UP Board Class 12 Mathematics Question Paper 2023 with Answer Key and Solutions PDF (February 27, Code 324 AY)

Step 1: Differentiate \(y\) with respect to \(t\).
\[ y = 4t \quad \Rightarrow \quad \frac{dy}{dt} = 4 \]

Step 2: Differentiate \(x\) with respect to \(t\).
\[ x = \frac{4}{t} \quad \Rightarrow \quad \frac{dx}{dt} = -\frac{4}{t^{2}} \]

Step 3: Apply the chain rule.
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4}{-\frac{4}{t^{2}}} = -t^{2} \]

Step 4: Conclusion.

The correct answer is (A) \(-t^{2}\).
Quick Tip: When \(y\) and \(x\) are given in terms of a parameter \(t\), use \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\).

Step 1: Write the given matrices.
\[ A = \begin{bmatrix} 0 & 1
1 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0
0 & -1 \end{bmatrix} \]

Step 2: Compute \(BA\).
\[ BA = \begin{bmatrix} 1 & 0
0 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1
1 & 0 \end{bmatrix} \]

Performing multiplication: \[ BA = \begin{bmatrix} (1 \cdot 0 + 0 \cdot 1) & (1 \cdot 1 + 0 \cdot 0)
(0 \cdot 0 + (-1) \cdot 1) & (0 \cdot 1 + (-1) \cdot 0) \end{bmatrix} = \begin{bmatrix} 0 & 1
-1 & 0 \end{bmatrix} \]

Step 3: Compare with given options.

This matches option (B).


Step 4: Conclusion.

The correct answer is (B) \(\begin{bmatrix} 0 & -1
1 & 0 \end{bmatrix}\).
Quick Tip: Always check the order when multiplying matrices: \(BA \neq AB\) in general.

Step 1: Differentiate the curve.

Given: \[ y = 2x^{2} + 3\sin x \]
Differentiate with respect to \(x\): \[ \frac{dy}{dx} = 4x + 3\cos x \]

Step 2: Find slope of tangent at \(x=0\).
\[ \frac{dy}{dx}\Big|_{x=0} = 4(0) + 3\cos(0) = 3 \]
So, slope of tangent at \(x=0\) is \(3\).


Step 3: Find slope of normal.

The slope of the normal is the negative reciprocal of the slope of the tangent. \[ m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{3} \]

Step 4: Match with options.

The correct answer is option (C) \(-\dfrac{1}{3}\).
Quick Tip: The slope of the normal to a curve at a point is the negative reciprocal of the slope of the tangent at that point.

Step 1: Recall the standard integral.
\[ \int \frac{dx}{1+x^{2}} = \tan^{-1}x + C \]

Step 2: Apply limits.
\[ \int_{1}^{\sqrt{3}} \frac{dx}{1+x^{2}} = \left[ \tan^{-1}x \right]_{1}^{\sqrt{3}} \]
\[ = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) \]

Step 3: Simplify.
\[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}, \quad \tan^{-1}(1) = \frac{\pi}{4} \]
\[ \therefore \int_{1}^{\sqrt{3}} \frac{dx}{1+x^{2}} = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \]

Step 4: Conclusion.

The correct answer is (B) \(\dfrac{\pi}{12}\).
Quick Tip: Always remember \(\int \dfrac{dx}{1+x^{2}} = \tan^{-1}x + C\) and use trigonometric values of \(\tan^{-1}(1)\) and \(\tan^{-1}(\sqrt{3})\).

Step 1: Understanding the modulus function.

The modulus function \(f(x) = |x|\) maps every real number \(x\) to a non-negative real number. Its codomain is \(\mathbb{R}^{+}\) (non-negative real numbers).


Step 2: Check one-one property.

If \(f(a) = f(b)\), then \(|a| = |b|\). This means either \(a = b\) or \(a = -b\).
Thus, different inputs (e.g., \(2\) and \(-2\)) give the same output. Hence, \(f(x)\) is **not one-one**, but **many-one**.


Step 3: Check onto property.

For every \(y \in \mathbb{R}^{+}\), there exists an \(x \in \mathbb{R}\) such that \(f(x) = |x| = y\).
Thus, the function is **onto** \(\mathbb{R}^{+}\).


Step 4: Conclusion.

The function is **many-one and onto**, so the correct answer is (B).
Quick Tip: The modulus function \(f(x) = |x|\) is many-one because \(f(a) = f(-a)\), and it is onto \(\mathbb{R}^{+}\) since every non-negative real number has a preimage.

Step 1: Substitution.
Let \(t = x^{3} \implies dt = 3x^{2}dx \implies x^{2}dx = \dfrac{dt}{3}\).

Step 2: Transform the integral. \[ \int x^{2}\sin(x^{3})\,dx = \int \sin(t)\,\dfrac{dt}{3} \]

Step 3: Solve. \[ = \dfrac{1}{3}\int \sin(t)\,dt = -\dfrac{1}{3}\cos(t) + C \]

Step 4: Back-substitution. \[ = -\dfrac{1}{3}\cos(x^{3}) + C \]


Final Answer: \[ \boxed{-\dfrac{1}{3}\cos(x^{3}) + C} \] Quick Tip: Always check if the derivative of the inner function exists in the integrand when using substitution.

Step 1: Recall condition for perpendicular vectors.
Two vectors \(\vec{u}\) and \(\vec{v}\) are perpendicular if: \[ \vec{u}\cdot \vec{v} = 0 \]

Step 2: Write the given vectors. \[ \vec{u} = 2\hat{i} + \hat{j} - a\hat{k}, \quad \vec{v} = \hat{i} + 4\hat{j} + \hat{k} \]

Step 3: Take the dot product. \[ \vec{u}\cdot \vec{v} = (2)(1) + (1)(4) + (-a)(1) \] \[ = 2 + 4 - a = 6 - a \]

Step 4: Apply perpendicular condition. \[ 6 - a = 0 \implies a = 6 \]


Final Answer: \[ \boxed{a = 6} \] Quick Tip: For perpendicular vectors → always use \(\vec{u}\cdot \vec{v} = 0\).

Step 1: Rearranging.

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

Step 2: Recognizing derivative form.

This suggests: \[ \frac{d}{dx}\left( \frac{y^2}{2} \right) = x + e^x \]

Step 3: Integration.

