UP Board Class 12 Mathematics Question Paper 2024 (Code 324 FC) Available- Download Here with Solution PDF

To find the derivative of \( \sin^{-1}(e^{-x}) \): \[ Let y = \sin^{-1}(e^{-x}). \]
Differentiating both sides with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (e^{-x})^2}} \cdot \frac{d}{dx}(e^{-x}). \] \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - e^{-2x}}} \cdot (-e^{-x}). \]
Simplifying: \[ \frac{dy}{dx} = -\frac{1}{\sqrt{e^{2x} - 1}}. \] Quick Tip: When differentiating inverse trigonometric functions, remember to use the chain rule and simplify carefully.

Using integration by parts, let: \[ u = x \quad and \quad dv = \sin x \, dx. \]
Then: \[ du = dx \quad and \quad v = -\cos x. \]
Applying the formula \( \int u \, dv = uv - \int v \, du \): \[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx. \] \[ \int x \sin x \, dx = -x \cos x + \sin x + C. \] Quick Tip: When solving integrals using parts, carefully choose \( u \) and \( dv \) for easier differentiation and integration.

The modulus function \( f(x) = |x| \) maps every \( x \in \mathbb{R} \) to a non-negative real number \( \mathbb{R}^+ \).
For any positive \( y \), both \( x = y \) and \( x = -y \) map to the same output, making \( f(x) \) many-one. Quick Tip: Modulus functions are many-one due to symmetry around the origin and map all negatives to positive values.

Equating the elements of the matrices: \[ 2x - y = 1 \quad and \quad x + 2y = 3. \]
Solving these equations simultaneously:
From the first equation: \[ y = 2x - 1. \]
Substitute \( y \) into the second equation: \[ x + 2(2x - 1) = 3 \quad \Rightarrow \quad x + 4x - 2 = 3 \quad \Rightarrow \quad 5x = 5 \quad \Rightarrow \quad x = 1. \]
Substitute \( x = 1 \) into \( y = 2x - 1 \): \[ y = 2(1) - 1 = \frac{1}{2}. \] Quick Tip: For matrix equations, equate corresponding elements and solve the resulting system of linear equations.

The cosine of the angle between two vectors is given by: \[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{\|\vec{A}\| \|\vec{B}\|}. \] \[ \vec{A} = 3\hat{i} - 2\hat{j} + \hat{k}, \quad \vec{B} = 2\hat{i} + \hat{j} + 3\hat{k}. \]
Compute \( \vec{A} \cdot \vec{B} \): \[ \vec{A} \cdot \vec{B} = (3)(2) + (-2)(1) + (1)(3) = 6 - 2 + 3 = 7. \]
Compute \( \|\vec{A}\| \) and \( \|\vec{B}\| \): \[ \|\vec{A}\| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}. \] \[ \|\vec{B}\| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}. \] \[ \cos \theta = \frac{7}{\sqrt{14} \cdot \sqrt{14}} = \frac{7}{14} = \frac{1}{14}. \] \[ \theta = \cos^{-1}\left(\frac{1}{14}\right). \] Quick Tip: To find the angle between two vectors, calculate the dot product and magnitudes carefully.

Step 1: The direction cosines are calculated as: \[ l = \frac{1}{\sqrt{1^2 + 1^2 + (-2)^2}}, \, m = \frac{1}{\sqrt{1^2 + 1^2 + (-2)^2}}, \, n = \frac{-2}{\sqrt{1^2 + 1^2 + (-2)^2}}. \]
Simplify: \[ l = m = \frac{1}{\sqrt{6}}, \, n = \frac{-2}{\sqrt{6}}. \] Quick Tip: Direction cosines are calculated by dividing each component by the magnitude of the vector.

Using the trigonometric identity: \[ \sin^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{6} \quad and \quad \tan^{-1} x = \frac{\pi}{6}. \]
Therefore, \( x = \tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}} \). Quick Tip: Use trigonometric identities like \(\tan^{-1} x = \tan(\theta)\) to simplify and compare values.

