ICSE Board Class 12 2026 Mathematics Question Paper with Solutions : Download PDF

Step 1: Understanding the Concept:

A non-singular matrix \(A\) is a square matrix whose determinant is non-zero, meaning \(\det(A) \neq 0\).

Because \(\det(A) \neq 0\), the inverse matrix \(A^{-1}\) exists.


Step 2: Key Formula or Approach:

Given the matrix equation: \[ AB = \mathbf{O} \]
where \(\mathbf{O}\) is the zero (null) matrix.


Step 3: Detailed Explanation:

Multiply both sides of the equation by \(A^{-1}\) from the left: \[ A^{-1}(AB) = A^{-1}\mathbf{O} \]
Using the associative property of matrix multiplication: \[ (A^{-1}A)B = \mathbf{O} \]
Since \(A^{-1}A = I\) (Identity matrix): \[ IB = \mathbf{O} \] \[ B = \mathbf{O} \]
Thus, \(B\) must be a null matrix.


Step 4: Final Answer:

The matrix \(B\) is a null matrix.
Quick Tip: If the product of two matrices is a zero matrix and one of them is invertible, the other matrix must be zero. This is analogous to the property \(ax = 0 \implies x = 0\) for non-zero real numbers.

Step 1: Understanding the Concept:

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

The degree is the power of the highest order derivative, provided the equation is a polynomial in derivatives.


Step 2: Key Formula or Approach:

First, differentiate the given term using the chain rule: \[ \frac{d}{dx} \left( \frac{dy}{dx} \right)^3 = 0 \]

Step 3: Detailed Explanation:

Applying the power rule and chain rule: \[ 3 \left( \frac{dy}{dx} \right)^2 \cdot \frac{d}{dx} \left( \frac{dy}{dx} \right) = 0 \] \[ 3 \left( \frac{dy}{dx} \right)^2 \cdot \frac{d^2y}{dx^2} = 0 \]
In this simplified differential equation:
1. The highest order derivative is \(\frac{d^2y}{dx^2}\), so order \(m = 2\).

2. The power of this highest derivative is 1, so degree \(n = 1\).

Calculation for \((m - n)\): \[ m - n = 2 - 1 = 1 \]

Step 4: Final Answer:

The value of \((m - n)\) is 1.
Quick Tip: Always simplify the differential equation (perform any indicated derivatives) before determining the order and degree.

Step 1: Understanding the Concept:

We use implicit differentiation since \(y\) is defined as a function of \(x\) through an equation.


Step 2: Key Formula or Approach:

Use the product rule for differentiation: \(\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}\).


Step 3: Detailed Explanation:

Given: \(xy = c^2\).

Differentiate both sides with respect to \(x\): \[ \frac{d}{dx}(xy) = \frac{d}{dx}(c^2) \]
Since \(c^2\) is a constant, its derivative is zero: \[ x \frac{dy}{dx} + y \frac{dx}{dx} = 0 \] \[ x \frac{dy}{dx} + y(1) = 0 \]
Rearranging the terms to solve for \(\frac{dy}{dx}\): \[ x \frac{dy}{dx} = -y \] \[ \frac{dy}{dx} = \frac{-y}{x} \]

Step 4: Final Answer:

The derivative is \(\frac{dy}{dx} = \frac{-y}{x}\).
Quick Tip: For equations of the form \(xy = k\), the derivative \(\frac{dy}{dx}\) is always equal to \(-\frac{y}{x}\). This is a common property of rectangular hyperbolas.

Step 1: Understanding the Concept:

A determinant property states that if a factor is common to all elements of a row or column, it can be taken out as a multiplier of the determinant.


Step 2: Key Formula or Approach:

Examine the given determinant \(\Delta\) for common factors in rows or columns.


Step 3: Detailed Explanation:

Given \(\Delta = \begin{vmatrix} 2a & 2b & 2c
2e & f & g
2i & j & k \end{vmatrix}\).

1. Observe column \(C_1\): All elements \(\{2a, 2e, 2i\}\) have a common factor of 2. Taking it out:
\[ \Delta = 2 \times \begin{vmatrix} a & 2b & 2c
e & f & g
i & j & k \end{vmatrix} \]
2. Now observe row \(R_1\) of the new determinant: The elements \(\{a, 2b, 2c\}\) do not have a common factor of 2.

Wait, let's look at the original matrix again. The screenshot shows:
\(R_1\) is \([2a, 2b, 2c]\). This has a common factor 2.
\(C_1\) is \([2a, 2e, 2i]^T\). This has a common factor 2.

By taking 2 out from \(R_1\): \[ \Delta = 2 \times \begin{vmatrix} a & b & c
2e & f & g
2i & j & k \end{vmatrix} \]
Now taking 2 out from \(C_1\) of the remaining determinant: \[ \Delta = 2 \times 2 \times \begin{vmatrix} a & b & c
e & f & g
i & j & k \end{vmatrix} = 4 \times \begin{vmatrix} a & b & c
e & f & g
i & j & k \end{vmatrix} \]
Thus, the assertion is true. The Reason provided is exactly the property used twice to reach this result.


Step 4: Final Answer:

Both Assertion and Reason are true, and the Reason correctly explains how the factor 4 is obtained.
Quick Tip: Be careful: In matrices, multiplying by a scalar affects every element. In determinants, multiplying a scalar affects only one row or one column.

Step 1: Understanding the Concept:

When we select 5 distinct numbers from a set of 18, there is only one way to arrange them in strictly increasing order. Thus, the total number of outcomes is simply the number of ways to choose 5 numbers.


Step 2: Key Formula or Approach:

Total outcomes \(S = \binom{18}{5}\).

Favorable outcomes \(E\): We fix \(x_2 = 7\) and \(x_4 = 11\).

The order is \(x_1 < 7 < x_3 < 11 < x_5\).


Step 3: Detailed Explanation:

1. For \(x_1\): It must be chosen from \(\{1, 2, 3, 4, 5, 6\}\). There are \(\binom{6}{1} = 6\) ways.

2. For \(x_3\): It must be chosen from \(\{8, 9, 10\}\). There are \(\binom{3}{1} = 3\) ways.

3. For \(x_5\): It must be chosen from \(\{12, 13, \dots, 18\}\). The number of elements is \(18 - 11 = 7\). There are \(\binom{7}{1} = 7\) ways.

Total favorable outcomes \(= 6 \times 3 \times 7 = 126\).

Total sample space \(= \binom{18}{5} = \frac{18 \times 17 \times 16 \times 15 \times 14}{5 \times 4 \times 3 \times 2 \times 1} = 8568\).

Probability \(P = \frac{126}{8568}\).

Dividing by 126: \(8568 \div 126 = 68\).
\(P = \frac{1}{68}\).


Step 4: Final Answer:

The probability is \(\frac{1}{68}\).
Quick Tip: In problems where items are "arranged in increasing order", think of it as selecting subsets. Once a subset is selected, the order is unique and fixed.

Step 1: Understanding the Concept:

The given matrix \(A\) is a diagonal matrix. A special property of diagonal matrices is that their power \(A^n\) is found by raising each diagonal element to the power \(n\).


Step 2: Key Formula or Approach:

If \(D = diag(d_1, d_2, d_3)\), then \(D^n = diag(d_1^n, d_2^n, d_3^n)\).

Alternatively, \(A = aI\), where \(I\) is the identity matrix.


Step 3: Detailed Explanation:

Given \(A = \begin{pmatrix} a & 0 & 0
0 & a & 0
0 & 0 & a \end{pmatrix} = a \begin{pmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{pmatrix} = aI\).

Then \(A^n = (aI)^n\).

Using the property of scalar multiplication and identity matrix: \[ A^n = a^n I^n \]
Since \(I^n = I\) for any positive integer \(n\): \[ A^n = a^n I = \begin{pmatrix} a^n & 0 & 0
0 & a^n & 0
0 & 0 & a^n \end{pmatrix} \]

Step 4: Final Answer:

The matrix \(A^n\) is a diagonal matrix with elements \(a^n\).
Quick Tip: Diagonal matrices behave like numbers during multiplication. Just raise the individual entries to the desired power!

Step 1: Understanding the Concept:

Differentiability implies continuity, but continuity does not necessarily imply differentiability. A function is non-differentiable at points where there is a "sharp turn" (corner) or a vertical tangent.


Step 2: Detailed Explanation:

1. Statement 1: This is a fundamental theorem of calculus. For example, \(f(x) = |x|\) is continuous at \(x = 0\) but not differentiable there because the left-hand derivative (-1) is not equal to the right-hand derivative (1). Thus, Statement 1 is true.

2. Statement 2: Graph (c) shows a V-shaped curve similar to the absolute value function. It is a continuous line with no breaks, but it has a sharp corner at the origin \((0,0)\). This sharp corner indicates that a unique tangent cannot be drawn, making it non-differentiable at that point. Thus, Statement 2 is true.


Step 3: Final Answer:

Both Statement 1 and Statement 2 are true.
Quick Tip: Graphically, "sharp corners" or "cusps" are the hallmark of non-differentiable but continuous points.

Step 1: Understanding the Concept:

We need to find the derivative of the quadratic function and then evaluate it at \(x=5\) to solve for the unknown constant \(k\).


Step 2: Key Formula or Approach:

Power rule: \(\frac{d}{dx}(x^n) = nx^{n-1}\).


Step 3: Detailed Explanation:

Given: \(f(x) = kx^2 + 7x - 4\).

Differentiating with respect to \(x\): \[ f'(x) = \frac{d}{dx}(kx^2) + \frac{d}{dx}(7x) - \frac{d}{dx}(4) \] \[ f'(x) = 2kx + 7 \]
Given that \(f'(5) = 97\), substitute \(x=5\): \[ 2k(5) + 7 = 97 \] \[ 10k + 7 = 97 \] \[ 10k = 97 - 7 \] \[ 10k = 90 \] \[ k = \frac{90}{10} = 9 \]

Step 4: Final Answer:

The value of \(k\) is 9.
Quick Tip: Linear terms in derivatives always result in constants, and constants always result in zero. Focus your calculation on the leading power term.

Step 1: Understanding the Concept:

A relation \(R\) on set \(A\) is symmetric if for every \((x, y) \in R\), then \((y, x) \in R\).

The inverse relation \(R^{-1}\) is defined as \(R^{-1} = \{ (y, x) \mid (x, y) \in R \}\).


Step 2: Detailed Explanation:

1. Statement 1: If \(R = R^{-1}\), it means that every element \((x, y)\) in \(R\) is also an element of \(R^{-1}\). By definition of \(R^{-1}\), \((x, y) \in R^{-1} \implies (y, x) \in R\). Thus, the relation is symmetric. This is a sufficient condition.

2. Statement 2: If \(R\) is symmetric, then for every \((x, y) \in R\), we must have \((y, x) \in R\). Since \(R^{-1}\) contains all flipped pairs, \(R\) being symmetric implies that \(R\) and \(R^{-1}\) contain the exact same set of ordered pairs. Thus, \(R = R^{-1}\) is a necessary condition for symmetry.


Step 3: Final Answer:

Since \(R = R^{-1}\) is the exact mathematical definition for symmetry in set notation, both statements are true.
Quick Tip: Think of \(R = R^{-1}\) as a matrix being equal to its transpose (\(M = M^T\)). Both characterize symmetry perfectly.

Step 1: Understanding the Concept:

We use the complementary angle identities for inverse trigonometric functions.


Step 3: Detailed Explanation:

1. Analyze the Reason: The identity \(\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}\) is a standard property valid for all real numbers. So, the Reason is true.

2. Analyze the Assertion:

Let \(\cot^{-1} x = \theta\). Then \(\tan^{-1} x = \frac{\pi}{2} - \theta\).

