \]
Replacing \(x\) and \(y\) by \(\lambda x\) and \(\lambda y\): \[ F(\lambda x,\lambda y) = 3\lambda x+2\lambda y \] \[ =\lambda(3x+2y) \] \[ =\lambda F(x,y). \]
Thus, \(F(x,y)\) is homogeneous of degree \(1\), not degree \(0\).

Therefore, (i) does not give a homogeneous differential equation of the required form.

Step 2: Test function (ii).

Given: \[ F(x,y)=\sin\frac{y}{x}+\log y-\log x. \]
Using the logarithm property, \[ \log y-\log x=\log\left(\frac{y}{x}\right). \]
Therefore, \[ F(x,y) = \sin\frac{y}{x} + \log\left(\frac{y}{x}\right). \]
This is entirely a function of \(y/x\). Alternatively, \[ F(\lambda x,\lambda y) = \sin\left(\frac{\lambda y}{\lambda x}\right) + \log\left(\frac{\lambda y}{\lambda x}\right) \] \[ = \sin\frac{y}{x} + \log\frac{y}{x} = F(x,y). \]
Hence, (ii) is homogeneous of degree zero.

Step 3: Test function (iii).

Given: \[ F(x,y)=e^{y/x}+1. \]
Replacing \(x\) and \(y\) by \(\lambda x\) and \(\lambda y\): \[ F(\lambda x,\lambda y) = e^{\frac{\lambda y}{\lambda x}}+1 \] \[ =e^{y/x}+1 \] \[ =F(x,y). \]
Thus, \(F(x,y)\) is homogeneous of degree zero.

Hence, (iii) also gives a homogeneous differential equation.

Step 4: Test function (iv).

Given: \[ F(x,y)=\sqrt{x^2+y^2}-y. \]
Replacing \(x\) and \(y\) by \(\lambda x\) and \(\lambda y\): \[ F(\lambda x,\lambda y) = \sqrt{(\lambda x)^2+(\lambda y)^2}-\lambda y \] \[ = \sqrt{\lambda^2(x^2+y^2)}-\lambda y. \]
For positive scaling factor \(\lambda\), \[ F(\lambda x,\lambda y) = \lambda\sqrt{x^2+y^2}-\lambda y \] \[ = \lambda\left(\sqrt{x^2+y^2}-y\right) \] \[ =\lambda F(x,y). \]
Therefore, this function is homogeneous of degree \(1\), not degree \(0\).

Hence, (iv) does not satisfy the condition for a homogeneous differential equation of the form \[ \frac{dy}{dx}=F(x,y). \]

Therefore, only (ii) and (iii) satisfy the required condition.
\[ \boxed{(ii) and (iii)} \]

Thus, the correct option is \(\boxed{(D)}\). Quick Tip: For a differential equation of the form \( \frac{dy}{dx}=F(x,y) \), check whether \(F(\lambda x,\lambda y)=F(x,y)\). If this condition holds, \(F\) is homogeneous of degree zero. Expressions involving only \(y/x\) or \(x/y\) are standard examples.

Concept:
This question tests fundamental vector dot product and cross product inequalities, specifically the Cauchy-Schwarz Inequality for vectors, which establishes a relation between the dot product of two vectors and the product of their magnitudes.

Step 1: Analyze the definition of the dot product.

By definition, the dot product of two vectors \( \vec{a} \) and \( \vec{b} \) is given by: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
where \( \theta \) is the angle between the two vectors \( \vec{a} \) and \( \vec{b} \), such that \( 0 \le \theta \le \pi \).

Step 2: Apply the range of the cosine function.

We know that for any real angle \( \theta \): \[ \cos \theta \le 1 \]
Multiplying both sides by the non-negative scalar quantity \( |\vec{a}| |\vec{b}| \), we get: \[ |\vec{a}| |\vec{b}| \cos \theta \le |\vec{a}| |\vec{b}| \cdot 1 \]
Substituting the definition of the dot product back into the inequality: \[ \vec{a} \cdot \vec{b} \le |\vec{a}| |\vec{b}| \]
This statement is universally true for any pair of vectors, matching option (A). Quick Tip: The Cauchy-Schwarz inequality states that \( |\vec{a} \cdot \vec{b}| \le |\vec{a}||\vec{b}| \). Since any real number is less than or equal to its absolute value (\( x \le |x| \)), it follows directly that \( \vec{a} \cdot \vec{b} \le |\vec{a}||\vec{b}| \).

Concept:
This problem uses the distributive property of the dot product over vector addition and subtraction, along with the magnitude relationship \( \vec{x} \cdot \vec{x} = |\vec{x}|^2 \).

Step 1: Expand the dot product equation.

We are given: \[ (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 198 \]
Expanding using the distributive law: \[ \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 198 \]
Since the dot product is commutative (\( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \)), the middle terms cancel out: \[ |\vec{a}|^2 - |\vec{b}|^2 = 198 \]

Step 2: Substitute the given magnitude relationship.

We are given that \( |\vec{a}| = 10 |\vec{b}| \). Substituting this into our simplified equation: \[ (10 |\vec{b}|)^2 - |\vec{b}|^2 = 198 \] \[ 100 |\vec{b}|^2 - |\vec{b}|^2 = 198 \] \[ 99 |\vec{b}|^2 = 198 \]

Step 3: Solve for \( |\vec{b}| \) and \( |\vec{a}| \).

Divide both sides by 99: \[ |\vec{b}|^2 = \frac{198}{99} = 2 \]
Taking the positive square root (since magnitude is always non-negative): \[ |\vec{b}| = \sqrt{2} \]
This matches option (B). Quick Tip: The vector identity \( (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2 - |\vec{b}|^2 \) works identically to the algebraic difference of squares formula \( (x+y)(x-y) = x^2 - y^2 \).

Concept:
The angle between two lines in three-dimensional space can be determined using the direction cosines of the lines. If two lines have direction cosines \( (l_1, m_1, n_1) \) and \( (l_2, m_2, n_2) \), the cosine of the angle \( \theta \) between them is given by their dot-product analogue. Since \( \theta \) is explicitly stated to be an acute angle, \( \cos \theta \) must be non-negative, necessitating the absolute value signs.

Step 1: Standard formula definition.

The unit vectors along the lines \( L_1 \) and \( L_2 \) can be written as: \[ \hat{u}_1 = l_1\hat{i} + m_1\hat{j} + n_1\hat{k} \quad and \quad \hat{u}_2 = l_2\hat{i} + m_2\hat{j} + n_2\hat{k} \]
The cosine of the angle \( \theta \) between these vectors is: \[ \cos \theta = \hat{u}_1 \cdot \hat{u}_2 = l_1l_2 + m_1m_2 + n_1n_2 \]

Step 2: Accounting for the acute angle constraint.

Since lines extend infinitely in both directions, they form two supplementary angles. The problem explicitly specifies that \( \theta \) is the acute angle, meaning \( 0 \le \theta \le \frac{\pi}{2} \). In this quadrant, cosine values are always positive. To guarantee a positive result regardless of the orientation choice of direction cosines, we introduce absolute value brackets: \[ \cos \theta = |l_1l_2 + m_1m_2 + n_1n_2| \]
This precisely matches option (D). Quick Tip: Whenever a question specifies an "acute angle" between lines in 3D geometry, always look for the option containing the absolute value modulus sign to ensure the output value is positive.

Concept:
For two lines in three-dimensional space to be parallel, their direction ratios must be proportional to one another. If the direction ratios of the first line are \( \langle a_1, b_1, c_1 \rangle \) and the second line are \( \langle a_2, b_2, c_2 \rangle \), then the condition for parallelism is: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

Step 1: Extract the components and set up the proportion.

Given direction ratios:
- Line 1: \( a_1 = 1, \, b_1 = -2, \, c_1 = 3 \)
- Line 2: \( a_2 = -2, \, b_2 = p, \, c_2 = -6 \)
Applying the condition for parallel lines: \[ \frac{1}{-2} = \frac{-2}{p} = \frac{3}{-6} \]

Step 2: Simplify the known ratios and solve for \(p\).

Notice that the first and third ratios both simplify to \( -\frac{1}{2} \): \[ -\frac{1}{2} = \frac{-2}{p} \]
Cross-multiplying to solve for \( p \): \[ 1 \cdot p = (-2) \cdot (-2) \] \[ p = 4 \]
This value corresponds to option (B). Quick Tip: Parallel lines possess direction vectors that are scalar multiples of each other. Inspecting the numbers here, multiplying line 1's vector by \(-2\) directly yields line 2's vector: \(-2 \times (-2) = 4\).

Concept:
This problem involves independent events and conditional probability. Key properties used:

For independent events, \( P(E \cap F) = P(E) \cdot P(F) \)
Conditional probabilities simplify to: \( P(E|F) = P(E) \) and \( P(F|E) = P(F) \)
Probability of union: \( P(E \cup F) = P(E) + P(F) - P(E \cap F) \)


Step 1: Determine the value of \( P(F) \).

Using the formula for the union of two events: \[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]
Since \( E \) and \( F \) are independent, substitute \( P(E \cap F) = P(E) \cdot P(F) \): \[ P(E \cup F) = P(E) + P(F) - P(E) \cdot P(F) \]
Substitute the given values \( P(E) = \frac{3}{10} \) and \( P(E \cup F) = \frac{1}{2} \): \[ \frac{1}{2} = \frac{3}{10} + P(F) - \frac{3}{10}P(F) \] \[ \frac{1}{2} - \frac{3}{10} = P(F) \left(1 - \frac{3}{10}\right) \] \[ \frac{5 - 3}{10} = P(F) \cdot \frac{7}{10} \quad \Rightarrow \quad \frac{2}{10} = \frac{7}{10} P(F) \] \[ P(F) = \frac{2}{7} \]

Step 2: Evaluate the conditional probability expression.

Because \( E \) and \( F \) are independent events, the occurrence of one does not affect the probability of the other: \[ P(E|F) = P(E) = \frac{3}{10} \] \[ P(F|E) = P(F) = \frac{2}{7} \]
Now compute the required difference: \[ P(E|F) - P(F|E) = \frac{3}{10} - \frac{2}{7} = \frac{3 \times 7 - 2 \times 10}{70} = \frac{21 - 20}{70} = \frac{1}{70} \]
This value directly matches option (C). Quick Tip: For independent events, conditional probabilities collapse immediately: \( P(A|B) = P(A) \). This avoids wasting time setting up complex fractions for the final subtraction step.

Concept:
The domain of the standard inverse sine function \( \sin^{-1}(\theta) \) is restricted to the interval \( [-1, 1] \). Therefore, for the function \( \sin^{-1}(1 - 2x) \) to be well-defined in the real number system, its argument must lie within these closed boundaries: \[ -1 \le 1 - 2x \le 1 \]

Step 1: Set up and solve the compound inequality.

Subtract 1 from all parts of the inequality chain: \[ -1 - 1 \le -2x \le 1 - 1 \] \[ -2 \le -2x \le 0 \]

Step 2: Divide by the negative coefficient.

Divide the entire inequality by \( -2 \). Remember that dividing or multiplying an inequality by a negative number reverses the direction of the inequality signs: \[ \frac{-2}{-2} \ge \frac{-2x}{-2} \ge \frac{0}{-2} \] \[ 1 \ge x \ge 0 \]
Rewriting this in the standard low-to-high interval notation: \[ 0 \le x \le 1 \quad \Rightarrow \quad x \in [0, 1] \]
This range matches option (D). Quick Tip: Always remember to flip the inequality signs whenever you divide or multiply by a negative quantity. A quick check of boundary values (like plugging in \( x=0 \) and \( x=1 \)) can confirm correctness instantly.

Concept:
This problem can be efficiently solved by manipulating matrix equations using matrix multiplication properties. Specifically, multiplying a matrix equation by its inverse \( A^{-1} \) allows us to lower the powers of the matrix and isolate \( A^{-1} \) explicitly in terms of \( A \) and the identity matrix \( I \).

Step 1: Set up the matrix equation and pre-multiply by \( A^{-1} \).

The given matrix equation is: \[ A^2 = 4A + 3I \]
Assuming \( A \) is invertible, we multiply both sides of the equation by \( A^{-1} \): \[ A^{-1} \cdot A^2 = A^{-1} \cdot (4A + 3I) \]
Using the associative and distributive properties of matrix multiplication: \[ (A^{-1} A) A = 4(A^{-1} A) + 3(A^{-1} I) \]

Step 2: Simplify using identity matrix properties.