Integrating both sides w.r.t. \(x\): \[ \frac{y^2}{2} = \int (x + e^x) \, dx \]
\[ \frac{y^2}{2} = \frac{x^2}{2} + e^x + C \]

Step 4: Simplify.
\[ y^2 = x^2 + 2e^x + C' \]


Final Answer: \[ \boxed{y^2 = x^2 + 2e^x + C} \] Quick Tip: Always try to rewrite differential equations in recognizable derivative forms for easier integration.

Step 1: Simplify the inverse sine.
\[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \]

So, \[ \tan^{-1}x = \frac{\pi}{6} \]

Step 2: Apply tangent function.

Taking tangent on both sides: \[ x = \tan\left(\frac{\pi}{6}\right) \]
\[ x = \frac{1}{\sqrt{3}} \]


Final Answer: \[ \boxed{\dfrac{1}{\sqrt{3}}} \] Quick Tip: Use standard values of trigonometric functions: \(\sin^{-1}\left(\tfrac{1}{2}\right) = \tfrac{\pi}{6}\), \(\cos^{-1}\left(\tfrac{1}{2}\right) = \tfrac{\pi}{3}\), etc.

We are given two matrices: \[ A = \begin{bmatrix} 2 & 4
3 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 3
-2 & 5 \end{bmatrix} \]

Step 1: Find \((A + B)\). \[ A + B = \begin{bmatrix} 2 & 4
3 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3
-2 & 5 \end{bmatrix} = \begin{bmatrix} (2+1) & (4+3)
(3+(-2)) & (2+5) \end{bmatrix} = \begin{bmatrix} 3 & 7
1 & 7 \end{bmatrix} \]

Step 2: Find \((A - B)\). \[ A - B = \begin{bmatrix} 2 & 4
3 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 3
-2 & 5 \end{bmatrix} = \begin{bmatrix} (2-1) & (4-3)
(3-(-2)) & (2-5) \end{bmatrix} = \begin{bmatrix} 1 & 1
5 & -3 \end{bmatrix} \]


Final Answer: \[ A + B = \begin{bmatrix} 3 & 7
1 & 7 \end{bmatrix}, \quad A - B = \begin{bmatrix} 1 & 1
5 & -3 \end{bmatrix} \] Quick Tip: Matrix addition and subtraction are performed by adding or subtracting corresponding elements of the two matrices.

Step 1: Differentiate once. \[ y = A\cos\theta + B\sin\theta \] \[ \dfrac{dy}{d\theta} = -A\sin\theta + B\cos\theta \]

Step 2: Differentiate again. \[ \dfrac{d^{2}y}{d\theta^{2}} = -A\cos\theta - B\sin\theta \]

Step 3: Compare with original function. \[ \dfrac{d^{2}y}{d\theta^{2}} = -(A\cos\theta + B\sin\theta) = -y \]


Final Answer: \[ \boxed{\dfrac{d^{2}y}{d\theta^{2}} = -y} \] Quick Tip: Trigonometric functions \(\sin\theta\) and \(\cos\theta\) always satisfy the differential equation \(y'' + y = 0\).

We have the system of inequalities: \[ 3x + 4y \leq 12, \quad 4x + 3y \leq 12, \quad x \geq 0, \quad y \geq 0 \]

Step 1: Convert inequalities into equations. \[ 3x + 4y = 12 \quad (1) \] \[ 4x + 3y = 12 \quad (2) \]

Step 2: Find intercepts.

For (1):
- If \(x=0 \implies y = 3\)
- If \(y=0 \implies x = 4\)

For (2):
- If \(x=0 \implies y = 4\)
- If \(y=0 \implies x = 3\)

Step 3: Plot lines and shade feasible region.
- Line (1): passes through \((0,3)\) and \((4,0)\).
- Line (2): passes through \((0,4)\) and \((3,0)\).
- Feasible region lies in the first quadrant bounded by these lines and coordinate axes.

Step 4: Find corner points of feasible region.
By solving (1) and (2): \[ 3x + 4y = 12, \quad 4x + 3y = 12 \]
Multiply first by 3: \(9x + 12y = 36\)
Multiply second by 4: \(16x + 12y = 48\)
Subtract: \(7x = 12 \implies x = \dfrac{12}{7}\)

Substitute in (1): \[ 3\left(\dfrac{12}{7}\right) + 4y = 12 \implies \dfrac{36}{7} + 4y = 12 \] \[ 4y = \dfrac{84 - 36}{7} = \dfrac{48}{7} \implies y = \dfrac{12}{7} \]

So, point of intersection is \(\left(\dfrac{12}{7}, \dfrac{12}{7}\right)\).

Step 5: Vertices of feasible region. \[ (0,0), \; (4,0), \; (0,4), \; \left(\dfrac{12}{7}, \dfrac{12}{7}\right) \]

Conclusion:
The shaded feasible region is the quadrilateral formed by these points in the first quadrant.


Final Answer:
Feasible region vertices are \[ \boxed{(0,0), \; (4,0), \; (0,4), \; \left(\dfrac{12}{7}, \dfrac{12}{7}\right)} \] Quick Tip: In graphical inequalities → always test intercepts, plot boundary lines, and then shade the common feasible region.

Step 1: Expand determinant.

Let \[ \Delta = \begin{vmatrix} a+b+2c & a & b
c & b+c+2a & b
c & a & c+a+2b \end{vmatrix} \]

Step 2: Apply column transformation.

Perform \(C_1 \to C_1 + C_2 + C_3\): \[ \Delta = \begin{vmatrix} (a+b+2c)+a+b & a & b
c+(b+c+2a)+b & b+c+2a & b
c+a+(c+a+2b) & a & c+a+2b \end{vmatrix} \]
\[ \Delta = \begin{vmatrix} 2(a+b+c) & a & b
2(a+b+c) & b+c+2a & b
2(a+b+c) & a & c+a+2b \end{vmatrix} \]

Step 3: Factor common term.

Take \(2(a+b+c)\) common from \(C_1\): \[ \Delta = 2(a+b+c) \begin{vmatrix} 1 & a & b
1 & b+c+2a & b
1 & a & c+a+2b \end{vmatrix} \]

Step 4: Expand determinant.

Apply \(R_2 \to R_2 - R_1\), \(R_3 \to R_3 - R_1\): \[ = 2(a+b+c) \begin{vmatrix} 1 & a & b
0 & b+c+a & 0
0 & 0 & c+a+b \end{vmatrix} \]

Step 5: Simplify.
\[ = 2(a+b+c)\big[(b+c+a)(c+a+b)\big] \]
\[ = 2(a+b+c)(a+b+c)(a+b+c) \]
\[ = 2(a+b+c)^3 \]


Final Answer: \[ \boxed{\Delta = 2(a+b+c)^3} \] Quick Tip: Column/row transformations often simplify determinants to triangular form.