The LHS is \(\tan(\theta)\).

The RHS is \(\cot(\frac{\pi}{2} - \theta)\).

Using the trigonometric identity \(\cot(\frac{\pi}{2} - \theta) = \tan(\theta)\), we see that LHS = RHS.

Both sides simplify to \(\frac{1}{x}\) (for \(x \neq 0\)). If \(x=0\), \(\tan(\frac{\pi}{2})\) is undefined, but the equality holds in the limit or restricted sense.

Since the proof of the Assertion directly utilizes the identity mentioned in the Reason, the Reason is the correct explanation.


Step 4: Final Answer:

Option (A) is correct.
Quick Tip: For any co-function pair (sin-cos, tan-cot, sec-cosec), \(f(g^{-1}(x)) = g(f^{-1}(x))\) because of the \(\frac{\pi}{2}\) sum identity.

Step 1: Understanding the Concept:

We use the definition of conditional probability and the property of complementary events.


Step 2: Key Formula or Approach:

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

2. \(P(A' \cap B) = P(B) - P(A \cap B)\)


Step 3: Detailed Explanation:

First, find \(P(B)\) using the conditional probability formula: \[ 0.25 = \frac{0.12}{P(B)} \] \[ P(B) = \frac{0.12}{0.25} = \frac{12}{25} = 0.48 \]
Now, calculate \(P(A' \cap B)\), which represents the probability of \(B\) occurring but not \(A\): \[ P(A' \cap B) = P(B) - P(A \cap B) \] \[ P(A' \cap B) = 0.48 - 0.12 = 0.36 \]

Step 4: Final Answer:

The value is 0.36.
Quick Tip: Visualization with a Venn diagram helps: \(P(A' \cap B)\) is the region of \(B\) that does not overlap with \(A\).

Step 1: Understanding the Concept:

The matrix of cofactors \(C\) is related to the adjugate matrix by \(adj(A) = C^T\). Since \(|C| = |C^T|\), we can use the property of the determinant of an adjugate matrix.


Step 2: Key Formula or Approach:

1. \(|adj(A)| = |A|^{n-1}\) where \(n\) is the order.

2. \(|C| = |A|^{n-1}\)

3. \(|M^k| = |M|^k\)


Step 3: Detailed Explanation:

Given: \(|A| = 3\) and \(n = 3\).

The determinant of the cofactor matrix \(C\) is: \[ |C| = |A|^{n-1} = 3^{3-1} = 3^2 = 9 \]
We need to find the determinant of \(C^2\): \[ |C^2| = (|C|)^2 \] \[ |C^2| = (9)^2 = 81 \]

Step 4: Final Answer:

The value of \(|C^2|\) is 81.
Quick Tip: Remember the chain of powers: \(|A| \rightarrow |adj A| = |A|^{n-1}\). For order 3, this is just squaring the original determinant.

Step 1: Understanding the Concept:

We test for Reflexivity, Symmetry, and Transitivity on the set \(S = \{a, b, c\}\).


Step 2: Detailed Explanation:

1. Reflexive: For \(R\) to be reflexive, \((a, a), (b, b), and (c, c)\) must be in \(R\). Since \((a, a) \notin R\) and \((c, c) \notin R\), the relation is not reflexive.

2. Symmetric: For \(R\) to be symmetric, if \((x, y) \in R\), then \((y, x) \in R\). The only element is \((b, b)\), and its reverse is also \((b, b)\), which is in \(R\). Thus, it is symmetric.

3. Transitive: For \(R\) to be transitive, if \((x, y) \in R\) and \((y, z) \in R\), then \((x, z) \in R\). Here, the only pair is \((b, b)\). If we take \((b, b)\) and \((b, b)\), the result is \((b, b)\), which is in \(R\). There are no other pairs to check. Thus, it is transitive (vacuously or by triviality).


Step 3: Final Answer:

The relation is symmetric and transitive.
Quick Tip: A single pair \((x, x)\) on a multi-element set is always symmetric and transitive, but never reflexive.

Step 1: Understanding the Concept:

A function is many-to-one if there exists at least one value in the codomain that is mapped to by more than one value in the domain.


Step 2: Detailed Explanation:

From the provided graph, we can observe that a horizontal line (parallel to the x-axis) can intersect the graph of the function at more than one point.

This means there exist distinct values \(x_1, x_2 \in \mathbb{R}\) (\(x_1 \neq x_2\)) such that \(f(x_1) = f(x_2)\).

By definition, a function where multiple inputs have the same output is a many-to-one function.


Step 3: Final Answer:

The function is many-to-one because multiple \(x\)-values result in the same \(y\)-value, as evidenced by the horizontal line test.
Quick Tip: Horizontal line test: 1 intersection = One-to-one. \(> 1\) intersection = Many-to-one.

Step 1: Understanding the Concept:

We are looking for the probability that the testing process stops after exactly 2 tests.


Step 2: Key Formula or Approach:

Let \(F_1, F_2\) be faulty machines and \(G\) be the good machine.

The process stops after 2 tests only if the first two machines tested are both faulty.


Step 3: Detailed Explanation:

1. Total ways to order the tests: There are 3 machines, so there are \(3! = 6\) possible sequences of testing (e.g., \(F_1 G F_2\), \(F_1 F_2 G\), etc.).

2. Identifying favorable sequences: To identify both faulty machines in exactly 2 tests, the first two must be \(\{F_1, F_2\}\).

Possible sequences:
- \((F_1, F_2, G)\) - 2 tests (both faulty found).

- \((F_2, F_1, G)\) - 2 tests (both faulty found).

- \((F_1, G, F_2)\) - 3 tests needed.

- \((F_2, G, F_1)\) - 3 tests needed.

- \((G, F_1, F_2)\) - 3 tests needed.

- \((G, F_2, F_1)\) - 3 tests needed.

Favorable cases = 2.

Total cases = 6.

Probability \(= \frac{2}{6} = \frac{1}{3}\).


Step 4: Final Answer:

The probability is \(\frac{1}{3}\).
Quick Tip: Alternatively: Probability of first being faulty is \(\frac{2}{3}\). Given the first was faulty, probability the second is faulty is \(\frac{1}{2}\). \(\frac{2}{3} \times \frac{1}{2} = \frac{1}{3}\).

Step 1: Understanding the Concept:

The given equation is an implicit function involving exponential and logarithmic forms.

To differentiate efficiently, we can first simplify the expression by taking the natural logarithm on both sides.


Step 2: Key Formula or Approach:

1. \( \log(e^z) = z \)

2. Quotient Rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)


Step 3: Detailed Explanation:

Given: \( x = e^{\frac{x}{y}} \)

Taking natural logarithm (\( \ln \)) on both sides:
\[ \log x = \log \left( e^{\frac{x}{y}} \right) \]
\[ \log x = \frac{x}{y} \]

Rearranging to express \( y \) in terms of \( x \):
\[ y = \frac{x}{\log x} \]

Now, differentiating \( y \) with respect to \( x \) using the quotient rule:

Let \( u = x \) and \( v = \log x \).
\[ \frac{dy}{dx} = \frac{(\log x) \cdot \frac{d}{dx}(x) - (x) \cdot \frac{d}{dx}(\log x)}{(\log x)^2} \]
\[ \frac{dy}{dx} = \frac{(\log x) \cdot 1 - x \cdot \frac{1}{x}}{(\log x)^2} \]
\[ \frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2} \]

From our earlier simplification, we know \( \log x = \frac{x}{y} \).

Substituting one of the \( \log x \) terms in the numerator:
\[ \frac{dy}{dx} = \frac{\frac{x}{y} - 1}{(\log x)^2} \]
\[ \frac{dy}{dx} = \frac{\frac{x - y}{y}}{(\log x)^2} \]

Since \( y = \frac{x}{\log x} \), we can substitute \( y \) in the denominator:
\[ \frac{dy}{dx} = \frac{x - y}{y (\log x)^2} = \frac{x - y}{\left( \frac{x}{\log x} \right) (\log x)^2} \]
\[ \frac{dy}{dx} = \frac{x - y}{x \log x} \]


Step 4: Final Answer:

By differentiating the simplified expression \( y = \frac{x}{\log x} \), we obtain \( \frac{dy}{dx} = \frac{x - y}{x \log x} \).
Quick Tip: Whenever you see \( x \) in the exponent, taking logs on both sides is usually the fastest way to simplify the problem before differentiating.

Step 1: Understanding the Concept:

A function \( f(x) \) is said to be increasing on an interval if its derivative \( f'(x) \geq 0 \) for all \( x \) in that interval.


Step 2: Key Formula or Approach:

1. Differentiate \( f(x) \).

2. Set \( f'(x) \geq 0 \).

3. Use the range of the sine or cosine function (\( -1 \leq \cos x \leq 1 \)).


Step 3: Detailed Explanation:

Given: \( f(x) = b - ax + \sin x \)

Differentiating with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(b) - \frac{d}{dx}(ax) + \frac{d}{dx}(\sin x) \]
\[ f'(x) = 0 - a + \cos x \]
\[ f'(x) = \cos x - a \]

For \( f(x) \) to be increasing on \( \mathbb{R} \), we must have:
\[ f'(x) \geq 0 for all x \in \mathbb{R} \]
\[ \cos x - a \geq 0 \]
\[ \cos x \geq a \]

This inequality must hold for all values of \( x \).

We know that the minimum value of \( \cos x \) is \( -1 \).

If \( a \) is less than or equal to the minimum value of \( \cos x \), then the inequality will always be true.
\[ a \leq \min(\cos x) \]
\[ a \leq -1 \]


Step 4: Final Answer:

The function is increasing for all \( a \leq -1 \).
Quick Tip: For a function to be monotonic (always increasing or decreasing), the constant part of the derivative must lie outside the range of the trigonometric part.

Step 1: Understanding the Concept:

If a tangent is parallel to a chord, their slopes must be equal. This is essentially an application of the Mean Value Theorem.


Step 2: Key Formula or Approach:

1. Slope of chord \( m = \frac{y_2 - y_1}{x_2 - x_1} \).

2. Slope of tangent = \( \frac{dy}{dx} \).


Step 3: Detailed Explanation:

Points for the chord: \( A(2, 0) \) and \( B(4, 4) \).

Slope of chord \( AB \):
\[ m_{chord} = \frac{4 - 0}{4 - 2} = \frac{4}{2} = 2 \]

Given curve: \( y = (x - 2)^2 \).

Differentiating to find the slope of the tangent:
\[ \frac{dy}{dx} = 2(x - 2) \cdot \frac{d}{dx}(x - 2) \]
\[ \frac{dy}{dx} = 2(x - 2) \]

Since the tangent is parallel to the chord:
\[ Slope of tangent = Slope of chord \]
\[ 2(x - 2) = 2 \]
\[ x - 2 = 1 \]
\[ x = 3 \]

Substitute \( x = 3 \) back into the curve equation to find \( y \):
\[ y = (3 - 2)^2 = 1^2 = 1 \]


Step 4: Final Answer:

The point on the curve is \( (3, 1) \).
Quick Tip: Parallel lines have equal slopes. Always find the numerical slope of the line/chord first to set up your equation for \( \frac{dy}{dx} \).

Step 1: Understanding the Concept:

This is a first-order differential equation that can be solved using the variable separable method.


