JKBOSE Class 12 2026 Mathematics Question Paper with Solutions : Download PDF

Step 1: Understanding the Concept:

A symmetric matrix is defined as a matrix \( A \) such that \( A^T = A \).

A skew-symmetric matrix is defined as a matrix \( A \) such that \( A^T = -A \).


Step 2: Key Formula or Approach:

We use the definitions of symmetric and skew-symmetric matrices simultaneously to solve for matrix \( A \).


Step 3: Detailed Explanation:

Given that matrix \( A \) is symmetric:
\[ A^T = A \quad ---(i) \]

Given that matrix \( A \) is skew-symmetric:
\[ A^T = -A \quad ---(ii) \]

From equations (i) and (ii), we can equate the values of \( A^T \):
\[ A = -A \]
\[ A + A = 0 \]
\[ 2A = 0 \]
\[ A = 0 \]

Therefore, \( A \) must be a zero matrix.


Step 4: Final Answer:

Since the matrix \( A \) must satisfy both conditions, every element of the matrix must be zero.

Hence, \( A \) is a zero matrix.
Quick Tip: Only a zero matrix can satisfy the condition of being equal to its own negative.
Remember: A skew-symmetric matrix must always have zero elements on its main diagonal.

Step 1: Understanding the Concept:

The anti-derivative of a function is its indefinite integral.


Step 2: Key Formula or Approach:

The standard integral for sine is:
\[ \int \sin(ax) dx = -\frac{\cos(ax)}{a} + c \]


Step 3: Detailed Explanation:

We need to find \( \int \sin 2x dx \).

Let \( 2x = t \). Then \( 2 dx = dt \implies dx = \frac{dt}{2} \).

Substituting these into the integral:
\[ \int \sin t \cdot \frac{dt}{2} = \frac{1}{2} \int \sin t dt \]
\[ = \frac{1}{2} (-\cos t) + c \]
\[ = -\frac{\cos 2x}{2} + c \]


Step 4: Final Answer:

The anti-derivative is \( -\frac{\cos 2x}{2} + c \).
Quick Tip: When integrating \( f(ax+b) \), always divide the result by the coefficient of \( x \), which is \( a \).
Anti-derivative of \( \sin \) is \( -\cos \), but derivative of \( \sin \) is \( \cos \). Don't mix them up!

Step 1: Understanding the Concept:

The vector joining two points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) directed from \( P \) to \( Q \) is given by the difference of their position vectors.


Step 2: Key Formula or Approach:
\[ \vec{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k} \]


Step 3: Detailed Explanation:

Given points:
\( P = (2, 3, 0) \) so \( \vec{OP} = 2\hat{i} + 3\hat{j} + 0\hat{k} \)
\( Q = (-1, -2, -4) \) so \( \vec{OQ} = -1\hat{i} - 2\hat{j} - 4\hat{k} \)

The vector directed from \( P \) to \( Q \) is:
\[ \vec{PQ} = \vec{OQ} - \vec{OP} \]
\[ \vec{PQ} = (-1 - 2)\hat{i} + (-2 - 3)\hat{j} + (-4 - 0)\hat{k} \]
\[ \vec{PQ} = -3\hat{i} - 5\hat{j} - 4\hat{k} \]


Step 4: Final Answer:

The required vector is \( -3\hat{i} - 5\hat{j} - 4\hat{k} \).
Quick Tip: Always subtract the "Initial" point coordinates from the "Final" point coordinates: \( Vector = Terminal - Initial \).

Step 1: Understanding the Concept:

In two dimensions, lines are either parallel or they intersect. In three dimensions, a third possibility exists.


Step 2: Detailed Explanation:

Skew lines are lines that do not lie in the same plane (non-coplanar).

Because they are not in the same plane:

1. They never intersect each other.

2. They are not parallel to each other.

The shortest distance between two skew lines is always a non-zero value and is measured along a line perpendicular to both.


Step 3: Final Answer:

Skew lines are defined as pairs of lines that are non-coplanar, meaning they are neither parallel nor do they intersect at any point.
Quick Tip: Think of a road overpass. The road on the bridge and the road below it are skew lines; they don't meet and they aren't going in the same direction.

