Class 12th Mathematics Sample Paper Set II (NCERT Exemplar)

Given:

The binary operation $*$ on R is defined as $a * b = a + b^2$.

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To Find:

The value of $-2*5$.

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Solution:

The given binary operation is $a * b = a + b^2$.

We need to find the value of $-2*5$.

Comparing $-2*5$ with $a*b$, we have $a = -2$ and $b = 5$.

Substitute these values into the definition of the operation:

$-2 * 5 = (-2) + (5)^2$

$-2 * 5 = -2 + 25$

$-2 * 5 = 23$

Thus, the value of $-2*5$ is $23$.

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The correct option is (B) 23.

Given:

A function $\sin^{–1} : [–1, 1] \to \left[\frac{\pi}{2},\frac{3\pi}{2}\right]$.

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To Find:

The value of $\sin^{-1} \left(-\frac{1}{2}\right)$.

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Solution:

The standard principal value branch of $\sin^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. In this range, $\sin^{-1} \left(-\frac{1}{2}\right) = -\frac{\pi}{6}$ because $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$ and $-\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

However, the problem defines the codomain of the given function as $\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$.

We need to find a value $\theta$ such that $\sin(\theta) = -\frac{1}{2}$ and $\theta$ lies in the interval $\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$.

The sine function is negative in the third and fourth quadrants.

The angle $\theta$ must satisfy $\sin(\theta) = -\frac{1}{2}$. The reference angle for $\sin \theta = \frac{1}{2}$ is $\frac{\pi}{6}$.

The interval $\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$ covers the second, third, and part of the fourth quadrant.

We look for an angle in this interval whose sine is $-\frac{1}{2}$. This angle must be in the third or fourth quadrant part of the interval.

Angles in the third quadrant are of the form $\pi + \alpha$, where $\alpha$ is the reference angle. For $\sin(\theta) = -\frac{1}{2}$, the reference angle is $\frac{\pi}{6}$.

A possible angle in the third quadrant is $\pi + \frac{\pi}{6} = \frac{7\pi}{6}$.

Let's check if $\frac{7\pi}{6}$ lies in the given codomain $\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$.

$\frac{\pi}{2} = \frac{3\pi}{6}$ and $\frac{3\pi}{2} = \frac{9\pi}{6}$.

Since $\frac{3\pi}{6} \le \frac{7\pi}{6} \le \frac{9\pi}{6}$, the angle $\frac{7\pi}{6}$ is in the interval $\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$.

Also, $\sin\left(\frac{7\pi}{6}\right) = \sin\left(\pi + \frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}$.

Therefore, the value of $\sin^{-1} \left(-\frac{1}{2}\right)$ in the codomain $\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$ is $\frac{7\pi}{6}$.

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Let's examine the options:

(A) $\frac{-\pi}{6}$: Not in the interval $\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$.

(B) $\frac{-\pi}{6}$: Same as (A).

(C) $\frac{5\pi}{6}$: $\frac{5\pi}{6}$ is in $\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$ (since $\frac{3\pi}{6} \le \frac{5\pi}{6} \le \frac{9\pi}{6}$), but $\sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, which is not $-\frac{1}{2}$.

(D) $\frac{7\pi}{6}$: As calculated, this value is in the interval $\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$ and $\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}$.

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The correct option is (D) $\frac{7\pi}{6}$.

Given:

The matrix equation $\left(\begin{matrix}9&6\\3&0\\\end{matrix}\right)=\left(\begin{matrix}2&3\\1&0\\\end{matrix}\right)\left(\begin{matrix}3&0\\1&2\\\end{matrix}\right)$.

Elementary row transformation R$_{1}$ → R$_{1}$ – 2 R$_{2}$ is applied to both sides.

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To Find:

The resulting equation after applying the given elementary row transformation to both sides.

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Solution:

Let the given equation be $A = BC$, where $A = \left(\begin{matrix}9&6\\3&0\\\end{matrix}\right)$, $B = \left(\begin{matrix}2&3\\1&0\\\end{matrix}\right)$, and $C = \left(\begin{matrix}3&0\\1&2\\\end{matrix}\right)$.