Step 1: Recall definition.

For any non-singular matrix \(M\), \[ M \cdot M^{-1} = I \]

Step 2: Take \(M = AB\).
\[ (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} \]
\[ = AIA^{-1} \]
\[ = AA^{-1} = I \]

Step 3: Conclude.

Since \((AB)(B^{-1}A^{-1}) = I\), it follows that \[ (AB)^{-1} = B^{-1}A^{-1} \]


Final Answer: \[ \boxed{(AB)^{-1} = B^{-1}A^{-1}} \] Quick Tip: Note: The order of multiplication reverses when taking the inverse of product of matrices.

We are given the parametric equations: \[ x = a \sin^{3}t, \quad y = b \cos^{3}t \]

Step 1: Differentiate \(x\) and \(y\) with respect to \(t\). \[ \frac{dx}{dt} = 3a \sin^{2}t \cos t, \quad \frac{dy}{dt} = -3b \cos^{2}t \sin t \]

Step 2: Find slope of tangent. \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-3b \cos^{2}t \sin t}{3a \sin^{2}t \cos t} = -\frac{b \cos t}{a \sin t} \]

Step 3: At \(t = \dfrac{\pi}{2}\). \[ x = a \sin^{3}\left(\frac{\pi}{2}\right) = a(1)^{3} = a \] \[ y = b \cos^{3}\left(\frac{\pi}{2}\right) = b(0)^{3} = 0 \] \[ \frac{dy}{dx} = -\frac{b \cos(\pi/2)}{a \sin(\pi/2)} = -\frac{b \cdot 0}{a \cdot 1} = 0 \]

Step 4: Equation of tangent.
Point: \((a, 0)\), slope \(m = 0\).
Equation: \[ y - 0 = 0 \cdot (x - a) \quad \Rightarrow \quad y = 0 \]


Final Answer: \[ \boxed{y = 0} \] Quick Tip: For parametric curves, slope of tangent = \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\).

Step 1: Total possible outcomes.
When two dice are thrown, total outcomes = \(36\).
Given condition: sum of numbers = 6.

Step 2: Outcomes where sum = 6. \[ (1,5), (2,4), (3,3), (4,2), (5,1) \]
So, total favourable cases for sum = 6 → \(5\).

Step 3: Outcomes where at least one 4 appears (within sum = 6). \[ (2,4), (4,2) \]
So, number of favourable outcomes = \(2\).

Step 4: Conditional probability. \[ P(at least one 4 \mid sum = 6) = \frac{Favourable outcomes}{Total outcomes with sum = 6} = \frac{2}{5} \]


Final Answer: \[ \boxed{\dfrac{2}{5}} \] Quick Tip: Conditional Probability = \(\dfrac{n(Favourable outcomes satisfying condition)}{n(Total outcomes under given condition)}\).

Step 1: Compare with symmetric form.
A line in symmetric form is: \[ \frac{x-x_{1}}{a} = \frac{y-y_{1}}{b} = \frac{z-z_{1}}{c} \]

Here, \[ \frac{x+3}{2} = \frac{y-5}{4} = \frac{z+6}{2} \]
So, \[ (x_{1}, y_{1}, z_{1}) = (-3, 5, -6), \quad \vec{d} = (2, 4, 2) \]

Step 2: Write vector equation of line.
General form: \[ \vec{r} = \vec{a} + \lambda \vec{d} \]
where \(\vec{a}\) is position vector of a point, \(\vec{d}\) is direction vector.

Step 3: Substitute values. \[ \vec{a} = -3\hat{i} + 5\hat{j} - 6\hat{k}, \quad \vec{d} = 2\hat{i} + 4\hat{j} + 2\hat{k} \]

So, \[ \vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k}) \]


Final Answer: \[ \boxed{\vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k})} \] Quick Tip: Vector equation of a line: \(\vec{r} = \vec{a} + \lambda \vec{d}\), where \(\vec{a}\) is a point and \(\vec{d}\) is direction vector.

Step 1: Rewrite equation. \[ \log\left(\dfrac{dy}{dx}\right) = 3x + 4y \]

Take exponential both sides: \[ \dfrac{dy}{dx} = e^{3x + 4y} \]

Step 2: Separate variables. \[ e^{-4y} dy = e^{3x} dx \]

Step 3: Integrate both sides. \[ \int e^{-4y} dy = \int e^{3x} dx \]
\[ \dfrac{e^{-4y}}{-4} = \dfrac{e^{3x}}{3} + C \]

Step 4: Simplify. \[ e^{-4y} = -\dfrac{4}{3} e^{3x} + C' \]
(where \(C' = 4C\) is constant of integration).


Final Answer: \[ \boxed{e^{-4y} + \dfrac{4}{3}e^{3x} = C} \] Quick Tip: Always use separation of variables for equations of the form \(\dfrac{dy}{dx} = f(x)\,g(y)\).

We need to prove that \(R\) is reflexive, symmetric, and transitive.

Step 1: Reflexivity.

For any \(a \in \mathbb{Z}\), we have \(a-a = 0\), which is divisible by \(2\).
So, \((a,a) \in R\). Hence, \(R\) is reflexive.

Step 2: Symmetry.

Suppose \((a,b) \in R\). Then, \((a-b)\) is divisible by \(2\).
So, \(a-b = 2k\) for some \(k \in \mathbb{Z}\). \(\Rightarrow b-a = -2k\), which is also divisible by \(2\).
Thus, \((b,a) \in R\). Hence, \(R\) is symmetric.

Step 3: Transitivity.

Suppose \((a,b) \in R\) and \((b,c) \in R\).
Then, \(a-b = 2k_1\), and \(b-c = 2k_2\) for some integers \(k_1, k_2\).
Adding, \(a-c = (a-b) + (b-c) = 2k_1 + 2k_2 = 2(k_1+k_2)\).
Thus, \(a-c\) is divisible by \(2\) \(\Rightarrow (a,c) \in R\).
Hence, \(R\) is transitive.

Step 4: Conclusion.

Since \(R\) is reflexive, symmetric, and transitive, \(R\) is an equivalence relation.