Since \( A^{-1}A = I \) and \( A^{-1}I = A^{-1} \), the expression becomes: \[ I \cdot A = 4I + 3A^{-1} \] \[ A = 4I + 3A^{-1} \]

Step 3: Isolate the matrix inverse term \( A^{-1} \).

Rearrange the terms to make \( 3A^{-1} \) the subject: \[ 3A^{-1} = A - 4I \]
Divide by 3: \[ A^{-1} = \frac{1}{3}A - \frac{4}{3}I \]

Step 4: Compare coefficients and find the sum \( x + y \).

The problem states that \( A^{-1} = xA + yI \). Comparing this with our derived equation: \[ x = \frac{1}{3} \quad and \quad y = -\frac{4}{3} \]
Now calculate the requested sum \( x + y \): \[ x + y = \frac{1}{3} + \left(-\frac{4}{3}\right) = \frac{1 - 4}{3} = \frac{-3}{3} = -1 \]
This directly evaluates to option (A). Quick Tip: Instead of computing complex characteristic polynomials, simply isolate \( I \) in the polynomial equation: \( 3I = A^2 - 4A \). Multiplying by \( A^{-1} \) directly gives \( 3A^{-1} = A - 4I \).

Concept:
A square matrix \( M \) is symmetric if \( M' = M \) (where \( M' \) denotes the transpose of \( M \)) and skew-symmetric if \( M' = -M \). Important properties of transposes used here include:

\( (X + Y)' = X' + Y' \)
\( (XY)' = Y'X' \)
\( (X')' = X \)


Step 1: Apply the given conditions for matrices A and B.

Since \( A \) and \( B \) are skew-symmetric matrices, by definition we have: \[ A' = -A \quad and \quad B' = -B \]

Step 2: Take the transpose of the given matrix expression.

Let the given matrix expression be denoted by \( P \): \[ P = AB' + BA' \]
Taking the transpose on both sides: \[ P' = (AB' + BA')' \]
Using the sum property of transposes: \[ P' = (AB')' + (BA')' \]

Step 3: Apply the reversal law of transposes.

Using the property \( (XY)' = Y'X' \): \[ P' = (B')'A' + (A')'B' \]
Since the transpose of a transpose returns the original matrix (\( (X')' = X \)): \[ P' = BA' + AB' \]

Step 4: Use commutativity of matrix addition to compare with the original matrix.

By rearranging the terms using matrix addition commutativity: \[ P' = AB' + BA' = P \]
Since \( P' = P \), the matrix \( AB' + BA' \) is a symmetric matrix. This corresponds to option (A). Quick Tip: For any two matrices \( A \) and \( B \), an expression of the form \( XY + YX \) where \( X=A \) and \( Y=B' \) retains its structure under transpose because the reversal law swaps the multiplication order, mirroring the terms back into themselves.

Concept:
The rule for matrix multiplication states that if matrix \( X \) has order \( m \times n \) and matrix \( Y \) has order \( n \times p \), then the resulting matrix \( XY \) is defined and has an order of \( m \times p \).

Step 1: Identify the orders of the given matrices.

Let the unknown order of matrix \( B \) be \( m \times n \).
The row matrix given is \( C = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \). It has 1 row and 3 columns, so its order is \( 1 \times 3 \).
The resulting product matrix on the right-hand side is a square matrix with 3 rows and 3 columns, so its order is \( 3 \times 3 \).

Step 2: Apply the matrix multiplication compatibility rule.

The product \( B \cdot C \) is defined, which implies that the number of columns in \( B \) must equal the number of rows in \( C \): \[ n = 1 \]

Step 3: Determine the rows using the product order.

The order of the product matrix \( BC \) is given by the number of rows of \( B \) and the number of columns of \( C \), which is \( m \times 3 \).
We are given that the product matrix has order \( 3 \times 3 \). Therefore: \[ m = 3 \]

Step 4: Combine the dimensions to find the order of B.

Since \( m = 3 \) and \( n = 1 \), the order of matrix \( B \) is \( 3 \times 1 \). This corresponds exactly to option (B). Quick Tip: Order matching equation: \( (m \times n) \times (1 \times 3) = (3 \times 3) \). The inside numbers must match (\( n = 1 \)), and the outside numbers must equal the product's dimensions (\( m = 3 \)).

Concept:
The condition \[ A^2=A \]
shows that \(A\) is an idempotent matrix. For an idempotent matrix, all positive integral powers of \(A\) reduce to \(A\). In particular, \[ A^3=A. \]

Since \(I\) commutes with every square matrix \(A\), the ordinary binomial expansion can be used for expressions involving \(I-A\).

Step 1: Expand \( (I-A)^3 \).

Using \[ (X-Y)^3=X^3-3X^2Y+3XY^2-Y^3, \]
with \(X=I\) and \(Y=A\), we get: \[ (I-A)^3 = I^3-3I^2A+3IA^2-A^3. \]

Since \[ I^2=I,\qquad I^3=I,\qquad IA=A,\qquad IA^2=A^2, \]
we obtain: \[ (I-A)^3 = I-3A+3A^2-A^3. \]

Step 2: Use the given condition \(A^2=A\).

From \[ A^2=A, \]
multiplying both sides by \(A\) gives: \[ A^3=A^2=A. \]
Therefore, \[ (I-A)^3 = I-3A+3A-A. \]
Combining the terms containing \(A\): \[ (I-A)^3=I-A. \]

Step 3: Compare with the given expression.

We are given: \[ (I-A)^3=xA+I. \]
But from the idempotent property, we have: \[ (I-A)^3=I-A. \]
Hence, \[ I-A=xA+I. \]
Subtracting \(I\) from both sides: \[ -A=xA. \]
Thus, \[ (x+1)A=0. \]

Since the given relation must hold for the matrix \(A\), the required coefficient is: \[ x+1=0. \]
Therefore, \[ \boxed{x=-1}. \]

Hence, the correct option is \(\boxed{(D)}\). Quick Tip: If \(A^2=A\), then \(A^n=A\) for every positive integer \(n\). Thus, expressions such as \( (I-A)^3 \) can be simplified quickly by expanding and replacing every positive power of \(A\) by \(A\).

Concept:
This question relies on important determinant properties of adjoint matrices:

\( A(adj A) = |A|I_n \)
\( |adj A| = |A|^{n-1} \), where \( n \) is the order of the square matrix.


Step 1: Factor out the scalar from the matrix expression.

The given matrix expression is: \[ A(adj A) = \begin{bmatrix} 2026 & 0 & 0
0 & 2026 & 0
0 & 0 & 2026 \end{bmatrix} \]
Factoring out the scalar 2026 from the matrix yields: \[ A(adj A) = 2026 \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} = 2026 I_3 \]
where \( I_3 \) is the identity matrix of order 3.

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

Comparing \( A(adj A) = 2026 I_3 \) with the theorem formula \( A(adj A) = |A|I_3 \), we get: \[ |A| = 2026 \]

Step 3: Calculate \( |adj A| \).

The matrix has order \( n = 3 \). Using the determinant property: \[ |adj A| = |A|^{3-1} = |A|^2 \]
Substituting \( |A| = 2026 \): \[ |adj A| = (2026)^2 \]
This matches option (D). Quick Tip: For any scalar matrix \( A(adj A) = kI_n \), the determinant value is \( |A| = k \), and the determinant of its adjoint is always \( k^{n-1} \). Here, \( n=3 \), so the answer is immediately \( k^2 \).

Concept:
For a function \( f(x) \) to be continuous at a specific point \( x = a \), the limit of the function as \( x \) approaches \( a \) must exist and be equal to the value of the function at that point: \[ \lim_{x \to a} f(x) = f(a) \]

Step 1: Evaluate the limit of \( f(x) \) as \( x \to 0 \).

We need to find: \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( x^2 \sin \frac{1}{x} \right) \]
We can evaluate this limit using the Sandwich (Squeeze) Theorem. We know that the sine function is bounded between \(-1\) and \(1\) for all real arguments: \[ -1 \le \sin \frac{1}{x} \le 1 \]

Step 2: Apply the inequality bounded limits.

Multiply the entire inequality chain by the non-negative term \( x^2 \) (since \( x^2 \ge 0 \) for all real \( x \)): \[ -x^2 \le x^2 \sin \frac{1}{x} \le x^2 \]
Taking limits as \( x \to 0 \): \[ \lim_{x \to 0} (-x^2) = 0 \quad and \quad \lim_{x \to 0} (x^2) = 0 \]
Since the lower and upper bounds both approach 0, by the Squeeze Theorem: \[ \lim_{x \to 0} \left( x^2 \sin \frac{1}{x} \right) = 0 \]

Step 3: Equate the limit to the value of the function at \( x = 0 \).

From the function definition, the value at \( x = 0 \) is: \[ f(0) = k(0 + 1) = k \]
For continuity at \( x = 0 \): \[ \lim_{x \to 0} f(x) = f(0) \quad \Rightarrow \quad 0 = k \]
Thus, \( k = 0 \), which matches option (D). Quick Tip: Any function composed of \( x^n \times (bounded function) \) as \( x \to 0 \) will always have a limit of 0 as long as \( n > 0 \). Consequently, simply evaluate the other branch at 0 and set it equal to 0.

Concept:

Collinear vectors are vectors that are parallel to the same line, meaning one can be expressed as a scalar multiple of the other (\( \vec{b} = \lambda \vec{a} \)).
The dot product of two vectors is zero if and only if the vectors are perpendicular (orthogonal) to each other, or if at least one of them is a zero vector.


Step 1: Evaluate Assertion (A).

The two vectors given are \( \vec{a} \) and \( -2\vec{a} \).
Let \( \vec{b} = -2\vec{a} \). Here, \( \vec{b} \) is a scalar multiple of \( \vec{a} \) with \( \lambda = -2 \).
Since one vector is a scalar multiple of the other, they are parallel (acting along the same line in opposite directions), which makes them collinear. Therefore, Assertion (A) is true.

Step 2: Evaluate Reason (R).

Let us calculate the dot product given in the reason: \[ \vec{a} \cdot (-2\vec{a}) = -2(\vec{a} \cdot \vec{a}) = -2|\vec{a}|^2 \]
Since it is given that \( \vec{a} \neq \vec{0} \), its magnitude \( |\vec{a}| > 0 \), meaning \( -2|\vec{a}|^2 \neq 0 \).
Thus, the statement \( \vec{a} \cdot (-2\vec{a}) = 0 \) is completely false.

Conclusion:

Assertion (A) is true, and Reason (R) is false. This matches option (C). Quick Tip: Collinear vectors have a dot product of \( \pm 2|\vec{a}|^2 \), not zero. A zero dot product indicates orthogonal vectors, which are at an angle of \( 90^\circ \), making them perpendicular, not parallel/collinear.

Concept:
To find the solution of the given differential equation, we use the variable-separable method. A general solution contains an arbitrary constant \( C \), while a particular solution is obtained by assigning a specific numerical value to \( C \).

Step 1: Solve the differential equation by separating variables.

The given equation is: \[ \frac{dy}{dx} = e^{x+y} = e^x \cdot e^y \]
Separating the variables \( x \) and \( y \) onto opposite sides: \[ \frac{1}{e^y} \, dy = e^x \, dx \quad \Rightarrow \quad e^{-y} \, dy = e^x \, dx \]

Step 2: Integrate both sides.

Integrating both sides of the separated equation: \[ \int e^{-y} \, dy = \int e^x \, dx \] \[ -e^{-y} = e^x + C_1 \]
Rearranging the terms to align with the standard equation forms: \[ e^x + e^{-y} = -C_1 \]
Let \( -C_1 = C \) (where \( C \) is an arbitrary constant): \[ e^x + e^{-y} = C \]
This represents the general solution, so Reason (R) is true.

Step 3: Evaluate Assertion (A).

A particular solution is obtained by substituting a specific value for the constant \( C \). If we select \( C = -2 \), the equation becomes: \[ e^x + e^{-y} = -2 \]
This matches the assertion statement perfectly. Thus, Assertion (A) is true.

Conclusion:

Both statements are true, and the general solution equation directly explains how the particular solution is constructed. Thus, option (A) is the correct choice. Quick Tip: Always separate indices first when exponential functions contain sums: \( e^{x+y} \to e^x e^y \). This makes it easy to spot that the equation is solvable via direct variable separation.

Concept:
The expression contains nested trigonometric and inverse trigonometric functions. The most convenient approach is to evaluate the innermost expression first and then proceed outward step-by-step.