An elementary row transformation on a matrix is equivalent to pre-multiplying the matrix by the corresponding elementary matrix. Let $E$ be the elementary matrix corresponding to the transformation R$_{1}$ → R$_{1}$ – 2 R$_{2}$.

Applying the transformation to both sides of $A = BC$ means we perform $EA = E(BC)$.

Since matrix multiplication is associative, $E(BC) = (EB)C$.

Thus, the transformation is applied to the matrix $A$ on the left side and to the matrix $B$ (the leftmost matrix) on the right side.

Applying R$_{1}$ → R$_{1}$ – 2 R$_{2}$ to the matrix $A = \left(\begin{matrix}9&6\\3&0\\\end{matrix}\right)$:

The new Row 1 is (9 - 2*3, 6 - 2*0) = (9 - 6, 6 - 0) = (3, 6).

The new matrix on the left side is $\left(\begin{matrix}3&6\\3&0\\\end{matrix}\right)$.

Applying R$_{1}$ → R$_{1}$ – 2 R$_{2}$ to the matrix $B = \left(\begin{matrix}2&3\\1&0\\\end{matrix}\right)$:

The new Row 1 is (2 - 2*1, 3 - 2*0) = (2 - 2, 3 - 0) = (0, 3).

The new matrix $EB$ is $\left(\begin{matrix}0&3\\1&0\\\end{matrix}\right)$.

The matrix $C = \left(\begin{matrix}3&0\\1&2\\\end{matrix}\right)$ remains unchanged.

The new equation is $EA = (EB)C$, which is:

$\left(\begin{matrix}3&6\\3&0\\\end{matrix}\right)=\left(\begin{matrix}0&3\\1&0\\\end{matrix}\right)\left(\begin{matrix}3&0\\1&2\\\end{matrix}\right)$

Let's verify this by multiplying the matrices on the right side:

$\left(\begin{matrix}0&3\\1&0\\\end{matrix}\right)\left(\begin{matrix}3&0\\1&2\\\end{matrix}\right) = \left(\begin{matrix}0 \cdot 3 + 3 \cdot 1 & 0 \cdot 0 + 3 \cdot 2 \\ 1 \cdot 3 + 0 \cdot 1 & 1 \cdot 0 + 0 \cdot 2 \\\end{matrix}\right) = \left(\begin{matrix}0 + 3 & 0 + 6 \\ 3 + 0 & 0 + 0 \\\end{matrix}\right) = \left(\begin{matrix}3&6\\3&0\\\end{matrix}\right)$

This matches the left side matrix obtained after the transformation.

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Comparing our resulting equation with the given options, we find that it matches option (B).

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The correct option is (B) $\left(\begin{matrix}3&6\\3&0\\\end{matrix}\right)=\left(\begin{matrix}0&3\\1&0\\\end{matrix}\right)\left(\begin{matrix}3&0\\1&2\\\end{matrix}\right)$.

Given:

A is a square matrix of order 3.

The determinant of A, $|A| = 5$.

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To Find:

The value of $|Adj. A|$.

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Solution:

For any square matrix A of order $n$, the determinant of its adjoint is given by the formula:

$|Adj. A| = |A|^{n-1}$

In this problem, the order of the matrix A is $n=3$, and the determinant of A is $|A|=5$.

Substituting these values into the formula:

$|Adj. A| = |A|^{3-1}$

$|Adj. A| = |A|^2$

$|Adj. A| = (5)^2$

$|Adj. A| = 25$

Thus, the value of $|Adj. A|$ is $25$.

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Given:

A and B are square matrices of order 3.

The determinant of A, $|A| = -1$.

The determinant of B, $|B| = 4$.

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To Find:

The value of $|3(AB)|$.

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Solution:

Given that A and B are square matrices of order 3.

We are given $|A| = -1$ and $|B| = 4$.

We need to find the value of $|3(AB)|$.

We use the properties of determinants:

1. For square matrices X and Y of the same order, $|XY| = |X||Y|$.

2. For a scalar $k$ and a square matrix X of order $n$, $|kX| = k^n |X|$.

In this case, the order of the matrices A and B is $n=3$.

First, consider the matrix product AB. Since A and B are both of order 3, AB is also a square matrix of order 3.