Step 1: Understanding the Concept:

Optimization refers to making something as functional or effective as possible.


Step 2: Detailed Explanation:

In mathematics and linear programming, an optimization problem consists of:

1. Objective Function: A linear function (like \( Z = ax + by \)) that needs to be maximized (e.g., profit) or minimized (e.g., cost).

2. Constraints: A set of linear inequalities or equations that restrict the values of the variables.

3. Feasible Region: The set of all possible points that satisfy all the given constraints.

The goal is to find a point in the feasible region that gives the optimal (maximum or minimum) value of the objective function.


Step 3: Final Answer:

An optimization problem is a problem which seeks to maximize or minimize a specific quantity (the objective function), subject to a set of constraints expressed as linear inequalities or equations.
Quick Tip: In board exams, optimization problems usually appear as Linear Programming Problems (LPP) where you use the corner point method to find the maximum or minimum.

Step 1: Understanding the Concept:

A relation \( R \) on a set \( A \) is:

1. Reflexive if \( (a, a) \in R \) for every \( a \in A \).

2. Symmetric if \( (a, b) \in R \implies (b, a) \in R \).

3. Transitive if \( (a, b) \in R \) and \( (b, c) \in R \implies (a, c) \in R \).


Step 2: Key Formula or Approach:

Define the set \( R \) explicitly based on the given conditions \( y = x + 5 \) and \( x < 4 \) where \( x, y \in \mathbb{N} \).

The values for \( x \) can only be \( \{1, 2, 3\} \).

Substituting \( x = 1, 2, 3 \) into \( y = x + 5 \):

When \( x = 1, y = 6 \).

When \( x = 2, y = 7 \).

When \( x = 3, y = 8 \).

So, \( R = \{ (1, 6), (2, 7), (3, 8) \} \).


Step 3: Detailed Explanation:

Reflexivity:

For \( R \) to be reflexive, \( (1, 1), (2, 2) \dots \) must belong to \( R \).

Since \( (1, 1) \notin R \), the relation is not reflexive.

Symmetry:

We have \( (1, 6) \in R \). For symmetry, \( (6, 1) \) must be in \( R \).

But \( (6, 1) \notin R \), so the relation is not symmetric.

Transitivity:

A relation is transitive if whenever \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \).

In this set, there are no elements \( (a, b) \) and \( (b, c) \) such that the second element of the first pair is the first element of the second pair.

(i.e., there is no \( y \) such that \( (x, y) \in R \) and \( (y, z) \in R \)).

When the hypothesis "if \( (a, b) \in R \) and \( (b, c) \in R \)" is never met, the statement is vacuously true.

Therefore, the relation is transitive.


Step 4: Final Answer:

The relation \( R \) is transitive but neither reflexive nor symmetric.
Quick Tip: A relation is "vacuously transitive" if you cannot find a pair of elements \( (a, b) \) and \( (b, c) \) in the set to test the condition. In such cases, the relation is always considered transitive by default.

Step 1: Understanding the Concept:

The principal value of an inverse trigonometric function is the value that falls within its standard restricted range.


Step 2: Key Formula or Approach:

The range of the principal value of \( \sin^{-1} x \) is \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).


Step 3: Detailed Explanation:

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

This implies \( \sin \theta = \frac{1}{\sqrt{2}} \).

We know that \( \sin \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \).

Since \( \frac{\pi}{4} \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \), the principal value is \( \frac{\pi}{4} \).


Step 4: Final Answer:

The principal value of \( \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) \) is \( \frac{\pi}{4} \).
Quick Tip: Always check if your calculated angle lies within the principal range:
\( \sin^{-1}: [-\pi/2, \pi/2] \)
\( \cos^{-1}: [0, \pi] \)
\( \tan^{-1}: (-\pi/2, \pi/2) \)

Step 1: Understanding the Concept:

The cross product of two vectors results in a vector perpendicular to both. Its magnitude represents the area of a parallelogram formed by the two vectors.