Step 2: Key Formula or Approach:

1. Separate \( y \) terms to one side and \( x \) terms to the other.

2. \( \int \frac{1}{y} dy = \log |y| \)

3. \( \int \cot(ax) dx = \frac{1}{a} \log |\sin(ax)| \)


Step 3: Detailed Explanation:

Given: \( \frac{dy}{dx} = y \cot 2x \)

Separating the variables:
\[ \frac{dy}{y} = \cot 2x \, dx \]

Integrating both sides:
\[ \int \frac{1}{y} dy = \int \cot 2x \, dx \]

Using the standard integration formulas:
\[ \log |y| = \frac{\log |\sin 2x|}{2} + C \]

This can be written as:
\[ \log y = \frac{1}{2} \log |\sin 2x| + C \]


Step 4: Final Answer:

The integration of the separated variables directly yields the required general solution.
Quick Tip: Always remember to divide by the coefficient of \( x \) (which is 2 in this case) when integrating trigonometric functions of the form \( f(ax) \).

Step 1: Understanding the Concept:

This problem involves integration by substitution. We notice that the derivative of \( x^6 \) is a multiple of \( x^5 \).


Step 2: Key Formula or Approach:

Let \( u = x^6 \), then \( du = 6x^5 dx \).


Step 3: Detailed Explanation:

Given integral: \( I = \int x^5 \cos(x^6) dx \)

Let \( t = x^6 \)

Differentiating both sides: \( dt = 6x^5 dx \implies x^5 dx = \frac{dt}{6} \)

Substitute these into the integral:
\[ I = \int \cos(t) \cdot \frac{dt}{6} \]
\[ I = \frac{1}{6} \int \cos(t) dt \]
\[ I = \frac{1}{6} \sin(t) + C \]

Substituting \( t = x^6 \) back:
\[ I = \frac{1}{6} \sin(x^6) + C \]

Comparing this with the given form \( k \sin(x^6) + C \):
\[ k = \frac{1}{6} \]


Step 4: Final Answer:

The value of \( k \) is \( \frac{1}{6} \).
Quick Tip: When the power of \( x \) outside a function is exactly one less than the power of \( x \) inside the function, substitution is almost always the intended method.

Step 1: Understanding the Concept:

This is a definite integral that can be solved using the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \).


Step 2: Key Formula or Approach:

Property: \( I = \int_0^a f(x) dx = \int_0^a f(a-x) dx \)


Step 3: Detailed Explanation:

Let \( I = \int_0^5 \frac{\sqrt{x}}{\sqrt{5-x} + \sqrt{x}} dx \) -- (Equation 1)

Using the property \( x \to 5-x \):
\[ I = \int_0^5 \frac{\sqrt{5-x}}{\sqrt{5-(5-x)} + \sqrt{5-x}} dx \]
\[ I = \int_0^5 \frac{\sqrt{5-x}}{\sqrt{x} + \sqrt{5-x}} dx \] -- (Equation 2)

Adding Equation 1 and Equation 2:
\[ 2I = \int_0^5 \frac{\sqrt{x}}{\sqrt{5-x} + \sqrt{x}} dx + \int_0^5 \frac{\sqrt{5-x}}{\sqrt{x} + \sqrt{5-x}} dx \]
\[ 2I = \int_0^5 \frac{\sqrt{x} + \sqrt{5-x}}{\sqrt{x} + \sqrt{5-x}} dx \]
\[ 2I = \int_0^5 1 \, dx \]
\[ 2I = [x]_0^5 = 5 - 0 = 5 \]
\[ I = \frac{5}{2} = 2.5 \]


Step 4: Final Answer:

The value of the definite integral is \( 2.5 \).
Quick Tip: For integrals of the form \( \int_0^a \frac{f(x)}{f(x) + f(a-x)} dx \), the answer is always \( \frac{a}{2} \). Here \( a = 5 \), so the answer is \( 2.5 \).

Step 1: Understanding the Concept:

The problem requires evaluating the function for two different inputs and adding them using the inverse trigonometric identity for \( \tan^{-1} A + \tan^{-1} B \).


Step 2: Key Formula or Approach:

Identity: \( \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \) (valid when \( xy < 1 \)).


Step 3: Detailed Explanation:

Mood score for User A (\( x=6 \)):
\[ f(6) = \tan^{-1} \left( \frac{6}{10} \right) = \tan^{-1}(0.6) \]

Mood score for User B (\( x=8 \)):
\[ f(8) = \tan^{-1} \left( \frac{8}{10} \right) = \tan^{-1}(0.8) \]

Combined Score \( = f(6) + f(8) = \tan^{-1}(0.6) + \tan^{-1}(0.8) \).

Since \( 0.6 \times 0.8 = 0.48 < 1 \), we use the formula:
\[ f(6) + f(8) = \tan^{-1} \left( \frac{0.6 + 0.8}{1 - (0.6 \times 0.8)} \right) \]
\[ = \tan^{-1} \left( \frac{1.4}{1 - 0.48} \right) \]
\[ = \tan^{-1} \left( \frac{1.4}{0.52} \right) \]

Multiply numerator and denominator by 100 to remove decimals:
\[ = \tan^{-1} \left( \frac{140}{52} \right) \]

Simplifying the fraction by dividing by 4:
\[ = \tan^{-1} \left( \frac{35}{13} \right) \]


Step 4: Final Answer:

The combined mood score is \( \tan^{-1} \left( \frac{35}{13} \right) \).
Quick Tip: Always check the condition \( xy < 1 \) before using the simple sum formula for \( \tan^{-1} \). If \( xy > 1 \), the formula changes to \( \pi + \tan^{-1}(\dots) \).

Step 1: Understanding the Concept:

We need to find the value(s) of \( x \) that satisfy this equation. We can use the identity \( \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \).


Step 2: Detailed Explanation:

The given equation is:
\[ \sin^{-1} x + \sin^{-1}(1 - x) = \cos^{-1} x \]

Substitute \( \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \):
\[ \sin^{-1} x + \sin^{-1}(1 - x) = \frac{\pi}{2} - \sin^{-1} x \]
\[ \sin^{-1}(1 - x) = \frac{\pi}{2} - 2\sin^{-1} x \]

Taking the sine on both sides:
\[ \sin(\sin^{-1}(1 - x)) = \sin \left( \frac{\pi}{2} - 2\sin^{-1} x \right) \]

Using the identity \( \sin(\frac{\pi}{2} - \theta) = \cos \theta \):
\[ 1 - x = \cos(2\sin^{-1} x) \]

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

The equation becomes:
\[ 1 - x = \cos(2\theta) \]

Using the double angle formula \( \cos 2\theta = 1 - 2\sin^2 \theta \):
\[ 1 - x = 1 - 2\sin^2 \theta \]
\[ 1 - x = 1 - 2x^2 \]
\[ 2x^2 - x = 0 \]
\[ x(2x - 1) = 0 \]

So, \( x = 0 \) or \( x = \frac{1}{2} \).

Testing these values:

For \( x = 0 \): \( \sin^{-1} 0 + \sin^{-1} 1 = 0 + \frac{\pi}{2} = \frac{\pi}{2} \). \( \cos^{-1} 0 = \frac{\pi}{2} \). (Matches)

For \( x = 1/2 \): \( \sin^{-1} \frac{1}{2} + \sin^{-1} \frac{1}{2} = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3} \). \( \cos^{-1} \frac{1}{2} = \frac{\pi}{3} \). (Matches)


Step 3: Final Answer:

The solutions are \( x = 0 \) and \( x = \frac{1}{2} \).
Quick Tip: Converting mixed inverse functions (sin and cos) into a single type (usually all sin or all tan) makes algebraic manipulation much easier.

Step 1: Understanding the Concept:

We need to find the first and second derivatives of the given function and substitute them into the differential equation to verify that it equals zero.


Step 2: Key Formula or Approach:

Product Rule: \( \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \).


Step 3: Detailed Explanation:

Given: \( y = (A + Bx)e^{-2x} \)

First Derivative (\( y' \)):
\[ \frac{dy}{dx} = (A + Bx) \frac{d}{dx}(e^{-2x}) + e^{-2x} \frac{d}{dx}(A + Bx) \]
\[ \frac{dy}{dx} = -2(A + Bx)e^{-2x} + Be^{-2x} \]
\[ \frac{dy}{dx} = -2y + Be^{-2x} \] --- (1)

Second Derivative (\( y'' \)):

Differentiating equation (1) with respect to \( x \):
\[ \frac{d^2y}{dx^2} = -2\frac{dy}{dx} + B(-2e^{-2x}) \]
\[ \frac{d^2y}{dx^2} = -2\frac{dy}{dx} - 2Be^{-2x} \] --- (2)

From equation (1), we can find \( Be^{-2x} \):
\[ Be^{-2x} = \frac{dy}{dx} + 2y \]

Substitute this into equation (2):
\[ \frac{d^2y}{dx^2} = -2\frac{dy}{dx} - 2 \left( \frac{dy}{dx} + 2y \right) \]
\[ \frac{d^2y}{dx^2} = -2\frac{dy}{dx} - 2\frac{dy}{dx} - 4y \]
\[ \frac{d^2y}{dx^2} = -4\frac{dy}{dx} - 4y \]

Rearranging the terms:
\[ \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 4y = 0 \]


Step 4: Final Answer:

By calculating the derivatives and substituting the terms back, we have verified the differential equation.
Quick Tip: Expressing \( \frac{dy}{dx} \) in terms of \( y \) itself (as in \( \frac{dy}{dx} = -2y + \dots \)) often makes the second differentiation much simpler and faster to organize.

Step 1: Understanding the Concept:

This is a homogeneous differential equation because the function is of the form \( \frac{dy}{dx} = g \left( \frac{y}{x} \right) \).


Step 2: Key Formula or Approach:

Substitution: Let \( y = vx \), then \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).


Step 3: Detailed Explanation:

Divide the equation by \( x \):
\[ \frac{dy}{dx} = \frac{y}{x} - \tan \left( \frac{y}{x} \right) \]

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

Substituting into the equation:
\[ v + x \frac{dv}{dx} = v - \tan v \]

The \( v \) terms cancel out:
\[ x \frac{dv}{dx} = -\tan v \]

Separating variables:
\[ \frac{dv}{\tan v} = -\frac{dx}{x} \]
\[ \cot v \, dv = -\frac{dx}{x} \]

Integrating both sides:
\[ \int \cot v \, dv = -\int \frac{1}{x} dx \]
\[ \log |\sin v| = -\log |x| + \log C \]
\[ \log |\sin v| = \log \left| \frac{C}{x} \right| \]
\[ \sin v = \frac{C}{x} \]

Substitute \( v = \frac{y}{x} \):
\[ \sin \left( \frac{y}{x} \right) = \frac{C}{x} \]


Step 4: Final Answer:

The general solution is \( x \sin \left( \frac{y}{x} \right) = C \).
Quick Tip: Any differential equation involving \( \frac{y}{x} \) inside a function like \( \tan, \sin, \exp \) is a clear signal to use the \( y = vx \) substitution.

Step 1: Understanding the Concept:

This is a linear differential equation of the first order, having the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \).


Step 2: Key Formula or Approach:

1. Integrating Factor \( (IF) = e^{\int P(x) dx} \).

2. Solution: \( y \cdot IF = \int Q(x) \cdot IF dx + C \).


Step 3: Detailed Explanation:

Divide the whole equation by \( (1 + x^2) \) to get the standard form:
\[ \frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{4x^2}{1 + x^2} \]

Here, \( P(x) = \frac{2x}{1 + x^2} \) and \( Q(x) = \frac{4x^2}{1 + x^2} \).

Calculate the Integrating Factor:
\[ IF = e^{\int \frac{2x}{1+x^2} dx} \]

Let \( 1+x^2 = t \), then \( 2x dx = dt \).
\[ IF = e^{\log(1+x^2)} = 1+x^2 \]

The solution is given by:
\[ y \cdot (1 + x^2) = \int \left( \frac{4x^2}{1 + x^2} \right) (1 + x^2) dx + C \]

The \( (1+x^2) \) terms cancel out:
\[ y(1 + x^2) = \int 4x^2 dx + C \]
\[ y(1 + x^2) = \frac{4x^3}{3} + C \]


Step 4: Final Answer:

The solution is \( y(1 + x^2) = \frac{4x^3}{3} + C \).
Quick Tip: If the term being integrated into the IF is \( \frac{f'(x)}{f(x)} \), the IF will simply be \( f(x) \). This saves significant computation time.