Using property 1:

$|AB| = |A||B|$

Substituting the given values of $|A|$ and $|B|$:

$|AB| = (-1)(4)$

$|AB| = -4$

Now, we need to find the determinant of $3(AB)$. Here, the scalar is $k=3$ and the matrix is AB, which is of order $n=3$.

Using property 2:

$|3(AB)| = 3^n |AB|$

$|3(AB)| = 3^3 |AB|$

We know that $3^3 = 3 \times 3 \times 3 = 27$.

So, $|3(AB)| = 27 |AB|$.

Substitute the value of $|AB| = -4$ into this equation:

$|3(AB)| = 27 (-4)$

$|3(AB)| = -108$

Thus, the value of $|3(AB)|$ is $-108$.

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Given:

The differential equation $\left[1+\left(\frac{dy}{dx}\right)^3\right]=\left(\frac{d^2y}{{dx}^2}\right)^2$.

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To Find:

The degree of the given differential equation.

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Solution:

The order of a differential equation is the order of the highest derivative appearing in the equation.

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

The given differential equation is:

$\left[1+\left(\frac{dy}{dx}\right)^3\right]=\left(\frac{d^2y}{{dx}^2}\right)^2$

First, we identify the highest order derivative in the equation.

The derivatives present are $\frac{dy}{dx}$ (first order) and $\frac{d^2y}{dx^2}$ (second order).

The highest order derivative is $\frac{d^2y}{dx^2}$.

The order of the differential equation is therefore 2.

Next, we check if the equation is a polynomial in terms of its derivatives. The given equation is already in polynomial form with respect to the derivatives $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

The highest order derivative is $\frac{d^2y}{dx^2}$. The power of this highest order derivative in the equation is 2 (from the term $\left(\frac{d^2y}{{dx}^2}\right)^2$).

Therefore, the degree of the differential equation is 2.

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Given:

The linear differential equation $x \frac{dy}{dx} - y = x^2$.

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To Find:

The integrating factor for the given differential equation.

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Solution:

The standard form of a first-order linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$.

The given differential equation is $x \frac{dy}{dx} - y = x^2$.

To convert it to the standard form, we divide the entire equation by $x$ (assuming $x \neq 0$):

$\frac{dy}{dx} - \frac{1}{x}y = \frac{x^2}{x}$

$\frac{dy}{dx} - \frac{1}{x}y = x$

Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we can identify $P(x)$ and $Q(x)$.

Here, $P(x) = -\frac{1}{x}$ and $Q(x) = x$.

The integrating factor (IF) is given by the formula:

$IF = e^{\int P(x) dx}$

Let's calculate the integral of $P(x)$:

$\int P(x) dx = \int -\frac{1}{x} dx = -\int \frac{1}{x} dx$

$\int P(x) dx = -\log|x| + C$

We take the integral without the constant of integration for the integrating factor.

$\int P(x) dx = -\log|x|$

Now, substitute this into the formula for the integrating factor:

$IF = e^{-\log|x|}$

Using the property of logarithms and exponentials, $e^{a \log b} = e^{\log b^a} = b^a$, we have:

$IF = e^{\log|x|^{-1}}$

$IF = |x|^{-1}$

$IF = \frac{1}{|x|}$

For practical purposes in solving the differential equation, we often use $\frac{1}{x}$ as the integrating factor, assuming a specific domain where $x > 0$ or $x < 0$. The question asks for "the" integrating factor, and $\frac{1}{x}$ is the commonly accepted form derived from $e^{\int -\frac{1}{x} dx}$.

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The integrating factor for solving the linear differential equation $x \frac{dy}{dx} - y = x^2$ is $\frac{1}{x}$.

Given:

The expression $\left|\hat{i}-\hat{j}\right|^2$.

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To Find:

The value of $\left|\hat{i}-\hat{j}\right|^2$.

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Solution:

We are asked to find the value of $\left|\hat{i}-\hat{j}\right|^2$.