Final Answer: \[ \boxed{R is an equivalence relation.} \] Quick Tip: To prove equivalence relation: check reflexive, symmetric, and transitive properties.

Step 1: Compute \(AB\).
\[ AB = \begin{bmatrix} 2 & 3
1 & -4 \end{bmatrix} \begin{bmatrix} 1 & -2
-1 & 3 \end{bmatrix} = \begin{bmatrix} 2(1)+3(-1) & 2(-2)+3(3)
1(1)+(-4)(-1) & 1(-2)+(-4)(3) \end{bmatrix} \]
\[ AB = \begin{bmatrix} -1 & 5
5 & -14 \end{bmatrix} \]

Step 2: Find \((AB)^{-1}\).

For a \(2 \times 2\) matrix \(\begin{bmatrix} p & q
r & s \end{bmatrix}\), \[ M^{-1} = \frac{1}{ps-qr}\begin{bmatrix} s & -q
-r & p \end{bmatrix} \]

Here, \(\det(AB) = (-1)(-14) - (5)(5) = 14 - 25 = -11\).
\[ (AB)^{-1} = \frac{1}{-11} \begin{bmatrix} -14 & -5
-5 & -1 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 14 & 5
5 & 1 \end{bmatrix} \]

Step 3: Find \(A^{-1}\) and \(B^{-1}\).

For \(A = \begin{bmatrix} 2 & 3
1 & -4 \end{bmatrix}\), \(\det(A) = (2)(-4) - (3)(1) = -8 - 3 = -11\).
\[ A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3
-1 & 2 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 4 & 3
1 & -2 \end{bmatrix} \]

For \(B = \begin{bmatrix} 1 & -2
-1 & 3 \end{bmatrix}\), \(\det(B) = (1)(3) - (-2)(-1) = 3 - 2 = 1\).
\[ B^{-1} = \begin{bmatrix} 3 & 2
1 & 1 \end{bmatrix} \]

Step 4: Compute \(B^{-1}A^{-1}\).
\[ B^{-1}A^{-1} = \begin{bmatrix} 3 & 2
1 & 1 \end{bmatrix} \cdot \frac{1}{11}\begin{bmatrix} 4 & 3
1 & -2 \end{bmatrix} \]
\[ = \frac{1}{11}\begin{bmatrix} 3(4)+2(1) & 3(3)+2(-2)
1(4)+1(1) & 1(3)+1(-2) \end{bmatrix} \]
\[ = \frac{1}{11}\begin{bmatrix} 14 & 5
5 & 1 \end{bmatrix} \]

Step 5: Compare results.
\[ (AB)^{-1} = \frac{1}{11}\begin{bmatrix} 14 & 5
5 & 1 \end{bmatrix}, \quad B^{-1}A^{-1} = \frac{1}{11}\begin{bmatrix} 14 & 5
5 & 1 \end{bmatrix} \]
\[ \therefore (AB)^{-1} = B^{-1}A^{-1} \]


Final Answer: \[ \boxed{(AB)^{-1} = B^{-1}A^{-1}} \] Quick Tip: Always remember: inverse of product reverses the order → \((AB)^{-1} = B^{-1}A^{-1}\).

We have two terms: \(y = (\cos x)^{\tan x} + x^{x}\)

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Part 1: Differentiate \((\cos x)^{\tan x}\)

Let \(u = (\cos x)^{\tan x}\) \[ \ln u = \tan x \cdot \ln(\cos x) \]

Differentiate w.r.t \(x\): \[ \frac{1}{u}\frac{du}{dx} = \sec^{2}x \cdot \ln(\cos x) + \tan x \cdot \frac{1}{\cos x}(-\sin x) \]
\[ \frac{1}{u}\frac{du}{dx} = \sec^{2}x \cdot \ln(\cos x) - \tan^{2}x \]

So, \[ \frac{du}{dx} = (\cos x)^{\tan x}\left(\sec^{2}x \cdot \ln(\cos x) - \tan^{2}x\right) \]

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Part 2: Differentiate \(x^{x}\)

Let \(v = x^{x}\) \[ \ln v = x\ln x \]

Differentiate: \[ \frac{1}{v}\frac{dv}{dx} = \ln x + 1 \]
\[ \frac{dv}{dx} = x^{x}(\ln x + 1) \]

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Final derivative: \[ \frac{dy}{dx} = (\cos x)^{\tan x}\left(\sec^{2}x \cdot \ln(\cos x) - \tan^{2}x\right) + x^{x}(\ln x + 1) \]


Final Answer: \[ \boxed{\dfrac{dy}{dx} = (\cos x)^{\tan x}\Big(\sec^{2}x \cdot \ln(\cos x) - \tan^{2}x\Big) + x^{x}(\ln x + 1)} \] Quick Tip: For \(f(x)^{g(x)}\), use \(\ln\) differentiation: \(\ln y = g(x)\ln f(x)\).

Condition for continuity at \(x=3\): \[ \lim_{x\to 3^-} f(x) = f(3) = \lim_{x\to 3^+} f(x) \]

Step 1: Left-hand limit (\(x \leq 3\)). \[ \lim_{x\to 3^-} f(x) = a(3) + 1 = 3a + 1 \]

Step 2: Value at \(x=3\).
Since \(x=3\) is included in first case: \[ f(3) = 3a + 1 \]

Step 3: Right-hand limit (\(x > 3\)). \[ \lim_{x\to 3^+} f(x) = b(3) + 3 = 3b + 3 \]

Step 4: Apply continuity condition. \[ 3a + 1 = 3b + 3 \]
\[ 3a - 3b = 2 \quad \implies \quad a - b = \tfrac{2}{3} \]

Conclusion:
The values of \(a\) and \(b\) are related by: \[ \boxed{a - b = \tfrac{2}{3}} \] Quick Tip: For piecewise functions → apply \(\lim_{x\to c^-} f(x) = \lim_{x\to c^+} f(x) = f(c)\) for continuity.

Step 1: Differentiate the function.
\[ f(x) = 4x^3 - 6x^2 - 72x + 30 \]
\[ f'(x) = 12x^2 - 12x - 72 \]

Step 2: Simplify derivative.
\[ f'(x) = 12(x^2 - x - 6) = 12(x-3)(x+2) \]

Step 3: Critical points.

Set \(f'(x) = 0\): \[ 12(x-3)(x+2) = 0 \quad \Rightarrow \quad x = 3, \; x = -2 \]

Step 4: Sign analysis of \(f'(x)\).