Step 2: Key Formula or Approach:

Use the determinant method for the cross product:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
a_1 & a_2 & a_3
b_1 & b_2 & b_3 \end{vmatrix} \]


Step 3: Detailed Explanation:

Calculate \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -7 & 7
3 & -2 & 2 \end{vmatrix} \]
\[ = \hat{i}[(-7 \times 2) - (7 \times -2)] - \hat{j}[(1 \times 2) - (7 \times 3)] + \hat{k}[(1 \times -2) - (-7 \times 3)] \]
\[ = \hat{i}[-14 + 14] - \hat{j}[2 - 21] + \hat{k}[-2 + 21] \]
\[ = 0\hat{i} + 19\hat{j} + 19\hat{k} \]

Now, find the magnitude:
\[ |\vec{a} \times \vec{b}| = \sqrt{0^2 + 19^2 + 19^2} \]
\[ = \sqrt{2 \times 19^2} = 19\sqrt{2} \]


Step 4: Final Answer:

The magnitude of the cross product is \( 19\sqrt{2} \).
Quick Tip: When finding the magnitude of a vector like \( k\hat{j} + k\hat{k} \), the result is always \( |k|\sqrt{2} \). This saves time in calculations.

Step 1: Understanding the Concept:
\( P(A/B) \) denotes the conditional probability of event A occurring given that event B has already occurred.


Step 2: Key Formula or Approach:
\[ P(A/B) = \frac{P(A \cap B)}{P(B)} \]


Step 3: Detailed Explanation:

Given:
\( P(A) = \frac{7}{13} \)
\( P(B) = \frac{9}{13} \)
\( P(A \cap B) = \frac{4}{13} \)

Applying the formula:
\[ P(A/B) = \frac{4/13}{9/13} \]

The denominator 13 cancels out:
\[ P(A/B) = \frac{4}{9} \]


Step 4: Final Answer:

The value of \( P(A/B) \) is \( \frac{4}{9} \).
Quick Tip: In conditional probability problems with fractions having common denominators, you can simply divide the numerators: \( \frac{Numerator of P(A \cap B)}{Numerator of P(B)} \).

Step 1: Understanding the Concept:

We use the general addition rule of probability and the definition of conditional probability.


Step 2: Key Formula or Approach:

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

2. \( P(A \cap B) = P(B) \cdot P(A/B) \)


Step 3: Detailed Explanation:

From the given information:
\( P(B) = \frac{5}{13} \)
\( 2P(A) = \frac{5}{13} \implies P(A) = \frac{5}{26} \)
\( P(A/B) = \frac{2}{5} \)

First, find \( P(A \cap B) \):
\[ P(A \cap B) = P(A/B) \cdot P(B) = \frac{2}{5} \cdot \frac{5}{13} = \frac{2}{13} \]

Now, calculate \( P(A \cup B) \):
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13} \]

Converting to a common denominator of 26:
\[ P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26} \]
\[ P(A \cup B) = \frac{11}{26} \]


Step 4: Final Answer:

The probability \( P(A \cup B) \) is \( \frac{11}{26} \).
Quick Tip: Always convert all probabilities to a common denominator before adding or subtracting to avoid calculation errors.

Step 1: Understanding the Concept:

A \( 3 \times 4 \) matrix has 3 rows and 4 columns. We calculate each element \( a_{ij} \) by substituting the row index \( i \) and column index \( j \) into the given formula.


Step 2: Key Formula or Approach:

Row indices: \( i \in \{1, 2, 3\} \).

Column indices: \( j \in \{1, 2, 3, 4\} \).

Formula: \( a_{ij} = |2i - j| \).


