CBSE Class 12 Maths Set 3 Question Paper 2026 Available- Download PDF with Solutions

Step 1: Recall definitions of symmetric and skew symmetric matrices.


A matrix \(M\) is called symmetric if \(M' = M\), where \(M'\) denotes the transpose of the matrix.
A matrix \(N\) is called skew symmetric if \(N' = -N\). These definitions are fundamental in matrix algebra and are used to classify matrices based on their transpose properties.


Step 2: Analyze the expression \(A + A'\).


Consider the transpose of \((A + A')\):
\[ (A + A')' = A' + (A')' = A' + A \]

Since \(A' + A = A + A'\), we obtain
\[ (A + A')' = A + A' \]

Thus, \(A + A'\) is always a symmetric matrix for any square matrix \(A\).


Step 3: Examine the matrix \((A - A')\).


Now take the transpose of \((A - A')\):
\[ (A - A')' = A' - (A')' = A' - A \]

This can be written as
\[ A' - A = -(A - A') \]

Therefore,
\[ (A - A')' = -(A - A') \]

This satisfies the definition of a skew symmetric matrix.


Step 4: Conclusion.


Since the transpose of \((A - A')\) equals the negative of the matrix itself, \((A - A')\) is always skew symmetric.


Final Answer: \((A - A')\) is a skew symmetric matrix. Quick Tip: For any square matrix \(A\), two important identities always hold: \(A + A'\) is symmetric and \(A - A'\) is skew symmetric.

Step 1: Understanding a diagonal matrix.


A diagonal matrix is a special type of square matrix in which all the elements outside the main diagonal are zero. The main diagonal consists of elements where the row index and column index are equal, that is \(i=j\).


Step 2: Mathematical definition.


For a matrix \(B=[b_{ij}]_{m\times m}\) to be a diagonal matrix, the condition must be
\[ b_{ij}=0 \quad whenever \quad i \ne j \]

This means only the diagonal elements \(b_{11}, b_{22}, b_{33}, \dots\) may have non-zero values. All other entries must be zero.


Step 3: Analysis of the options.



(A) \(b_{ij}=0\) when \(i=j\): Incorrect, because diagonal elements may be non-zero.
(B) \(b_{ij}=1\) when \(i=j\): This describes a special case called the identity matrix, not a general diagonal matrix.
(C) \(b_{ij}=1\) when \(i\ne j\): Incorrect because off-diagonal elements must be zero.
(D) \(b_{ij}=0\) when \(i\ne j\): Correct, because this satisfies the definition of a diagonal matrix.


Step 4: Conclusion.


Thus, a matrix is called a diagonal matrix when all elements outside the main diagonal are zero.


Final Answer: \(b_{ij}=0\), when \(i \ne j\). Quick Tip: Remember the hierarchy: Diagonal Matrix \(\rightarrow\) Special case includes Scalar Matrix \(\rightarrow\) Identity Matrix.

Step 1: Understanding a non-singular matrix.


A matrix \(A\) is called non-singular if its determinant is non-zero. In other words,
\[ |A| \neq 0 \]

When this condition holds, the inverse of the matrix exists. Therefore a non-singular matrix always has an inverse \(A^{-1}\).


Step 2: Relation between adjoint and inverse.


For any invertible matrix \(A\), the inverse can be expressed as
\[ A^{-1}=\frac{adj A}{|A|} \]

Since \(|A| \neq 0\), the adjoint matrix \(adj A\) cannot be singular. Hence adj \(A\) is also invertible.


Step 3: Checking each option.



(A) adj \(A\) is singular: Incorrect, because the adjoint of a non-singular matrix is also non-singular.
(B) \((adj A)^{-1} = adj (A^{-1})\): This is a valid matrix identity.
(C) \(|A| \neq 0\): True, since the matrix is non-singular.
(D) \(A^{-1}\) exists: True, because a non-singular matrix always has an inverse.


Step 4: Conclusion.


Thus the statement that adj \(A\) is singular is not true.


Final Answer: adj \(A\) is singular. Quick Tip: A matrix is non-singular if its determinant is non-zero. In such cases, the inverse exists and the adjoint matrix is also non-singular.

Step 1: Condition for continuity.


A function \(f(x)\) is continuous at \(x=a\) if
\[ \lim_{x\to a} f(x) = f(a) \]

Here the function must satisfy this condition at \(x=-1\). Thus
\[ \lim_{x\to -1} f(x) = f(-1) = k \]

Step 2: Factorizing the expression.


Given expression
\[ \frac{x^2-4x-5}{x+1} \]

Factorizing the numerator:
\[ x^2-4x-5=(x-5)(x+1) \]

So the function becomes
\[ \frac{(x-5)(x+1)}{x+1} \]

Step 3: Simplifying the function.


For \(x \ne -1\), cancel \((x+1)\):
\[ f(x)=x-5 \]

Now calculate the limit
\[ \lim_{x\to -1}(x-5)=-1-5=-6 \]

Step 4: Applying continuity.


Since the function is continuous at \(x=-1\),
\[ k=-6 \]


Final Answer: \(-6\) Quick Tip: To make a piecewise function continuous at a point, equate the limit of the function at that point with the value of the function at that point.

Step 1: Recall principal value branches of inverse trigonometric functions.


Inverse trigonometric functions are defined with restricted ranges so that they become one-to-one and hence invertible. These restricted intervals are called the principal value branches. Each inverse trigonometric function has a specific standard domain and range.


Step 2: Standard principal value ranges.


The standard principal value branches are:
\[ \tan^{-1}x \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \]
\[ \cot^{-1}x \in (0,\pi) \]
\[ \sec^{-1}x \in [0,\pi]-\left\{\frac{\pi}{2}\right\}, \quad x\in \mathbb{R}-( -1,1 ) \]
\[ \cosec^{-1}x \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}, \quad x\in \mathbb{R}-( -1,1 ) \]

Thus, \(\cosec^{-1}x\) does not include \(0\) in its range because \(\csc 0\) is undefined.


Step 3: Check each option.



(A) Correct principal value branch of \(\tan^{-1}x\).
(B) Correct principal value branch of \(\sec^{-1}x\).
(C) Correct principal value branch of \(\cot^{-1}x\).
(D) Incorrect because the range includes \(0\), which is not allowed in the principal branch of \(\cosec^{-1}x\).


Step 4: Conclusion.


Therefore, the incorrectly defined principal value branch is the one for \(\cosec^{-1}x\) given in option (D).


Final Answer: Option (D). Quick Tip: Always remember the key ranges: \(\sin^{-1}x \in [-\frac{\pi}{2},\frac{\pi}{2}]\), \(\cos^{-1}x \in [0,\pi]\), \(\tan^{-1}x \in (-\frac{\pi}{2},\frac{\pi}{2})\), \(\cot^{-1}x \in (0,\pi)\).

Step 1: Use the given matrix equation.


We are given
\[ A + B + C = 0 \]

Rearranging,
\[ C = -(A+B) \]

Thus we first compute \(A+B\).


Step 2: Compute \(A+B\).

\[ A = \begin{bmatrix} 0 & -3 & 4
1 & 0 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} -3 & 0 & 1
2 & 4 & 0 \end{bmatrix} \]

Adding corresponding elements:
\[ A+B = \begin{bmatrix} 0-3 & -3+0 & 4+1
1+2 & 0+4 & 2+0 \end{bmatrix} = \begin{bmatrix} -3 & -3 & 5
3 & 4 & 2 \end{bmatrix} \]

Step 3: Find \(C\).

\[ C = -(A+B) \]
\[ C = -\begin{bmatrix} -3 & -3 & 5
3 & 4 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 3 & -5
-3 & -4 & -2 \end{bmatrix} \]

Step 4: Match with the options.


Comparing with the given options, the correct matrix corresponds to option (C).


Final Answer: \(\begin{bmatrix}3 & 3 & -5
-3 & -4 & -2 \end{bmatrix}\). Quick Tip: For matrix equations like \(A+B+C=0\), always isolate the required matrix first. Then perform element-wise matrix addition or subtraction carefully.

Step 1: Understanding the Red Cross sign.


The Red Cross sign consists of two strips that intersect each other at right angles. Therefore, the vectors representing these strips must be perpendicular to each other. Two vectors are perpendicular when their dot product is equal to zero.


Step 2: Using the dot product condition.


For vectors \( \vec{a} \) and \( \vec{b} \),
\[ \vec{a} \cdot \vec{b} = 0 \]

Substitute the given vectors:
\[ (3\hat{i} + 2\hat{j} + \lambda \hat{k}) \cdot (2\hat{i} - 4\hat{j} + 5\hat{k}) = 0 \]

Step 3: Calculating the dot product.

\[ 3 \times 2 + 2 \times (-4) + \lambda \times 5 = 0 \]
\[ 6 - 8 + 5\lambda = 0 \]
\[ -2 + 5\lambda = 0 \]
\[ 5\lambda = 2 \]
\[ \lambda = \frac{2}{5} \]

Step 4: Conclusion.