Check intervals \((-\infty,-2)\), \((-2,3)\), and \((3,\infty)\):


For \(x < -2\): Choose \(x = -3\), then \((x-3)(x+2) = (-6)(-1) = +6 > 0\).
\(\Rightarrow f'(x) > 0\) → Increasing.
For \(-2 < x < 3\): Choose \(x = 0\), then \((x-3)(x+2) = (-3)(2) = -6 < 0\).
\(\Rightarrow f'(x) < 0\) → Decreasing.
For \(x > 3\): Choose \(x = 4\), then \((x-3)(x+2) = (1)(6) = +6 > 0\).
\(\Rightarrow f'(x) > 0\) → Increasing.


Step 5: Conclusion.


Function is increasing in intervals \((-\infty, -2)\) and \((3, \infty)\).
Function is decreasing in interval \((-2, 3)\).



Final Answer: \[ \boxed{Increasing in (-\infty, -2) \cup (3,\infty), \quad Decreasing in (-2,3)} \] Quick Tip: Use first derivative test: \(f'(x) > 0\) → Increasing, \(f'(x) < 0\) → Decreasing.

We have: \[ (\tan^{-1}y - x)dy = (1+y^{2})dx \]

Step 1: Rearrange. \[ \frac{dy}{dx} = \frac{1+y^{2}}{\tan^{-1}y - x} \]

Step 2: Substitution.
Let \(z = \tan^{-1}y \implies \frac{dz}{dy} = \frac{1}{1+y^{2}} \implies dy = (1+y^{2})dz\)

Step 3: Rewrite.
Substitute into original equation: \[ (z - x)(1+y^{2})dz = (1+y^{2})dx \]

Cancel \((1+y^{2})\): \[ (z - x)dz = dx \]

Step 4: Rearrange terms. \[ \frac{dx}{dz} + x = z \]

Step 5: Solve linear ODE.
This is linear in \(x\): \[ \frac{dx}{dz} + x = z \]

Integrating factor: \(e^{\int 1dz} = e^{z}\)
\[ \frac{d}{dz}(xe^{z}) = ze^{z} \]

Step 6: Integrate. \[ xe^{z} = \int ze^{z}dz \]

Integration by parts: \[ \int ze^{z}dz = (z-1)e^{z} + C \]

So, \[ xe^{z} = (z-1)e^{z} + C \]
\[ x = z - 1 + Ce^{-z} \]

Step 7: Back substitution. \(z = \tan^{-1}y\)
\[ \boxed{x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}} \] Quick Tip: When \(\arctan y\) appears, use substitution \(z = \tan^{-1}y\) to simplify.

Step 1: Write in standard form.
First line passes through \(A(1,2,-4)\), direction vector \(\vec{d} = (2,3,6)\).
Second line passes through \(B(3,3,-5)\), direction vector \(\vec{d} = (2,3,6)\).

Step 2: Check relation of lines.
Since both lines have the same direction vector \((2,3,6)\), they are parallel.

Step 3: Formula for shortest distance between parallel lines. \[ d = \frac{|\overrightarrow{AB} \times \vec{d}|}{|\vec{d}|} \]

Step 4: Find \(\overrightarrow{AB}\). \[ \overrightarrow{AB} = (3-1,\,3-2,\,-5-(-4)) = (2,1,-1) \]

Step 5: Compute cross product. \[ \overrightarrow{AB} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 1 & -1
2 & 3 & 6 \end{vmatrix} \]
\[ = \hat{i}(1\cdot 6 - (-1)\cdot 3) - \hat{j}(2\cdot 6 - (-1)\cdot 2) + \hat{k}(2\cdot 3 - 1\cdot 2) \]
\[ = \hat{i}(6+3) - \hat{j}(12+2) + \hat{k}(6-2) \]
\[ = 9\hat{i} - 14\hat{j} + 4\hat{k} \]

Step 6: Magnitudes. \[ |\overrightarrow{AB} \times \vec{d}| = \sqrt{9^{2} + (-14)^{2} + 4^{2}} = \sqrt{81+196+16} = \sqrt{293} \]
\[ |\vec{d}| = \sqrt{2^{2}+3^{2}+6^{2}} = \sqrt{4+9+36} = \sqrt{49} = 7 \]

Step 7: Distance. \[ d = \frac{\sqrt{293}}{7} \]


Final Answer: \[ \boxed{d = \dfrac{\sqrt{293}}{7}} \] Quick Tip: For parallel lines, use \(d = \dfrac{|\overrightarrow{AB}\times \vec{d}|}{|\vec{d}|}\).

Step 1: Define the experiment.

A die is thrown three times. Each throw has \(6\) possible outcomes.
Total number of outcomes: \[ n(S) = 6^3 = 216 \]

Step 2: Use complementary probability.

We want the probability of at least one odd number. \[ P(at least one odd) = 1 - P(no odd) \]

Step 3: Probability of no odd number.

Odd numbers on a die: \(\{1,3,5\}\) (3 outcomes).
Even numbers on a die: \(\{2,4,6\}\) (3 outcomes).

Probability of even in one throw = \(\dfrac{3}{6} = \dfrac{1}{2}\).

For three throws, probability of all even = \[ \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]

Step 4: Final probability.
\[ P(at least one odd) = 1 - \frac{1}{8} = \frac{7}{8} \]


Final Answer: \[ \boxed{\dfrac{7}{8}} \] Quick Tip: In probability problems involving "at least one", it is easier to use the complement rule: \(P(at least one) = 1 - P(none)\).

Step 1: Rewrite the equations. \[ Circle: x^{2} + y^{2} = 8x \quad \Rightarrow \quad (x-4)^{2} + y^{2} = 16 \]
This is a circle with centre \((4,0)\) and radius \(4\).
\[ Parabola: y^{2} = 4x \quad \Rightarrow \quad x = \frac{y^{2}}{4} \]

Step 2: Points of intersection.
Substitute \(x = \frac{y^{2}}{4}\) in circle equation: \[ \left(\frac{y^{2}}{4}\right)^{2} + y^{2} = 8\left(\frac{y^{2}}{4}\right) \] \[ \frac{y^{4}}{16} + y^{2} = 2y^{2} \quad \Rightarrow \quad \frac{y^{4}}{16} = y^{2} \] \[ y^{2}\left(\frac{y^{2}}{16} - 1\right) = 0 \quad \Rightarrow \quad y = 0, \; y = \pm 4 \]
On \(x\)-axis, \(y=0 \Rightarrow x=0\). For \(y=4\), \(x = \frac{16}{4} = 4\).