Use the formula for conditional probability: \[ P(A / B) = \frac{P(A \cap B)}{P(B)} \quad \Rightarrow \quad \frac{2}{5} = \frac{P(A \cap B)}{\frac{5}{13}}. \]
Simplify: \[ P(A \cap B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13}. \] Quick Tip: For conditional probability, always use the formula \(P(A / B) = \frac{P(A \cap B)}{P(B)}\).

Rewrite the equation: \[ \frac{dy}{1 - y} = (1 + x) dx. \]
Integrate both sides: \[ -\ln|1 - y| = x + \frac{x^2}{2} + C. \] Quick Tip: Separate variables and integrate both sides carefully for differential equations.

Differentiate the given curve: \[ \frac{dy}{dx} = \frac{d}{dx}(2x^2 - 3\cos x) = 4x + 3\sin x. \]
At \( x = 0 \): \[ \frac{dy}{dx} = 4(0) + 3\sin(0) = 0. \] Quick Tip: For slope, differentiate the function and substitute the given point.

Taking the logarithm of both sides: \[ y \log_e x = x - y. \]
Differentiating with respect to \( x \): \[ \log_e x \frac{dy}{dx} + \frac{y}{x} = 1 - \frac{dy}{dx}. \]
Rearranging terms: \[ \frac{dy}{dx}(1 + \log_e x) = 1 - \frac{y}{x}. \]
From the original equation \( y = \frac{x}{1 + \log_e x} \), substitute \( y \): \[ \frac{dy}{dx} = \frac{\log_e x}{(1 + \log_e x)^2}. \] Quick Tip: Use logarithmic differentiation for equations involving powers and exponential terms.

Rewrite the equation: \[ \frac{\tan y}{\sec^2 y} \, dy = -\frac{\tan x}{\sec^2 x} \, dx. \]
Integrating both sides: \[ \int \sin y \, dy = -\int \sin x \, dx. \] \[ -\cos y = \cos x + C. \]
Simplify: \[ \cos x + \cos y = C. \] Quick Tip: Separate variables for trigonometric differential equations before integration.

The properties of cube roots of unity are: \[ \omega^3 = 1, \, 1 + \omega + \omega^2 = 0. \]
Expanding the determinant:

Simplify using properties of \( \omega \): \[ = 0. \] Quick Tip: Use properties of roots of unity such as \( 1 + \omega + \omega^2 = 0 \) to simplify expressions.

(A) For independent events: \[ P(A \cap B) = P(A) \cdot P(B) = 0.5 \cdot 0.4 = 0.2. \]
(B) Using the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B). \]
Substitute values: \[ P(A \cup B) = 0.5 + 0.4 - 0.2 = 0.7. \] Quick Tip: For independent events, multiply probabilities for \( P(A \cap B) \) and use the union formula for \( P(A \cup B) \).

Differentiating \( ay^2 = x^3 \) with respect to \( x \): \[ 2ay \frac{dy}{dx} = 3x^2. \]
At \( (am^2, am^3) \), substitute \( x = am^2 \) and \( y = am^3 \): \[ 2a(am^3) \frac{dy}{dx} = 3(am^2)^2. \] \[ 2am^3 \frac{dy}{dx} = 3a^2m^4. \] \[ \frac{dy}{dx} = \frac{3am^4}{2m^3} = \frac{3am}{2}. \] Quick Tip: Substitute the given point into the derivative after differentiating implicitly.

Find vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \): \[ \overrightarrow{AB} = (-1 - 2, -2 - 3, 1 - 4) = (-3, -5, -3). \] \[ \overrightarrow{BC} = (5 - (-1), 8 - (-2), 7 - 1) = (6, 10, 6). \]
Check if \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \) are scalar multiples: \[ \frac{-3}{6} = \frac{-5}{10} = \frac{-3}{6} = -\frac{1}{2}. \]
Since the ratios are equal, the points are collinear. Quick Tip: To check collinearity, verify if vectors are scalar multiples of each other.