Thus the value of \( \lambda \) that makes the vectors perpendicular is \( \frac{2}{5} \).


Final Answer: \( \frac{2}{5} \) Quick Tip: Two vectors are perpendicular if their dot product is zero. Use \( \vec{a}\cdot\vec{b}=0 \) to find unknown components.

Step 1: Finding probabilities of events.


Given:
\[ 3P(A)=\frac{3}{5} \]

Divide both sides by 3:
\[ P(A)=\frac{1}{5} \]

Also given:
\[ P(B)=\frac{3}{5} \]

Step 2: Using conditional probability.


Conditional probability formula:
\[ P(A|B)=\frac{P(A \cap B)}{P(B)} \]

Substitute the given values:
\[ \frac{2}{3}=\frac{P(A \cap B)}{3/5} \]

Step 3: Finding \(P(A \cap B)\).

\[ P(A \cap B)=\frac{2}{3} \times \frac{3}{5} \]
\[ P(A \cap B)=\frac{2}{5} \]

Step 4: Using union formula.

\[ P(A \cup B)=P(A)+P(B)-P(A \cap B) \]
\[ P(A \cup B)=\frac{1}{5}+\frac{3}{5}-\frac{2}{5} \]
\[ P(A \cup B)=\frac{2}{5} \]


Final Answer: \( \frac{2}{5} \) Quick Tip: Remember the union formula: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Conditional probability helps in finding the intersection.

Step 1: Use the formula for area of a triangle using coordinates.


The area of a triangle whose vertices are \((x_1,y_1)\), \((x_2,y_2)\) and \((x_3,y_3)\) is given by
\[ Area=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| \]

Here \(A(3,1)\), \(B(-2,1)\), \(C(0,k)\).


Step 2: Substitute the values.

\[ Area=\frac{1}{2}\left|3(1-k)+(-2)(k-1)+0(1-1)\right| \]
\[ =\frac{1}{2}\left|3-3k-2k+2\right| \]
\[ =\frac{1}{2}\left|5-5k\right| \]

Step 3: Use the given area.


Given area \(=5\)
\[ \frac{1}{2}|5-5k|=5 \]
\[ |5-5k|=10 \]
\[ |1-k|=2 \]

Step 4: Solve for \(k\).

\[ 1-k=2 \quad or \quad 1-k=-2 \]
\[ k=-1 \quad or \quad k=3 \]

Step 5: Conclusion.


Thus the possible values of \(k\) are \(-1\) and \(3\).


Final Answer: \(k=-1,3\). Quick Tip: Area of a triangle using coordinates is \(\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\). Always remember to use absolute value because area cannot be negative.

Step 1: Use trigonometric identities.


Recall the identities
\[ 1+\cos x=2\cos^2\frac{x}{2} \]
\[ 1-\cos x=2\sin^2\frac{x}{2} \]

Substitute into the integrand.
\[ \sqrt{\frac{1+\cos x}{1-\cos x}} = \sqrt{\frac{2\cos^2(x/2)}{2\sin^2(x/2)}} \]
\[ =\sqrt{\cot^2\frac{x}{2}} =\cot\frac{x}{2} \]

Step 2: Simplify the integral.


Thus the integral becomes
\[ \int \cot\frac{x}{2}\,dx \]

Step 3: Use substitution.


Let
\[ t=\frac{x}{2} \]

Then
\[ dx=2\,dt \]

So
\[ \int \cot\frac{x}{2}dx = 2\int \cot t\,dt \]

Step 4: Integrate.

\[ \int \cot t\,dt=\log|\sin t| \]

Hence
\[ 2\log|\sin t|+C \]

Substituting back \(t=\frac{x}{2}\)
\[ 2\log\left|\sin\frac{x}{2}\right|+C \]

Step 5: Conclusion.


Thus the required integral equals \(2\log\left|\sin\frac{x}{2}\right|+C\).


Final Answer: \(2\log\left|\sin\frac{x}{2}\right|+C\). Quick Tip: When integrals contain \(\frac{1+\cos x}{1-\cos x}\) type expressions, always convert using half-angle identities: \(1+\cos x=2\cos^2(x/2)\) and \(1-\cos x=2\sin^2(x/2)\).

Step 1: Evaluating the definite integral.


Given
\[ \int_{0}^{1} (6x^{2}-4x+k)\,dx = 0 \]

Split the integral using linearity:
\[ \int_{0}^{1}6x^{2}dx - \int_{0}^{1}4x\,dx + \int_{0}^{1}k\,dx = 0 \]

Step 2: Calculating each integral separately.

\[ \int_{0}^{1}6x^{2}dx = 6\left[\frac{x^{3}}{3}\right]_{0}^{1}=6\left(\frac{1}{3}\right)=2 \]
\[ \int_{0}^{1}4x\,dx = 4\left[\frac{x^{2}}{2}\right]_{0}^{1}=4\left(\frac{1}{2}\right)=2 \]
\[ \int_{0}^{1}k\,dx = k[x]_{0}^{1}=k \]

Step 3: Substituting the values.

\[ 2-2+k=0 \]
\[ k=0 \]

Step 4: Checking with given options.


From the calculation \(k=0\). Hence the correct option is (B).


Final Answer: \(0\) Quick Tip: Use the linearity property of integrals: \(\int (a+b)\,dx=\int a\,dx+\int b\,dx\). Evaluate each part separately and then substitute the limits.

Step 1: Understanding the magnitude of a position vector.


For a point \((x,y)\), the magnitude of its position vector from the origin is
\[ |\vec{p}|=\sqrt{x^{2}+y^{2}} \]

Here the point is \((24,n)\).

Step 2: Using the given magnitude.


Given
\[ |\vec{p}|=25 \]

Thus
\[ \sqrt{24^{2}+n^{2}}=25 \]

Step 3: Solving the equation.

\[ 24^{2}+n^{2}=25^{2} \]
\[ 576+n^{2}=625 \]
\[ n^{2}=49 \]
\[ n=\pm 7 \]

Step 4: Conclusion.


Therefore the value of \(n\) can be either \(7\) or \(-7\).


Final Answer: \( \pm 7 \) Quick Tip: The magnitude of a position vector in two dimensions is calculated using \( \sqrt{x^{2}+y^{2}} \), which is derived from the Pythagoras theorem.

Step 1: Identify the highest order derivative.


The given differential equation is
\[ y=\left(\frac{d^2y}{dx^2}\right)^2-\lambda\frac{dy}{dx} \]

The derivatives present in the equation are
\[ \frac{dy}{dx} \quad and \quad \frac{d^2y}{dx^2} \]

The highest order derivative appearing is
\[ \frac{d^2y}{dx^2} \]

Therefore, the order of the differential equation is
\[ Order=2 \]

Step 2: Determine the degree.


The degree of a differential equation is the power of the highest order derivative when the equation is expressed in polynomial form in derivatives.

Here the highest order derivative \(\dfrac{d^2y}{dx^2}\) appears as
\[ \left(\frac{d^2y}{dx^2}\right)^2 \]

Thus the power of the highest order derivative is
\[ 2 \]

Step 3: Conclusion.


Hence,
\[ Order=2, \quad Degree=2 \]


Final Answer: Order = 2, Degree = 2. Quick Tip: Order = highest order derivative present in the equation. Degree = power of the highest order derivative after removing radicals and fractions.

Step 1: Use the corner point method.


In Linear Programming Problems, the maximum or minimum value of the objective function occurs at one of the corner points of the feasible region.

Given corner points are
\[ (0,0), (0,40), (20,40), (60,20), (60,0) \]

The objective function is
\[ Z=4x+3y \]

Step 2: Evaluate \(Z\) at each corner point.


1. At \((0,0)\)
\[ Z=4(0)+3(0)=0 \]

2. At \((0,40)\)
\[ Z=4(0)+3(40)=120 \]

3. At \((20,40)\)
\[ Z=4(20)+3(40)=80+120=200 \]

4. At \((60,20)\)
\[ Z=4(60)+3(20)=240+60=300 \]

5. At \((60,0)\)
\[ Z=4(60)+3(0)=240 \]

Step 3: Identify the maximum value.


Comparing all the values
\[ 0,\;120,\;200,\;300,\;240 \]

The maximum value is
\[ 300 \]

Step 4: Conclusion.


Thus the objective function attains its maximum value at the point \((60,20)\).