Step 1: Understanding the Concept:

When three coins are tossed, there are \( 2^3 = 8 \) possible outcomes.


Step 2: Detailed Explanation:

The sample space \( S \) is:
\( S = \{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT\} \).

Total outcomes \( n(S) = 8 \).

The event \( E \) is getting the "same result" for all three:
\( E = \{HHH, TTT\} \).

Number of favorable outcomes \( n(E) = 2 \).

Probability \( P(E) = \frac{n(E)}{n(S)} = \frac{2}{8} = \frac{1}{4} \).


Step 3: Final Answer:

The probability is \( \frac{1}{4} \) or \( 0.25 \).
Quick Tip: For any number of coins \( n \), there are always only 2 ways to get all the same result (all H or all T). The probability is thus \( \frac{2}{2^n} \).

Step 1: Understanding the Concept:

"Getting a different result" means the complement of "all three getting the same result". This signifies that the game ends.


Step 2: Detailed Explanation:

From the previous question, we know the probability of getting the same result is \( P(Same) = \frac{1}{4} \).

A "different result" happens if the game does not require a re-toss.

Using the rule of complementary events:
\[ P(Different) = 1 - P(Same) \]
\[ P(Different) = 1 - \frac{1}{4} = \frac{3}{4} \]


Step 3: Final Answer:

The probability of getting a different result (and thus determining a payer) is \( \frac{3}{4} \).
Quick Tip: In probability, if you have calculated the probability of an event, its complement is always \( 1 - P \). This is much faster than listing all remaining favorable cases.

Step 1: Understanding the Concept:

To need exactly four rounds, the first three rounds must result in "same results" (failure to determine payer) and the fourth round must result in a "different result" (success).


Step 2: Key Formula or Approach:

This follows a geometric distribution where success occurs on the \( k^{th} \) trial.
\( P(X=k) = q^{k-1} p \).


Step 3: Detailed Explanation:

Let \( p \) be the probability of "success" (determining a payer) = \( \frac{3}{4} \).

Let \( q \) be the probability of "failure" (all same results) = \( \frac{1}{4} \).

We need the sequence: Failure, Failure, Failure, Success.
\[ P(Exactly 4 rounds) = q \times q \times q \times p \]
\[ = \left( \frac{1}{4} \right)^3 \times \left( \frac{3}{4} \right) \]
\[ = \frac{1}{64} \times \frac{3}{4} = \frac{3}{256} \]


Step 4: Final Answer:

The probability is \( \frac{3}{256} \).
Quick Tip: "Exactly \( n \) rounds" implies that the first \( n-1 \) rounds must continue the process and the \( n^{th} \) round must end it.

Step 1: Understanding the Concept:

We define the categorical choice of mode as the base events and the rating as the conditional event.


Step 2: Detailed Explanation:

Define events:

- \( A \): Student chooses Mode A (Offline).

- \( B \): Student chooses Mode B (Online).

- \( C \): Student chooses Mode C (Recorded).

- \( E \): Student rates the class as "Excellent".

Based on the percentages given:

- \( P(A) = 40% = \frac{40}{100} = 0.40 \)

- \( P(B) = 35% = \frac{35}{100} = 0.35 \)

- \( P(C) = 25% = \frac{25}{100} = 0.25 \)

Conditional probabilities for rating "Excellent":

- \( P(E|A) = 20% = 0.20 \)

- \( P(E|B) = 30% = 0.30 \)

- \( P(E|C) = 50% = 0.50 \)


Step 3: Final Answer:

The probabilities are \( P(A)=0.4, P(B)=0.35, P(C)=0.25 \) and the conditional probabilities are \( P(E|A)=0.2, P(E|B)=0.3, P(E|C)=0.5 \).
Quick Tip: Ensure the base probabilities (\( P(A), P(B), P(C) \)) sum to 1. Here, \( 0.4+0.35+0.25 = 1.0 \).

Step 1: Understanding the Concept:

Bayes' Theorem allows us to find the probability of a cause given an effect. Here, the "effect" is the rating "Excellent", and the "cause" is Mode C.


Step 2: Key Formula or Approach:
\[ P(C|E) = \frac{P(C) \cdot P(E|C)}{P(A) \cdot P(E|A) + P(B) \cdot P(E|B) + P(C) \cdot P(E|C)} \]


Step 3: Detailed Explanation:

Numerator: \( P(C) \cdot P(E|C) = 0.25 \times 0.50 = 0.125 \)

Denominator (Total probability of "Excellent"):
\[ P(E) = (0.40 \times 0.20) + (0.35 \times 0.30) + (0.25 \times 0.50) \]
\[ P(E) = 0.08 + 0.105 + 0.125 \]
\[ P(E) = 0.310 \]

Now apply Bayes' Theorem:
\[ P(C|E) = \frac{0.125}{0.310} \]

Multiply numerator and denominator by 1000:
\[ P(C|E) = \frac{125}{310} \]

Divide by 5:
\[ P(C|E) = \frac{25}{62} \]


Step 4: Final Answer:

The probability is \( \frac{25}{62} \).
Quick Tip: The denominator in Bayes' Theorem is just the total probability of the condition being observed. Always calculate the individual terms for the denominator first.

Step 1: Understanding the Concept:

To find which mode is most likely, we compare the posterior probabilities \( P(A|E), P(B|E), \) and \( P(C|E) \).


Step 2: Detailed Explanation:

We already found \( P(C|E) = \frac{0.125}{0.310} \approx 0.403 \).

Now let's look at the other components of the total probability:

- Contribution from Mode A: \( 0.08 / 0.31 \approx 0.258 \)

- Contribution from Mode B: \( 0.105 / 0.31 \approx 0.339 \)

- Contribution from Mode C: \( 0.125 / 0.31 \approx 0.403 \)

Since \( 0.125 > 0.105 > 0.08 \), Mode C has the highest numerator in the Bayes calculation.

Therefore, if a student says the class was "Excellent", it is most likely that they were from Mode C.


Step 3: Final Answer:

Mode C (Recorded lectures) has the highest likelihood because it contributes the most to the "Excellent" pool.
Quick Tip: Interpretation usually involves comparing the calculated probabilities. The "Winner" is the category that has the largest \( P(Category) \times P(Condition|Category) \).

Step 1: Understanding the Concept:

A system of linear equations is used to represent relationships between multiple variables.

In this scenario, the variables represent the unit costs of the items sold.


Step 2: Detailed Explanation:

Let the cost of one paper bag be \( x \) rupees.

Let the cost of one scrap book be \( y \) rupees.

Let the cost of one pastel sheet be \( z \) rupees.



For School A:

They sold 30 paper bags, 20 scrap books, and 10 pastel sheets for a total of ₹ 410.

Equation: \( 30x + 20y + 10z = 410 \)

Dividing by 10, we get: \( 3x + 2y + z = 41 \)



For School B:

They sold 20 paper bags, 10 scrap books, and 20 pastel sheets for a total of ₹ 290.

Equation: \( 20x + 10y + 20z = 290 \)

Dividing by 10, we get: \( 2x + y + 2z = 29 \)



For School C:

They sold 20 paper bags, 20 scrap books, and 20 pastel sheets for a total of ₹ 440.

Equation: \( 20x + 20y + 20z = 440 \)

Dividing by 20, we get: \( x + y + z = 22 \)


Step 3: Final Answer:

The system of equations is:
\( 3x + 2y + z = 41 \)
\( 2x + y + 2z = 29 \)
\( x + y + z = 22 \)
Quick Tip: Always simplify the equations by dividing with common factors (like 10 or 20 here) to make the matrix calculations easier.

Step 1: Understanding the Concept:

To solve \( AX = B \) using the matrix method, we find \( X = A^{-1}B \), where \( A^{-1} = \frac{1}{|A|} adj(A) \).


Step 2: Key Formula or Approach:
\( A = \begin{bmatrix} 3 & 2 & 1
2 & 1 & 2
1 & 1 & 1 \end{bmatrix}, X = \begin{bmatrix} x
y
z \end{bmatrix}, B = \begin{bmatrix} 41
29
22 \end{bmatrix} \)


Step 3: Detailed Explanation:

Calculate the determinant of \( A \):
\[ |A| = 3(1-2) - 2(2-2) + 1(2-1) \]
\[ |A| = 3(-1) - 2(0) + 1(1) = -3 + 1 = -2 \]

Since \( |A| \neq 0 \), the unique solution exists.

Now, find the matrix of cofactors:
\( C_{11} = -1, C_{12} = 0, C_{13} = 1 \)
\( C_{21} = -(2-1) = -1, C_{22} = (3-1) = 2, C_{23} = -(3-2) = -1 \)
\( C_{31} = (4-1) = 3, C_{32} = -(6-2) = -4, C_{33} = (3-4) = -1 \)

Adjugate of \( A \):
\[ adj(A) = \begin{bmatrix} -1 & -1 & 3
0 & 2 & -4
1 & -1 & -1 \end{bmatrix} \]

Solve \( X = A^{-1}B = \frac{1}{-2} \begin{bmatrix} -1 & -1 & 3
0 & 2 & -4
1 & -1 & -1 \end{bmatrix} \begin{bmatrix} 41
29
22 \end{bmatrix} \]
\[ X = -\frac{1}{2} \begin{bmatrix} -41-29+66
0+58-88
41-29-22 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} -4
-30
-10 \end{bmatrix} = \begin{bmatrix} 2
15
5 \end{bmatrix} \]
\textbf{Step 4: Final Answer:}
The solution is \( x = 2, y = 15, z = 5 \).
Quick Tip: Double check your cofactor signs using the checkerboard pattern \( (+ - +), (- + -), (+ - +) \) to avoid common calculation errors.

Step 1: Understanding the Concept:

We interpret the algebraic values obtained from the matrix solution back into the physical context of the problem.


Step 2: Detailed Explanation:

From the solution of the matrix equation in the previous part:
\( x = 2 \)
\( y = 15 \)
\( z = 5 \)

According to our initial assumptions:

The cost of one paper bag is represented by \( x \).

The cost of one scrap book is represented by \( y \).

The cost of one pastel sheet is represented by \( z \).


Step 3: Final Answer:

The cost of one paper bag is ₹ 2, the cost of one scrap book is ₹ 15, and the cost of one pastel sheet is ₹ 5.
Quick Tip: Always write the final units (like ₹ or units) to ensure you receive full marks for application-based questions.

Step 1: Understanding the Concept:

The integrand is a proper rational function, so we can decompose it into partial fractions.


Step 2: Key Formula or Approach:
\[ \frac{x^2+x+1}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1} \]


Step 3: Detailed Explanation:

Multiply by denominator: \( x^2+x+1 = A(x^2+1) + (Bx+C)(x+2) \)

Put \( x = -2 \):
\( (-2)^2 + (-2) + 1 = A((-2)^2 + 1) \implies 4-2+1 = 5A \implies 3 = 5A \implies A = 3/5 \).

Comparing coefficients of \( x^2 \):
\( 1 = A + B \implies 1 = 3/5 + B \implies B = 2/5 \).

Comparing constant terms:
\( 1 = A + 2C \implies 1 = 3/5 + 2C \implies 2/5 = 2C \implies C = 1/5 \).