Complete the square in the expression \( 5 + 2x + x^2 \): \[ 5 + 2x + x^2 = (x+1)^2 + 4. \]
Substitute \( u = x + 1 \), then \( du = dx \): \[ \int \sqrt{5 + 2x + x^2} \, dx = \int \sqrt{u^2 + 4} \, du. \]
Solve using trigonometric substitution: \[ u = 2 \tan \theta \quad \Rightarrow \quad \sqrt{u^2 + 4} = 2 \sec \theta. \]
Evaluate and simplify the integral. Quick Tip: Complete the square before integrating expressions under a square root.

Take the logarithm of both sides: \[ \ln y = x \ln x. \]
Differentiate with respect to \( x \): \[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1. \]
Multiply through by \( y = x^x \): \[ \frac{dy}{dx} = x^x (\ln x + 1). \] Quick Tip: For expressions like \( y = x^x \), use logarithmic differentiation.

To minimize \( Z = 3x + 9y \), plot the constraints on a graph and find the feasible region:
1. \( x + 3y \leq 60 \) is a line with intercepts at \( x = 60, y = 20 \).
2. \( x + y \geq 10 \) is a line passing through \( (10, 0) \) and \( (0, 10) \).
3. \( x \leq y \) represents the region below \( x = y \).
4. \( x \geq 0, y \geq 0 \) restricts to the first quadrant.

The feasible region is bounded, and \( Z = 3x + 9y \) is evaluated at corner points. Solve for \( Z \) at these vertices:

\( (0, 10) \): \( Z = 3(0) + 9(10) = 90 \),
\( (0, 20) \): \( Z = 3(0) + 9(20) = 180 \),
\( (30, 10) \): \( Z = 3(30) + 9(10) = 90 + 30 = 120 \).


The minimum value is \( Z = 90 \) at \( (0, 10) \). Quick Tip: For optimization problems, identify the feasible region and evaluate the objective function at each vertex to determine the minimum or maximum.

The shortest distance \( d \) between two skew lines is given by: \[ d = \frac{|(\vec{r_1} - \vec{r_2}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}, \]
where \( \vec{r_1} = 3\hat{i} + 3\hat{j} - 5\hat{k}, \vec{r_2} = \hat{i} + 2\hat{j} - 4\hat{k} \),
and \( \vec{d_1} = 2\hat{i} + 3\hat{j} + 6\hat{k}, \vec{d_2} = 2\hat{i} + 3\hat{j} + 6\hat{k} \).

1. Compute \( \vec{r_1} - \vec{r_2} = 2\hat{i} + \hat{j} - \hat{k} \).
2. Compute \( \vec{d_1} \times \vec{d_2} = \vec{0} \) since \( \vec{d_1} \) and \( \vec{d_2} \) are parallel.

Since \( \vec{d_1} \parallel \vec{d_2} \), the lines are not skew but parallel, and the shortest distance is the perpendicular distance between planes. The distance is \( d = 0 \). Quick Tip: For shortest distance, check for skew or parallel lines. Use cross products for skew, and perpendicular distance for parallel lines.

To prove \( R \) is an equivalence relation, we verify the three properties:
1. Reflexive: For any \( a \in Z \), \( a - a = 0 \), which is a multiple of 5. Hence, \( (a, a) \in R \).
2. Symmetric: If \( (a, b) \in R \), then \( a - b \) is a multiple of 5. Thus, \( b - a = -(a - b) \), which is also a multiple of 5. Hence, \( (b, a) \in R \).
3. Transitive: If \( (a, b) \in R \) and \( (b, c) \in R \), then \( a - b \) and \( b - c \) are multiples of 5. Adding these gives \( a - c = (a - b) + (b - c) \), which is a multiple of 5. Hence, \( (a, c) \in R \).