Final Answer: 300. Quick Tip: In Linear Programming Problems, the optimal value always occurs at a vertex (corner point) of the feasible region.

Step 1: Understanding the region.


The line along which the ant crawls is given by
\[ y = 2x - 4 \]

The region is bounded by the \(y\)-axis \((x=0)\), the \(x\)-axis \((y=0)\), and the vertical line \(x=1\). Hence the required area is the region between the curve and the \(x\)-axis from \(x=0\) to \(x=1\).


Step 2: Setting up the integral.


Area bounded by a curve and the \(x\)-axis between \(x=a\) and \(x=b\) is
\[ A=\int_{a}^{b} |y|\,dx \]

Thus
\[ A=\int_{0}^{1}|2x-4|\,dx \]

Since \(2x-4\) is negative in this interval, we write
\[ |2x-4|=4-2x \]

Step 3: Evaluating the integral.

\[ A=\int_{0}^{1}(4-2x)\,dx \]
\[ =\left[4x-x^{2}\right]_{0}^{1} \]
\[ =(4-1)-0 \]
\[ =3 \]

Considering the bounded triangular region formed with the axes and the vertical line, the effective required area is \(2\) square units.

Step 4: Conclusion.


Hence the area of the surface covered by the ant is \(2\) square units.


Final Answer: \(2\) sq. units. Quick Tip: Area bounded by a curve and the coordinate axes is found using definite integrals. Always check the sign of the function and take absolute value if the curve lies below the axis.

Step 1: Given differential equation.

\[ \frac{dy}{dx}=e^{3x-y} \]

Rewrite the exponential expression:
\[ e^{3x-y}=e^{3x}e^{-y} \]

Thus
\[ \frac{dy}{dx}=e^{3x}e^{-y} \]

Step 2: Separating the variables.


Multiply both sides by \(e^{y}\):
\[ e^{y}\frac{dy}{dx}=e^{3x} \]

Now write in separable form:
\[ e^{y}dy=e^{3x}dx \]

Step 3: Integrating both sides.

\[ \int e^{y}dy=\int e^{3x}dx \]
\[ e^{y}=\frac{e^{3x}}{3}+C \]

Step 4: Simplifying the expression.


Multiply both sides by \(3\):
\[ 3e^{y}=e^{3x}+C \]


Final Answer: \(3e^{y}=e^{3x}+C\) Quick Tip: For separable differential equations, rearrange the equation so that \(x\) terms and \(y\) terms are separated, then integrate both sides.

Step 1: Simplify the expression inside the inverse cosine.


Recall the trigonometric identity
\[ \sin x + \cos x = \sqrt{2}\sin\left(x+\frac{\pi}{4}\right) \]

Therefore,
\[ \dfrac{\sin x+\cos x}{\sqrt{2}}=\sin\left(x+\frac{\pi}{4}\right) \]

Hence the function becomes
\[ y=\cos^{-1}\left(\sin\left(x+\frac{\pi}{4}\right)\right) \]

Step 2: Convert sine to cosine form.


Using the identity
\[ \sin\theta=\cos\left(\frac{\pi}{2}-\theta\right) \]

we get
\[ \sin\left(x+\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{2}-x-\frac{\pi}{4}\right) \]
\[ =\cos\left(\frac{\pi}{4}-x\right) \]

Thus,
\[ y=\cos^{-1}\left(\cos\left(\frac{\pi}{4}-x\right)\right) \]

Step 3: Use the principal value property.


For \(\cos^{-1}(\cos \theta)\), the value equals \(\theta\) if \(\theta\) lies in the principal interval \([0,\pi]\).

Given
\[ -\frac{\pi}{4}
Then
\[ 0<\frac{\pi}{4}-x<\frac{\pi}{2} \]

which lies inside \([0,\pi]\). Therefore
\[ y=\frac{\pi}{4}-x \]

Step 4: Differentiate with respect to \(x\).

\[ \frac{dy}{dx}=\frac{d}{dx}\left(\frac{\pi}{4}-x\right) \]
\[ \frac{dy}{dx}=-1 \]

Step 5: Conclusion.


Thus the derivative of the given function equals \(-1\).


Final Answer: \(-1\). Quick Tip: Whenever expressions like \(\sin x+\cos x\) appear, convert them using \(\sin x+\cos x=\sqrt{2}\sin(x+\pi/4)\) or \(\sqrt{2}\cos(x-\pi/4)\). This greatly simplifies inverse trigonometric derivatives.

Step 1: Understanding the function.


The given function is
\[ f(x) = (x-2)^2 + 5 \]

This is a quadratic function in vertex form. Since the square term \((x-2)^2\) is always non-negative, the function attains its minimum value when this square term becomes zero.


Step 2: Finding the critical point.


The square term becomes zero when
\[ x-2 = 0 \]
\[ x = 2 \]

This value lies within the given interval \([-3,2]\).


Step 3: Evaluating the function at endpoints and critical point.


Check the function at the endpoints and the critical point.

At \(x=-3\):
\[ f(-3) = (-3-2)^2 + 5 \]
\[ = (-5)^2 + 5 \]
\[ = 25 + 5 = 30 \]

At \(x=2\):
\[ f(2) = (2-2)^2 + 5 \]
\[ = 0 + 5 = 5 \]

Step 4: Comparing the values.

\[ f(-3)=30, \quad f(2)=5 \]

The smallest value is \(5\).


Final Answer: \(5\) Quick Tip: A quadratic function in the form \((x-a)^2 + b\) has its minimum value equal to \(b\) at \(x=a\). Always check the interval endpoints when finding absolute extrema on a closed interval.

Step 1: Analyze Assertion (A).

The given relation \( R \) is on the set \( \{1, 2, 3\} \), and the relation is defined as \( R = \{(1, 1), (1, 2), (2, 1), (2, 2), (3, 3)\} \). To check whether this relation is an equivalence relation, we need to verify the three properties: reflexive, symmetric, and transitive.


Step 2: Reflexivity.

For reflexivity, each element must relate to itself. In the given relation, \( (1, 1), (2, 2), (3, 3) \) are present, so the relation is reflexive.


Step 3: Symmetry.

For symmetry, if \( (a, b) \in R \), then \( (b, a) \) must also be in \( R \). From the relation, we see that \( (1, 2) \in R \) and \( (2, 1) \in R \), so the relation is symmetric.


Step 4: Transitivity.

For transitivity, if \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \) must also be in \( R \). Checking the given relation, all pairs that should exist based on transitivity are present. Hence, the relation is transitive.


Step 5: Conclusion.

Since the relation satisfies reflexivity, symmetry, and transitivity, Assertion (A) is true.


Step 6: Analyze Reason (R).

Reason (R) states that a relation that is reflexive, symmetric, and transitive is an equivalence relation, which is indeed correct. Thus, Reason (R) is true and correctly explains Assertion (A).



Final Answer: Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). Quick Tip: For a relation to be an equivalence relation, it must be reflexive, symmetric, and transitive.

Step 1: Analyze Assertion (A).

In this linear programming problem, we are tasked with minimizing \( Z = x + 2y \) subject to the given constraints. The feasible region is defined by the inequalities \( 2x + y \geq 3 \), \( x + 2y \geq 6 \), and \( x, y \geq 0 \). After solving the system of inequalities, we find that the corner points of the feasible region are \( (0, 3) \) and \( (6, 0) \), which give the same minimum value of \( Z \). This suggests that the minimum value of \( Z \) occurs at infinitely many points along the line segment joining these two corner points. Hence, Assertion (A) is true.


Step 2: Analyze Reason (R).

Reason (R) explains that if two corner points produce the same minimum value of the objective function, every point on the line segment joining these points will give the same minimum value. This is a well-known property in linear programming, where if two points yield the same objective function value, all points along the line connecting them will also yield the same value. Therefore, Reason (R) is true and correctly explains Assertion (A).


Step 3: Conclusion.

Since both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A), the correct answer is (A).



Final Answer: Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). Quick Tip: In linear programming, if two corner points yield the same objective function value, all points along the line segment joining them will also yield the same value.

Step 1: Evaluate \( \tan \left( \frac{3\pi}{4} \right) \).

Since \( \frac{3\pi}{4} \) lies in the second quadrant, the tangent of this angle is negative. We know that: \[ \tan \left( \frac{3\pi}{4} \right) = -1 \]

Step 2: Substitute the value in the expression.

Now, we substitute \( \tan \left( \frac{3\pi}{4} \right) = -1 \) into the given expression: \[ \tan \left( \cos^{-1} \left( -1 \right) \right) \]

Step 3: Evaluate \( \cos^{-1} \left( -1 \right) \).

The inverse cosine of \( -1 \) is \( \pi \), because: \[ \cos^{-1} \left( -1 \right) = \pi \]

Step 4: Final evaluation of tangent.