The given expression is: \[ \sin\left[\cot^{-1}\left\{\sqrt{2}\left(\cos(\tan^{-1}1)\right)\right\}\right]. \]

Step 1: Evaluate \( \tan^{-1}1 \).

We know that: \[ \tan\frac{\pi}{4}=1. \]
Therefore, \[ \tan^{-1}(1)=\frac{\pi}{4}. \]

Step 2: Evaluate \( \cos(\tan^{-1}1) \).

Substituting the value obtained above: \[ \cos(\tan^{-1}1) = \cos\frac{\pi}{4} \] \[ = \frac{1}{\sqrt{2}}. \]

Step 3: Simplify the expression inside \( \cot^{-1} \).

Multiplying by \(\sqrt{2}\): \[ \sqrt{2}\left(\cos(\tan^{-1}1)\right) = \sqrt{2}\left(\frac{1}{\sqrt{2}}\right) \] \[ =1. \]
Hence, the original expression becomes: \[ \sin(\cot^{-1}1). \]

Step 4: Evaluate \( \cot^{-1}1 \).

Since \[ \cot\frac{\pi}{4}=1, \]
we have: \[ \cot^{-1}(1)=\frac{\pi}{4}. \]
Therefore, \[ \sin(\cot^{-1}1) = \sin\frac{\pi}{4}. \]

Using \[ \sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}, \]
we obtain: \[ \boxed{\frac{1}{\sqrt{2}}}. \]

Thus, the required value is \[ \boxed{\frac{1}{\sqrt{2}}}. \] Quick Tip: For nested trigonometric expressions, evaluate the innermost function first and proceed outward. Here, \( \tan^{-1}1=\frac{\pi}{4} \), which gives \( \cos(\tan^{-1}1)=\frac{1}{\sqrt{2}} \), causing the factor \( \sqrt{2} \) to cancel immediately.

Concept:

A relation is symmetric if \( (a,b) \in R \implies (b,a) \in R \).
An equivalence relation must be reflexive, symmetric, and transitive simultaneously.


Step 1: Check for symmetry in R.

We are given \( R = \{(1, 1), (3, 3), (1, 2)\} \) on set \( A = \{1, 2, 3\} \).
Notice that the ordered pair \( (1, 2) \in R \), but its reverse pair \( (2, 1) \notin R \). Therefore, \( R \) is not symmetric.

Step 2: Identify the requirements to make \( R \cup R_1 \) an equivalence relation.

Let the equivalence relation be \( E = R \cup R_1 \).

Reflexivity Requirement: For \( E \) to be reflexive, it must contain \( (a,a) \) for every element \( a \in A \). Thus, we must have \( (1,1), (2,2), (3,3) \in E \). Since \( (1,1) \) and \( (3,3) \) are already in \( R \), we must add \( (2,2) \) via \( R_1 \).
Symmetry Requirement: Since \( (1,2) \in R \), its reverse pair \( (2,1) \) must be included in \( E \), meaning \( (2,1) \) must be added via \( R_1 \).
Transitivity Requirement: Let's check pairs with the new additions: \( \{(1,1), (2,2), (3,3), (1,2), (2,1)\} \). Here \( (1,2) \) and \( (2,1) \implies (1,1) \), and \( (2,1) \) and \( (1,2) \implies (2,2) \). The set is already transitive.


Step 3: Construct the smallest relation set \( R_1 \).

The missing essential pairs that must be provided by \( R_1 \) are: \[ R_1 = \{(2, 2), (2, 1)\} \] Quick Tip: To build an equivalence relation from a minimal set, sequentially satisfy reflexivity first, then symmetry, and lastly check transitivity to find the minimal required elements.

Concept:
For unit vectors, their magnitudes are equal to 1 (\( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \)). To eliminate the vector magnitude bars, we square both sides of the equation and expand using the vector identity \( |\vec{x}|^2 = \vec{x} \cdot \vec{x} \).

Step 1: Square both sides of the given equation.

Given: \[ |\vec{a} + 2\vec{b}| = |2\vec{a} - \vec{b}| \]
Squaring both sides: \[ |\vec{a} + 2\vec{b}|^2 = |2\vec{a} - \vec{b}|^2 \]

Step 2: Expand using the dot product formula.

Expanding both sides: \[ |\vec{a}|^2 + 4(\vec{a} \cdot \vec{b}) + 4|\vec{b}|^2 = 4|\vec{a}|^2 - 4(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \]

Step 3: Substitute the magnitudes of the unit vectors.

Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, substitute \( |\vec{a}|^2 = 1 \) and \( |\vec{b}|^2 = 1 \): \[ 1 + 4(\vec{a} \cdot \vec{b}) + 4(1) = 4(1) - 4(\vec{a} \cdot \vec{b}) + 1 \] \[ 5 + 4(\vec{a} \cdot \vec{b}) = 5 - 4(\vec{a} \cdot \vec{b}) \]

Step 4: Solve for the dot product and the angle.

Subtract 5 from both sides: \[ 4(\vec{a} \cdot \vec{b}) = -4(\vec{a} \cdot \vec{b}) \quad \Rightarrow \quad 8(\vec{a} \cdot \vec{b}) = 0 \quad \Rightarrow \quad \vec{a} \cdot \vec{b} = 0 \]
The dot product definition states \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = 0 \).
Since \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \): \[ \cos \theta = 0 \quad \Rightarrow \quad \theta = \frac{\pi}{2} or 90^\circ \] Quick Tip: Squaring vector magnitude equations is the single most common technique used to break them down into basic dot product and scalar value expansions.

Concept:
For a line in three-dimensional space written in the form \[ \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}, \]
the quantities \(a,b,c\) are its direction ratios.

If two lines have direction ratios \[ (a_1,b_1,c_1) \]
and \[ (a_2,b_2,c_2), \]
then they are perpendicular if and only if the dot product of their direction vectors is zero: \[ a_1a_2+b_1b_2+c_1c_2=0. \]

Step 1: Determine the direction ratios of the first line.

The first line is: \[ \frac{x-3}{1}=\frac{1-y}{1}=\frac{z+2}{p}. \]
The second numerator can be rewritten as: \[ 1-y=-(y-1). \]
Therefore, \[ \frac{1-y}{1}=\frac{y-1}{-1}. \]
Hence, the line can be written as: \[ \frac{x-3}{1}=\frac{y-1}{-1}=\frac{z+2}{p}. \]
Therefore, the direction ratios of the first line are: \[ (a_1,b_1,c_1)=(1,-1,p). \]

Step 2: Determine the direction ratios of the second line.

The second line is: \[ \frac{2-x}{3}=\frac{y+1}{5}=\frac{z+56}{2p}. \]
Since \[ 2-x=-(x-2), \]
we have: \[ \frac{2-x}{3}=\frac{x-2}{-3}. \]
Thus, the line becomes: \[ \frac{x-2}{-3}=\frac{y+1}{5}=\frac{z+56}{2p}. \]
Therefore, the direction ratios of the second line are: \[ (a_2,b_2,c_2)=(-3,5,2p). \]

Step 3: Apply the perpendicularity condition.

Since the two lines are perpendicular, their direction vectors must have zero dot product: \[ (1)(-3)+(-1)(5)+(p)(2p)=0. \]
Simplifying: \[ -3-5+2p^2=0 \] \[ 2p^2-8=0 \] \[ 2(p^2-4)=0 \] \[ p^2-4=0. \]
Therefore, \[ p^2=4. \]
Taking square roots: \[ p=\pm2. \]

Hence, the possible values of \(p\) are: \[ \boxed{p=\pm2}. \]

Therefore, both \(p=2\) and \(p=-2\) make the two lines perpendicular. Quick Tip: For two lines in three-dimensional space, first identify their direction ratios carefully. If the lines are perpendicular, the dot product of their direction vectors must be zero. Remember that changing the signs of both numerator and denominator of a fraction does not change its value, but it may change the signs of the direction ratios when putting the line into standard form.

Concept:
The vector equation of a straight line passing through a point with position vector \( \vec{a} \) and parallel to a direction vector \( \vec{b} \) is given by: \[ \vec{r} = \vec{a} + \kappa \vec{b} \]
Here, the line passes through the origin, so \( \vec{a} = \vec{0} \). The required line is perpendicular to two given lines, meaning its direction vector \( \vec{b} \) must be perpendicular to the direction vectors of both given lines. We can find this direction vector by calculating the cross product of the direction vectors of the two lines.

Step 1: Extract the direction vectors of the given lines.

From the line equations, their respective parallel direction vectors are: \[ \vec{b}_1 = 3\hat{i} + 4\hat{j} + 2\hat{k} \] \[ \vec{b}_2 = \hat{i} - \hat{j} + \hat{k} \]

Step 2: Compute the cross product \( \vec{b} = \vec{b}_1 \times \vec{b}_2 \).

Using the determinant method for cross products: \[ \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & 4 & 2
1 & -1 & 1 \end{vmatrix} \]
Expanding along the first row: \[ \vec{b} = \hat{i}((4)(1) - (2)(-1)) - \hat{j}((3)(1) - (2)(1)) + \hat{k}((3)(-1) - (4)(1)) \] \[ \vec{b} = \hat{i}(4 + 2) - \hat{j}(3 - 2) + \hat{k}(-3 - 4) \] \[ \vec{b} = 6\hat{i} - \hat{j} - 7\hat{k} \]

Step 3: Write down the final vector line equation.

Since the line passes through the origin \( \vec{a} = 0\hat{i} + 0\hat{j} + 0\hat{k} \), the vector equation simplifies to: \[ \vec{r} = \kappa(6\hat{i} - \hat{j} - 7\hat{k}) \]
where \( \kappa \) is a scalar parameter. Quick Tip: Whenever a line is specified as being perpendicular to two other lines simultaneously, its direction vector can always be found immediately by computing the cross product of their individual direction vectors.

Concept:
Since both \(x\) and \(y\) are given in terms of the parameter \(t\), the derivative \(dy/dx\) is obtained using parametric differentiation: \[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}. \]
We differentiate both expressions with respect to \(t\), simplify the resulting quotient, and then substitute \(t=-2\).

Step 1: Differentiate \(x\) with respect to \(t\).

Given: \[ x=e^{t+\frac{1}{t}}. \]
Using the chain rule: \[ \frac{dx}{dt} = e^{t+\frac{1}{t}} \frac{d}{dt}\left(t+\frac{1}{t}\right). \]
Since \[ \frac{d}{dt}\left(t+\frac{1}{t}\right) = 1-\frac{1}{t^2}, \]
we get: \[ \frac{dx}{dt} = e^{t+\frac{1}{t}} \left(1-\frac{1}{t^2}\right). \]

Step 2: Differentiate \(y\) with respect to \(t\).

Given: \[ y=e^{t-\frac{1}{t}}. \]
Using the chain rule: \[ \frac{dy}{dt} = e^{t-\frac{1}{t}} \frac{d}{dt}\left(t-\frac{1}{t}\right). \]
Now, \[ \frac{d}{dt}\left(t-\frac{1}{t}\right) = 1+\frac{1}{t^2}. \]
Therefore, \[ \frac{dy}{dt} = e^{t-\frac{1}{t}} \left(1+\frac{1}{t^2}\right). \]

Step 3: Find \( \frac{dy}{dx} \).

Using the parametric differentiation formula: \[ \frac{dy}{dx} = \frac{e^{t-\frac{1}{t}}\left(1+\frac{1}{t^2}\right)} {e^{t+\frac{1}{t}}\left(1-\frac{1}{t^2}\right)}. \]
Using \[ \frac{e^A}{e^B}=e^{A-B}, \]
we obtain: \[ \frac{dy}{dx} = e^{\left(t-\frac{1}{t}\right)-\left(t+\frac{1}{t}\right)} \frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}. \]
Simplifying the exponent: \[ \left(t-\frac{1}{t}\right)-\left(t+\frac{1}{t}\right) = -\frac{2}{t}. \]
Also, \[ \frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}} = \frac{\frac{t^2+1}{t^2}}{\frac{t^2-1}{t^2}} = \frac{t^2+1}{t^2-1}. \]
Hence, \[ \frac{dy}{dx} = e^{-\frac{2}{t}} \frac{t^2+1}{t^2-1}. \]

Step 4: Evaluate at \(t=-2\).