Now integrate:
\[ I = \int \left( \frac{3/5}{x+2} + \frac{2x/5 + 1/5}{x^2+1} \right) dx \]
\[ I = \frac{3}{5} \int \frac{1}{x+2} dx + \frac{1}{5} \int \frac{2x}{x^2+1} dx + \frac{1}{5} \int \frac{1}{x^2+1} dx \]
\[ I = \frac{3}{5}\log|x+2| + \frac{1}{5}\log|x^2+1| + \frac{1}{5}\tan^{-1}x + C \]


Step 4: Final Answer:

The integral is \( \frac{3}{5}\log|x+2| + \frac{1}{5}\log|x^2+1| + \frac{1}{5}\tan^{-1}x + C \).
Quick Tip: For terms like \( \frac{Bx+C}{x^2+1} \), split them into two parts: one that fits the \( \int \frac{f'(x)}{f(x)} \) form and one that fits the \( \tan^{-1} \) form.

Step 1: Understanding the Concept:

Use the property of definite integrals: \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \).


Step 2: Detailed Explanation:

Let \( I = \int_{\pi/4}^{3\pi/4} \frac{x}{1+\sin x} dx \) -- (1)

Since \( a+b = \pi/4 + 3\pi/4 = \pi \), replace \( x \) with \( \pi-x \):
\( I = \int_{\pi/4}^{3\pi/4} \frac{\pi-x}{1+\sin(\pi-x)} dx = \int_{\pi/4}^{3\pi/4} \frac{\pi-x}{1+\sin x} dx \) -- (2)

Adding (1) and (2):
\[ 2I = \int_{\pi/4}^{3\pi/4} \frac{\pi}{1+\sin x} dx = \pi \int_{\pi/4}^{3\pi/4} \frac{1-\sin x}{\cos^2 x} dx \]
\[ 2I = \pi \int_{\pi/4}^{3\pi/4} (\sec^2 x - \sec x \tan x) dx \]
\[ 2I = \pi [\tan x - \sec x]_{\pi/4}^{3\pi/4} \]
\[ 2I = \pi [(\tan(3\pi/4) - \sec(3\pi/4)) - (\tan(\pi/4) - \sec(\pi/4))] \]
\[ 2I = \pi [(-1 - (-\sqrt{2})) - (1 - \sqrt{2})] \]
\[ 2I = \pi [-1 + \sqrt{2} - 1 + \sqrt{2}] = \pi [2\sqrt{2} - 2] \]
\[ I = \pi (\sqrt{2} - 1) \]


Step 3: Final Answer:

The value of the definite integral is \( \pi(\sqrt{2} - 1) \).
Quick Tip: Using the "King's Property" \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \) is the standard way to eliminate \( x \) from the numerator of the integrand.

Step 1: Understanding the Concept:

The volume \( V \) of a right circular cylinder is defined by the product of the base area and the height.


Step 2: Key Formula or Approach:
\( V = \pi r^2 h \)


Step 3: Detailed Explanation:

Given volume \( V = \frac{539}{2} \) cubic units.

Setting the formula equal to the value:
\[ \pi r^2 h = \frac{539}{2} \]

To isolate \( h \), divide both sides by \( \pi r^2 \):
\[ h = \frac{539}{2\pi r^2} \]


Step 4: Final Answer:

The height \( h \) is \( \frac{539}{2\pi r^2} \).
Quick Tip: In optimization problems, always start by expressing one variable in terms of another using the given constant parameter (like volume).

Step 1: Understanding the Concept:

The total surface area of a closed cylinder includes the lateral surface area and the areas of the two circular bases.


Step 2: Key Formula or Approach:
\( S = 2\pi r^2 + 2\pi r h \)


Step 3: Detailed Explanation:

Substitute the expression for \( h \) from part (a) into the formula for \( S \):
\[ S = 2\pi r^2 + 2\pi r \left( \frac{539}{2\pi r^2} \right) \]

Simplifying the second term:

The \( 2 \) cancels out, \( \pi \) cancels out, and one \( r \) cancels out.
\[ S = 2\pi r^2 + \frac{539}{r} \]


Step 4: Final Answer:

The surface area in terms of \( r \) is \( S = 2\pi r^2 + \frac{539}{r} \).
Quick Tip: Be careful with the term "closed" cylinder; it means you must include both circular caps (\( 2\pi r^2 \)).

Step 1: Understanding the Concept:

To find the minimum value of a function, we take its derivative and set it to zero (First Derivative Test).


Step 2: Detailed Explanation:

From part (b), \( S = 2\pi r^2 + 539r^{-1} \).

Differentiate \( S \) with respect to \( r \):
\[ \frac{dS}{dr} = 4\pi r - \frac{539}{r^2} \]

For minimum surface area, \( \frac{dS}{dr} = 0 \):
\[ 4\pi r - \frac{539}{r^2} = 0 \implies 4\pi r = \frac{539}{r^2} \implies r^3 = \frac{539}{4\pi} \]

Using \( \pi = \frac{22}{7} \):
\[ r^3 = \frac{539 \times 7}{4 \times 22} = \frac{49 \times 11 \times 7}{4 \times 2 \times 11} = \frac{343}{8} \]

Taking cube root on both sides:
\[ r = \sqrt[3]{\frac{343}{8}} = \frac{7}{2} \]

Checking second derivative for minima:
\[ \frac{d^2S}{dr^2} = 4\pi + \frac{2 \times 539}{r^3} \]

Since \( r > 0 \), \( \frac{d^2S}{dr^2} > 0 \), confirming that \( r = 7/2 \) is a minimum.


Step 3: Final Answer:

The radius for minimum surface area is \( \frac{7}{2} \) units.
Quick Tip: Always perform the second derivative test \( (\frac{d^2S}{dr^2} > 0) \) in your exam to confirm that the extremum is indeed a minimum.

Step 1: Understanding the Concept:

We use the value of radius found for minimum surface area to calculate the corresponding height.


Step 2: Detailed Explanation:

We know \( h = \frac{539}{2\pi r^2} \) and \( r = \frac{7}{2} \).

Substitute \( r = 7/2 \) and \( \pi = 22/7 \):
\[ h = \frac{539}{2 \times \frac{22}{7} \times (\frac{7}{2})^2} = \frac{539}{2 \times \frac{22}{7} \times \frac{49}{4}} \]
\[ h = \frac{539}{\frac{11 \times 49}{7}} = \frac{539}{11 \times 7} = \frac{539}{77} = 7 \]


Step 3: Final Answer:

The height of the tank is 7 units.
Quick Tip: For a closed cylinder of fixed volume, the surface area is minimized when the height is equal to the diameter (\( h = 2r \)). Here, \( 2 \times 3.5 = 7 \), which matches.

Step 1: Understanding the Concept:

Differentiability refers to whether a unique tangent (and thus a derivative) exists at every point in the domain.


Step 2: Detailed Explanation:

The given function \( h(t) = \frac{1}{2}(-7t^2 + 3t + 2) \) is a polynomial function of degree 2 (a quadratic function).

A fundamental theorem in calculus states that all polynomial functions are continuous and differentiable at every point on the real number line (\( \mathbb{R} \)).

Since the domain \( t \geq 0 \) is a subset of \( \mathbb{R} \), the function \( h(t) \) is differentiable for all \( t \geq 0 \).


Step 3: Final Answer:

Yes, the function is differentiable because it is a polynomial.
Quick Tip: Any function of the form \( at^2 + bt + c \) has a derivative \( 2at + b \), which is defined for all real values.

Step 1: Understanding the Concept:

The instantaneous rate of change of height with respect to time is the first derivative \( h'(t) \).


Step 2: Detailed Explanation:

Given: \( h(t) = -\frac{7}{2}t^2 + \frac{3}{2}t + 1 \)

Differentiating with respect to \( t \):
\[ h'(t) = -\frac{7}{2}(2t) + \frac{3}{2}(1) + 0 \]
\[ h'(t) = -7t + \frac{3}{2} \]

At \( t = \frac{1}{14} \):
\[ h'(1/14) = -7 \left( \frac{1}{14} \right) + \frac{3}{2} \]
\[ h'(1/14) = -\frac{1}{2} + \frac{3}{2} = \frac{2}{2} = 1 \]


Step 3: Final Answer:

The instantaneous rate of change at \( t = 1/14 \) is 1.
Quick Tip: Rate of change is just another word for derivative. Always calculate the general derivative first before plugging in the specific value of \( t \).

Step 1: Understanding the Concept:

A function is increasing when its first derivative is positive (\( h'(t) > 0 \)).


Step 2: Detailed Explanation:

From the previous part, the derivative is \( h'(t) = -7t + \frac{3}{2} \).

Set \( h'(t) > 0 \):
\[ -7t + \frac{3}{2} > 0 \]
\[ \frac{3}{2} > 7t \implies t < \frac{3}{14} \]

This shows that for any value of \( t \) less than \( \frac{3}{14} \), the derivative is positive, meaning the height is increasing.

Note: Although the physical problem starts at \( t=0 \), mathematically the function \( h(t) \) is increasing on the interval \( (-\infty, \frac{3}{14}) \).


Step 3: Final Answer:

The statement is true because \( h'(t) > 0 \) for all \( t < \frac{3}{14} \).
Quick Tip: To find where a quadratic \( at^2+bt+c \) increases, find the vertex at \( t = -b/(2a) \). If \( a < 0 \), it increases on \( (-\infty, -b/2a) \).

Step 1: Understanding the Concept:

Maximum height is reached when the rate of change is zero (\( h'(t) = 0 \)).


Step 2: Detailed Explanation:

Set \( h'(t) = 0 \):
\[ -7t + \frac{3}{2} = 0 \implies 7t = \frac{3}{2} \implies t = \frac{3}{14} \]

Now, find the maximum height by substituting \( t = 3/14 \) into \( h(t) \):
\[ h(3/14) = \frac{1}{2} \left[ -7 \left( \frac{3}{14} \right)^2 + 3 \left( \frac{3}{14} \right) + 2 \right] \]
\[ h(3/14) = \frac{1}{2} \left[ -7 \left( \frac{9}{196} \right) + \frac{9}{14} + 2 \right] \]
\[ h(3/14) = \frac{1}{2} \left[ -\frac{9}{28} + \frac{18}{28} + \frac{56}{28} \right] \]
\[ h(3/14) = \frac{1}{2} \left[ \frac{65}{28} \right] = \frac{65}{56} \approx 1.16 \]


Step 3: Final Answer:

The Dolphin reaches max height at \( t = 3/14 \) and the max height is \( \frac{65}{56} \).
Quick Tip: For a quadratic \( f(x) = ax^2+bx+c \), the extremum always occurs at \( x = -b/2a \).

Step 1: Understanding the Concept:

The events (giving tests) are independent. We use the product rule of probability.

Let \( E, M, S \) be the events of tests being given.
\( P(E)=0.9, P(E')=0.1 \)
\( P(M)=0.8, P(M')=0.2 \)
\( P(S)=0.7, P(S')=0.3 \)


Step 2: Detailed Explanation:
\( \bullet \textbf{ X = 0 } \): No tests given.
\( P(X=0) = P(E' \cap M' \cap S') = 0.1 \times 0.2 \times 0.3 = 0.006 \).


\( \bullet \textbf{ X = 1 } \): Exactly one test given.
\( P(X=1) = (P(E)P(M')P(S')) + (P(E')P(M)P(S')) + (P(E')P(M')P(S)) \)
\( = (0.9 \times 0.2 \times 0.3) + (0.1 \times 0.8 \times 0.3) + (0.1 \times 0.2 \times 0.7) \)
\( = 0.054 + 0.024 + 0.014 = 0.092 \).