Since \( R \) satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation. Quick Tip: For equivalence relations, always verify reflexivity, symmetry, and transitivity step-by-step.

Given: \[ x \sqrt{1 + y} + y \sqrt{1 + x} + x = 0. \]
Differentiate both sides with respect to \( x \): \[ \frac{d}{dx} \left( x \sqrt{1 + y} + y \sqrt{1 + x} + x \right) = 0. \]
Using the product rule: \[ \sqrt{1 + y} + x \frac{1}{2\sqrt{1 + y}} \frac{dy}{dx} + \sqrt{1 + x} \frac{dy}{dx} + y \frac{1}{2\sqrt{1 + x}} + 1 = 0. \]
Combine terms involving \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left( x \frac{1}{2\sqrt{1 + y}} + \sqrt{1 + x} \right) = -\left( \sqrt{1 + y} + y \frac{1}{2\sqrt{1 + x}} + 1 \right). \]
Substitute \( x \sqrt{1 + y} + y \sqrt{1 + x} + x = 0 \) and simplify to get: \[ \frac{dy}{dx} = -\frac{1}{(1 + x)^2}. \] Quick Tip: For implicit differentiation, carefully apply the product rule and isolate \( \frac{dy}{dx} \) terms.

Expand the determinant along the first row:

Calculate each minor:

Substitute back: \[ Determinant = (1 + a)(bc + b + c) - c - b. \]
Simplify and verify: \[ Determinant = abc \left( 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right). \] Quick Tip: Expand determinants systematically and use minor expansions to simplify step-by-step.

The given differential equation is \[ \frac{dy}{dx} + y \cot x = 2x. \]
The integrating factor (IF) is calculated as: \[ IF = e^{\int \cot x \, dx} = \sin x. \]
Multiplying through by \( \sin x \), the equation becomes: \[ \sin x \frac{dy}{dx} + y \sin x \cot x = 2x \sin x \quad \Rightarrow \quad \frac{d}{dx}(y \sin x) = 2x \sin x. \]
Integrating both sides: \[ y \sin x = \int 2x \sin x \, dx. \]
Using integration by parts: \[ \int 2x \sin x \, dx = -2x \cos x + 2 \int \cos x \, dx = -2x \cos x + 2 \sin x. \]
Thus: \[ y \sin x = -2x \cos x + 2 \sin x + C. \]
Substitute \( x = \frac{\pi}{2}, y = 0 \) to find \( C \): \[ 0 \cdot \sin\left(\frac{\pi}{2}\right) = -2 \cdot \frac{\pi}{2} \cdot \cos\left(\frac{\pi}{2}\right) + 2 \cdot \sin\left(\frac{\pi}{2}\right) + C \quad \Rightarrow \quad C = -2. \]
The particular solution is: \[ y = \frac{-2x \cos x + 2 \sin x - 2}{\sin x}. \] Quick Tip: For first-order linear differential equations, always find the integrating factor and use it to simplify the equation.

The given integral can be solved using partial fraction decomposition: \[ \frac{x}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}. \]
Multiply through by \( (x-a)(x-b)(x-c) \) and solve for \( A, B, C \) by substituting appropriate values of \( x \). After finding the coefficients, integrate each term individually: \[ \int \frac{A}{x-a} \, dx + \int \frac{B}{x-b} \, dx + \int \frac{C}{x-c} \, dx. \]
The result is: \[ \ln|x-a|, \ln|x-b|, \ln|x-c|, \]
combined with the coefficients to give the final solution. Quick Tip: Partial fractions simplify rational functions into easily integrable terms.

Rewriting the equation: \[ x^2 - 2x + y^2 = 8 \quad \Rightarrow \quad (x-1)^2 + y^2 = 9. \]
This represents a circle centered at \( (1, 0) \) with radius \( r = 3 \). The area of the circle is: \[ Area = \pi r^2 = \pi (3)^2 = 9\pi. \] Quick Tip: Rewrite equations of circles into standard form to identify key parameters like center and radius.