Substituting \(t=-2\): \[ e^{-\frac{2}{-2}}=e^1=e. \]
Further, \[ \frac{(-2)^2+1}{(-2)^2-1} = \frac{4+1}{4-1} = \frac{5}{3}. \]
Therefore, \[ \left.\frac{dy}{dx}\right|_{t=-2} = e\left(\frac{5}{3}\right) \] \[ \boxed{\left.\frac{dy}{dx}\right|_{t=-2}=\frac{5e}{3}}. \]

Hence, the correct answer is: \[ \boxed{\frac{5e}{3}}. \] Quick Tip: For parametric equations \(x=x(t)\) and \(y=y(t)\), always use \( \frac{dy}{dx}=\frac{dy/dt}{dx/dt} \). When exponential functions occur, simplify the quotient using \(e^A/e^B=e^{A-B}\) before substituting the value of the parameter.

Concept:
For a continuous function \( f(x) \) to be increasing on an interval, its first derivative must be strictly greater than zero (\( f'(x) > 0 \)) within that interval. We will find \( f'(x) \) using the product rule of differentiation: \[ \frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x) \]

Step 1: Compute the first derivative \( f'(x) \).

The function is given by: \[ f(x) = x^2 e^{-x} \]
Let \( u(x) = x^2 \) and \( v(x) = e^{-x} \). Differentiating both with respect to \( x \): \[ u'(x) = 2x \quad and \quad v'(x) = -e^{-x} \]
Applying the product rule: \[ f'(x) = (2x)(e^{-x}) + (x^2)(-e^{-x}) \]
Factoring out the common exponential term \( e^{-x} \) and \( x \): \[ f'(x) = e^{-x}(2x - x^2) = x e^{-x}(2 - x) \]

Step 2: Set up the inequality for an increasing function.

For \( f(x) \) to be strictly increasing, we set: \[ f'(x) > 0 \quad \Rightarrow \quad x e^{-x}(2 - x) > 0 \]

Step 3: Analyze the signs of the individual factors over the domain \( (0, \infty) \).

We are given that \( x \in (0, \infty) \), which means:

\( x > 0 \) is always positive.
\( e^{-x} > 0 \) is strictly positive for all real values of \( x \).

Since \( x \) and \( e^{-x} \) are both positive, the sign of the overall derivative depends solely on the remaining factor \( (2 - x) \): \[ 2 - x > 0 \quad \Rightarrow \quad 2 > x \quad \Rightarrow \quad x < 2 \]

Step 4: Intersect the inequality with the given domain.

Combining our result \( x < 2 \) with the specified domain constraint \( x > 0 \), we obtain the open interval: \[ 0 < x < 2 \quad \Rightarrow \quad x \in (0, 2) \]
Thus, the function is increasing on the sub-interval \( (0, 2) \). Quick Tip: Exponential terms like \( e^{-x} \) or \( e^{x} \) can never be negative or zero for any real \( x \). You can divide them out of inequalities safely without changing the inequality sign direction.

Concept:
This problem can be solved effectively by the method of substitution followed by partial fraction decomposition. Since the derivative of \( \sin x \) is \( \cos x \), choosing \( t = \sin x \) converts the trigonometric integrand into a rational algebraic function.

Step 1: Apply integration by substitution.

Let: \[ t = \sin x \]
Differentiating both sides with respect to \( x \): \[ dt = \cos x \, dx \]
Substituting these values back into the given integral: \[ I = \int \frac{\cos x \, dx}{(2 + \sin x)(4 + \sin x)} = \int \frac{dt}{(2 + t)(4 + t)} \]

Step 2: Decomposition using partial fractions.

We can split the integrand using partial fractions: \[ \frac{1}{(2 + t)(4 + t)} = \frac{A}{2 + t} + \frac{B}{4 + t} \]
Multiplying through by the common denominator \( (2 + t)(4 + t) \): \[ 1 = A(4 + t) + B(2 + t) \]
To find the coefficients \( A \) and \( B \), we substitute the roots of the linear factors:

Let \( t = -2 \): \( 1 = A(4 - 2) + B(0) \Rightarrow 1 = 2A \Rightarrow A = \frac{1}{2} \)
Let \( t = -4 \): \( 1 = A(0) + B(2 - 4) \Rightarrow 1 = -2B \Rightarrow B = -\frac{1}{2} \)

Thus, our integrand can be rewritten as: \[ \frac{1}{(2 + t)(4 + t)} = \frac{1}{2(2 + t)} - \frac{1}{2(4 + t)} \]

Step 3: Perform individual integrations.

Substitute the partial fractions back into the integral: \[ I = \int \left[ \frac{1}{2(2 + t)} - \frac{1}{2(4 + t)} \right] dt = \frac{1}{2} \log|2 + t| - \frac{1}{2} \log|4 + t| + C \]
Using logarithm properties \( \log M - \log N = \log\left(\frac{M}{N}\right) \): \[ I = \frac{1}{2} \log\left| \frac{2 + t}{4 + t} \right| + C \]

Step 4: Substitute back original variables.

Replacing \( t \) with \( \sin x \) gives the final result: \[ I = \frac{1}{2} \log\left| \frac{2 + \sin x}{4 + \sin x} \right| + C \] Quick Tip: When linear terms in the denominator differ by a constant, you can skip full partial fractions by using the difference observation: \( \frac{1}{(t+2)(t+4)} = \frac{1}{2} \left[ \frac{1}{t+2} - \frac{1}{t+4} \right] \).

Concept:
To integrate a linear expression over a quadratic expression, \( \int \frac{px + q}{ax^2 + bx + c} \, dx \), we express the numerator as a linear combination involving the derivative of the denominator plus a constant: \[ Numerator = A \cdot \frac{d}{dx}(Denominator) + B \]

Step 1: Set up the numerator decomposition.

The denominator is \( x^2 + 4x + 5 \). Its derivative is: \[ \frac{d}{dx}(x^2 + 4x + 5) = 2x + 4 \]
We express the numerator \( x + 3 \) as: \[ x + 3 = A(2x + 4) + B \]
Expanding the right-hand side: \[ x + 3 = 2Ax + (4A + B) \]

Step 2: Equate coefficients to calculate \( A \) and \( B \).

Comparing coefficients of like terms on both sides:

For \( x \): \( 1 = 2A \Rightarrow A = \frac{1}{2} \)
For constant terms: \( 3 = 4A + B \)

Substitute \( A = \frac{1}{2} \) into the constant equation: \[ 3 = 4\left(\frac{1}{2}\right) + B \quad \Rightarrow \quad 3 = 2 + B \quad \Rightarrow \quad B = 1 \]
Therefore, we rewrite the numerator as: \[ x + 3 = \frac{1}{2}(2x + 4) + 1 \]

Step 3: Split the integral into two distinct manageable parts.
\[ I = \int \frac{\frac{1}{2}(2x + 4) + 1}{x^2 + 4x + 5} \, dx = \frac{1}{2} \int \frac{2x + 4}{x^2 + 4x + 5} \, dx + \int \frac{1}{x^2 + 4x + 5} \, dx \]
Let these be \( I = \frac{1}{2}I_1 + I_2 \).

Step 4: Evaluate both integrals \( I_1 \) and \( I_2 \).

For \( I_1 \), the numerator is the exact derivative of the denominator: \[ I_1 = \int \frac{2x + 4}{x^2 + 4x + 5} \, dx = \log|x^2 + 4x + 5| \]
For \( I_2 \), complete the square in the quadratic denominator: \[ x^2 + 4x + 5 = (x + 2)^2 - 4 + 5 = (x + 2)^2 + 1 \]
Using the standard integration formula \( \int \frac{du}{u^2 + 1} = \tan^{-1}(u) \): \[ I_2 = \int \frac{dx}{(x + 2)^2 + 1} = \tan^{-1}(x + 2) \]

Step 5: Combine the final integrated components.
\[ I = \frac{1}{2}\log|x^2 + 4x + 5| + \tan^{-1}(x + 2) + C \] Quick Tip: Always look to see if simple inspection works: \( x + 3 = \frac{1}{2}(2x + 4) + 1 \). Splitting the linear numerator mentally can save you from writing down long algebraic parameter linear system setups.

Concept:
The given differential equation can be written as \( \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \). Since the total degree of every term in both the numerator and denominator is equal to 2, this is a homogeneous differential equation. We solve it using the standard substitution method: \[ y = vx \quad \Rightarrow \quad \frac{dy}{dx} = v + x\frac{dv}{dx} \]

Step 1: Rewrite the differential equation in standard derivative form.
\[ \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \]

Step 2: Substitute \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \).

Substituting these variables into our equation: \[ v + x\frac{dv}{dx} = \frac{x(vx)}{x^2 + (vx)^2} = \frac{vx^2}{x^2(1 + v^2)} \]
Canceling \( x^2 \) from numerator and denominator gives: \[ v + x\frac{dv}{dx} = \frac{v}{1 + v^2} \]

Step 3: Separate the variables \( v \) and \( x \).

Isolate the term containing \( \frac{dv}{dx} \): \[ x\frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v - v(1 + v^2)}{1 + v^2} = \frac{v - v - v^3}{1 + v^2} = \frac{-v^3}{1 + v^2} \]
Separating terms by moving all \( v \) components to the left and \( x \) components to the right: \[ \frac{1 + v^2}{v^3} \, dv = -\frac{1}{x} \, dx \]
Splitting the fraction on the left: \[ \left( \frac{1}{v^3} + \frac{1}{v} \right) dv = -\frac{1}{x} \, dx \]

Step 4: Integrate both sides of the equation.
\[ \int \left( v^{-3} + \frac{1}{v} \right) dv = -\int \frac{1}{x} \, dx \] \[ \frac{v^{-2}}{-2} + \log|v| = -\log|x| + C \] \[ -\frac{1}{2v^2} + \log|v| + \log|x| = C \]
Using logarithm properties \( \log|v| + \log|x| = \log|vx| \): \[ -\frac{1}{2v^2} + \log|vx| = C \]

Step 5: Substitute back \( v = \frac{y}{x} \) to obtain the final answer.

Since \( y = vx \), we can substitute \( vx = y \) and \( v = \frac{y}{x} \): \[ -\frac{1}{2\left(\frac{y}{x}\right)^2} + \log|y| = C \quad \Rightarrow \quad -\frac{x^2}{2y^2} + \log|y| = C \] Quick Tip: In homogeneous equations, look out for log properties during simplification. Combining \( \log|v| + \log|x| = \log|vx| \) simplifies back to \( \log|y| \) directly, saving major back-substitution algebra steps.

Concept:
The given differential equation is of the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), which describes a first-order linear differential equation. We can solve this system using the Integrating Factor method: \[ I.F. = e^{\int P(x) \, dx} \]
The general solution is then given by: \[ y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \, dx + C \]

Step 1: Identify components and compute the Integrating Factor (I.F.).

Comparing with the standard linear form: \[ P(x) = -3\cot x \quad and \quad Q(x) = \sin 2x \]
Calculate the integral of \( P(x) \): \[ \int P(x) \, dx = \int -3\cot x \, dx = -3\log|\sin x| = \log|(\sin x)^{-3}| = \log\left(\frac{1}{\sin^3 x}\right) \]
Now calculate the Integrating Factor: \[ I.F. = e^{\log\left(\frac{1}{\sin^3 x}\right)} = \frac{1}{\sin^3 x} = \csc^3 x \]

Step 2: Write out the solution format.
\[ y \cdot \frac{1}{\sin^3 x} = \int (\sin 2x) \cdot \frac{1}{\sin^3 x} \, dx \]
Expand using the double-angle identity \( \sin 2x = 2\sin x \cos x \): \[ y \cdot \frac{1}{\sin^3 x} = \int \frac{2\sin x \cos x}{\sin^3 x} \, dx = 2 \int \frac{\cos x}{\sin^2 x} \, dx \]

Step 3: Integrate the right-hand side using substitution.

For the integral \( \int \frac{\cos x}{\sin^2 x} \, dx \), substitute \( u = \sin x \), which gives \( du = \cos x \, dx \): \[ 2 \int \frac{du}{u^2} = 2 \left(-\frac{1}{u}\right) = -\frac{2}{\sin x} \]
Thus, our general solution equation is: \[ \frac{y}{\sin^3 x} = -\frac{2}{\sin x} + C \]
Multiplying through by \( \sin^3 x \): \[ y = -2\sin^2 x + C\sin^3 x \]

Step 4: Use the initial conditions to find the particular solution parameter \( C \).