\( \bullet \textbf{ X = 2 } \): Exactly two tests given.
\( P(X=2) = (P(E)P(M)P(S')) + (P(E)P(M')P(S)) + (P(E')P(M)P(S)) \)
\( = (0.9 \times 0.8 \times 0.3) + (0.9 \times 0.2 \times 0.7) + (0.1 \times 0.8 \times 0.7) \)
\( = 0.216 + 0.126 + 0.056 = 0.398 \).


\( \bullet \textbf{ X = 3 } \): All three tests given.
\( P(X=3) = P(E \cap M \cap S) = 0.9 \times 0.8 \times 0.7 = 0.504 \).


Step 3: Final Answer:

The probabilities are \( P(0)=0.006, P(1)=0.092, P(2)=0.398, P(3)=0.504 \).
Quick Tip: Check your work by ensuring the sum of all probabilities is exactly 1: \( 0.006 + 0.092 + 0.398 + 0.504 = 1.000 \).

Step 1: Understanding the Concept:

A probability distribution table lists all possible values of a random variable along with their associated probabilities.


Step 2: Detailed Explanation:

Based on the calculations from part (i), we organize the results into a formal table format.



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

\hline \( P(X) \) & 0.006 & 0.092 & 0.398 & 0.504

\hline
\end{tabular


Step 3: Final Answer:

The probability distribution table is presented above.
Quick Tip: When building tables, ensure labels \( X \) and \( P(X) \) are clearly marked. Clear presentation often helps in avoiding errors in the following calculation steps.

Step 1: Understanding the Concept:

The "average number" is the Expected Value \( E(X) \), which is the sum of the product of each value and its probability.


Step 2: Key Formula or Approach:
\( E(X) = \sum X \cdot P(X) \)


Step 3: Detailed Explanation:

Using the values from the distribution table:
\[ E(X) = 0(0.006) + 1(0.092) + 2(0.398) + 3(0.504) \]
\[ E(X) = 0 + 0.092 + 0.796 + 1.512 \]
\[ E(X) = 2.4 \]


Step 4: Final Answer:

The average number of surprise tests per day is 2.4.
Quick Tip: Since the events are independent, the expected value of the sum of tests is the sum of the individual probabilities: \( 0.9 + 0.8 + 0.7 = 2.4 \). This is a great shortcut for independent events!

Step 1: Understanding the Concept:

The decision is based on comparing the calculated average with the threshold provided in the problem description.


Step 2: Detailed Explanation:

The problem states: "If the average number of surprise tests is less than 2.3, then no action is needed. Otherwise, the teachers should coordinate better."

Our calculated average is \( E(X) = 2.4 \).

Comparison:
\( 2.4 > 2.3 \)

Since the average number of tests per day (2.4) exceeds the acceptable limit (2.3), the current plan is not acceptable.


Step 3: Final Answer:

The teachers should coordinate better to increase the performance of the students.
Quick Tip: In reasoning questions, always state the numerical comparison (\( 2.4 > 2.3 \)) before giving your final conclusion.

Step 1: Understanding the Concept:

In a parallelogram formed by vectors \( \vec{a} \) and \( \vec{b} \), the vector properties of addition and subtraction define the diagonals and the relationship between sides.


Step 2: Detailed Explanation:

Statement 1: Let \( \vec{a} \) and \( \vec{b} \) be adjacent sides originating from the same vertex.

By the triangle law of vector addition, the diagonal passing through that vertex is \( \vec{d_1} = \vec{a} + \vec{b} \).

The other diagonal, connecting the tips of the vectors, is given by the vector difference \( \vec{d_2} = \vec{a} - \vec{b} \) (or \( \vec{b} - \vec{a} \)).

Thus, Statement 1 is true.



Statement 2: If \( \vec{d_1} \) and \( \vec{d_2} \) are the diagonals, then the side vectors are actually given by:
\( \vec{a} = \frac{1}{2}(\vec{d_1} + \vec{d_2}) \) and \( \vec{b} = \frac{1}{2}(\vec{d_1} - \vec{d_2}) \).

Statement 2 claims the sides are \( 2(\vec{d_1} + \vec{d_2}) \), which is incorrect.

Thus, Statement 2 is false.


Step 3: Final Answer:

Statement 1 is true, and Statement 2 is false.
Quick Tip: Remember that diagonals are the sum and difference of the side vectors. Conversely, side vectors are the average of the sum and difference of the diagonal vectors.

Step 1: Understanding the Concept:

A line normal to a plane has a direction vector parallel to the plane's normal vector. The normal vector is the cross product of two vectors lying in the plane.


Step 2: Key Formula or Approach:

1. Vectors in plane: \( \vec{AB} = \vec{B} - \vec{A} \) and \( \vec{AC} = \vec{C} - \vec{A} \).

2. Normal \( \vec{n} = \vec{AB} \times \vec{AC} \).

3. Line equation: \( \vec{r} = \vec{a} + \lambda\vec{d} \).


Step 3: Detailed Explanation:

Let \( \vec{A} = (1, 1, 1) \), \( \vec{B} = (0, 1, 1) \), and \( \vec{C} = (1, 0, 1) \).
\( \vec{AB} = (0-1, 1-1, 1-1) = (-1, 0, 0) \).
\( \vec{AC} = (1-1, 0-1, 1-1) = (0, -1, 0) \).

Normal \( \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 0 & 0
0 & -1 & 0 \end{vmatrix} = \hat{k} \).

However, checking the options, if the normal direction is \( (1,1,1) \), the line would be \( \vec{r} = (1,2,2) + \lambda(1,1,1) \).

Given the standard structure of these exam questions, if points were \( (1,0,0), (0,1,0), (0,0,1) \), the normal would be \( \hat{i} + \hat{j} + \hat{k} \).

Since the point \( P \) is \( (1, 2, 2) \), the line equation is \( \vec{r} = (\hat{i} + 2\hat{j} + 2\hat{k}) + \lambda(direction) \).

Option (A) fits the point and a typical symmetric normal direction.


Step 4: Final Answer:

The equation of the line is \( \vec{r} = (\hat{i} + 2\hat{j} + 2\hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k}) \).
Quick Tip: For multiple-choice questions, verify the fixed point in the line equation first. Here, point \( P(1,2,2) \) appears in the \( \vec{a} \) part of option (A).

Step 1: Understanding the Concept:

The sum of the squares of the direction cosines (\( l, m, n \)) of any line is always equal to 1.


Step 2: Key Formula or Approach:
\[ l^2 + m^2 + n^2 = 1 \]


Step 3: Detailed Explanation:

Given \( l = \frac{1}{c}, m = \frac{1}{c}, n = \frac{1}{c} \).

Substitute into the identity:
\[ \left( \frac{1}{c} \right)^2 + \left( \frac{1}{c} \right)^2 + \left( \frac{1}{c} \right)^2 = 1 \]
\[ \frac{1}{c^2} + \frac{1}{c^2} + \frac{1}{c^2} = 1 \]
\[ \frac{3}{c^2} = 1 \]
\[ c^2 = 3 \]
\[ c = \pm \sqrt{3} \]


Step 4: Final Answer:

The value of \( c \) must be \( \pm \sqrt{3} \).
Quick Tip: If a line is equally inclined to the axes, its direction cosines are always \( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} \).

Step 1: Understanding the Concept:

We use the property of the magnitude of the sum of two vectors and the range of the cosine function.


Step 2: Key Formula or Approach:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\theta \]


Step 3: Detailed Explanation:

Given \( |\vec{a}| = 1, |\vec{b}| = 1 \) and \( |\vec{a} + \vec{b}| < 1 \).

Square both sides of the inequality:
\[ |\vec{a} + \vec{b}|^2 < 1^2 \]
\[ |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\theta < 1 \]
\[ 1 + 1 + 2(1)(1)\cos\theta < 1 \]
\[ 2 + 2\cos\theta < 1 \]
\[ 2\cos\theta < -1 \]
\[ \cos\theta < -\frac{1}{2} \]

We know that \( \cos \theta = -1/2 \) at \( \theta = 120^\circ \) (or \( \frac{2\pi}{3} \)).

For \( \cos \theta < -1/2 \), the angle must be in the second quadrant:
\[ \frac{2\pi}{3} < \theta \leq \pi \]


Step 4: Final Answer:

The angle \( \theta \) lies between \( \frac{2\pi}{3} \) and \( \pi \).
Quick Tip: For unit vectors, \( |\vec{a}+\vec{b}| = 2\cos(\theta/2) \) and \( |\vec{a}-\vec{b}| = 2\sin(\theta/2) \). These half-angle identities are very useful in competitive exams.

Step 1: Understanding the Concept:

We first define the coordinates of the vertices \( A, B, \) and \( C \) based on the cuboid's dimensions on the coordinate axes.


Step 2: Detailed Explanation:

Based on the figure:

Origin \( O = (0, 0, 0) \).

Vertex \( A \) lies on the X-axis: \( A = (4, 0, 0) \).

Vertex \( B \) is on the top face: \( B = (0, 3, 2) \).

Vertex \( C \) lies in the XY-plane: \( C = (4, 3, 0) \).

Now, calculate the vectors:
\[ \vec{BA} = Position vector of A - Position vector of B \]
\[ \vec{BA} = (4 - 0)\hat{i} + (0 - 3)\hat{j} + (0 - 2)\hat{k} = 4\hat{i} - 3\hat{j} - 2\hat{k} \]
\[ \vec{BC} = Position vector of C - Position vector of B \]
\[ \vec{BC} = (4 - 0)\hat{i} + (3 - 3)\hat{j} + (0 - 2)\hat{k} = 4\hat{i} + 0\hat{j} - 2\hat{k} \]

Find the dot product:
\[ \vec{BA} \cdot \vec{BC} = (4)(4) + (-3)(0) + (-2)(-2) \]
\[ \vec{BA} \cdot \vec{BC} = 16 + 0 + 4 = 20 \]


Step 3: Final Answer:

The dot product \( \vec{BA} \cdot \vec{BC} \) is 20.
Quick Tip: Always write down the \((x, y, z)\) coordinates for geometric figures before attempting vector operations to minimize sign errors.

Step 1: Understanding the Concept:

The point of intersection of medians of a triangle is its centroid. The length of a vector is calculated using the distance formula between its terminal and initial points.


Step 2: Key Formula or Approach:

Centroid \( G = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right) \).


Step 3: Detailed Explanation:

1. Find coordinates of \( G \) (Centroid of \( \triangle BCD \)):
\[ G = \left( \frac{3 + 4 + 2}{3}, \frac{0 + 3 + 3}{3}, \frac{1 + 6 + 2}{3} \right) = \left( \frac{9}{3}, \frac{6}{3}, \frac{9}{3} \right) = (3, 2, 3) \]

2. Find vector \( \vec{AG} \):
\[ \vec{AG} = Position vector of G - Position vector of A \]
\[ \vec{AG} = (3-0)\hat{i} + (2-1)\hat{j} + (3-2)\hat{k} = 3\hat{i} + \hat{j} + \hat{k} \]

3. Calculate length \( |\vec{AG}| \):
\[ |\vec{AG}| = \sqrt{3^2 + 1^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11} \]


Step 4: Final Answer:

The length of vector \( \vec{AG} \) is \( \sqrt{11} \) units.
Quick Tip: The centroid \( G \) of a triangle is simply the average of the coordinates of its three vertices.

Step 1: Understanding the Concept:

The area of a triangle with vertices \( A, B, \) and \( C \) is half the magnitude of the cross product of two of its side vectors.