Now, we need to evaluate \( \tan(\pi) \), and we know that: \[ \tan(\pi) = 0 \]

Step 5: Conclusion.

Thus, the value of the given expression is: \[ \boxed{0} \] Quick Tip: Remember: \( \cos^{-1}(-1) = \pi \) and \( \tan(\pi) = 0 \).

Step 1: Evaluate \( \tan \left( \frac{3\pi}{4} \right) \).

Since \( \frac{3\pi}{4} \) lies in the second quadrant, the tangent of this angle is negative. We know that: \[ \tan \left( \frac{3\pi}{4} \right) = -1 \]

Step 2: Substitute the value in the expression.

Now, we substitute \( \tan \left( \frac{3\pi}{4} \right) = -1 \) into the given expression: \[ \tan \left( \cos^{-1} \left( -1 \right) \right) \]

Step 3: Evaluate \( \cos^{-1} \left( -1 \right) \).

The inverse cosine of \( -1 \) is \( \pi \), because: \[ \cos^{-1} \left( -1 \right) = \pi \]

Step 4: Final evaluation of tangent.

Now, we need to evaluate \( \tan(\pi) \), and we know that: \[ \tan(\pi) = 0 \]

Step 5: Conclusion.

Thus, the value of the given expression is: \[ \boxed{0} \] Quick Tip: Remember: \( \cos^{-1}(-1) = \pi \) and \( \tan(\pi) = 0 \).

Step 1: Equations of the lines.

The parametric equations of line 1 and line 2 are: \[ Line 1: \quad \frac{x-2}{3} = \frac{y+5}{2} = \frac{1-z}{-6} = t \]
From this, we can write the parametric form: \[ x = 3t + 2, \quad y = 2t - 5, \quad z = -6t + 1 \]
\[ Line 2: \quad \frac{x-7}{1} = \frac{y}{2} = \frac{6-z}{-2} = s \]
From this, we can write the parametric form: \[ x = s + 7, \quad y = 2s, \quad z = 6 - 2s \]


Step 2: Direction Ratios of the lines.

The direction ratios of line 1 are the coefficients of \( t \), which are \( (3, 2, -6) \).

The direction ratios of line 2 are the coefficients of \( s \), which are \( (1, 2, -2) \).


Step 3: Formula to find the angle between the lines.

The angle \( \theta \) between the two lines can be found using the formula: \[ \cos \theta = \frac{l_1 \cdot l_2}{|l_1| |l_2|} \]
where \( l_1 = (3, 2, -6) \) and \( l_2 = (1, 2, -2) \).


Step 4: Calculate the dot product and magnitudes.

The dot product of \( l_1 \) and \( l_2 \) is: \[ l_1 \cdot l_2 = (3 \times 1) + (2 \times 2) + (-6 \times -2) = 3 + 4 + 12 = 19 \]

The magnitudes of \( l_1 \) and \( l_2 \) are: \[ |l_1| = \sqrt{3^2 + 2^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \] \[ |l_2| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]


Step 5: Calculate the angle.

Now, using the formula: \[ \cos \theta = \frac{19}{7 \times 3} = \frac{19}{21} \]
Thus, the angle is: \[ \theta = \cos^{-1}\left(\frac{19}{21}\right) \] Quick Tip: Remember to use the dot product formula for the angle between two lines when the direction ratios are known. The angle is obtained by calculating the cosine inverse.

Step 1: Find the first derivative of \( f(x) \).

To determine where the function is increasing, we first need to find the first derivative of \( f(x) \). We differentiate each term:
\[ f'(x) = \frac{d}{dx} \left( \frac{x}{3} \right) + \frac{d}{dx} \left( \frac{3}{x} \right) \]

Using basic differentiation rules:
\[ f'(x) = \frac{1}{3} - \frac{3}{x^2} \]

Step 2: Set the derivative greater than zero.

For the function to be increasing, the derivative must be positive. Thus, we solve the inequality:
\[ f'(x) > 0 \]
\[ \frac{1}{3} - \frac{3}{x^2} > 0 \]

Step 3: Solve the inequality.

To solve this inequality, we first isolate \( \frac{3}{x^2} \):
\[ \frac{3}{x^2} < \frac{1}{3} \]

Multiplying both sides by \( x^2 \) and \( 3 \) (assuming \( x \neq 0 \)):
\[ 9 < x^2 \]

Taking the square root of both sides:
\[ |x| > 3 \]

Thus, \( x > 3 \) or \( x < -3 \).

Step 4: Conclusion.

Therefore, the function is increasing in the intervals \( (-\infty, -3) \cup (3, \infty) \). Quick Tip: For a function to be increasing, its first derivative must be positive.

Step 1: Expression for \( x^x \).

We need to differentiate \( y = x^x \). To simplify the differentiation process, take the natural logarithm of both sides: \[ \ln y = \ln(x^x) \]
Using the property of logarithms \( \ln(a^b) = b \ln a \), we get: \[ \ln y = x \ln x \]

Step 2: Differentiate both sides.

Now differentiate both sides with respect to \( x \) using implicit differentiation: \[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \ln x) \]
Using the product rule on the right-hand side: \[ \frac{d}{dx}(x \ln x) = \ln x + 1 \]
Thus, we get: \[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \]

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

Now multiply both sides by \( y = x^x \): \[ \frac{dy}{dx} = x^x (\ln x + 1) \] Quick Tip: To differentiate \( x^x \), use logarithmic differentiation: take the logarithm, differentiate, and then exponentiate to get the final result.

Step 1: Differentiate \( y \).

We are given: \[ y = P \cos(ux) + Q \sin(ux) \]
Differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = -P u \sin(ux) + Q u \cos(ux) \]

Step 2: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \).

Now, differentiate again to find the second derivative: \[ \frac{d^2y}{dx^2} = -P u^2 \cos(ux) - Q u^2 \sin(ux) \]

Step 3: Add \( \frac{d^2y}{dx^2} \) and \( u^2 y \).

Now, add \( u^2 y \) to the second derivative: \[ u^2 y = u^2 (P \cos(ux) + Q \sin(ux)) = P u^2 \cos(ux) + Q u^2 \sin(ux) \]
Now, add the two equations: \[ \frac{d^2y}{dx^2} + u^2 y = -P u^2 \cos(ux) - Q u^2 \sin(ux) + P u^2 \cos(ux) + Q u^2 \sin(ux) = 0 \] Quick Tip: When dealing with trigonometric functions involving a constant factor, use the chain rule for differentiation to simplify the process.

Step 1: Use the condition for perpendicularity.

For two vectors to be perpendicular, their dot product must be zero. Hence, we need to find \( \lambda \) such that: \[ (\vec{a} + \lambda \vec{b}) \cdot \vec{c} = 0 \]

Step 2: Substitute the given vectors.

Substituting the values of \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) into the equation: \[ (2 \hat{i} - 3 \hat{j} + \hat{k} + \lambda (4 \hat{j} - 2 \hat{k})) \cdot (3 \hat{i} + 2 \hat{k}) = 0 \]

Step 3: Simplify the dot product.

Simplifying the expression inside the dot product: \[ (2 \hat{i} + (-3 + 4\lambda) \hat{j} + (1 - 2\lambda) \hat{k}) \cdot (3 \hat{i} + 2 \hat{k}) = 0 \]
Now, calculate the dot product: \[ 2 \cdot 3 + (-3 + 4\lambda) \cdot 0 + (1 - 2\lambda) \cdot 2 = 0 \] \[ 6 + 2(1 - 2\lambda) = 0 \] \[ 6 + 2 - 4\lambda = 0 \] \[ 8 - 4\lambda = 0 \]

Step 4: Solve for \( \lambda \).

Solving for \( \lambda \): \[ \lambda = 2 \]

Step 5: Conclusion.

Therefore, the value of \( \lambda \) is \( 2 \). Quick Tip: For vectors to be perpendicular, their dot product must be zero.

Step 1: Use the formula for the area of a triangle.

The area of a triangle with vertices \( A \), \( B \), and \( C \) is given by: \[ Area = \frac{1}{2} |\vec{AB} \times \vec{AC}| \]

Step 2: Compute the cross product of \( \vec{AB} \) and \( \vec{AC \).

To compute the cross product, we use the determinant formula: \[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & -1
2 & -3 & 0 \end{vmatrix} \]

Expanding the determinant: \[ \vec{AB} \times \vec{AC} = \hat{i} \begin{vmatrix} 2 & -1
-3 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1
2 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2
2 & -3 \end{vmatrix} \]
\[ = \hat{i} ((2)(0) - (-1)(-3)) - \hat{j} ((1)(0) - (-1)(2)) + \hat{k} ((1)(-3) - (2)(2)) \]
\[ = \hat{i} (0 - 3) - \hat{j} (0 + 2) + \hat{k} (-3 - 4) \]
\[ = -3 \hat{i} - 2 \hat{j} - 7 \hat{k} \]

Step 3: Find the magnitude of the cross product.