We are given that \( y = 2 \) when \( x = \frac{\pi}{2} \). Substitute these boundary values: \[ 2 = -2\sin^2\left(\frac{\pi}{2}\right) + C\sin^3\left(\frac{\pi}{2}\right) \]
Since \( \sin\left(\frac{\pi}{2}\right) = 1 \): \[ 2 = -2(1)^2 + C(1)^3 \quad \Rightarrow \quad 2 = -2 + C \quad \Rightarrow \quad C = 4 \]
Substituting \( C = 4 \) back into the general equation yields our unique particular solution: \[ y = -2\sin^2 x + 4\sin^3 x \] Quick Tip: Remember the log property rule: \( e^{\log(f(x))} = f(x) \). Bring numerical coefficients inside the log as exponents first, otherwise they prevent correct cancellation of the base \( e \).

Concept:
When a variable base is raised to a variable exponent power, \( [f(x)]^{g(y)} \), the derivative cannot be computed via direct simple power rules. Instead, we must use logarithmic differentiation by taking the natural logarithm (\( \log \)) of both sides to convert exponents into standard operational products, and then differentiate implicitly.

Step 1: Take the natural logarithm on both sides.

The given equation is: \[ (\sin x)^y = y^{\cos x} \]
Taking \( \log \) on both sides: \[ \log\left((\sin x)^y\right) = \log\left(y^{\cos x}\right) \]
Using power logarithm rules \( \log(M^N) = N\log M \): \[ y \log(\sin x) = \cos x \log y \]

Step 2: Differentiate both sides with respect to \( x \) using the product rule.

Differentiating the left side: \[ \frac{d}{dx}[y \cdot \log(\sin x)] = \frac{dy}{dx} \cdot \log(\sin x) + y \cdot \frac{1}{\sin x} \cdot \cos x = \frac{dy}{dx}\log(\sin x) + y\cot x \]
Differentiating the right side: \[ \frac{d}{dx}[\cos x \cdot \log y] = (-\sin x)\log y + \cos x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = -\sin x\log y + \frac{\cos x}{y}\frac{dy}{dx} \]
Equating both differentiated results: \[ \frac{dy}{dx}\log(\sin x) + y\cot x = -\sin x\log y + \frac{\cos x}{y}\frac{dy}{dx} \]

Step 3: Group terms containing \( \frac{dy}{dx} \) together to isolate it.

Collect all terms involving \( \frac{dy}{dx} \) on the left-hand side and remaining terms on the right-hand side: \[ \frac{dy}{dx}\log(\sin x) - \frac{\cos x}{y}\frac{dy}{dx} = -\sin x\log y - y\cot x \]
Factoring out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left[ \log(\sin x) - \frac{\cos x}{y} \right] = -(\sin x\log y + y\cot x) \]

Step 4: Solve for \( \frac{dy}{dx} \) cleanly by taking a common denominator.

Simplify the expression inside the brackets: \[ \frac{dy}{dx} \left[ \frac{y\log(\sin x) - \cos x}{y} \right] = -(\sin x\log y + y\cot x) \]
Now multiply both sides by \( y \) and divide by the brackets factor: \[ \frac{dy}{dx} = \frac{-y(\sin x\log y + y\cot x)}{y\log(\sin x) - \cos x} \]
Distributing the negative sign through the denominator to make it clean: \[ \frac{dy}{dx} = \frac{y^2\cot x + y\sin x\log y}{\cos x - y\log(\sin x)} \] Quick Tip: Be extra vigilant when applying the chain rule to implicit logs like \( \log y \). Never forget to append a trailing factor of \( \frac{dy}{dx} \) due to function composition dependencies.

Concept:
Since the trials (surgeries) are independent and have a constant probability of success, this scenario follows a Binomial Distribution model, denoted by \( B(n, p) \). The probability of obtaining exactly \( r \) successes out of \( n \) independent trials is given by the formula: \[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where:

\( n = 3 \) (total number of patients/surgeries)
\( p = 90% = 0.9 \) (probability of a successful surgery)
\( q = 1 - p = 1 - 0.9 = 0.1 \) (probability of a failed surgery)


Part (i): Probability that exactly one surgery is successful (\( r = 1 \)).

Substitute \( n = 3 \), \( r = 1 \), \( p = 0.9 \), and \( q = 0.1 \) into the binomial formula: \[ P(X = 1) = \binom{3}{1} (0.9)^1 (0.1)^{3-1} \]
Since \( \binom{3}{1} = 3 \): \[ P(X = 1) = 3 \times 0.9 \times (0.1)^2 = 3 \times 0.9 \times 0.01 = 2.7 \times 0.01 = 0.027 \]
Thus, the probability that exactly one surgery is successful is \( 0.027 \) or \( \frac{27}{1000} \).

Part (ii): Probability that at most two surgeries are successful (\( X \le 2 \)).

"At most two" means the number of successful surgeries can be 0, 1, or 2. It is computationally quicker to solve this using the complement rule: \[ P(X \le 2) = 1 - P(X = 3) \]
Let's first find \( P(X = 3) \), which represents the probability that all three surgeries are successful: \[ P(X = 3) = \binom{3}{3} (0.9)^3 (0.1)^{3-3} = 1 \times 0.729 \times 1 = 0.729 \]
Now substitute this back into the complement equation: \[ P(X \le 2) = 1 - 0.729 = 0.271 \]
Thus, the probability that at most two surgeries are successful is \( 0.271 \) or \( \frac{271}{1000} \). Quick Tip: Whenever an inequality asks for "at most \( n-1 \)" or "at least 1", checking the complement side (\( 1 - P(opposite) \)) almost always circumvents adding several tedious individual binomial combinations together.

Concept:
When an integrand involves fractional powers of \( x \), we can eliminate the radical symbols completely by selecting a substitution variable power equal to the least common multiple (LCM) of the denominators of the fractional exponents. Here, the exponents are \( \frac{1}{2} \) and \( \frac{1}{3} \). The LCM of 2 and 3 is 6.

Step 1: Set up the algebraic substitution.

Let: \[ x = t^6 \quad \Rightarrow \quad dx = 6t^5 \, dt \]
Now find expressions for the terms inside the denominator: \[ x^{1/2} = (t^6)^{1/2} = t^3 \] \[ x^{1/3} = (t^6)^{1/3} = t^2 \]

Step 2: Substitute components into the integral and simplify.

Substituting these values back into the expression: \[ I = \int \frac{6t^5 \, dt}{t^3 + t^2} \]
Factor out \( t^2 \) from the denominator to cancel terms with the numerator: \[ I = \int \frac{6t^5}{t^2(t + 1)} \, dt = 6 \int \frac{t^3}{t + 1} \, dt \]

Step 3: Perform polynomial division on the integrand.

To integrate a rational expression where the numerator's degree is higher than the denominator, apply algebraic division or use a clever manipulation: \[ t^3 = (t^3 + 1) - 1 = (t + 1)(t^2 - t + 1) - 1 \]
Dividing each part by \( (t + 1) \): \[ \frac{t^3}{t + 1} = \frac{(t + 1)(t^2 - t + 1) - 1}{t + 1} = t^2 - t + 1 - \frac{1}{t + 1} \]

Step 4: Integrate the terms step-by-step.

Substitute the broken down terms back into the integral: \[ I = 6 \int \left( t^2 - t + 1 - \frac{1}{t + 1} \right) dt \] \[ I = 6 \left[ \frac{t^3}{3} - \frac{t^2}{2} + t - \log|t + 1| \right] + C \]
Distributing the constant factor 6: \[ I = 2t^3 - 3t^2 + 6t - 6\log|t + 1| + C \]

Step 5: Back-substitute original variable parameters.

Since \( x = t^6 \), we have \( t = x^{1/6} \). Let's convert high powers back:

\( t^3 = (x^{1/6})^3 = x^{1/2} \)
\( t^2 = (x^{1/6})^2 = x^{1/3} \)
\( t = x^{1/6} \)

Substituting these back yields the final solution: \[ I = 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6\log|x^{1/6} + 1| + C \] Quick Tip: Adding and subtracting 1 inside the numerator polynomial (\( t^3 \to t^3+1-1 \)) allows you to use the sum of cubes factorization formula \( a^3+b^3 = (a+b)(a^2-ab+b^2) \), avoiding long polynomial division entirely.

Concept:
Before applying integration techniques like Integration by Parts, we can simplify the inverse trigonometric expression using standard trigonometric identity substitutions or inverse tangent properties: \[ \tan^{-1}\left(\frac{A - B}{1 + AB}\right) = \tan^{-1}A - \tan^{-1}B \]

Step 1: Simplify the inverse trigonometric function expression.

Let the integrand factor be split using the property formula, where \( A = 1 \) and \( B = x \): \[ \tan^{-1}\left(\frac{1 - x}{1 + 1 \cdot x}\right) = \tan^{-1}(1) - \tan^{-1}(x) \]
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), the integral can be rewritten as: \[ I = \int \left( \frac{\pi}{4} - \tan^{-1}x \right) dx = \frac{\pi}{4}\int dx - \int \tan^{-1}x \, dx \] \[ I = \frac{\pi x}{4} - \int \tan^{-1}x \, dx \]

Step 2: Solve the remaining integral using Integration by Parts.

To compute \( I_1 = \int \tan^{-1}x \, dx \), we treat the integrand as a product with 1: \( \int (\tan^{-1}x \cdot 1) \, dx \).
Using the ILATE rule, choose:

First function \( u = \tan^{-1}x \implies du = \frac{1}{1 + x^2} \, dx \)
Second function \( v = 1 \implies \int v \, dx = x \)

Applying the Integration by Parts formula \( \int u \, dv = uv - \int v \, du \): \[ I_1 = x\tan^{-1}x - \int \frac{x}{1 + x^2} \, dx \]

Step 3: Integrate the algebraic fraction component.

For the remaining integral \( \int \frac{x}{1 + x^2} \, dx \), multiply and divide by 2 so the numerator perfectly reflects the denominator derivative: \[ \int \frac{x}{1 + x^2} \, dx = \frac{1}{2}\int \frac{2x}{1 + x^2} \, dx = \frac{1}{2}\log|1 + x^2| \]
Thus, the value of \( I_1 \) is: \[ I_1 = x\tan^{-1}x - \frac{1}{2}\log|1 + x^2| \]

Step 4: Combine parts to write the final integral solution.

Substitute \( I_1 \) back into our primary expression setup: \[ I = \frac{\pi x}{4} - \left( x\tan^{-1}x - \frac{1}{2}\log|1 + x^2| \right) + C \] \[ I = \frac{\pi x}{4} - x\tan^{-1}x + \frac{1}{2}\log|1 + x^2| + C \] Quick Tip: Alternatively, substituting \( x = \tan\theta \) directly simplifies the inner fraction to \( \tan(\frac{\pi}{4} - \theta) \), transforming the entire inverse function expression into a simple linear expression \( \frac{\pi}{4} - \theta \) right from the start.

Concept:
This definite integral can be resolved using King's Property of definite integrals, which states that: \[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]

Step 1: Write down the primary integral expression.

Let our initial given integral be represented as equation (1): \[ I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{2^x + 1} \, dx \quad --- (1) \]

Step 2: Apply King's Property to generate a companion equation.

Here, lower bound \( a = -\frac{\pi}{2} \) and upper bound \( b = \frac{\pi}{2} \). The variable substitute parameter is: \[ a + b - x = -\frac{\pi}{2} + \frac{\pi}{2} - x = -x \]
Replacing \( x \) with \( -x \) inside the integrand: \[ I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2(-x)}{2^{-x} + 1} \, dx \]
We know that \( \cos(-x) = \cos x \), so \( \cos^2(-x) = \cos^2 x \). Also rewrite the exponential index: \[ 2^{-x} = \frac{1}{2^x} \]
Substituting these simplifications back into the definite integral: \[ I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{\frac{1}{2^x} + 1} \, dx = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{\frac{1 + 2^x}{2^x}} \, dx = \int_{-\pi/2}^{\pi/2} \frac{2^x \cos^2 x}{2^x + 1} \, dx \quad --- (2) \]

Step 3: Add equations (1) and (2) together.

Since both equations have matching limits of integration, we can merge their integrands: \[ 2I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{2^x + 1} \, dx + \int_{-\pi/2}^{\pi/2} \frac{2^x \cos^2 x}{2^x + 1} \, dx \] \[ 2I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x (1 + 2^x)}{2^x + 1} \, dx \]
The matching factor \( (2^x + 1) \) cancels out cleanly from the numerator and denominator: \[ 2I = \int_{-\pi/2}^{\pi/2} \cos^2 x \, dx \]

Step 4: Integrate the simplified function using even/odd properties.