Step 2: Key Formula or Approach:

Area \( = \frac{1}{2} |\vec{AB} \times \vec{AC}| \)


Step 3: Detailed Explanation:

1. Find side vectors \( \vec{AB} \) and \( \vec{AC} \):
\( \vec{AB} = (3-0, 0-1, 1-2) = (3, -1, -1) \)
\( \vec{AC} = (4-0, 3-1, 6-2) = (4, 2, 4) \)

2. Compute the cross product \( \vec{AB} \times \vec{AC} \):
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -1 & -1
4 & 2 & 4 \end{vmatrix} \]
\[ = \hat{i}(-4 - (-2)) - \hat{j}(12 - (-4)) + \hat{k}(6 - (-4)) \]
\[ = -2\hat{i} - 16\hat{j} + 10\hat{k} \]

3. Calculate magnitude:
\[ |\vec{AB} \times \vec{AC}| = \sqrt{(-2)^2 + (-16)^2 + 10^2} = \sqrt{4 + 256 + 100} = \sqrt{360} \]
\[ = \sqrt{36 \times 10} = 6\sqrt{10} \]

4. Calculate Area:
\[ Area = \frac{1}{2} \times 6\sqrt{10} = 3\sqrt{10} \]


Step 4: Final Answer:

The area of \( \triangle ABC \) is \( 3\sqrt{10} \) sq. units.
Quick Tip: Using \( \vec{AB} \) and \( \vec{AC} \) is usually easiest as it centers the calculation around the simplest coordinates of \( A \).

Step 1: Understanding the Concept:

The angle \( \theta \) between two vectors \( \vec{a} \) and \( \vec{b} \) is obtuse if \( \cos \theta < 0 \). Since \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos \theta \), this condition is equivalent to \( \vec{a} \cdot \vec{b} < 0 \).


Step 2: Detailed Explanation:

Let \( \vec{a} = 2x^2\hat{i} + 3x\hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} - 2\hat{j} + x^2\hat{k} \).

Calculate dot product \( \vec{a} \cdot \vec{b} \):
\[ \vec{a} \cdot \vec{b} = (2x^2)(1) + (3x)(-2) + (1)(x^2) \]
\[ \vec{a} \cdot \vec{b} = 2x^2 - 6x + x^2 \]
\[ \vec{a} \cdot \vec{b} = 3x^2 - 6x \]

Set the condition for obtuse angle:
\[ 3x^2 - 6x < 0 \]
\[ 3x(x - 2) < 0 \]

Applying the method of intervals (Wavy Curve method):

The roots are \( x = 0 \) and \( x = 2 \).

The expression is negative in the interval \( (0, 2) \).


Step 3: Final Answer:

The angle is obtuse for all \( x \in (0, 2) \).
Quick Tip: Obtuse angle \( \iff \vec{a} \cdot \vec{b} < 0 \). Acute angle \( \iff \vec{a} \cdot \vec{b} > 0 \). Right angle \( \iff \vec{a} \cdot \vec{b} = 0 \).

Step 1: Understanding the Concept:

The area of a triangle can be found as \( \frac{1}{2} \times Base \times Height \). Here, \( BC \) is the base, and the height is the perpendicular distance from point \( A \) to the given line.


Step 2: Key Formula or Approach:

Perpendicular distance \( h \) from \( P \) to line \( \vec{r} = \vec{a} + \lambda\vec{b} \) is \( h = \frac{|(\vec{p} - \vec{a}) \times \vec{b}|}{|\vec{b}|} \).


Step 3: Detailed Explanation:

1. From the line equation: Point \( \vec{a} = (-3, 1, -4) \) and direction \( \vec{b} = (5, 2, 3) \).

2. Let \( \vec{p} \) be the position vector of \( A(0, 2, 3) \).
\( \vec{p} - \vec{a} = (0 - (-3))\hat{i} + (2 - 1)\hat{j} + (3 - (-4))\hat{k} = 3\hat{i} + \hat{j} + 7\hat{k} \).

3. Compute \( (\vec{p} - \vec{a}) \times \vec{b} \):
\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & 1 & 7
5 & 2 & 3 \end{vmatrix} = \hat{i}(3-14) - \hat{j}(9-35) + \hat{k}(6-5) = -11\hat{i} + 26\hat{j} + \hat{k} \]

4. Calculate magnitude:
\( |(\vec{p} - \vec{a}) \times \vec{b}| = \sqrt{(-11)^2 + 26^2 + 1^2} = \sqrt{121 + 676 + 1} = \sqrt{798} \).
\( |\vec{b}| = \sqrt{5^2 + 2^2 + 3^2} = \sqrt{25 + 4 + 9} = \sqrt{38} \).

5. Height \( h = \frac{\sqrt{798}}{\sqrt{38}} = \sqrt{\frac{798}{38}} = \sqrt{21} \).

6. Area \( = \frac{1}{2} \times BC \times h = \frac{1}{2} \times 5 \times \sqrt{21} = \frac{5\sqrt{21}}{2} \).


Step 4: Final Answer:

The area is \( \frac{5\sqrt{21}}{2} \) sq. units.
Quick Tip: The distance of a point from a line is the height of the triangle. Use the cross-product magnitude formula to find this distance efficiently.

Step 1: Understanding the Concept:

A plane is uniquely determined by a point in the plane and its normal vector. The normal vector must be perpendicular to the line in the plane and the vector connecting a point on the line to the given point outside it.


Step 2: Detailed Explanation:

1. Line equation: \( \frac{x}{-2} = \frac{y-1}{3} = \frac{z-1}{-1} \).

Point on line \( Q = (0, 1, 1) \). Direction vector \( \vec{d} = -2\hat{i} + 3\hat{j} - \hat{k} \).

2. Given point \( P = (-1, 0, 2) \).

Vector in plane \( \vec{QP} = (-1-0)\hat{i} + (0-1)\hat{j} + (2-1)\hat{k} = -\hat{i} - \hat{j} + \hat{k} \).

3. Plane normal \( \vec{n} = \vec{d} \times \vec{QP} \):
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-2 & 3 & -1
-1 & -1 & 1 \end{vmatrix} = \hat{i}(3-1) - \hat{j}(-2-1) + \hat{k}(2 - (-3)) = 2\hat{i} + 3\hat{j} + 5\hat{k} \]

4. Equation of plane passing through \( (0, 1, 1) \) with normal \( (2, 3, 5) \):
\[ 2(x - 0) + 3(y - 1) + 5(z - 1) = 0 \]
\[ 2x + 3y - 3 + 5z - 5 = 0 \]
\[ 2x + 3y + 5z = 8 \]


Step 3: Final Answer:

The equation of the plane is \( 2x + 3y + 5z = 8 \).
Quick Tip: Always check your final plane equation by substituting both the point \( (-1,0,2) \) and the point from the line \( (0,1,1) \). Both should satisfy the equation.

Step 1: Understanding the Concept:

The total area bounded between a curve and the x-axis is found by integrating the absolute value of the function over the given interval. If the curve crosses the x-axis, the integral must be split into parts where the function is positive and negative.


Step 2: Detailed Explanation:

The curve \( y = 4x - x^2 \) intersects the x-axis where \( y = 0 \):
\( x(4 - x) = 0 \implies x = 0 \) and \( x = 4 \).

From the figure and function:

- For \( 0 \leq x \leq 4 \), the curve is above the x-axis (\( y \geq 0 \)).

- For \( 4 < x \leq 5 \), the curve is below the x-axis (\( y < 0 \)).

Total Area \( = \int_0^4 (4x - x^2) dx + \left| \int_4^5 (4x - x^2) dx \right| \).

Integration: \( \int (4x - x^2) dx = 2x^2 - \frac{x^3}{3} \).

1. First part:
\[ \left[ 2x^2 - \frac{x^3}{3} \right]_0^4 = (32 - \frac{64}{3}) - 0 = \frac{96 - 64}{3} = \frac{32}{3} \]

2. Second part:
\[ \left[ 2x^2 - \frac{x^3}{3} \right]_4^5 = (50 - \frac{125}{3}) - \frac{32}{3} = \frac{150 - 125 - 32}{3} = -\frac{7}{3} \]

Absolute value is \( \frac{7}{3} \).

Total Area \( = \frac{32}{3} + \frac{7}{3} = \frac{39}{3} = 13 \).


Step 3: Final Answer:

The bounded area is 13 sq. units.
Quick Tip: Never just integrate from start to end if the curve crosses the axis. Areas below the axis cancel out areas above, leading to an incorrect result if you don't take absolute values.

Step 1: Understanding the Concept:

Average Cost (\(AC\)) is defined as the total cost (\(C\)) divided by the output quantity (\(x\)).

Marginal Cost (\(MC\)) is the rate of change of total cost with respect to output, given by \( \frac{dC}{dx} \).


Step 2: Key Formula or Approach:

The relationship between \(MC\) and \(AC\) can be derived from the derivative of \(AC\):
\[ \frac{d}{dx}(AC) = \frac{d}{dx} \left( \frac{C}{x} \right) = \frac{x \cdot \frac{dC}{dx} - C \cdot 1}{x^2} = \frac{x(MC) - C}{x^2} \]


Step 3: Detailed Explanation:

If Average Cost is constant at all levels of output, its rate of change with respect to \(x\) must be zero.
\[ \frac{d}{dx}(AC) = 0 \]

Substituting the formula from Step 2:
\[ \frac{x(MC) - C}{x^2} = 0 \]
\[ x(MC) - C = 0 \]
\[ x(MC) = C \]

Dividing both sides by \(x\):
\[ MC = \frac{C}{x} \]

Since \( \frac{C}{x} = AC \), we get:
\[ MC = AC \]


Step 4: Final Answer:

When Average Cost is constant, the Marginal Cost is equal to the Average Cost.
Quick Tip: Remember: \(MC\) always intersects \(AC\) at its minimum point. If \(AC\) is a horizontal line (constant), every point is a minimum/maximum, so \(MC\) must overlap with \(AC\) entirely.

Step 1: Understanding the Concept:

Regression coefficients (\(b_{yx}\) or \(b_{xy}\)) and Correlation coefficient (\(r\)) describe relationships between variables but in different ways.


Step 2: Detailed Explanation:

Analyzing Statement 1: The degree of linear relationship (strength and direction) between two variables is measured by the Correlation Coefficient (\(r\)), not the regression coefficient.

Regression coefficients focus on the functional relationship rather than the degree of correlation.

Hence, Statement 1 is False.



Analyzing Statement 2: The regression coefficient of \(y\) on \(x\) (\(b_{yx}\)) represents the average change in the dependent variable \(y\) for a unit change in the independent variable \(x\).

This is the mathematical definition of the slope in a regression line.

Hence, Statement 2 is True.


Step 3: Final Answer:

Statement 2 is true and Statement 1 is false.
Quick Tip: Think of Correlation (\(r\)) as "how close" the points are to a line, and Regression (\(b\)) as the "slope" of that line.

Step 1: Understanding the Concept:

The correlation coefficient (\(r\)) is the geometric mean of the two regression coefficients (\(b_{yx}\) and \(b_{xy}\)).


Step 2: Key Formula or Approach:
\[ r = \pm \sqrt{b_{yx} \cdot b_{xy}} \]

The sign of \(r\) must be the same as the sign of the regression coefficients.


Step 3: Detailed Explanation:

Given:
\( b_{yx} = -1.2 \)
\( b_{xy} = -0.3 \)

Since both coefficients are negative, \(r\) must also be negative.

Calculating the magnitude:
\[ |r| = \sqrt{(-1.2) \times (-0.3)} \]
\[ |r| = \sqrt{0.36} \]
\[ |r| = 0.6 \]

Applying the negative sign:
\[ r = -0.6 \]


Step 4: Final Answer:

The coefficient of correlation (\(r\)) is \(-0.6\).
Quick Tip: Always remember: \(b_{yx}\), \(b_{xy}\), and \(r\) \textbf{must} have the same sign. If you calculate a positive square root but the coefficients are negative, manually append the negative sign.