So, points of intersection: \((0,0)\) and \((4,4)\).

Step 3: Required area.
We want area bounded by parabola, circle, and \(x\)-axis above.

In terms of \(y\): \[ Parabola: x = \frac{y^{2}}{4}, \quad Circle: x = 4 + \sqrt{16 - y^{2}} \]

Thus, area is: \[ A = \int_{0}^{4} \left[ \big(4 + \sqrt{16 - y^{2}}\big) - \frac{y^{2}}{4} \right] dy \]

Step 4: Solve integration. \[ A = \int_{0}^{4} \left(4 - \frac{y^{2}}{4}\right) dy + \int_{0}^{4} \sqrt{16 - y^{2}} \, dy \]

First part: \[ \int_{0}^{4} \left(4 - \frac{y^{2}}{4}\right) dy = \left[4y - \frac{y^{3}}{12}\right]_{0}^{4} = 16 - \frac{64}{12} = 16 - \frac{16}{3} = \frac{32}{3} \]

Second part: \[ \int_{0}^{4} \sqrt{16 - y^{2}} \, dy \]
This represents a quarter circle of radius \(4\), so area = \(\frac{1}{4} \pi r^{2} = 4\pi\).

Total Area: \[ A = \frac{32}{3} + 4\pi \]


Final Answer: \[ \boxed{\; \dfrac{32}{3} + 4\pi \;} \] Quick Tip: For area bounded by two curves, always take \(\int (x_{right} - x_{left}) dy\).

Step 1: Separate the integral. \[ I = \int \log(\log x) \, dx + \int \frac{1}{(\log x)^{2}} \, dx \]

Step 2: Substitution.
Let \(t = \log x \;\Rightarrow\; dx = e^{t} dt = x dt\).
\[ I = \int \log t \cdot e^{t} dt + \int \frac{e^{t}}{t^{2}} dt \]

Step 3: First integral by parts. \[ \int e^{t} \log t \, dt = e^{t} \log t - \int \frac{e^{t}}{t} dt \]

Step 4: Combine both integrals. \[ I = e^{t} \log t - \int \frac{e^{t}}{t} dt + \int \frac{e^{t}}{t^{2}} dt \]

Notice that: \[ \frac{d}{dt}\left(-\frac{e^{t}}{t}\right) = -\frac{e^{t}}{t} + \frac{e^{t}}{t^{2}} \]

So, \[ - \int \frac{e^{t}}{t} dt + \int \frac{e^{t}}{t^{2}} dt = -\frac{e^{t}}{t} \]

Step 5: Final result. \[ I = e^{t} \log t - \frac{e^{t}}{t} + C \]

Back substitute \(t = \log x\): \[ I = x \log(\log x) - \frac{x}{\log x} + C \]


Final Answer: \[ \boxed{\; x \log(\log x) - \dfrac{x}{\log x} + C \;} \] Quick Tip: When you see \(\log(\log x)\) or \(\dfrac{1}{(\log x)^{2}}\), always try substitution \(t = \log x\).

Step 1: Compute \(A^{2}\). \[ A^{2} = \begin{bmatrix} 2 & 3
1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3
1 & 2 \end{bmatrix} = \begin{bmatrix} 4+3 & 6+6
2+2 & 3+4 \end{bmatrix} = \begin{bmatrix} 7 & 12
4 & 7 \end{bmatrix} \]

Step 2: Compute \(A^{2} - 4A + I_{2}\). \[ A^{2} - 4A + I_{2} = \begin{bmatrix} 7 & 12
4 & 7 \end{bmatrix} - 4 \begin{bmatrix} 2 & 3
1 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix} \]
\[ = \begin{bmatrix} 7 & 12
4 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12
4 & 8 \end{bmatrix} + \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0
0 & 0 \end{bmatrix} \]

Hence proved. ✅

Step 3: Find \(A^{-1}\).
From Cayley-Hamilton theorem: \[ A^{2} - 4A + I = 0 \]
\[ \implies A^{2} - 4A = -I \]
\[ \implies A(A - 4I) = -I \]
\[ \implies A^{-1} = -(A - 4I) \]
\[ = -\left(\begin{bmatrix} 2 & 3
1 & 2 \end{bmatrix} - \begin{bmatrix} 4 & 0
0 & 4 \end{bmatrix}\right) = -\begin{bmatrix} -2 & 3
1 & -2 \end{bmatrix} = \begin{bmatrix} 2 & -3
-1 & 2 \end{bmatrix} \]


Final Answer: \[ \boxed{A^{-1} = \begin{bmatrix} 2 & -3
-1 & 2 \end{bmatrix}} \] Quick Tip: Cayley-Hamilton theorem provides a direct method to compute \(A^{-1}\) without using the adjoint and determinant.

Step 1: Write system in matrix form. \[ AX = B \]

where \[ A = \begin{bmatrix} 3 & -2 & 3
2 & 1 & -1
4 & -3 & 2 \end{bmatrix},\quad X = \begin{bmatrix} x
y
z \end{bmatrix},\quad B = \begin{bmatrix} 8
1
4 \end{bmatrix} \]

Step 2: Solve using inverse method. \[ X = A^{-1}B \]

Step 3: Find \(\det(A)\). \[ \det(A) = 3\begin{vmatrix} 1 & -1
-3 & 2 \end{vmatrix} - (-2)\begin{vmatrix} 2 & -1
4 & 2 \end{vmatrix} + 3\begin{vmatrix} 2 & 1
4 & -3 \end{vmatrix} \]
\[ = 3(1\cdot 2 - (-1)(-3)) + 2(2\cdot 2 - (-1)(4)) + 3(2(-3)-1(4)) \]
\[ = 3(2-3) + 2(4+4) + 3(-6-4) = 3(-1) + 16 + (-30) = -17 \]

So, \(\det(A) \neq 0\), hence solution exists.