Since \( \cos^2 x \) is an even function (\( f(-x) = f(x) \)), we can change the integration baseline using the property \( \int_{-a}^{a} f(x)dx = 2\int_{0}^{a} f(x)dx \): \[ 2I = 2 \int_{0}^{\pi/2} \cos^2 x \, dx \quad \Rightarrow \quad I = \int_{0}^{\pi/2} \cos^2 x \, dx \]
Using the trigonometric identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \): \[ I = \int_{0}^{\pi/2} \left( \frac{1 + \cos 2x}{2} \right) dx = \frac{1}{2} \left[ x + \frac{\sin 2x}{2} \right]_{0}^{\pi/2} \]

Step 5: Compute boundaries to isolate the final numeric answer.

Evaluate at the upper limit \( \frac{\pi}{2} \) and lower limit 0: \[ I = \frac{1}{2} \left[ \left( \frac{\pi}{2} + \frac{\sin(\pi)}{2} \right) - \left( 0 + \frac{\sin(0)}{2} \right) \right] \]
Since \( \sin(\pi) = 0 \) and \( \sin(0) = 0 \): \[ I = \frac{1}{2} \left[ \frac{\pi}{2} \right] = \frac{\pi}{4} \] Quick Tip: An integrand containing a factor of \( \frac{1}{a^x + 1} \) alongside symmetric bounds \( [-L, L] \) is a classic indicator that applying King's property will eliminate the exponential tracking variable entirely.

Concept:
A system of linear equations can be represented in matrix form as \(AX = B\). If the matrix \(A\) is non-singular (\(|A| \neq 0\)), its unique solution can be found using the inverse matrix: \(X = A^{-1}B\), where: \[ A^{-1} = \frac{1}{|A|} adj(A) \]

Step 1: Set up variables and translate the word problem into equations.

Let the number of ₹ 1,000 vouchers given out be \(x\).

Let the number of ₹ 500 vouchers given out be \(y\).

According to the given information:

A total of 60 vouchers were given:
\[ x + y = 60 \]
The number of ₹ 1,000 vouchers (\(x\)) added to 3 times the number of ₹ 500 vouchers (\(3y\)) equals 100:
\[ x + 3y = 100 \]


Step 2: Express the equations in matrix form \(AX = B\).
\[ \begin{bmatrix} 1 & 1
1 & 3 \end{bmatrix} \begin{bmatrix} x
y \end{bmatrix} = \begin{bmatrix} 60
100 \end{bmatrix} \]
Here, matrix \(A = \begin{bmatrix} 1 & 1
1 & 3 \end{bmatrix}\), matrix of unknowns \(X = \begin{bmatrix} x
y \end{bmatrix}\), and constant column matrix \(B = \begin{bmatrix} 60
100 \end{bmatrix}\).

Step 3: Check for invertibility by calculating the determinant \(|A|\).
\[ |A| = (1 \times 3) - (1 \times 1) = 3 - 1 = 2 \]
Since \(|A| = 2 \neq 0\), the matrix \(A\) is invertible and a unique solution exists.

Step 4: Find the adjoint of matrix \(A\) and compute \(A^{-1}\).

For a \(2 \times 2\) matrix \(\begin{bmatrix} a & b
c & d \end{bmatrix}\), the adjoint is found by swapping the diagonal elements and changing the signs of the off-diagonal elements: \(\begin{bmatrix} d & -b
-c & a \end{bmatrix}\). \[ adj(A) = \begin{bmatrix} 3 & -1
-1 & 1 \end{bmatrix} \]
Thus, the inverse matrix \(A^{-1}\) is: \[ A^{-1} = \frac{1}{2} \begin{bmatrix} 3 & -1
-1 & 1 \end{bmatrix} \]

Step 5: Multiply \(A^{-1}\) by \(B\) to find vector \(X\).
\[ X = A^{-1}B = \frac{1}{2} \begin{bmatrix} 3 & -1
-1 & 1 \end{bmatrix} \begin{bmatrix} 60
100 \end{bmatrix} \]
Multiplying the row and column pairs: \[ \begin{bmatrix} x
y \end{bmatrix} = \frac{1}{2} \begin{bmatrix} (3 \times 60) + (-1 \times 100)
(-1 \times 60) + (1 \times 100) \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 180 - 100
-60 + 100 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 80
40 \end{bmatrix} = \begin{bmatrix} 40
20 \end{bmatrix} \]
Therefore, \(x = 40\) and \(y = 20\).
There were 40 vouchers of ₹ 1,000 and 20 vouchers of ₹ 500 given away. Quick Tip: To quickly verify your results in a matrix word problem, plug your numbers back into the original statements: \(40 + 20 = 60\) total vouchers, and \(40 + 3(20) = 100\). This instantly confirms your answer is completely correct.

Concept:
We are given the matrix equation \(PQ - RS = O\), where \(O\) is the null matrix. Rearranging the matrix expression gives: \[ RS = PQ \]
To isolate matrix \(S\), we pre-multiply both sides of the equation by \(R^{-1}\) (provided \(R\) is non-singular): \[ R^{-1}RS = R^{-1}PQ \implies S = R^{-1}PQ \]

Step 1: Compute the matrix product \(PQ\).
\[ PQ = \begin{bmatrix} 2 & -1
3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 2
7 & 4 \end{bmatrix} = \begin{bmatrix} (2)(5) + (-1)(7) & (2)(2) + (-1)(4)
(3)(5) + (4)(7) & (3)(2) + (4)(4) \end{bmatrix} \] \[ PQ = \begin{bmatrix} 10 - 7 & 4 - 4
15 + 28 & 6 + 16 \end{bmatrix} = \begin{bmatrix} 3 & 0
43 & 22 \end{bmatrix} \]

Step 2: Find the determinant and verify invertibility of matrix \(R\).

The matrix \(R\) is given as \(\begin{bmatrix} 2 & 5
3 & 8 \end{bmatrix}\). \[ |R| = (2 \times 8) - (5 \times 3) = 16 - 15 = 1 \]
Since \(|R| = 1 \neq 0\), the inverse matrix \(R^{-1}\) exists.

Step 3: Determine the inverse matrix \(R^{-1}\).

Using the standard inversion formula for a \(2 \times 2\) matrix: \[ R^{-1} = \frac{1}{|R|} adj(R) = \frac{1}{1} \begin{bmatrix} 8 & -5
-3 & 2 \end{bmatrix} = \begin{bmatrix} 8 & -5
-3 & 2 \end{bmatrix} \]

Step 4: Calculate matrix \(S\) by evaluating \(R^{-1}(PQ)\).
\[ S = R^{-1}PQ = \begin{bmatrix} 8 & -5
-3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 0
43 & 22 \end{bmatrix} \]
Multiplying the rows of \(R^{-1}\) by the columns of \(PQ\): \[ S = \begin{bmatrix} (8)(3) + (-5)(43) & (8)(0) + (-5)(22)
(-3)(3) + (2)(43) & (-3)(0) + (2)(22) \end{bmatrix} \] \[ S = \begin{bmatrix} 24 - 215 & 0 - 110
-9 + 86 & 0 + 44 \end{bmatrix} = \begin{bmatrix} -191 & -110
77 & 44 \end{bmatrix} \] Quick Tip: Since \(|R| = 1\), finding the inverse requires no fractional scaling. When solving equations of the form \(RS = B\), make sure to pre-multiply by \(R^{-1}\) on the left hand side, because matrix multiplication is generally non-commutative (\(R^{-1}B \neq BR^{-1}\)).

Concept:
A line in three-dimensional space can be represented in vector form as: \[ \vec r=\vec a+\lambda\vec b, \]
where \(\vec a\) is the position vector of a point on the line and \(\vec b\) is its direction vector.

To determine whether two lines intersect, we write their parametric equations and check whether there exist parameter values that give the same \(x\), \(y\), and \(z\) coordinates.

Step 1: Write line \(l_1\) in vector form.

The first line is: \[ \frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}. \]
Comparing with \[ \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}, \]
we obtain the point: \[ (-3,1,5) \]
and the direction vector: \[ (-3,1,5). \]
Therefore, the vector equation of \(l_1\) is: \[ \vec r_1 = (-3\hat i+\hat j+5\hat k) + \lambda(-3\hat i+\hat j+5\hat k). \]

The corresponding parametric equations are: \[ x=-3-3\lambda, \] \[ y=1+\lambda, \] \[ z=5+5\lambda. \]

Step 2: Write line \(l_2\) in vector form.

The second line is: \[ \frac{x+1}{-1}=\frac{2-y}{-2}=\frac{z-5}{5}. \]
For the second fraction: \[ \frac{2-y}{-2} = \frac{y-2}{2}. \]
Therefore, the line can be rewritten as: \[ \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{5}. \]
Hence, a point on \(l_2\) is: \[ (-1,2,5), \]
and its direction vector is: \[ (-1,2,5). \]
Therefore, the vector equation of \(l_2\) is: \[ \vec r_2 = (-\hat i+2\hat j+5\hat k) + \mu(-\hat i+2\hat j+5\hat k). \]

The corresponding parametric equations are: \[ x=-1-\mu, \] \[ y=2+2\mu, \] \[ z=5+5\mu. \]

Step 3: Find the condition for intersection.

For the two lines to intersect, their coordinates must be equal for some values of \(\lambda\) and \(\mu\).

From the \(x\)-coordinates: \[ -3-3\lambda=-1-\mu. \]
Rearranging: \[ 3\lambda-\mu=-2. \tag{1} \]

From the \(y\)-coordinates: \[ 1+\lambda=2+2\mu. \]
Therefore: \[ \lambda-2\mu=1. \tag{2} \]

From the \(z\)-coordinates: \[ 5+5\lambda=5+5\mu. \]
Hence: \[ \lambda=\mu. \tag{3} \]

Step 4: Solve for the parameters.

Using \[ \lambda=\mu \]
in equation (1): \[ 3\lambda-\lambda=-2 \] \[ 2\lambda=-2 \] \[ \lambda=-1. \]
Therefore, \[ \mu=-1. \]

We can verify equation (2): \[ \lambda-2\mu = -1-2(-1) = -1+2 = 1, \]
which satisfies equation (2).

Thus, the parameter values are consistent: \[ \lambda=\mu=-1. \]

Step 5: Find the point of intersection.

Substitute \(\lambda=-1\) into the parametric equations of \(l_1\): \[ x=-3-3(-1)=0, \] \[ y=1+(-1)=0, \] \[ z=5+5(-1)=0. \]
Therefore, the point of intersection is: \[ \boxed{(0,0,0)}. \]

We can also verify this point directly in both line equations.

For \(l_1\): \[ \frac{0+3}{-3}=-1,\qquad \frac{0-1}{1}=-1,\qquad \frac{0-5}{5}=-1. \]
Thus, the origin lies on \(l_1\).

For \(l_2\): \[ \frac{0+1}{-1}=-1,\qquad \frac{2-0}{-2}=-1,\qquad \frac{0-5}{5}=-1. \]
Thus, the origin also lies on \(l_2\).

Hence, the two lines intersect at the origin.
\[ \boxed{The lines are intersecting and their point of intersection is (0,0,0).} \] Quick Tip: For two lines in three dimensions, write both in parametric form and equate the corresponding \(x\), \(y\), and \(z\) coordinates. If the resulting equations have a common solution for the parameters, the lines intersect. Always substitute the obtained parameters back into all three coordinates to verify the intersection point.

Concept:

The distance between two parallel lines \(\vec{r} = \vec{a}_1 + \lambda \vec{b}\) and \(\vec{r} = \vec{a}_2 + \mu \vec{b}\) represents the side length (\(d\)) of the square:
\[ d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|} \]
Since a square has perpendicular adjacent sides, the direction vector of the first pair (\(\vec{b}\)) must be perpendicular to the direction vector of the second pair (\(\vec{c}\)), meaning their dot product is zero: \(\vec{b} \cdot \vec{c} = 0\).


Step 1: Calculate the value of \(p\) using orthogonality.

The direction vector of the given parallel lines is \(\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}\).
The direction ratios of the other pair of sides are given as \(\langle -3, 6, p \rangle\), which translates to a direction vector \(\vec{c} = -3\hat{i} + 6\hat{j} + p\hat{k}\).
Since adjacent sides of a square are perpendicular: \[ \vec{b} \cdot \vec{c} = 0 \implies (2)(-3) + (3)(6) + (6)(p) = 0 \] \[ -6 + 18 + 6p = 0 \implies 12 + 6p = 0 \implies 6p = -12 \implies p = -2 \]

Step 2: Set up parameters to find the distance between the parallel lines.