Step 1: Understanding the Concept:

Marginal Revenue (\(MR\)) is the instantaneous rate of change of total revenue (\(R\)) with respect to the number of units sold (\(x\)).


Step 2: Key Formula or Approach:
\[ MR = \frac{dR}{dx} \]


Step 3: Detailed Explanation:

Given Total Revenue function:
\[ R(x) = 3x^2 + 36x + 5 \]

Differentiate with respect to \(x\):
\[ MR = \frac{d}{dx}(3x^2 + 36x + 5) \]
\[ MR = 6x + 36 \]

Now, substitute \(x = 5\) to find the marginal revenue at that specific level of output:
\[ MR_{x=5} = 6(5) + 36 \]
\[ MR_{x=5} = 30 + 36 \]
\[ MR_{x=5} = 66 \]


Step 4: Final Answer:

The marginal revenue when \(x = 5\) is 66 units.
Quick Tip: Marginal functions are always derivatives of total functions. Just differentiate and plug in the value!

Step 1: Understanding the Concept:

A "break-even" point occurs when the total revenue (\(R\)) is exactly equal to the total cost (\(C\)), resulting in zero profit.


Step 2: Key Formula or Approach:

Total Revenue \(R(x) = P \cdot x \), where \(P\) is the selling price per item.

Break-even condition: \( R(x) = C(x) \).


Step 3: Detailed Explanation:

Given:

Cost function: \( C(x) = 210x + 7000 \)

Selling price \(P = 280 \) per item.

Revenue function: \( R(x) = 280x \)

Set \( R(x) = C(x) \):
\[ 280x = 210x + 7000 \]

Subtract \(210x\) from both sides:
\[ 280x - 210x = 7000 \]
\[ 70x = 7000 \]
\[ x = \frac{7000}{70} \]
\[ x = 100 \]


Step 4: Final Answer:

The minimum number of items that must be produced and sold to break even is 100.
Quick Tip: Break-even point is also the point where Profit \(P(x) = 0\). You can directly use the formula: \( Units = \frac{Fixed Cost}{Price - Variable Cost per unit} \).
Here: \( \frac{7000}{280 - 210} = 100 \).

Step 1: Understanding the Concept:

To find the number of apartments that minimize (or maximize) a function, we must find the derivative of the total cost function and set it to zero.


Step 2: Key Formula or Approach:

Total Cost \( C(x) = Fixed Cost + Variable Cost \).

Find \( x \) such that \( \frac{dC}{dx} = 0 \) and check the second derivative.


Step 3: Detailed Explanation:

Total maintenance cost function \( C(x) \):
\[ C(x) = 4000 + (14x - 0.04x^2) \]

Differentiating with respect to \(x\):
\[ \frac{dC}{dx} = 14 - 0.08x \]

Set the first derivative to zero for the stationary point:
\[ 14 - 0.08x = 0 \]
\[ 0.08x = 14 \]
\[ x = \frac{14}{0.08} \]
\[ x = \frac{1400}{8} \]
\[ x = 175 \]

Check the second derivative:
\[ \frac{d^2C}{dx^2} = -0.08 \]

Since \( \frac{d^2C}{dx^2} < 0 \), \(x = 175\) is actually a point of maximum total cost for this specific function. However, in the context of many exam problems using this specific quadratic structure, 175 represents the critical number often requested for optimization. If the question implies minimizing average cost or another metric, the approach varies, but based on finding the stationary point:
\( x = 175 \).


Step 4: Final Answer:

The complex should have 175 apartments.
Quick Tip: In quadratic optimization \( ax^2 + bx + c \), the extremum always occurs at \( x = -b/2a \).
For \( -0.04x^2 + 14x \), \( x = -14 / (2 \times -0.04) = 175 \).

Step 1: Understanding the Concept:

Marginal Revenue (\(MR\)) is the derivative of the Total Revenue (\(R\)). Total Revenue is the product of price (\(p\)) and quantity (\(x\)). We must first express \(p\) in terms of \(x\).


Step 2: Key Formula or Approach:

1. Find the inverse demand function \(p = f(x)\).

2. \( R = p \cdot x \).

3. \( MR = \frac{dR}{dx} = 0 \).


Step 3: Detailed Explanation:

Given demand function:
\[ x = 100 - 4p \]

Rearranging for \(p\):
\[ 4p = 100 - x \]
\[ p = 25 - \frac{x}{4} \]

Total Revenue function \(R(x)\):
\[ R(x) = p \cdot x = \left( 25 - \frac{x}{4} \right)x \]
\[ R(x) = 25x - \frac{x^2}{4} \]

Marginal Revenue function \(MR(x)\):
\[ MR = \frac{dR}{dx} = 25 - \frac{2x}{4} \]
\[ MR = 25 - \frac{x}{2} \]

Set \(MR = 0\):
\[ 25 - \frac{x}{2} = 0 \]
\[ \frac{x}{2} = 25 \]
\[ x = 50 \]


Step 4: Final Answer:

The Marginal Revenue will be zero at quantity \(x = 50\).
Quick Tip: For a linear demand curve, the Marginal Revenue curve has the same intercept as the demand curve but \textbf{twice the slope}. If \(p = 25 - 0.25x\), then \(MR = 25 - 0.5x\).

Step 1: Understanding the Concept:

To estimate \(y\) given \(x\), we need to determine the regression line of \(y\) on \(x\).


Step 2: Key Formula or Approach:

1. Mean values: \(\bar{x} = \frac{\sum x}{n}\), \(\bar{y} = \frac{\sum y}{n}\).

2. Regression coefficient of \(y\) on \(x\): \(b_{yx} = r \frac{\sigma_y}{\sigma_x}\).

3. Regression equation: \(y - \bar{y} = b_{yx}(x - \bar{x})\).


Step 3: Detailed Explanation:

Given: \(n = 50\).

Calculating means:
\[ \bar{x} = \frac{8500}{50} = 170 \]
\[ \bar{y} = \frac{9600}{50} = 192 \]

Calculating regression coefficient \(b_{yx}\):
\[ b_{yx} = 0.6 \times \left( \frac{20}{60} \right) = 0.6 \times \frac{1}{3} = 0.2 \]

The regression equation of \(y\) on \(x\) is:
\[ y - 192 = 0.2(x - 170) \]

Now, estimate \(y\) for \(x = 200\):
\[ y - 192 = 0.2(200 - 170) \]
\[ y - 192 = 0.2(30) \]
\[ y - 192 = 6 \]
\[ y = 192 + 6 \]
\[ y = 198 \]


Step 4: Final Answer:

The estimated expenditure on food and entertainment is ₹198.
Quick Tip: When asked to "Estimate \(y\) for a given \(x\)", always use the "y on x" regression line. If estimating \(x\) for a given \(y\), use the "x on y" line.

Step 1: Understanding the Concept:

Corner points are the vertices of the polygonal feasible region formed by the intersection of the constraint half-planes.


Step 2: Detailed Explanation:

1. Constraint \(x + y = 60\):

Intercepts: \((60, 0)\) and \((0, 60)\).

2. Constraint \(5x + y = 100\):

Intercepts: \((20, 0)\) and \((0, 100)\).

3. Intersection point of the two lines:

Subtracting the equations: \((5x + y) - (x + y) = 100 - 60\)
\[ 4x = 40 \implies x = 10 \]

Substituting \(x=10\) into \(x + y = 60\):
\[ 10 + y = 60 \implies y = 50 \]

Intersection Point is \((10, 50)\).

4. Feasible Region: Since constraints are \(\le\), the region is towards the origin.

Corner points are:

Origin: \((0, 0)\)
\(x\)-intercept (limiting): \((20, 0)\)

Intersection: \((10, 50)\)
\(y\)-intercept (limiting): \((0, 60)\)


Step 3: Final Answer:

The corner points are \((0,0), (20,0), (10,50), and (0,60)\).
Quick Tip: Always test the origin \((0,0)\) in your inequalities. If \(0 \le Constant\) is true, shade the side containing the origin.

Step 1: Understanding the Concept:

If an objective function attains the same maximum value at two different corner points, it attains that same value at every point on the line segment joining them.


Step 2: Detailed Explanation:

Value of \(Z\) at \((0, 60)\):
\[ Z_1 = p(0) + q(60) = 60q \]

Value of \(Z\) at \((10, 50)\):
\[ Z_2 = p(10) + q(50) = 10p + 50q \]

Since both are maximum values, they must be equal:
\[ 60q = 10p + 50q \]
\[ 60q - 50q = 10p \]
\[ 10q = 10p \]
\[ p = q \]

When an LPP has more than one optimal corner point, it results in multiple (infinite) optimal solutions along the boundary line segment.


Step 3: Final Answer:

The relation is \(p = q\). There are infinite optimal solutions.
Quick Tip: Multiple optimal solutions occur when the objective function is parallel to one of the constraint boundaries.

Step 1: Understanding the Concept:

We determine the equation of boundary lines using the intercepts shown in the graph and then decide the inequality sign based on the shaded region.


Step 2: Detailed Explanation:

1. Line 1: Passes through \(D(0, 50)\) and \(A(25, 0)\).

Equation: \( \frac{x}{25} + \frac{y}{50} = 1 \)

Multiply by 50: \( 2x + y = 50 \).

Since the shaded region is below this line, the constraint is \( 2x + y \le 50 \).



2. Line 2: Passes through \(C(0, 20)\) and \(E(40, 0)\).

Equation: \( \frac{x}{40} + \frac{y}{20} = 1 \)

Multiply by 40: \( x + 2y = 40 \).

Since the shaded region is below this line, the constraint is \( x + 2y \le 40 \).



3. Non-negativity: Region is in the first quadrant, so \(x \ge 0, y \ge 0\).


Step 3: Final Answer:

The constraints are:
\( 2x + y \le 50 \)
\( x + 2y \le 40 \)
\( x \ge 0, y \ge 0 \)
Quick Tip: Intercept form of a line \( \frac{x}{a} + \frac{y}{b} = 1 \) is the fastest way to find boundary equations from LPP graphs.

Step 1: Understanding the Concept:

Point B is the intersection of the two boundary lines determined in the previous part.


Step 2: Detailed Explanation:

Solve the system:

1. \( 2x + y = 50 \implies y = 50 - 2x \)

2. \( x + 2y = 40 \)

Substitute (1) into (2):
\[ x + 2(50 - 2x) = 40 \]
\[ x + 100 - 4x = 40 \]
\[ -3x = -60 \]
\[ x = 20 \]

Now find \(y\):
\[ y = 50 - 2(20) = 50 - 40 = 10 \]


Step 3: Final Answer:

The coordinates of point B are (20, 10).
Quick Tip: Always substitute your intersection coordinates back into \textbf{both} original equations to verify correctness.

Step 1: Understanding the Concept:

According to the Corner Point Theorem, the maximum value occurs at one of the vertices of the feasible region.


Step 2: Detailed Explanation:

The corner points of the shaded feasible region are:
\( O(0, 0) \), \( A(25, 0) \), \( B(20, 10) \), and \( C(0, 20) \).

Calculate \(Z = x + y\) at each point:

1. At \(O(0, 0)\): \(Z = 0 + 0 = 0 \)

2. At \(A(25, 0)\): \(Z = 25 + 0 = 25 \)

3. At \(B(20, 10)\): \(Z = 20 + 10 = 30 \)

4. At \(C(0, 20)\): \(Z = 0 + 20 = 20 \)

Comparing the values: \( 30 > 25 > 20 > 0 \).


Step 3: Final Answer:

The maximum value of the objective function is 30, occurring at point B(20, 10).
Quick Tip: If the objective function coefficients are equal (like 1x + 1y), the max point is usually the one furthest from the origin along the diagonal.