Step 3: Detailed Explanation:

For Row 1 (i = 1):
\( a_{11} = |2(1) - 1| = |1| = 1 \)
\( a_{12} = |2(1) - 2| = |0| = 0 \)
\( a_{13} = |2(1) - 3| = |-1| = 1 \)
\( a_{14} = |2(1) - 4| = |-2| = 2 \)


For Row 2 (i = 2):
\( a_{21} = |2(2) - 1| = |3| = 3 \)
\( a_{22} = |2(2) - 2| = |2| = 2 \)
\( a_{23} = |2(2) - 3| = |1| = 1 \)
\( a_{24} = |2(2) - 4| = |0| = 0 \)


For Row 3 (i = 3):
\( a_{31} = |2(3) - 1| = |5| = 5 \)
\( a_{32} = |2(3) - 2| = |4| = 4 \)
\( a_{33} = |2(3) - 3| = |3| = 3 \)
\( a_{34} = |2(3) - 4| = |2| = 2 \)


Step 4: Final Answer:

The required matrix is:
\[ A = \begin{bmatrix} 1 & 0 & 1 & 2
3 & 2 & 1 & 0
5 & 4 & 3 & 2 \end{bmatrix} \]
Quick Tip: The absolute value \( |x| \) ensures that elements are always non-negative. Note how the values decrease as \( j \) increases until \( 2i = j \), then increase again.

Step 1: Understanding the Concept:

The area of a region bounded by a curve is calculated using definite integration.

Since an ellipse is symmetric about both coordinate axes, the total area can be found by calculating the area of one quadrant and multiplying it by four.


Step 2: Key Formula or Approach:

The area is given by \( A = 4 \int_{0}^{a} y \, dx \).

Standard integral formula: \( \int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \).


Step 3: Detailed Explanation:

The given equation is \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \).

Comparing with \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we get \( a^2 = 16 \implies a = 4 \) and \( b^2 = 9 \implies b = 3 \).

Solve for \( y \) in terms of \( x \):
\[ \frac{y^2}{9} = 1 - \frac{x^2}{16} = \frac{16 - x^2}{16} \]
\[ y^2 = \frac{9}{16} (16 - x^2) \implies y = \frac{3}{4} \sqrt{16 - x^2} \]

The total area \( A \) is:
\[ A = 4 \int_{0}^{4} \frac{3}{4} \sqrt{16 - x^2} \, dx = 3 \int_{0}^{4} \sqrt{16 - x^2} \, dx \]

Applying the standard integration formula:
\[ A = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4} \]

Substituting the upper and lower limits:
\[ A = 3 \left[ \left( \frac{4}{2}\sqrt{16 - 16} + 8\sin^{-1}(1) \right) - \left( \frac{0}{2}\sqrt{16 - 0} + 8\sin^{-1}(0) \right) \right] \]
\[ A = 3 \left[ (0 + 8 \cdot \frac{\pi}{2}) - (0 + 0) \right] = 3 \cdot 4\pi = 12\pi \]


Step 4: Final Answer:

The area of the region bounded by the ellipse is \( 12\pi \) square units.
Quick Tip: Shortcut: The area of any ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is always exactly \( \pi ab \).
For this problem: \( Area = \pi \times 4 \times 3 = 12\pi \). Use this to verify your steps in descriptive exams.

Step 1: Understanding the Concept:

The shortest distance between two skew lines \( \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu \vec{b}_2 \) is the length of the vector projection of the segment joining the two lines onto the direction perpendicular to both.


Step 2: Key Formula or Approach:

The distance \( d \) is given by:
\[ d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| \]


Step 3: Detailed Explanation:

Identify the components from the given equations:

Line 1: \( \vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k} \), \( \vec{b}_1 = \hat{i} - \hat{j} + \hat{k} \)

Line 2: \( \vec{a}_2 = 2\hat{i} - \hat{j} - \hat{k} \), \( \vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k} \)

Calculate the difference vector:
\[ \vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-1)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k} \]

Calculate the cross product of the direction vectors:
\[ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -1 & 1
2 & 1 & 2 \end{vmatrix} \]
\[ = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - (-2)) = -3\hat{i} + 3\hat{k} \]

Find the magnitude of the cross product:
\[ |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \]

Calculate the dot product \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \):
\[ (\hat{i} - 3\hat{j} - 2\hat{k}) \cdot (-3\hat{i} + 3\hat{k}) = 1(-3) + (-3)(0) + (-2)(3) = -3 - 6 = -9 \]

Substitute into the distance formula:
\[ d = \left| \frac{-9}{3\sqrt{2}} \right| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \]


Step 4: Final Answer:

The shortest distance between the two lines is \( \frac{3\sqrt{2}}{2} \) units.
Quick Tip: If the dot product \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \) turns out to be zero, it means the lines intersect at some point, and the shortest distance is 0.