From the line equations, extract the position vectors: \[ \vec{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k}, \quad \vec{a}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k} \]
Calculate the difference vector \((\vec{a}_2 - \vec{a}_1)\): \[ \vec{a}_2 - \vec{a}_1 = (3 - 1)\hat{i} + (3 - 2)\hat{j} + (-5 - (-4))\hat{k} = 2\hat{i} + \hat{j} - \hat{k} \]

Step 3: Compute the cross product \((\vec{a}_2 - \vec{a}_1) \times \vec{b}\).
\[ (\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 1 & -1
2 & 3 & 6 \end{vmatrix} \]
Expanding along the first row: \[ = \hat{i}((1)(6) - (-1)(3)) - \hat{j}((2)(6) - (-1)(2)) + \hat{k}((2)(3) - (1)(2)) \] \[ = \hat{i}(6 + 3) - \hat{j}(12 + 2) + \hat{k}(6 - 2) = 9\hat{i} - 14\hat{j} + 4\hat{k} \]

Step 4: Determine the magnitude of the cross product and the direction vector.
\[ |(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{9^2 + (-14)^2 + 4^2} = \sqrt{81 + 196 + 16} = \sqrt{293} \] \[ |\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]
The perpendicular distance (side length \(d\)) is: \[ d = \frac{\sqrt{293}}{7} \]

Step 5: Calculate the area of the square.

The area of a square is equal to the square of its side length (\(d^2\)): \[ Area = d^2 = \left(\frac{\sqrt{293}}{7}\right)^2 = \frac{293}{49} \approx 5.98 square units \] Quick Tip: Always double check vector components during cross-product evaluations. Finding the distance between parallel lines gives you the exact side length of the square, and squaring it gives the total area directly.

Concept:

A function is one-one (injective) if \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\) for all elements in the domain.
A function is onto (surjective) if the range of the function is equal to its codomain, meaning for every element \(y\) in the codomain, there exists at least one element \(x\) in the domain such that \(f(x) = y\).


Step 1: Prove that the function is One-One.

Let \(x_1, x_2 \in \mathbf{R}_+ = [0, \infty)\) such that \(f(x_1) = f(x_2)\): \[ 4x_1^2 + 4x_1 - 5 = 4x_2^2 + 4x_2 - 5 \]
Cancel out \(-5\) from both sides: \[ 4x_1^2 + 4x_1 = 4x_2^2 + 4x_2 \]
Rearranging all terms to one side: \[ 4(x_1^2 - x_2^2) + 4(x_1 - x_2) = 0 \]
Factoring using the difference of squares identity \(a^2 - b^2 = (a-b)(a+b)\): \[ 4(x_1 - x_2)(x_1 + x_2) + 4(x_1 - x_2) = 0 \]
Factor out the common term \(4(x_1 - x_2)\): \[ 4(x_1 - x_2)(x_1 + x_2 + 1) = 0 \]
This gives two possible cases:

\(x_1 - x_2 = 0 \implies x_1 = x_2\)
\(x_1 + x_2 + 1 = 0 \implies x_1 + x_2 = -1\)

Since the domain is restricted to non-negative real numbers (\(\mathbf{R}_+ = [0, \infty)\)), \(x_1 \ge 0\) and \(x_2 \ge 0\). Therefore, their sum plus one can never be zero (\(x_1 + x_2 + 1 \ge 1\)).

Thus, the second case is impossible, leaving only \(x_1 = x_2\). This confirms that \(f(x)\) is one-one.


Step 2: Prove that the function is Onto.

Let \(y \in [-5, \infty)\) be an arbitrary element in the codomain. Set \(f(x) = y\) and solve for \(x\): \[ 4x^2 + 4x - 5 = y \implies 4x^2 + 4x - (5 + y) = 0 \]
This is a quadratic equation in terms of \(x\). Applying the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \[ x = \frac{-4 \pm \sqrt{4^2 - 4(4)(-(5 + y))}}{2(4)} = \frac{-4 \pm \sqrt{16 + 16(5 + y)}}{8} \]
Factor out 16 inside the square root radical: \[ x = \frac{-4 \pm \sqrt{16(1 + 5 + y)}}{8} = \frac{-4 \pm 4\sqrt{6 + y}}{8} = \frac{-1 \pm \sqrt{6 + y}}{2} \]
Since the domain requires \(x \in [0, \infty)\), \(x\) must be non-negative. Therefore, we must discard the negative root sign: \[ x = \frac{\sqrt{6 + y} - 1}{2} \]
Now check if this value of \(x\) remains valid inside the domain for all values of \(y \in [-5, \infty)\):
If \(y \ge -5\), then \(6 + y \ge 1 \implies \sqrt{6 + y} \ge 1 \implies \sqrt{6 + y} - 1 \ge 0\).
Thus, \(x \ge 0\), which belongs to \(\mathbf{R}_+\). Since a valid preimage \(x\) exists for every element in the codomain, the function is onto.

Step 3: Find the value of \(p \in \mathbf{R}_+\) such that \(f(p) = 3\).

Set up the equation: \[ 4p^2 + 4p - 5 = 3 \implies 4p^2 + 4p - 8 = 0 \]
Dividing the entire equation by 4: \[ p^2 + p - 2 = 0 \]
Factoring by splitting the middle term: \[ p^2 + 2p - p - 2 = 0 \implies p(p + 2) - 1(p + 2) = 0 \implies (p - 1)(p + 2) = 0 \]
This yields \(p = 1\) or \(p = -2\). Since \(p\) must belong to the non-negative domain (\(\mathbf{R}_+\)), we select \(p = 1\). Quick Tip: To complete the square easily for quadratics like \(4x^2 + 4x - 5\), rewrite it as \((2x+1)^2 - 1 - 5 = (2x+1)^2 - 6\). Setting this equal to \(y\) makes isolating \(x\) direct and bypasses the full quadratic formula setup entirely.

Concept:
In a linear programming problem, the maximum or minimum value of a linear objective function over a bounded feasible region occurs at one of the corner points of the feasible region.

Therefore, we first determine the feasible region from the given inequalities, find all its vertices, and then evaluate the objective function \[ Z=12x+18y \]
at each vertex.

Step 1: Write the boundary equations.

The given constraints are: \[ x+y\leq1200, \] \[ x-2y\geq0, \] \[ x+3y\geq600, \] \[ x\geq0,\qquad y\geq0. \]

The corresponding boundary lines are: \[ L_1:x+y=1200, \] \[ L_2:x-2y=0, \] \[ L_3:x+3y=600. \]

The conditions \(x\geq0\) and \(y\geq0\) restrict the feasible region to the first quadrant.

Step 2: Determine the relevant regions.

For \[ x+y\leq1200, \]
the feasible side is the region below the line \(x+y=1200\).

For \[ x-2y\geq0, \]
we have: \[ x\geq2y \]
or equivalently, \[ y\leq\frac{x}{2}. \]
Thus, the feasible region lies below the line \(x=2y\) when expressed as \(y=x/2\).

For \[ x+3y\geq600, \]
the feasible region lies above the line \(x+3y=600\).

Together with \[ x\geq0,\qquad y\geq0, \]
these inequalities form a bounded feasible region in the first quadrant.

Step 3: Find the intersection of \(L_2\) and \(L_3\).

From \[ x-2y=0, \]
we get: \[ x=2y. \]
Substituting this into \[ x+3y=600: \] \[ 2y+3y=600 \] \[ 5y=600 \] \[ y=120. \]
Therefore, \[ x=2(120)=240. \]
Hence, one vertex is: \[ A=(240,120). \]

Step 4: Find the intersection of \(L_1\) and \(L_2\).

Again, \[ x=2y. \]
Substituting into \[ x+y=1200: \] \[ 2y+y=1200 \] \[ 3y=1200 \] \[ y=400. \]
Thus, \[ x=2(400)=800. \]
Hence, another vertex is: \[ B=(800,400). \]

Step 5: Find the vertices on the \(x\)-axis.

Since \(y=0\) on the \(x\)-axis, the constraint \[ x+3y\geq600 \]
becomes: \[ x\geq600. \]
The constraint \[ x+y\leq1200 \]
becomes: \[ x\leq1200. \]
Also, \[ x-2y=x\geq0 \]
is satisfied for these values.

Therefore, the feasible portion of the \(x\)-axis extends from \[ (600,0) \]
to \[ (1200,0). \]
Hence, the remaining two vertices are: \[ C=(1200,0) \]
and \[ D=(600,0). \]

Thus, the four corner points of the feasible region are: \[ A=(240,120),\quad B=(800,400),\quad C=(1200,0),\quad D=(600,0). \]

Step 6: Evaluate the objective function at each corner point.

The objective function is: \[ Z=12x+18y. \]

At \(A=(240,120)\): \[ Z=12(240)+18(120) \] \[ =2880+2160 \] \[ =5040. \]

At \(B=(800,400)\): \[ Z=12(800)+18(400) \] \[ =9600+7200 \] \[ =16800. \]

At \(C=(1200,0)\): \[ Z=12(1200)+18(0) \] \[ =14400. \]

At \(D=(600,0)\): \[ Z=12(600)+18(0) \] \[ =7200. \]

The results can be summarized as:



The largest value is: \[ 16800. \]

This occurs at: \[ (x,y)=(800,400). \]

Therefore, \[ \boxed{Maximum value of Z=16800} \]
at \[ \boxed{(x,y)=(800,400)}. \] Quick Tip: In a graphical linear programming problem, after identifying the feasible region, evaluate the objective function at every corner point. The largest value gives the maximum when the problem asks to maximise \(Z\). Always verify that every selected corner point satisfies all the original inequalities.

Concept:
The total group of students consists of dropout students and regular students. Since the percentage of dropout students and regular students together is \(100%\), the probability of selecting a regular student is obtained by subtracting the percentage of dropout students from \(100%\).

Step 1: Identify the percentage of dropout students.

According to the given information, \(40%\) of the students appearing in the examination were dropouts.

Therefore, \[ P(Dropout)=\frac{40}{100}=0.40. \]

Step 2: Find the percentage of regular students.

The remaining students are regular students of class XII. Hence, \[ P(Regular) = 1-P(Dropout). \]
Therefore, \[ P(Regular) = 1-0.40 \] \[ P(Regular)=0.60. \]

Thus, the required probability is: \[ \boxed{0.60} \]
or \[ \boxed{\frac{3}{5}}. \] Quick Tip: When two categories are complementary, their probabilities add up to \(1\). Thus, if \(40%\) of students are dropouts, the probability of selecting a regular student is \(1-0.40=0.60\).

Concept:
This is a conditional probability question because the student is selected specifically from the group of dropout students. We therefore consider only the dropout students.

The given information states that \(5%\) of the dropout students qualify the examination. Hence, the probability that a dropout student does not qualify is the complement of the probability of qualifying.

Step 1: Find the probability that a dropout student qualifies.

Given: \[ P(Qualifies\midDropout)=5%=0.05. \]

Step 2: Use the complement rule.

For any event, \[ P(Does not qualify)=1-P(Qualifies). \]
Therefore, \[ P(Does not qualify\midDropout) = 1-0.05 \] \[ =0.95. \]

Hence, the required probability is: \[ \boxed{0.95} \]
or \[ \boxed{\frac{19}{20}}. \] Quick Tip: Whenever the question asks for the probability of an event not occurring, use the complement rule \(P(A')=1-P(A)\). Here, only the dropout group is considered because the student is selected from that group.

Concept:
We need to find the probability that a student is not a dropout, given that the student has qualified the examination.

Let: \[ D=event that the student is a dropout, \] \[ R=event that the student is a regular student, \]
and \[ Q=event that the student qualifies the examination. \]

We need to find: \[ P(R\mid Q). \]

Using Bayes' theorem: \[ P(R\mid Q)=\frac{P(R)P(Q\mid R)}{P(Q)}. \]

Step 1: Write the given probabilities.

Since \(40%\) of the students are dropouts: \[ P(D)=0.40. \]
Therefore, \(60%\) are regular students: \[ P(R)=0.60. \]

The probability that a dropout qualifies is: \[ P(Q\mid D)=0.05. \]

The probability that a regular student qualifies is: \[ P(Q\mid R)=0.10. \]

Step 2: Find the probability that a randomly selected student qualifies.

Using the law of total probability: \[ P(Q)=P(D)P(Q\mid D)+P(R)P(Q\mid R). \]
Substituting the values: \[ P(Q) = (0.40)(0.05)+(0.60)(0.10). \]
Therefore, \[ P(Q)=0.02+0.06 \] \[ P(Q)=0.08. \]

Step 3: Apply Bayes' theorem.