Let \( A_1 \) and \( A_2 \) be the events of selecting the first and second bag, respectively. Let \( R \) be the event of drawing a red ball. Using Bayes' theorem: \[ P(A_1 \mid R) = \frac{P(R \mid A_1) P(A_1)}{P(R)}. \]
Here: \[ P(R \mid A_1) = \frac{4}{8}, \quad P(R \mid A_2) = \frac{2}{8}, \quad P(A_1) = P(A_2) = \frac{1}{2}. \] \[ P(R) = P(R \mid A_1)P(A_1) + P(R \mid A_2)P(A_2) = \frac{4}{8} \cdot \frac{1}{2} + \frac{2}{8} \cdot \frac{1}{2} = \frac{3}{8}. \]
Thus: \[ P(A_1 \mid R) = \frac{\frac{4}{8} \cdot \frac{1}{2}}{\frac{3}{8}} = \frac{2}{3}. \] Quick Tip: Apply Bayes' theorem to calculate conditional probabilities by systematically identifying prior and likelihood terms.

The area of the triangle is given by: \[ Area = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right|, \]
where \( \vec{AB} = \vec{b} - \vec{a}, \vec{AC} = \vec{c} - \vec{a} \). Expanding the cross product and simplifying yields the required expression. Quick Tip: Use vector cross products to compute areas of triangles in three-dimensional space.

Let \( I = \int_0^{\pi/2} \log \sin x \, dx \). Using the property of definite integrals: \[ \int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx, \]
we have: \[ I = \int_0^{\pi/2} \log \sin x \, dx = \int_0^{\pi/2} \log \sin(\pi/2 - x) \, dx. \]
Since \( \sin(\pi/2 - x) = \cos x \), the integral becomes: \[ I = \int_0^{\pi/2} \log \cos x \, dx. \]
Adding the two expressions for \( I \): \[ 2I = \int_0^{\pi/2} (\log \sin x + \log \cos x) \, dx = \int_0^{\pi/2} \log (\sin x \cos x) \, dx. \]
Using the identity \( \sin x \cos x = \frac{1}{2} \sin 2x \), we get: \[ 2I = \int_0^{\pi/2} \log \left(\frac{1}{2} \sin 2x\right) \, dx = \int_0^{\pi/2} \log \frac{1}{2} \, dx + \int_0^{\pi/2} \log \sin 2x \, dx. \]
The first term simplifies to: \[ \int_0^{\pi/2} \log \frac{1}{2} \, dx = \log \frac{1}{2} \cdot \frac{\pi}{2} = -\frac{\pi}{2} \log 2. \]
For the second term, using the substitution \( u = 2x \), we get: \[ \int_0^{\pi/2} \log \sin 2x \, dx = \frac{1}{2} \int_0^\pi \log \sin u \, du. \]
By symmetry of \( \sin u \), the integral evaluates to \( 0 \). Thus: \[ 2I = -\frac{\pi}{2} \log 2 \quad \Rightarrow \quad I = -\frac{\pi}{2} \log 2. \] Quick Tip: Use symmetry and integral properties to simplify expressions involving logarithmic trigonometric functions.

Let the integral be: \[ I = \int_0^{\pi/2} \frac{x}{a^2 \sin^2 x + b^2 \cos^2 x} \, dx. \]
Using the property of definite integrals: \[ \int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx, \]
we substitute \( x \to \pi/2 - x \). This gives: \[ I = \int_0^{\pi/2} \frac{\pi/2 - x}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx. \]
Adding these two forms of \( I \), we get: \[ 2I = \int_0^{\pi/2} \frac{x}{a^2 \sin^2 x + b^2 \cos^2 x} \, dx + \int_0^{\pi/2} \frac{\pi/2 - x}{a^2 \cos^2 x + b^2 \sin^2 x} \, dx. \]
Simplify: \[ 2I = \frac{\pi}{2} \int_0^{\pi/2} \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} \, dx. \]
Now, let: \[ J = \int_0^{\pi/2} \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} \, dx. \]
Using the standard result for this integral: \[ J = \frac{\pi}{2ab}. \]
Thus: \[ 2I = \frac{\pi}{2} \cdot \frac{\pi}{2ab} \quad \Rightarrow \quad I = \frac{\pi^2}{4ab}. \] Quick Tip: When solving integrals with trigonometric terms in the denominator, symmetry and standard results can simplify the process.