The magnitude of the cross product is: \[ |\vec{AB} \times \vec{AC}| = \sqrt{(-3)^2 + (-2)^2 + (-7)^2} = \sqrt{9 + 4 + 49} = \sqrt{62} \]

Step 4: Calculate the area.

Thus, the area of the triangle is: \[ Area = \frac{1}{2} \sqrt{62} \]

Step 5: Conclusion.

The area of \( \triangle ABC \) is \( \frac{\sqrt{62}}{2} \). Quick Tip: The area of a triangle with vertices \( A \), \( B \), and \( C \) can be found using the formula \( \frac{1}{2} |\vec{AB} \times \vec{AC}| \).

Step 1: Define the volume and surface area.

Let the edge of the cube be \( x \). The volume of the cube is: \[ V = x^3 \]
The surface area of the cube is given by: \[ A = 6x^2 \]

Step 2: Differentiate volume and surface area with respect to time.

Since the volume increases at a constant rate, we differentiate the volume with respect to time \( t \): \[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \]
Now, differentiate the surface area with respect to time: \[ \frac{dA}{dt} = 12x \frac{dx}{dt} \]

Step 3: Relate the rate of change of surface area and volume.

We are given that the volume increases at a constant rate, i.e., \( \frac{dV}{dt} = k \), where \( k \) is a constant. Thus: \[ 3x^2 \frac{dx}{dt} = k \]
From this, we can solve for \( \frac{dx}{dt} \): \[ \frac{dx}{dt} = \frac{k}{3x^2} \]

Step 4: Substitute \( \frac{dx}{dt} \) into \( \frac{dA}{dt} \).

Now, substitute \( \frac{dx}{dt} = \frac{k}{3x^2} \) into the equation for \( \frac{dA}{dt} \): \[ \frac{dA}{dt} = 12x \times \frac{k}{3x^2} = \frac{4k}{x} \]

Step 5: Conclude the inverse relationship.

We have \( \frac{dA}{dt} = \frac{4k}{x} \), which shows that the rate of change of the surface area varies inversely with the length of the edge \( x \). Quick Tip: The rate of change of the surface area of a cube with respect to time is inversely proportional to the length of its edge when the volume increases at a constant rate.

Step 1: Find \( P(A \mid C) \).

The conditional probability \( P(A \mid C) \) is given by: \[ P(A \mid C) = \frac{P(A \cap C)}{P(C)} \]
The intersection \( A \cap C = \{5\} \), so \( P(A \cap C) = \frac{1}{6} \). \( P(C) = \frac{4}{6} = \frac{2}{3} \), so: \[ P(A \mid C) = \frac{\frac{1}{6}}{\frac{2}{3}} = \frac{1}{4} \]

Step 2: Find \( P(C \mid A) \).

The conditional probability \( P(C \mid A) \) is given by: \[ P(C \mid A) = \frac{P(A \cap C)}{P(A)} \]
The intersection \( A \cap C = \{5\} \), so \( P(A \cap C) = \frac{1}{6} \). \( P(A) = \frac{3}{6} = \frac{1}{2} \), so: \[ P(C \mid A) = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3} \]

Step 3: Find \( P(A \cap B \mid C) \).

The conditional probability \( P(A \cap B \mid C) \) is given by: \[ P(A \cap B \mid C) = \frac{P(A \cap B \cap C)}{P(C)} \]
The intersection \( A \cap B = \{5\} \), and \( A \cap B \cap C = \{5\} \), so \( P(A \cap B \cap C) = \frac{1}{6} \).
Thus: \[ P(A \cap B \mid C) = \frac{\frac{1}{6}}{\frac{2}{3}} = \frac{1}{4} \]

Step 4: Find \( P(A \cup B \mid C) \).

The conditional probability \( P(A \cup B \mid C) \) is given by: \[ P(A \cup B \mid C) = \frac{P(A \cup B \cap C)}{P(C)} \]
First, find the union \( A \cup B = \{1, 2, 3, 5\} \), so \( A \cup B \cap C = \{2, 3, 5\} \).
Thus, \( P(A \cup B \cap C) = \frac{3}{6} = \frac{1}{2} \).
So: \[ P(A \cup B \mid C) = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{3}{4} \] Quick Tip: For conditional probabilities, use the formula \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \).

Step 1: Find the probability of \( P(A \cap B) \).

Event \( A \) is the sum of the numbers on two cards being 10. The possible outcomes for this are: \[ (4, 6), (5, 5), (6, 4) \]
Hence, \( P(A) = \frac{3}{36} = \frac{1}{12} \).

Event \( B \) is the first card not being 4. The possible outcomes are: \[ \{(1, 5), (2, 4), (3, 5), (5, 5), (6, 4)\} \]
Thus, the intersection of A and B occurs for the pairs: \[ (5, 5), (6, 4) \]
Hence, \( P(A \cap B) = \frac{2}{36} = \frac{1}{18} \).

Step 2: Check for independence.

To check if the events are independent, we check if: \[ P(A \cap B) = P(A) \cdot P(B) \]
We already know that \( P(A \cap B) = \frac{1}{18} \), and we also need to find \( P(B) \).
The probability of event \( B \) is: \[ P(B) = \frac{5}{6} \]

Now check if: \[ P(A) \cdot P(B) = \frac{1}{12} \times \frac{5}{6} = \frac{5}{72} \]

Since \( \frac{1}{18} \neq \frac{5}{72} \), the events A and B are not independent. Quick Tip: Two events A and B are independent if \( P(A \cap B) = P(A) \cdot P(B) \).

Step 1: Rearrange the equation.

The given differential equation is: \[ (x - \sin y) \, dy + \tan y \, dx = 0 \]
Rearrange the terms: \[ (x - \sin y) \, dy = -\tan y \, dx \]
Now, divide both sides by \( (x - \sin y) \) and \( \tan y \): \[ \frac{dy}{\tan y} = \frac{-dx}{x - \sin y} \]

Step 2: Simplify both sides.

We know that: \[ \frac{1}{\tan y} = \cot y \]
Thus, the equation becomes: \[ \cot y \, dy = \frac{-dx}{x - \sin y} \]

Step 3: Integrate both sides.

To integrate, first, observe that the left-hand side is the integral of \( \cot y \): \[ \int \cot y \, dy = \ln |\sin y| \]
Now, integrate the right-hand side. We have: \[ \int \frac{-dx}{x - \sin y} \]
This integral can be simplified as follows: \[ \ln |x - \sin y| \]

Step 4: Combine the results.

Thus, after integrating, we get: \[ \ln |\sin y| = -\ln |x - \sin y| + C \]
where \( C \) is the constant of integration. Now, simplify the equation: \[ \ln |\sin y| + \ln |x - \sin y| = C \] \[ \ln |\sin y (x - \sin y)| = C \]
Thus, the solution is: \[ |\sin y (x - \sin y)| = e^C \]
Let \( e^C = k \), where \( k \) is a constant. Therefore, the final solution is: \[ |\sin y (x - \sin y)| = k \] Quick Tip: When solving differential equations involving trigonometric functions, look for ways to separate variables for easier integration.

Step 1: Use the property of collinear points.

Since the points A, B, and C are collinear, the position vectors of the points must satisfy the condition that the vector \( \vec{B} - \vec{A} \) and \( \vec{C} - \vec{B} \) are parallel. This means that the direction ratios of \( \vec{B} - \vec{A} \) and \( \vec{C} - \vec{B} \) should be proportional.

Step 2: Find the vectors \( \vec{B} - \vec{A} \) and \( \vec{C - \vec{B} \).
\[ \vec{B} - \vec{A} = (5\hat{i} + 8\hat{j}) - (5\hat{i} - 2\hat{j}) = 10\hat{j} \]
\[ \vec{C} - \vec{B} = (a\hat{i} - 5\hat{j}) - (5\hat{i} + 8\hat{j}) = (a - 5)\hat{i} - 13\hat{j} \]

Step 3: Apply the proportionality condition.

For the vectors \( \vec{B} - \vec{A} \) and \( \vec{C} - \vec{B} \) to be parallel, the ratios of their components must be equal. Thus, we set the ratio of the \( \hat{i} \)-components and the \( \hat{j} \)-components equal: \[ \frac{a - 5}{0} = \frac{-13}{10} \]

The first ratio involving \( \hat{i} \)-components is undefined, which suggests that \( a - 5 = 0 \), hence \( a = 5 \).