Step 1: Understanding the Concept:

Two vectors are perpendicular if their scalar (dot) product is exactly zero.


Step 2: Key Formula or Approach:

Express the vector \( \vec{a} + \lambda \vec{b} \) in terms of components and then apply the condition \( (\vec{a} + \lambda \vec{b}) \cdot \vec{c} = 0 \).


Step 3: Detailed Explanation:

First, determine the vector \( \vec{a} + \lambda \vec{b} \):
\[ \vec{a} + \lambda \vec{b} = (2\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-\hat{i} + 2\hat{j} + \hat{k}) \]
\[ = (2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k} \]

Given that this vector is perpendicular to \( \vec{c} = 3\hat{i} + \hat{j} \):
\[ (\vec{a} + \lambda \vec{b}) \cdot \vec{c} = 0 \]
\[ [(2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}] \cdot (3\hat{i} + \hat{j} + 0\hat{k}) = 0 \]

Multiply corresponding components:
\[ 3(2 - \lambda) + 1(2 + 2\lambda) + 0(3 + \lambda) = 0 \]
\[ 6 - 3\lambda + 2 + 2\lambda = 0 \]
\[ 8 - \lambda = 0 \implies \lambda = 8 \]


Step 4: Final Answer:

The value of the scalar \( \lambda \) is 8.
Quick Tip: Perpendicular vectors = zero dot product.
Parallel vectors = cross product is zero OR one is a scalar multiple of the other.

Step 1: Understanding the Concept:

Simplifying inverse trigonometric expressions usually involves transforming the inner trigonometric part to match the outer function's type.


Step 2: Key Formula or Approach:

Divide the numerator and denominator by \( \cos x \) and use the tangent subtraction formula: \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \).


Step 3: Detailed Explanation:

Given: \( \tan^{-1} \left( \frac{\cos x - \sin x}{\cos x + \sin x} \right) \).

Divide both the numerator and denominator by \( \cos x \):
\[ = \tan^{-1} \left( \frac{1 - \tan x}{1 + \tan x} \right) \]

We know that \( 1 = \tan(\frac{\pi}{4}) \). Substitute this into the expression:
\[ = \tan^{-1} \left( \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x} \right) \]

Using the identity \( \tan(A - B) \):
\[ = \tan^{-1} \left( \tan \left( \frac{\pi}{4} - x \right) \right) \]

Check the range condition: \( -\frac{\pi}{4} < x < \frac{3\pi}{4} \).

Multiplying by -1: \( \frac{\pi}{4} > -x > -\frac{3\pi}{4} \).

Adding \( \frac{\pi}{4} \): \( \frac{\pi}{2} > \frac{\pi}{4} - x > -\frac{\pi}{2} \).

Since \( (\frac{\pi}{4} - x) \) is within the principal range of \( \tan^{-1} \), the expression simplifies directly.
\[ = \frac{\pi}{4} - x \]


Step 4: Final Answer:

The simplified form is \( \frac{\pi}{4} - x \).
Quick Tip: Remember: \( \frac{\cos x - \sin x}{\cos x + \sin x} = \tan(\frac{\pi}{4} - x) \) and \( \frac{\cos x + \sin x}{\cos x - \sin x} = \tan(\frac{\pi}{4} + x) \). These are very frequent identities.

Step 1: Understanding the Concept:

The question requires us to find a scalar \( K \) by evaluating powers of matrices and performing scalar multiplication and subtraction on matrices.


Step 2: Key Formula or Approach:

We will first calculate \( A^2 \) by performing matrix multiplication \( A \cdot A \), then substitute all values into the equation \( A^2 = KA - 2I \) and compare the corresponding elements.