We need: \[ P(R\mid Q) = \frac{P(R)P(Q\mid R)}{P(Q)}. \]
Thus, \[ P(R\mid Q) = \frac{(0.60)(0.10)}{0.08} \] \[ = \frac{0.06}{0.08} \] \[ =\frac{6}{8} \] \[ =\frac{3}{4}. \]

Therefore, \[ \boxed{P(R\mid Q)=\frac{3}{4}}. \]

Hence, the probability that a student who qualified was not a dropout is: \[ \boxed{\frac{3}{4}}. \] Quick Tip: For a question of the form ``given that the student qualified, find the probability that the student was regular'', use Bayes' theorem. First calculate the total probability of qualification from both groups, then divide the contribution from the required group by the total probability.

Concept:
We are given that the selected student did not qualify, and we need to find the probability that the student was a regular student.

Let: \[ D=event that the student is a dropout, \] \[ R=event that the student is a regular student, \]
and \[ Q'=event that the student does not qualify. \]

We need to find: \[ P(R\mid Q'). \]

By Bayes' theorem: \[ P(R\mid Q') = \frac{P(R)P(Q'\mid R)}{P(Q')}. \]

Step 1: Find the required basic probabilities.

We have: \[ P(D)=0.40, \qquad P(R)=0.60. \]
Also, \[ P(Q\mid D)=0.05, \qquad P(Q\mid R)=0.10. \]

Therefore, the probabilities of not qualifying are: \[ P(Q'\mid D)=1-0.05=0.95 \]
and \[ P(Q'\mid R)=1-0.10=0.90. \]

Step 2: Find the probability that a randomly selected student does not qualify.

By the law of total probability: \[ P(Q') = P(D)P(Q'\mid D)+P(R)P(Q'\mid R). \]
Substituting the values: \[ P(Q') = (0.40)(0.95)+(0.60)(0.90). \]
Thus, \[ P(Q')=0.38+0.54 \] \[ P(Q')=0.92. \]

Step 3: Apply Bayes' theorem.

Therefore, \[ P(R\mid Q') = \frac{(0.60)(0.90)}{0.92}. \]
Hence, \[ P(R\mid Q') = \frac{0.54}{0.92} = \frac{54}{92} = \frac{27}{46}. \]

Therefore, \[ \boxed{P(R\mid Q')=\frac{27}{46}}. \]

Thus, the probability that a student who did not qualify was a regular student is: \[ \boxed{\frac{27}{46}}. \] Quick Tip: For conditional probability involving ``did not qualify'', first find the complementary probabilities of non-qualification. Then apply Bayes' theorem using the total probability of non-qualification.

Concept:
The equation of a straight line passing through two points \((x_1,y_1)\) and \((x_2,y_2)\) can be obtained using the two-point form: \[ \frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1}. \]

The endpoints of the boundary line \(AB\) are: \[ A(0,4),\qquad B(-2,0). \]

Step 1: Find the slope of \(AB\).

The slope is: \[ m=\frac{0-4}{-2-0} \] \[ m=\frac{-4}{-2}=2. \]

Step 2: Use the point-slope form.

Using point \(A(0,4)\): \[ y-4=2(x-0). \]
Therefore, \[ y-4=2x. \]
Rearranging: \[ 2x-y+4=0. \]

However, checking the coordinates of \(B(-2,0)\): \[ 2(-2)-0+4=0, \]
which is satisfied. Hence the equation is: \[ \boxed{2x-y+4=0}. \]

Equivalently, \[ \boxed{y=2x+4}. \] Quick Tip: To find the equation of a boundary line from two vertices, calculate its slope and use the point-slope form. Always substitute both given points into the final equation to verify the result.

Concept:
The boundary line \(AC\) passes through the two vertices: \[ A(0,4) \]
and \[ C(3,0). \]
We can find its equation using the slope formula and the point-slope form.

Step 1: Find the slope of \(AC\).

The slope is: \[ m=\frac{0-4}{3-0} \] \[ m=-\frac{4}{3}. \]

Step 2: Use the point-slope form.

Using the point \(A(0,4)\): \[ y-4=-\frac{4}{3}(x-0). \]
Thus, \[ y-4=-\frac{4x}{3}. \]
Multiplying throughout by \(3\): \[ 3y-12=-4x. \]
Rearranging: \[ 4x+3y-12=0. \]

Therefore, the equation of the boundary line \(AC\) is: \[ \boxed{4x+3y-12=0}. \]

Equivalently, \[ \boxed{y=4-\frac{4x}{3}}. \] Quick Tip: For a line joining \(A(0,4)\) and \(C(3,0)\), the slope is negative because the line decreases as \(x\) increases. A quick check is to substitute both endpoints into the final equation.

Concept:
The region \(OAC\) is bounded by the \(y\)-axis, the \(x\)-axis, and the line \(AC\). From part (ii), the equation of \(AC\) is: \[ 4x+3y-12=0. \]
Solving for \(y\): \[ y=4-\frac{4x}{3}. \]

The region extends from \(x=0\) to \(x=3\). Therefore, its area can be found by integrating the function \(y\) with respect to \(x\): \[ A=\int_0^3 y\,dx. \]

Step 1: Set up the area integral.

Using \[ y=4-\frac{4x}{3}, \]
we have: \[ A=\int_0^3\left(4-\frac{4x}{3}\right)dx. \]

Step 2: Evaluate the integral.

Integrating term by term: \[ A= \left[ 4x-\frac{4}{3}\frac{x^2}{2} \right]_0^3. \]
Therefore, \[ A= \left[ 4x-\frac{2x^2}{3} \right]_0^3. \]
Substituting the upper limit: \[ A= 4(3)-\frac{2(3)^2}{3}. \]
Hence, \[ A=12-\frac{18}{3} \] \[ A=12-6 \] \[ A=6. \]

Therefore, the area of region \(OAC\) is: \[ \boxed{6 square units}. \] Quick Tip: For area under a straight line above the \(x\)-axis, use \(A=\int_a^b y\,dx\). Here, the line \(AC\) meets the axes at \(A(0,4)\) and \(C(3,0)\), so the limits are \(0\) and \(3\).

Concept:
The region \(AOB\) is bounded by the \(y\)-axis and the line segments \(AB\) and \(BO\). The line \(AB\) has equation: \[ y=2x+4. \]
Since \(B=(-2,0)\) and \(O=(0,0)\), the region extends from \(x=-2\) to \(x=0\).

Step 1: Write the equation of \(AB\).

From part (i): \[ y=2x+4. \]

For \(x=-2\): \[ y=2(-2)+4=0, \]
and for \(x=0\): \[ y=2(0)+4=4. \]
Thus, the required region lies between \(x=-2\) and \(x=0\).

Step 2: Set up the area integral.

The area is: \[ A=\int_{-2}^{0}(2x+4)\,dx. \]

Step 3: Evaluate the integral.

We have: \[ A= \left[ x^2+4x \right]_{-2}^{0}. \]
At \(x=0\): \[ 0^2+4(0)=0. \]
At \(x=-2\): \[ (-2)^2+4(-2)=4-8=-4. \]
Therefore, \[ A=0-(-4) \] \[ A=4. \]

Hence, the area of region \(AOB\) is: \[ \boxed{4 square units}. \] Quick Tip: When the interval lies on the negative \(x\)-axis, do not change the limits unnecessarily. Simply integrate from the smaller \(x\)-value to the larger \(x\)-value, here from \(-2\) to \(0\).

Concept:
The two poles have heights \(22\) m and \(16\) m, and they stand on opposite sides of a road of width \(20\) m. The common point \(R\) lies at a distance \(x\) m from the smaller \(16\) m pole.

The ladders form right-angled triangles with the road and the poles. We therefore use the Pythagorean theorem to express \(l_1^2\) and \(l_2^2\).

Step 1: Express \(l_1^2\).

The distance from \(R\) to the smaller pole is \(x\) m and the height of the smaller pole is \(16\) m. Hence: \[ l_1^2=x^2+16^2. \]
Therefore, \[ l_1^2=x^2+256. \]

Step 2: Express \(l_2^2\).

Since the total width of the road is \(20\) m and \(R\) is \(x\) m from the smaller pole, its distance from the taller pole is: \[ 20-x. \]
The taller pole has height \(22\) m. Hence: \[ l_2^2=(20-x)^2+22^2. \]
Therefore, \[ l_2^2=(20-x)^2+484. \]

Step 3: Find \(p(x)\).

Given: \[ p(x)=l_1^2+l_2^2. \]
Thus, \[ p(x)=x^2+256+(20-x)^2+484. \]
Therefore, \[ \boxed{p(x)=x^2+(20-x)^2+740}. \]

Expanding: \[ p(x)=x^2+(400-40x+x^2)+740 \] \[ p(x)=2x^2-40x+1140. \]

Hence, an equivalent form is: \[ \boxed{p(x)=2x^2-40x+1140}. \] Quick Tip: Use the Pythagorean theorem separately for each ladder. The horizontal distances from \(R\) to the two poles are \(x\) and \(20-x\), while the vertical distances are the corresponding pole heights.

Concept:
From part (i), the function representing the sum of the squares of the ladder lengths is: \[ p(x)=2x^2-40x+1140. \]
We differentiate this function with respect to \(x\) to obtain \(p'(x)\).

Step 1: Differentiate \(p(x)\).

Given: \[ p(x)=2x^2-40x+1140. \]
Using the power rule: \[ \frac{d}{dx}(x^n)=nx^{n-1}, \]
we get: \[ p'(x)=2(2x)-40. \]
Therefore, \[ \boxed{p'(x)=4x-40}. \] Quick Tip: Once \(p(x)\) has been simplified to a polynomial, differentiate term by term. Constants disappear upon differentiation, while \(2x^2\) differentiates to \(4x\).

Concept:
To minimize the sum \[ l_1^2+l_2^2=p(x), \]
we find the critical point by setting the first derivative equal to zero. We then verify that the critical point gives a minimum.

From the previous parts: \[ p(x)=2x^2-40x+1140 \]
and \[ p'(x)=4x-40. \]

Step 1: Find the critical point.

For the minimum value: \[ p'(x)=0. \]
Therefore, \[ 4x-40=0. \]
Hence, \[ 4x=40 \] \[ x=10. \]

Step 2: Verify that the value gives a minimum.

The second derivative is: \[ p''(x)=4. \]
Since \[ p''(x)>0, \]
the function \(p(x)\) is concave upward and the critical point corresponds to a minimum.

Therefore, \[ \boxed{x=10 m}. \]

Thus, the ladders should be placed at a distance of \(10\) m from the smaller pole. Quick Tip: For a differentiable function, a critical point satisfying \(p'(x)=0\) is a minimum when \(p''(x)>0\). Here, \(p'(x)=4x-40\) gives \(x=10\), and \(p''(x)=4>0\) confirms the minimum.

Concept:
If both poles have the same height of \(16\) m and are separated by a \(20\) m wide road, let \(x\) be the distance of the common point \(R\) from either pole.

Then the horizontal distance from \(R\) to the other pole is: \[ 20-x. \]
Using the Pythagorean theorem, we can express the sum of the squares of the two ladder lengths and minimize it.

Step 1: Express the squared lengths of the ladders.

For the first pole: \[ l_1^2=x^2+16^2 \] \[ l_1^2=x^2+256. \]

For the second pole, the horizontal distance is \(20-x\): \[ l_2^2=(20-x)^2+16^2 \] \[ l_2^2=(20-x)^2+256. \]

Therefore, \[ p(x)=l_1^2+l_2^2. \]
Thus, \[ p(x)=x^2+(20-x)^2+512. \]

Step 2: Simplify the expression.

Expanding: \[ p(x) = x^2+400-40x+x^2+512 \] \[ p(x)=2x^2-40x+912. \]

Step 3: Differentiate and find the critical point.

Differentiating: \[ p'(x)=4x-40. \]
For the minimum: \[ p'(x)=0. \]
Hence, \[ 4x-40=0 \] \[ x=10. \]

Step 4: Verify the minimum.

The second derivative is: \[ p''(x)=4. \]
Since \[ p''(x)>0, \]
the value \(x=10\) gives a minimum.

Therefore, the common point \(R\) should be located: \[ \boxed{10 m} \]
from either pole.

Thus, the ladders should be kept at a distance of \(10\) m from either pole. Quick Tip: When the two poles have equal heights and are \(20\) m apart, symmetry immediately suggests that the minimum occurs when \(R\) is midway between them. Differentiation confirms this result: \(p'(x)=0\) gives \(x=10\) m.