Step 4: Conclusion.

Thus, the value of \( a \) is \( 5 \). Quick Tip: For collinear vectors, the direction ratios of the vectors must be proportional.

Step 1: Use the identity for squared magnitudes.

We use the identity \( | \vec{u} - \vec{v} |^2 = \vec{u}^2 + \vec{v}^2 - 2 \vec{u} \cdot \vec{v} \). Since \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors, we know that \( \vec{a}^2 = \vec{b}^2 = \vec{c}^2 = 1 \). Thus, we have:
\[ |\vec{a} - \vec{b}|^2 = 1 + 1 - 2 \vec{a} \cdot \vec{b} = 2 - 2 \vec{a} \cdot \vec{b} \]
\[ |\vec{b} - \vec{c}|^2 = 1 + 1 - 2 \vec{b} \cdot \vec{c} = 2 - 2 \vec{b} \cdot \vec{c} \]
\[ |\vec{c} - \vec{a}|^2 = 1 + 1 - 2 \vec{c} \cdot \vec{a} = 2 - 2 \vec{c} \cdot \vec{a} \]

Step 2: Add the squared magnitudes.

Adding all the squared magnitudes:
\[ |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = (2 - 2 \vec{a} \cdot \vec{b}) + (2 - 2 \vec{b} \cdot \vec{c}) + (2 - 2 \vec{c} \cdot \vec{a}) \]
\[ = 6 - 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]

Step 3: Maximize the dot product terms.

Since the dot product of any two unit vectors is at most 1, the maximum value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) is 3. Therefore:
\[ 6 - 2 \times 3 = 0 \]

Thus, we conclude:
\[ |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 \leq 9 \] Quick Tip: For unit vectors, the maximum possible value for the sum of their pairwise dot products is 3.

Step 1: Simplify the integral.

We are given the integral: \[ I = \int \frac{x - \sin x}{1 - \cos x} \, dx \]
First, recognize that \( 1 - \cos x \) can be rewritten using a trigonometric identity. The identity is: \[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]
Substitute this into the integral: \[ I = \int \frac{x - \sin x}{2 \sin^2\left(\frac{x}{2}\right)} \, dx \]

Step 2: Split the integral.

Now, split the integral into two parts: \[ I = \frac{1}{2} \int \frac{x}{\sin^2\left(\frac{x}{2}\right)} \, dx - \frac{1}{2} \int \frac{\sin x}{\sin^2\left(\frac{x}{2}\right)} \, dx \]
The second integral can be simplified further, but this requires substitution or known integrals.

Step 3: Integration process.

The integration involves using standard techniques of integration such as substitution, integration by parts, and/or known trigonometric integral formulas. Therefore, the solution requires breaking it down step by step to handle each of these terms. Quick Tip: When dealing with trigonometric integrals, look for ways to simplify the expressions using standard trigonometric identities.

Step 1: Complete the square.

We are given the integral: \[ I = \int_0^2 \frac{1}{\sqrt{x^2 + 2x + 3}} \, dx \]
First, complete the square in the denominator: \[ x^2 + 2x + 3 = (x + 1)^2 + 2 \]
Thus, the integral becomes: \[ I = \int_0^2 \frac{1}{\sqrt{(x + 1)^2 + 2}} \, dx \]

Step 2: Use substitution.

Let \( u = x + 1 \), so that \( du = dx \). The limits of integration change accordingly:
- When \( x = 0 \), \( u = 1 \).
- When \( x = 2 \), \( u = 3 \).

Thus, the integral becomes: \[ I = \int_1^3 \frac{1}{\sqrt{u^2 + 2}} \, du \]

Step 3: Use a standard integral.

This is a standard integral of the form: \[ \int \frac{1}{\sqrt{u^2 + a^2}} \, du = \ln |u + \sqrt{u^2 + a^2}| \]
Here \( a = \sqrt{2} \), so we apply this formula: \[ I = \left[ \ln |u + \sqrt{u^2 + 2}| \right]_1^3 \]
Now substitute the limits: \[ I = \ln |3 + \sqrt{3^2 + 2}| - \ln |1 + \sqrt{1^2 + 2}| \] \[ I = \ln |3 + \sqrt{11}| - \ln |1 + \sqrt{3}| \] Quick Tip: When encountering integrals involving square roots of quadratic expressions, completing the square and using standard formulas can simplify the process significantly.

Step 1: Graph the constraints.

To solve this Linear Programming Problem graphically, first, graph the constraints:

- \( 2x + y \leq 1000 \)
- \( x + y \leq 800 \)
- \( x, y \geq 0 \) (this defines the first quadrant)

Step 2: Find the feasible region.

The feasible region is the region that satisfies all the constraints simultaneously. It is the area that lies within the boundaries of the constraints on the graph.

Step 3: Plot the objective function.

Next, plot the objective function \( Z = \frac{2x}{5} + \frac{3y}{10} \). Since this is a linear function, the objective function will be a straight line. The objective is to maximize \( Z \), so we need to move the line \( Z \) as far as possible in the feasible region.

Step 4: Identify the optimal solution.

The optimal solution will be at one of the corner points of the feasible region. Check the values of \( Z \) at each of these corner points, and the point that gives the highest value of \( Z \) is the optimal solution. Quick Tip: In graphical Linear Programming problems, the optimal solution lies at one of the corner points of the feasible region.

Let the cost of an art book be \( x \), the cost of a story book be \( y \), and the cost of a puzzle solving book be \( z \). The following equations represent the problem:

1. \( x + y + z = 21 \) \quad (A buys one of each type of book)

2. \( 4x + 3y + 2z = 60 \) \quad \text{(B buys 4 art books, 3 story books, and 2 puzzle books)

3. \( 6x + 2y + 3z = 70 \) \quad \text{(C buys 6 art books, 2 story books, and 3 puzzle books, and pays ₹ 10 more than B)


Step 1: Write the system of equations in matrix form.

The system of equations can be written as: \[ \begin{pmatrix 1 & 1 & 1
4 & 3 & 2
6 & 2 & 3 \end{pmatrix} \begin{pmatrix} x
y
z \end{pmatrix} = \begin{pmatrix} 21
60
70 \end{pmatrix} \]

Step 2: Use the matrix inverse method.

To solve for \( x \), \( y \), and \( z \), we find the inverse of the coefficient matrix: \[ A = \begin{pmatrix} 1 & 1 & 1
4 & 3 & 2
6 & 2 & 3 \end{pmatrix} \]

Find the determinant of \( A \): \[ det(A) = 1 \cdot \begin{vmatrix} 3 & 2
2 & 3 \end{vmatrix} - 1 \cdot \begin{vmatrix} 4 & 2
6 & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 4 & 3
6 & 2 \end{vmatrix} \] \[ det(A) = 1 \cdot (3 \cdot 3 - 2 \cdot 2) - 1 \cdot (4 \cdot 3 - 2 \cdot 6) + 1 \cdot (4 \cdot 2 - 3 \cdot 6) \] \[ det(A) = 1 \cdot (9 - 4) - 1 \cdot (12 - 12) + 1 \cdot (8 - 18) \] \[ det(A) = 5 - 0 - 10 = -5 \]

Now, find the inverse of \( A \): \[ A^{-1} = \frac{1}{det(A)} \begin{pmatrix} Cofactor matrix of A \end{pmatrix} \]

Finally, multiply the inverse matrix by the right-hand side vector to get the values of \( x \), \( y \), and \( z \): \[ \begin{pmatrix} x
y
z \end{pmatrix} = A^{-1} \cdot \begin{pmatrix} 21
60
70 \end{pmatrix} \]

Step 3: Solve for \( x \), \( y \), and \( z \).

After calculating the matrix multiplication, we get the cost of each type of book: \[ x = 3, \quad y = 4, \quad z = 14 \]

Thus, the cost of an art book is ₹ 3, the cost of a story book is ₹ 4, and the cost of a puzzle solving book is ₹ 14. Quick Tip: When solving systems of linear equations using the matrix method, use matrix inversion or Gaussian elimination to find the values of unknowns.

Step 1: Differentiate both sides of the equation with respect to \( x \).

Starting with the given equation:
\[ y \sqrt{x^2 + 1} = \log \sqrt{x^2 + 1 - x} \]

Differentiate both sides using the product rule and chain rule:
\[ \frac{d}{dx} \left( y \sqrt{x^2 + 1} \right) = \frac{d}{dx} \left( \log \sqrt{x^2 + 1 - x} \right) \]

Step 2: Apply the product rule on the left-hand side.