Step 3: Detailed Explanation:

First, calculate \( A^2 \):
\[ A^2 = \begin{bmatrix} 3 & -2
4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2
4 & -2 \end{bmatrix} \]
\[ A^2 = \begin{bmatrix} (3)(3) + (-2)(4) & (3)(-2) + (-2)(-2)
(4)(3) + (-2)(4) & (4)(-2) + (-2)(-2) \end{bmatrix} \]
\[ A^2 = \begin{bmatrix} 9 - 8 & -6 + 4
12 - 8 & -8 + 4 \end{bmatrix} = \begin{bmatrix} 1 & -2
4 & -4 \end{bmatrix} \]

Now, evaluate the right-hand side \( KA - 2I \):
\[ KA - 2I = K \begin{bmatrix} 3 & -2
4 & -2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix} \]
\[ KA - 2I = \begin{bmatrix} 3K & -2K
4K & -2K \end{bmatrix} - \begin{bmatrix} 2 & 0
0 & 2 \end{bmatrix} = \begin{bmatrix} 3K - 2 & -2K
4K & -2K - 2 \end{bmatrix} \]

Given \( A^2 = KA - 2I \), equate the matrices:
\[ \begin{bmatrix} 1 & -2
4 & -4 \end{bmatrix} = \begin{bmatrix} 3K - 2 & -2K
4K & -2K - 2 \end{bmatrix} \]

Equating corresponding elements:

1) \( 3K - 2 = 1 \implies 3K = 3 \implies K = 1 \)

2) \( -2K = -2 \implies K = 1 \)

3) \( 4K = 4 \implies K = 1 \)

4) \( -2K - 2 = -4 \implies -2K = -2 \implies K = 1 \)


Step 4: Final Answer:

The value of \( K \) that satisfies the equation is \( 1 \).
Quick Tip: When comparing two matrices, always verify the value of the variable from at least two different elements to ensure consistency.

Step 1: Understanding the Concept:

To integrate a rational function where the denominator is a product of linear factors, we use the method of partial fractions.


Step 2: Key Formula or Approach:

Express the integrand as:
\[ \frac{x}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3} \]


Step 3: Detailed Explanation:

Multiply through by the common denominator:
\[ x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) \]

Put \( x = 1 \): \( 1 = A(1-2)(1-3) \implies 1 = A(-1)(-2) \implies A = 1/2 \).

Put \( x = 2 \): \( 2 = B(2-1)(2-3) \implies 2 = B(1)(-1) \implies B = -2 \).

Put \( x = 3 \): \( 3 = C(3-1)(3-2) \implies 3 = C(2)(1) \implies C = 3/2 \).

The integral becomes:
\[ I = \int \left[ \frac{1}{2(x-1)} - \frac{2}{x-2} + \frac{3}{2(x-3)} \right] dx \]
\[ I = \frac{1}{2}\int \frac{1}{x-1} dx - 2\int \frac{1}{x-2} dx + \frac{3}{2}\int \frac{1}{x-3} dx \]
\[ I = \frac{1}{2}\log|x-1| - 2\log|x-2| + \frac{3}{2}\log|x-3| + C \]


Step 4: Final Answer:

The integral is \( \frac{1}{2}\log|x-1| - 2\log|x-2| + \frac{3}{2}\log|x-3| + C \).
Quick Tip: For non-repeated linear factors, use the "cover-up" method to find constants quickly: To find \( A \), cover \( (x-1) \) and put \( x=1 \) in the rest: \( 1/((1-2)(1-3)) = 1/2 \).

Step 1: Understanding the Concept:

The integral involves a quadratic expression under a square root. We solve this by completing the square to transform it into a standard integral form.


Step 2: Key Formula or Approach:

Standard form: \( \int \sqrt{x^2 + a^2} dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| + C \).


Step 3: Detailed Explanation:

Complete the square for \( x^2 + 2x + 5 \):
\[ x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x+1)^2 + 2^2 \]

Let \( I = \int \sqrt{(x+1)^2 + 2^2} dx \).