Step 4: Use Cramer’s Rule.
\[ x = \frac{\det(A_{1})}{\det(A)},\quad y = \frac{\det(A_{2})}{\det(A)},\quad z = \frac{\det(A_{3})}{\det(A)} \]

- Replace first column of \(A\) with \(B\) for \(A_{1}\): \[ A_{1} = \begin{bmatrix} 8 & -2 & 3
1 & 1 & -1
4 & -3 & 2 \end{bmatrix},\quad \det(A_{1}) = 17 \]

- Replace second column of \(A\) with \(B\) for \(A_{2}\): \[ A_{2} = \begin{bmatrix} 3 & 8 & 3
2 & 1 & -1
4 & 4 & 2 \end{bmatrix},\quad \det(A_{2}) = -34 \]

- Replace third column of \(A\) with \(B\) for \(A_{3}\): \[ A_{3} = \begin{bmatrix} 3 & -2 & 8
2 & 1 & 1
4 & -3 & 4 \end{bmatrix},\quad \det(A_{3}) = -51 \]

Step 5: Compute values. \[ x = \frac{17}{-17} = -1,\quad y = \frac{-34}{-17} = 2,\quad z = \frac{-51}{-17} = 3 \]


Final Answer: \[ \boxed{x = -1,\; y = 2,\; z = 3} \] Quick Tip: For \(3\times 3\) systems, Cramer’s Rule is the fastest way if \(\det(A)\neq 0\).

Step 1: Simplify constraints.
\[ 9x + 12y \leq 180 \;\;\Rightarrow\;\; 3x + 4y \leq 60 \quad (same as 2nd constraint) \]
So effective constraints are: \[ 3x + 4y \leq 60, \quad x + 3y \leq 30, \quad x \geq 0, \; y \geq 0 \]

Step 2: Find corner points.

- When \(x=0\): \(4y \leq 60 \Rightarrow y \leq 15\), and \(3y \leq 30 \Rightarrow y \leq 10\). So feasible point \((0,10)\).
- When \(y=0\): \(3x \leq 60 \Rightarrow x \leq 20\), and \(x \leq 30\). So feasible point \((20,0)\).
- Intersection of \(3x+4y=60\) and \(x+3y=30\): \[ 3x+4y=60 \quad (1), \quad x+3y=30 \quad (2) \]
From (2), \(x=30-3y\). Substitute into (1): \[ 3(30-3y) + 4y = 60 \;\;\Rightarrow\;\; 90-9y+4y=60 \;\;\Rightarrow\;\; -5y=-30 \;\;\Rightarrow\;\; y=6, \; x=12 \]
So intersection point \((12,6)\).

Step 3: Evaluate objective function at feasible points.
\[ Z = 8000x+12000y \]
- At \((0,0)\): \(Z=0\)
- At \((20,0)\): \(Z=160000\)
- At \((0,10)\): \(Z=120000\)
- At \((12,6)\): \(Z=8000(12)+12000(6)=96000+72000=168000\)

Step 4: Conclusion.

Maximum value of \(Z\) is \[ \boxed{168000 \;\;at\;\; (x,y)=(12,6)} \] Quick Tip: In LPP, the maximum/minimum value always occurs at a corner point of the feasible region.

Step 1: Use property of definite integrals.
\[ I = \int_0^{\pi/2} \frac{\sin^4x}{\sin^4x+\cos^4x} dx \]
Now, \[ I = \int_0^{\pi/2} \frac{\cos^4x}{\cos^4x+\sin^4x} dx \quad (by x \to \tfrac{\pi}{2}-x ) \]

Step 2: Add the two results.
\[ 2I = \int_0^{\pi/2} \frac{\sin^4x}{\sin^4x+\cos^4x} dx + \int_0^{\pi/2} \frac{\cos^4x}{\cos^4x+\sin^4x} dx \] \[ 2I = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \]

Step 3: Final value.
\[ I = \frac{\pi}{4} \]


Final Answer: \[ \boxed{\tfrac{\pi}{4}} \] Quick Tip: In integrals of the form \(\int_0^{\pi/2} f(\sin x,\cos x) dx\), use the property \(I=\int_0^{\pi/2} f(\cos x,\sin x) dx\).

Step 1: Use property of definite integrals.

Let \[ I = \int_0^{\pi/2} \ln(\sin x)\, dx \]
Using property: \[ I = \int_0^{\pi/2} \ln(\cos x)\, dx \]

Step 2: Add the two results.
\[ 2I = \int_0^{\pi/2} [\ln(\sin x) + \ln(\cos x)] dx \] \[ 2I = \int_0^{\pi/2} \ln(\sin x \cos x)\, dx \]

Step 3: Simplify.
\[ \sin x \cos x = \tfrac{1}{2}\sin 2x \]
So, \[ 2I = \int_0^{\pi/2} \ln\left(\tfrac{1}{2}\sin 2x\right)\, dx \]

Step 4: Substitution.

Put \(2x = t \;\Rightarrow\; dx = dt/2\), limits \(0 \to \pi\). \[ 2I = \frac{1}{2}\int_0^{\pi} \ln\left(\tfrac{1}{2}\sin t\right) dt \] \[ = \frac{1}{2}\left[\int_0^{\pi} \ln(\sin t)\, dt - \int_0^{\pi} \ln 2 \, dt\right] \]

Step 5: Evaluate.
\[ \int_0^{\pi} \ln(\sin t)\, dt = 2\int_0^{\pi/2} \ln(\sin t)\, dt = 2I \]
So, \[ 2I = \frac{1}{2}(2I - \pi \ln 2) \] \[ 4I = 2I - \pi \ln 2 \;\;\Rightarrow\;\; 2I = -\pi \ln 2 \] \[ I = -\frac{\pi}{2}\ln 2 \]


Final Answer: \[ \boxed{-\tfrac{\pi}{2}\ln 2} \] Quick Tip: Standard result: \(\int_0^{\pi/2} \ln(\sin x) dx = -\tfrac{\pi}{2}\ln 2\). Similarly, \(\int_0^{\pi/2} \ln(\cos x) dx = -\tfrac{\pi}{2}\ln 2\).

We are given: \[ y = \sin^{-1}x \]

Step 1: First derivative. \[ \frac{dy}{dx} = \frac{1}{\sqrt{1-x^{2}}} \]

Step 2: Second derivative. \[ \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left(\frac{1}{\sqrt{1-x^{2}}}\right) = \frac{1}{2}(1-x^{2})^{-\tfrac{3}{2}} \cdot (2x) = \frac{x}{(1-x^{2})^{\tfrac{3}{2}}} \]

Step 3: Substitute in given expression. \[ (1-x^{2}) \cdot \frac{d^{2}y}{dx^{2}} - x \cdot \frac{dy}{dx} \] \[ = (1-x^{2}) \cdot \frac{x}{(1-x^{2})^{3/2}} - x \cdot \frac{1}{\sqrt{1-x^{2}}} \] \[ = \frac{x(1-x^{2})}{(1-x^{2})^{3/2}} - \frac{x}{\sqrt{1-x^{2}}} \] \[ = \frac{x}{\sqrt{1-x^{2}}} - \frac{x}{\sqrt{1-x^{2}}} = 0 \]


Hence Proved. \[ \boxed{(1 - x^{2}) \frac{d^{2}y}{dx^{2}} - x \frac{dy}{dx} = 0} \] Quick Tip: For proving differential identities, always calculate \(\dfrac{dy}{dx}\) and \(\dfrac{d^{2}y}{dx^{2}}\), then simplify the given expression.