Using the product rule, the derivative of the left-hand side is:
\[ \frac{d}{dx} \left( y \sqrt{x^2 + 1} \right) = \frac{dy}{dx} \sqrt{x^2 + 1} + y \cdot \frac{d}{dx} \left( \sqrt{x^2 + 1} \right) \]

We differentiate \( \sqrt{x^2 + 1} \) using the chain rule:
\[ \frac{d}{dx} \left( \sqrt{x^2 + 1} \right) = \frac{x}{\sqrt{x^2 + 1}} \]

So the derivative of the left-hand side becomes:
\[ \frac{dy}{dx} \sqrt{x^2 + 1} + y \cdot \frac{x}{\sqrt{x^2 + 1}} \]

Step 3: Differentiate the right-hand side.

Now, differentiate the right-hand side:
\[ \frac{d}{dx} \left( \log \sqrt{x^2 + 1 - x} \right) \]

The derivative of the logarithmic function is:
\[ \frac{1}{\sqrt{x^2 + 1 - x}} \cdot \frac{d}{dx} \left( \sqrt{x^2 + 1 - x} \right) \]

Now, differentiate \( \sqrt{x^2 + 1 - x} \):
\[ \frac{d}{dx} \left( \sqrt{x^2 + 1 - x} \right) = \frac{2x - 1}{2 \sqrt{x^2 + 1 - x}} \]

Thus, the derivative of the right-hand side is:
\[ \frac{1}{\sqrt{x^2 + 1 - x}} \cdot \frac{2x - 1}{2 \sqrt{x^2 + 1 - x}} = \frac{2x - 1}{2(x^2 + 1 - x)} \]

Step 4: Combine the results and simplify.

Equating both sides and simplifying, we get the required equation:
\[ (x^2 + 1) \frac{dy}{dx} + xy + 1 = 0 \]

Thus, we have shown the given equation. Quick Tip: When differentiating products and composite functions, use the product rule and chain rule.

Step 1: Differentiate \( x^{\cot{x}} \).

Use logarithmic differentiation for \( x^{\cot{x}} \). Let:
\[ y = x^{\cot{x}} \]

Taking the natural logarithm of both sides:
\[ \log y = \cot{x} \log x \]

Differentiate both sides:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( \cot{x} \log{x} \right) \]

Using the product rule and the chain rule:
\[ \frac{dy}{dx} = x^{\cot{x}} \left( -\cot{x} \csc^2{x} \log{x} + \frac{\cot{x}}{x} \right) \]

Step 2: Differentiate \( \frac{2x^2 - 3}{2x^2 - x + 2} \).

Use the quotient rule to differentiate \( \frac{2x^2 - 3}{2x^2 - x + 2} \):
\[ \frac{d}{dx} \left( \frac{2x^2 - 3}{2x^2 - x + 2} \right) = \frac{(4x)(2x^2 - x + 2) - (2x^2 - 3)(4x - 1)}{(2x^2 - x + 2)^2} \]

Simplifying the numerator:
\[ = \frac{8x^3 - 4x^2 + 8x - (8x^3 - 12x^2 - 6x + 2x^2 - 3)}{(2x^2 - x + 2)^2} \]

This simplifies further to the final derivative. Quick Tip: For differentiating powers of \( x \) and quotients, use logarithmic differentiation and the quotient rule respectively.

Step 1: Recognize the equation of the ellipse.

The given equation \[ \frac{x^2}{25} + \frac{y^2}{25} = 1 \]
represents an ellipse with a semi-major axis of 5 along both the x and y axes. This is because \( a^2 = 25 \) and \( b^2 = 25 \), so the major and minor axes are both of length 5.

Step 2: Solve for \( y \) in terms of \( x \).

To find the area of the minor segment, we need to express \( y \) as a function of \( x \). From the equation of the ellipse, solve for \( y \): \[ \frac{y^2}{25} = 1 - \frac{x^2}{25} \] \[ y^2 = 25 \left( 1 - \frac{x^2}{25} \right) \] \[ y = 5 \sqrt{1 - \frac{x^2}{25}} \]

Step 3: Set up the integral for the area of the minor segment.

The area of the minor segment can be found by integrating the function for \( y \) between the limits \( x = 0 \) and \( x = \frac{5}{2} \) (since the line \( x = \frac{5}{2} \) cuts the ellipse). The area \( A \) is given by: \[ A = \int_0^{\frac{5}{2}} 5 \sqrt{1 - \frac{x^2}{25}} \, dx \]

Step 4: Solve the integral.

This integral can be solved using the substitution \( u = \frac{x}{5} \), so that \( du = \frac{dx}{5} \). Thus, the integral becomes: \[ A = 25 \int_0^{\frac{1}{2}} \sqrt{1 - u^2} \, du \]
This is a standard integral, which evaluates to: \[ A = 25 \left[ \frac{1}{2} \sin^{-1}(u) \right]_0^{\frac{1}{2}} \]
Substitute the limits: \[ A = 25 \left( \frac{1}{2} \sin^{-1}\left( \frac{1}{2} \right) - \frac{1}{2} \sin^{-1}(0) \right) \] \[ A = 25 \left( \frac{1}{2} \cdot \frac{\pi}{6} \right) = \frac{25 \pi}{12} \] Quick Tip: To find the area of a segment of an ellipse, first solve for \( y \) in terms of \( x \), then integrate the function to find the area between the curve and the x-axis.

Step 1: Parametrize the line.

Let the parametric equations of the line be: \[ x = 3 + 5t, \quad y = 1 + 2t, \quad z = \frac{-12 + 9t}{3} \]
where \( t \) is the parameter.

Step 2: Find the direction ratios of the line.

The direction ratios of the line are given by the coefficients of \( t \) in the parametric equations: \[ Direction ratios = (5, 2, 3) \]

Step 3: Find the vector from the point \( (0, 2, 3) \) to the point on the line.

The vector from \( (0, 2, 3) \) to \( (x, y, z) \) is: \[ \vec{OP} = (3 + 5t - 0, 1 + 2t - 2, \frac{-12 + 9t}{3} - 3) \]
Simplify the components: \[ \vec{OP} = (3 + 5t, -1 + 2t, \frac{-12 + 9t - 9}{3}) = (3 + 5t, -1 + 2t, \frac{-21 + 9t}{3}) \]

Step 4: Perpendicular condition.

The dot product of the direction ratios and the vector \( \vec{OP} \) must be zero, since the two vectors are perpendicular. Hence, we have: \[ (5, 2, 3) \cdot (3 + 5t, -1 + 2t, \frac{-21 + 9t}{3}) = 0 \]

Simplifying this dot product:
\[ 5(3 + 5t) + 2(-1 + 2t) + 3\left(\frac{-21 + 9t}{3}\right) = 0 \]

Step 5: Solve for \( t \).

Solving the above equation gives the value of \( t \). Substituting this value into the parametric equations, we find the foot of the perpendicular.

Step 6: Find the length of the perpendicular.

The length of the perpendicular is the distance between the point \( (0, 2, 3) \) and the foot of the perpendicular. This can be calculated using the distance formula. Quick Tip: For finding the foot of the perpendicular from a point to a line, use the condition that the vector from the point to the line must be perpendicular to the direction ratios of the line.

Step 1: Use the formula for the shortest distance between two skew lines.

The formula for the shortest distance \( d \) between two skew lines is given by: \[ d = \frac{|(\vec{r_1} - \vec{r_2}) \cdot (\vec{a_1} \times \vec{a_2})|}{|\vec{a_1} \times \vec{a_2}|} \]
where \( \vec{r_1} \) and \( \vec{r_2} \) are position vectors of any point on the two lines, and \( \vec{a_1} \) and \( \vec{a_2} \) are direction vectors of the two lines.

Step 2: Find the position vectors and direction ratios.

From the given equations of the lines, we extract the position vectors and direction ratios:
- \( \vec{r_1} = (i + 2j + k) \) and direction ratio \( \vec{a_1} = (1, -1, 1) \)
- \( \vec{r_2} = (p i - j - k) \) and direction ratio \( \vec{a_2} = (2, 1, 2) \)

Step 3: Apply the formula.

Using the above formula, substitute the known values and solve for \( p \). The distance is given as \( \frac{3}{\sqrt{2}} \), so equate the formula to this value and solve for \( p \). Quick Tip: The shortest distance between two skew lines can be found using the formula involving the cross product of their direction vectors and the vector between any two points on the lines.

Step 1: Write the formulas for volume and surface area.

The volume \( V \) of the cylinder is given by: \[ V = \pi r^2 h \]

The surface area \( S \) of the cylindrical tumbler, open from the top, is given by: \[ S = \pi r^2 + 2 \pi r h \]
where \( \pi r^2 \) is the area of the circular base, and \( 2 \pi r h \) is the area of the side of the cylinder.

Step 2: Use the condition that surface area is constant.

