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.