To find \( (AB)^{-1} \), we use the property of inverse matrices: \[ (AB)^{-1} = B^{-1} A^{-1}. \]
1. Compute \( A^{-1} \): Use the formula \( A^{-1} = \frac{1}{det(A)} adj(A) \), where:

2. Compute \( B^{-1} \): Similarly, calculate:
\[ B^{-1} = \frac{1}{det(B)} adj(B). \]
3. Multiply \( B^{-1} A^{-1} \) to get \( (AB)^{-1} \).

Perform these calculations to get the final result for \( (AB)^{-1} \). Quick Tip: The inverse of a product of matrices is the product of their inverses in reverse order: \( (AB)^{-1} = B^{-1} A^{-1} \).

Step 1: Represent the system of equations in matrix form.
The given system can be written as: \[ AX = B \]
where

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

Expanding along the first row:

Simplify each minor:

Substitute these values: \[ det(A) = 3(-1) - (-2)(8) + 3(-10) = -3 + 16 - 30 = -17. \]

Step 3: Compute the inverse of \( A \).
The inverse of \( A \) is given by: \[ A^{-1} = \frac{1}{det(A)} \cdot Adj(A), \]
where \( Adj(A) \) is the adjoint of \( A \). Compute the cofactors of \( A \) to find \( Adj(A) \).

Step 4: Solve for \( X \).
Using \( X = A^{-1}B \), compute the values of \( x \), \( y \), and \( z \). Quick Tip: To solve systems of equations using the matrix method, ensure that the determinant of the coefficient matrix is non-zero (\( det(A) \neq 0 \)).

Solution: For the first term:

y1 = (tan x)
cot x
.

Taking the logarithm:

ln y1 = cot x ln(tan x).

Differentiating:

1
y1
dy1
dx = − csc2 x ln(tan x) + cot x ·
sec2 x
tan x
.

Thus:

dy1
dx = (tan x)
cot x

− csc2 x ln(tan x) + cot x sec2 x
tan x

.

For the second term:

y2 = (sin x)
cos x

 

Taking the logarithm:

ln y2 = cos x ln(sin x).

Differentiating:

1
y2
dy2
dx = − sin x ln(sin x) + cos x cot x.

Thus:

dy2
dx = (sin x)
cos x
(− ln(sin x) sin x + cos x cot x).

Finally:

dy
dx =
dy1
dx +
dy2
dx .

Quick Tip: For differentiation of power functions with variable bases and exponents, use logarithmic differentiation effectively.

Compute the derivative:
f
′
(x) = 6
5
x
3 −
12
5
x
2 − 6x +
36
5
.

Simplify:

f
′
(x) = 6
5
(x
3 − 2x
2 − 5x + 6).

Solve f
′
(x) = 0 to find critical points:
x
3 − 2x
2 − 5x + 6 = 0.

Using factorization:

(x − 1)(x
2 − x − 6) = 0 ⇒ x = 1, x = −2, x = 3.
Test intervals (−∞, −2), (−2, 1), (1, 3), (3,∞) by substituting values in f
′
(x):

• f
′
(x) > 0: Function is increasing.
• f
′
(x) < 0: Function is decreasing.
Thus, f(x) is increasing on (−∞, −2) ∪ (1, 3) and decreasing on (−2, 1) ∪ (3,∞).

Quick Tip: To determine increasing and decreasing intervals, solve \( f'(x) = 0 \) and test the sign of \( f'(x) \) in each interval.