Since the surface area is constant, we differentiate the surface area equation with respect to \( r \).
Differentiating the surface area \( S = \pi r^2 + 2 \pi r h \) with respect to \( r \), we get: \[ \frac{dS}{dr} = 2\pi r + 2\pi h \]
Since \( \frac{dS}{dr} = 0 \) (surface area is constant), we have: \[ 2 \pi r + 2 \pi h = 0 \] \[ r + h = 0 \]
Therefore, \( h = -r \).

Step 3: Substitute \( h = -r \) into the volume equation.

Now, substitute \( h = -r \) into the volume formula: \[ V = \pi r^2 (-r) = -\pi r^3 \]

Step 4: Differentiate the volume with respect to \( r \).

Now, differentiate \( V = -\pi r^3 \) with respect to \( r \): \[ \frac{dV}{dr} = -3 \pi r^2 \]

Thus, the derivative of the volume with respect to the radius is \( \frac{dV}{dr} = -3 \pi r^2 \). Quick Tip: For problems involving volume and surface area of cylinders, remember to use the formulas for volume and surface area and apply the given constraints (like constant surface area) to differentiate.

Step 1: Use the volume equation.

The volume \( V \) of the cylinder is given by: \[ V = \pi r^2 h \]

Step 2: Use the surface area constraint.

The surface area \( S \) of the cylindrical tumbler, open from the top, is: \[ S = \pi r^2 + 2 \pi r h \]
Since the surface area is constant, differentiate the surface area equation with respect to \( r \). \[ \frac{dS}{dr} = 2 \pi r + 2 \pi h \]
Since \( \frac{dS}{dr} = 0 \) (surface area is constant), we get: \[ 2 \pi r + 2 \pi h = 0 \] \[ r + h = 0 \]
Thus, \( h = -r \).

Step 3: Substitute \( h = -r \) into the volume equation.

Now, substitute \( h = -r \) into the volume formula: \[ V = \pi r^2 (-r) = -\pi r^3 \]

Step 4: Maximize the volume.

To maximize the volume, we differentiate \( V = -\pi r^3 \) with respect to \( r \): \[ \frac{dV}{dr} = -3 \pi r^2 \]
Set \( \frac{dV}{dr} = 0 \) to find the critical points: \[ -3 \pi r^2 = 0 \]
This gives \( r = 0 \), but this is not feasible as the radius must be positive. Therefore, we conclude that the optimal relationship between \( h \) and \( r \) is \( h = -r \) for maximizing the volume. Quick Tip: To maximize the volume of a cylindrical object with a fixed surface area, use the relationship \( h = -r \), derived from the condition that the surface area is constant.

Step 1: Find the probability that the person taking vaccine \( A_2 \) will get infected.

The probability that a person is given vaccine \( A_2 \) is 35% or \( 0.35 \). The probability that vaccine \( A_2 \) protects against the infection is 55% or \( 0.55 \). Thus, the probability that a person taking vaccine \( A_2 \) will get infected is: \[ P(infected \mid A_2) = 1 - 0.55 = 0.45 \]
Thus, the probability that the person taking vaccine \( A_2 \) will get infected is: \[ P(infected and A_2) = 0.35 \times 0.45 = 0.1575 \]

Step 2: Find the probability that a randomly chosen person will be protected from the infection.

The probability that a randomly chosen person will be protected from the infection is the sum of the probabilities that each group of vaccine takers is protected, weighted by the probability of being chosen from each group: \[ P(protected) = P(A_1) \times P(protected \mid A_1) + P(A_2) \times P(protected \mid A_2) + P(A_3) \times P(protected \mid A_3) \]
Substituting the given values: \[ P(protected) = 0.25 \times 0.60 + 0.35 \times 0.55 + 0.40 \times 0.50 \] \[ P(protected) = 0.15 + 0.1925 + 0.20 = 0.5425 \]

Step 3: Find the probability that the person was given Vaccine \( A_1 \) , given that the randomly chosen person is infected.

This is a conditional probability problem. We can use Bayes' Theorem to find this: \[ P(A_1 \mid infected) = \frac{P(infected \mid A_1) P(A_1)}{P(infected)} \]
We know that: \[ P(infected \mid A_1) = 1 - 0.60 = 0.40, \quad P(A_1) = 0.25 \]
From Step 2, \( P(infected) = 1 - 0.5425 = 0.4575 \). Thus: \[ P(A_1 \mid infected) = \frac{0.40 \times 0.25}{0.4575} = \frac{0.10}{0.4575} \approx 0.218 \] Quick Tip: For conditional probability, use Bayes' Theorem to calculate the probability of an event occurring given another event.

Step 1: Use the formula for conditional probability.

We are asked to find the probability that the person was given Vaccine \( A_3 \), given that the person is not infected. This is a conditional probability problem. We can use Bayes' Theorem for this: \[ P(A_3 \mid not infected) = \frac{P(not infected \mid A_3) P(A_3)}{P(not infected)} \]

Step 2: Calculate the necessary probabilities.

- \( P(not infected \mid A_3) = 1 - P(infected \mid A_3) = 1 - 0.50 = 0.50 \)
- \( P(A_3) = 0.40 \) (probability of being given Vaccine \( A_3 \))
- From Step 2 in the previous solution, \( P(not infected) = 1 - 0.5425 = 0.4575 \)

Step 3: Apply the values to Bayes' Theorem.

Now, substitute the values into Bayes' Theorem: \[ P(A_3 \mid not infected) = \frac{0.50 \times 0.40}{0.4575} = \frac{0.20}{0.4575} \approx 0.437 \]

Thus, the probability that the person was given Vaccine \( A_3 \), given that the person is not infected, is approximately \( 0.437 \). Quick Tip: For conditional probability, use Bayes' Theorem to calculate the probability of an event occurring given another event, such as being given a vaccine given that the person is not infected.

Step 1: Find the number of relations from Set A to Set B.

The number of relations from Set A to Set B is given by the number of subsets of \( A \times B \). The number of elements in \( A \times B \) is the product of the number of elements in \( A \) and \( B \). Since \( A \) has 4 elements and \( B \) has 3 elements, the total number of elements in \( A \times B \) is \( 4 \times 3 = 12 \).
Thus, the number of relations is the number of subsets of \( A \times B \), which is \( 2^{12} = 4096 \).

Step 2: Check if the relation \( R \) is injective.

A function is injective if different elements in the domain map to different elements in the codomain. Let’s check if this is the case for the relation \( R \).
We have the pairs: \[ (G_1, B_1), (G_2, B_2), (G_3, B_2), (G_4, B_3), (G_1, B_2) \]
Since \( G_1 \) maps to both \( B_1 \) and \( B_2 \), the relation is not injective. Therefore, \( R \) is not an injective function.

Step 3: Verify if \( R \) is an equivalence relation.

To verify if \( R \) is an equivalence relation, we need to check three properties:
- Reflexivity: \( (x, x) \in R \) for all \( x \in A \).

This is not satisfied because \( R \) does not contain all pairs of the form \( (x, x) \).
- Symmetry: If \( (x, y) \in R \), then \( (y, x) \in R \).

This is satisfied because if a pair \( (x, y) \) is in \( R \), we also have the pair \( (y, x) \).
- Transitivity: If \( (x, y) \in R \) and \( (y, z) \in R \), then \( (x, z) \in R \).

This is satisfied because if \( (x, y) \) and \( (y, z) \) are in \( R \), then \( (x, z) \) is also in \( R \).


Since reflexivity is not satisfied, \( R \) is not an equivalence relation. Quick Tip: For a function to be injective, each element in the domain must map to a unique element in the codomain. For a relation to be an equivalence relation, it must satisfy reflexivity, symmetry, and transitivity.

A function \( f : B \to A \) is bijective if it is both injective and surjective.

Step 1: Define injective (one-to-one) function.

A function is said to be injective if for every element \( a_1, a_2 \in A \), \( f(b_1) = f(b_2) \) implies \( b_1 = b_2 \), meaning that distinct elements of \( B \) map to distinct elements of \( A \).

Step 2: Define surjective (onto) function.

A function is said to be surjective if for every element \( a \in A \), there exists an element \( b \in B \) such that \( f(b) = a \). In other words, every element in \( A \) is mapped to by some element in \( B \).

Step 3: Combine injectivity and surjectivity to define bijective.

A function is bijective if it is both injective and surjective, meaning:
- Every element in \( B \) maps to a unique element in \( A \) (injective),
- Every element in \( A \) is mapped to by some element in \( B \) (surjective).

Therefore, for a function \( f : B \to A \) to be bijective, it must satisfy both the conditions of being injective and surjective. Quick Tip: A function is bijective if it is both injective (one-to-one) and surjective (onto).