Let \( t = x+1 \), then \( dt = dx \).
\[ I = \int \sqrt{t^2 + 2^2} dt \]

Using the formula with \( a = 2 \):
\[ I = \frac{t}{2}\sqrt{t^2 + 2^2} + \frac{2^2}{2}\log|t + \sqrt{t^2 + 2^2}| + C \]

Substituting \( t = x+1 \) back:
\[ I = \frac{x+1}{2}\sqrt{x^2+2x+5} + 2\log|x+1+\sqrt{x^2+2x+5}| + C \]


Step 4: Final Answer:

The result of the integration is \( \frac{x+1}{2}\sqrt{x^2+2x+5} + 2\log|x+1+\sqrt{x^2+2x+5}| + C \).
Quick Tip: Remember the triplet of formulas for \( \int \sqrt{x^2 \pm a^2} \) and \( \int \sqrt{a^2 - x^2} \). They all start with \( \frac{x}{2}(same root) \).

Step 1: Understanding the Concept:

This is a problem of parametric differentiation. We find \( \frac{dy}{dx} \) using the chain rule: \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \).


Step 2: Key Formula or Approach:

1. Differentiate \( x \) with respect to \( \theta \).

2. Differentiate \( y \) with respect to \( \theta \).

3. Divide the results.


Step 3: Detailed Explanation:

Differentiating \( x \) w.r.t \( \theta \):
\[ \frac{dx}{d\theta} = a \left[ -\sin \theta + \frac{d}{d\theta}(\theta \sin \theta) \right] \]

Using product rule: \( \frac{d}{d\theta}(\theta \sin \theta) = \sin \theta + \theta \cos \theta \).
\[ \frac{dx}{d\theta} = a [-\sin \theta + \sin \theta + \theta \cos \theta] = a\theta \cos \theta \]

Differentiating \( y \) w.r.t \( \theta \):
\[ \frac{dy}{d\theta} = a \left[ \cos \theta - \frac{d}{d\theta}(\theta \cos \theta) \right] \]

Using product rule: \( \frac{d}{d\theta}(\theta \cos \theta) = \cos \theta - \theta \sin \theta \).
\[ \frac{dy}{d\theta} = a [\cos \theta - (\cos \theta - \theta \sin \theta)] \]
\[ \frac{dy}{d\theta} = a [\cos \theta - \cos \theta + \theta \sin \theta] = a\theta \sin \theta \]

Now, calculate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{a\theta \sin \theta}{a\theta \cos \theta} = \tan \theta \]


Step 4: Final Answer:

The derivative \( \frac{dy}{dx} \) is \( \tan \theta \).
Quick Tip: Be very careful with negative signs when applying the product rule inside brackets. Distributing the minus sign is a common place for errors.

Step 1: Understanding the Concept:

This problem involves second-order derivatives. We differentiate the given function twice and rearrange to form the desired differential equation.


Step 2: Key Formula or Approach:

Differentiate \( y \) with respect to \( x \), then rearrange the expression before differentiating again to simplify the process.


Step 3: Detailed Explanation:

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

Differentiating w.r.t \( x \):
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \]

Cross-multiplying:
\[ \sqrt{1 - x^2} \frac{dy}{dx} = 1 \]

Squaring both sides to remove the root:
\[ (1 - x^2) \left( \frac{dy}{dx} \right)^2 = 1 \]

Differentiating again w.r.t \( x \) using product rule and chain rule:
\[ (1 - x^2) \cdot \frac{d}{dx} \left( \frac{dy}{dx} \right)^2 + \left( \frac{dy}{dx} \right)^2 \cdot \frac{d}{dx}(1 - x^2) = 0 \]
\[ (1 - x^2) \cdot 2 \left( \frac{dy}{dx} \right) \cdot \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 \cdot (-2x) = 0 \]

Divide the entire equation by \( 2 \frac{dy}{dx} \) (assuming \( \frac{dy}{dx} \neq 0 \)):
\[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 0 \]


Step 4: Final Answer:

The given identity \( (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0 \) is proved.
Quick Tip: Squaring after the first derivative often makes second differentiation much easier than using the quotient rule directly.