Step 1: Slope of the given line.
Equation: \(x + 2y = 0 \;\Rightarrow\; y = -\tfrac{1}{2}x\).
So, slope = \(-\tfrac{1}{2}\).

Step 2: Differentiate the curve.
Curve: \[ y = \cos(x+y) \]

Differentiate both sides w.r.t. \(x\): \[ \frac{dy}{dx} = -\sin(x+y)\left(1 + \frac{dy}{dx}\right) \]
\[ \frac{dy}{dx} + \sin(x+y)\frac{dy}{dx} = -\sin(x+y) \] \[ \frac{dy}{dx}(1 + \sin(x+y)) = -\sin(x+y) \] \[ \frac{dy}{dx} = \frac{-\sin(x+y)}{1 + \sin(x+y)} \]

Step 3: Condition for parallelism.
We need \(\dfrac{dy}{dx} = -\tfrac{1}{2}\). \[ \frac{-\sin(x+y)}{1 + \sin(x+y)} = -\frac{1}{2} \] \[ \frac{\sin(x+y)}{1 + \sin(x+y)} = \frac{1}{2} \] \[ 2 \sin(x+y) = 1 + \sin(x+y) \quad \Rightarrow \quad \sin(x+y) = 1 \]

Step 4: Corresponding points.
If \(\sin(x+y) = 1 \;\Rightarrow\; x+y = \frac{\pi}{2} + 2n\pi\).

Also, from curve \(y = \cos(x+y) = \cos\left(\frac{\pi}{2} + 2n\pi\right) = 0\).
So, \(y=0\), \(x = \frac{\pi}{2} + 2n\pi\).

For \(-2\pi \leq x \leq 2\pi\), possible \(x = -\tfrac{3\pi}{2}, \; \tfrac{\pi}{2}\).

Step 5: Equation of tangent.
Equation of tangent: \(y - y_{1} = m(x - x_{1})\), with \(m = -\tfrac{1}{2}\).

At \(\left(\tfrac{\pi}{2},0\right)\): \[ y - 0 = -\tfrac{1}{2}(x - \tfrac{\pi}{2}) \] \[ x + 2y - \tfrac{\pi}{2} = 0 \]

At \(\left(-\tfrac{3\pi}{2},0\right)\): \[ y - 0 = -\tfrac{1}{2}(x + \tfrac{3\pi}{2}) \] \[ x + 2y + \tfrac{3\pi}{2} = 0 \]


Final Answer: \[ \boxed{\; x + 2y - \tfrac{\pi}{2} = 0 \quad or \quad x + 2y + \tfrac{3\pi}{2} = 0 \;} \] Quick Tip: For tangents parallel to a line: find slope from given line, equate with \(\dfrac{dy}{dx}\) of curve, then find points and equations.

Step 1: Equation of required plane.
The equation of a plane passing through the line of intersection of two planes is: \[ \vec{r} \cdot \vec{n}_1 = d_1 + \lambda (\vec{r} \cdot \vec{n}_2 - d_2) \]

Here, \[ \vec{n}_1 = (1,1,1), \quad d_1 = 6 \] \[ \vec{n}_2 = (2,3,4), \quad d_2 = -5 \]

So, \[ \vec{r} \cdot (1,1,1) - 6 + \lambda \big(\vec{r} \cdot (2,3,4) + 5 \big) = 0 \]

Step 2: General equation of the plane. \[ (x+y+z - 6) + \lambda (2x + 3y + 4z + 5) = 0 \]

Step 3: Condition of passing through \((1,1,1)\).
Substitute \(x=1, y=1, z=1\): \[ (1+1+1 - 6) + \lambda (2(1)+3(1)+4(1)+5) = 0 \]
\[ (-3) + \lambda (14) = 0 \quad \implies \quad \lambda = \frac{3}{14} \]

Step 4: Required plane.
Substitute \(\lambda = \frac{3}{14}\): \[ (x+y+z - 6) + \frac{3}{14}(2x + 3y + 4z + 5) = 0 \]

Multiply through by \(14\): \[ 14(x+y+z - 6) + 3(2x + 3y + 4z + 5) = 0 \]
\[ 14x + 14y + 14z - 84 + 6x + 9y + 12z + 15 = 0 \]
\[ 20x + 23y + 26z - 69 = 0 \]


Final Answer: \[ \boxed{20x + 23y + 26z - 69 = 0} \] Quick Tip: The plane through the intersection of two planes always involves a parameter \(\lambda\), which is determined using the given point.

Step 1: Standard form. \[ \frac{dy}{dx} - y = \cos x \]
This is a linear differential equation of the form: \[ \frac{dy}{dx} + P(x) y = Q(x), \quad P(x) = -1, \; Q(x) = \cos x \]

Step 2: Find the integrating factor (IF). \[ IF = e^{\int P(x)dx} = e^{\int -1 dx} = e^{-x} \]

Step 3: Multiply through by IF. \[ e^{-x}\frac{dy}{dx} - e^{-x}y = e^{-x}\cos x \]
\[ \implies \frac{d}{dx}(y e^{-x}) = e^{-x}\cos x \]

Step 4: Integrate both sides. \[ y e^{-x} = \int e^{-x}\cos x \, dx + C \]

Now, \[ \int e^{-x}\cos x \, dx = \frac{e^{-x}(-\cos x + \sin x)}{2} \]

Step 5: Final solution. \[ y e^{-x} = \frac{e^{-x}(\sin x - \cos x)}{2} + C \]
\[ y = \frac{\sin x - \cos x}{2} + Ce^{x} \]


Final Answer: \[ \boxed{y = \frac{\sin x - \cos x}{2} + Ce^{x}} \] Quick Tip: For first-order linear differential equations, always compute the Integrating Factor (IF) and reduce it to the derivative of \((y \cdot IF)\).