The vectors $\hat{i}$ and $\hat{j}$ are standard orthogonal unit vectors. We can represent them in component form. Assuming a 2D coordinate system (the calculation is the same in 3D):

$\hat{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$

$\hat{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$

First, let's find the vector $\hat{i}-\hat{j}$:

$\hat{i}-\hat{j} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1-0 \\ 0-1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$

Next, we find the magnitude of this vector, $|\hat{i}-\hat{j}|$. The magnitude of a vector $\begin{pmatrix} x \\ y \end{pmatrix}$ is given by $\sqrt{x^2 + y^2}$.

$|\hat{i}-\hat{j}| = \left|\begin{pmatrix} 1 \\ -1 \end{pmatrix}\right| = \sqrt{(1)^2 + (-1)^2}$

$|\hat{i}-\hat{j}| = \sqrt{1 + 1}$

$|\hat{i}-\hat{j}| = \sqrt{2}$

Finally, we need to find the square of the magnitude:

$|\hat{i}-\hat{j}|^2 = (\sqrt{2})^2$

$|\hat{i}-\hat{j}|^2 = 2$

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Alternatively, using the property that for any vector $\mathbf{v}$, $|\mathbf{v}|^2 = \mathbf{v} \cdot \mathbf{v}$:

$|\hat{i}-\hat{j}|^2 = (\hat{i}-\hat{j}) \cdot (\hat{i}-\hat{j})$

Using the distributive property of the dot product and the fact that $\hat{i}$ and $\hat{j}$ are orthogonal unit vectors ($\hat{i} \cdot \hat{i} = |\hat{i}|^2 = 1$, $\hat{j} \cdot \hat{j} = |\hat{j}|^2 = 1$, $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{i} = 0$):

$(\hat{i}-\hat{j}) \cdot (\hat{i}-\hat{j}) = \hat{i} \cdot \hat{i} - \hat{i} \cdot \hat{j} - \hat{j} \cdot \hat{i} + \hat{j} \cdot \hat{j}$

$= |\hat{i}|^2 - 0 - 0 + |\hat{j}|^2$

$= 1^2 + 1^2$

$= 1 + 1$

$= 2$

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The value of $\left|\hat{i}-\hat{j}\right|^2$ is 2.

Given:

The equations of the two planes are:

$3x + 4y - 7 = 0$

$6x + 8y + 6 = 0$

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To Find:

The distance between the two planes.

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Solution:

The equations of the planes are:

$P_1: 3x + 4y - 7 = 0$

$P_2: 6x + 8y + 6 = 0$

The normal vector to the first plane is $\mathbf{n}_1 = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix}$.

The normal vector to the second plane is $\mathbf{n}_2 = \begin{pmatrix} 6 \\ 8 \\ 0 \end{pmatrix}$.

Since $\mathbf{n}_2 = 2 \mathbf{n}_1$, the normal vectors are parallel, which indicates that the planes are parallel.

To find the distance between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$, we use the formula:

$d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$

We need to rewrite the given equations so that the coefficients of $x, y, z$ are the same.

The first equation is already in the form $Ax + By + Cz + D_1 = 0$ with $A=3$, $B=4$, $C=0$, and $D_1=-7$.

For the second equation, $6x + 8y + 6 = 0$, we can divide the entire equation by 2 to match the coefficients of the first plane:

$\frac{6x + 8y + 6}{2} = \frac{0}{2}$

$3x + 4y + 3 = 0$

This is in the form $Ax + By + Cz + D_2 = 0$ with $A=3$, $B=4$, $C=0$, and $D_2=3$.

Now, we apply the distance formula:

$d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$

$d = \frac{|-7 - 3|}{\sqrt{3^2 + 4^2 + 0^2}}$

$d = \frac{|-10|}{\sqrt{9 + 16 + 0}}$

$d = \frac{10}{\sqrt{25}}$

$d = \frac{10}{5}$

$d = 2$

The distance between the two planes is 2 units.

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The final answer is 2.

Given:

$\vec{a}$ is a unit vector, which means $|\vec{a}| = 1$.

The equation $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 99$ is given (the multiplication symbol implies a dot product between vectors).

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To Find:

The value of $|\vec{x}|$.

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Solution:

We are given the equation involving the dot product of two vector expressions:

$(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 99$

We can expand the dot product using the distributive property, similar to how we expand algebraic expressions:

$(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = \vec{x} \cdot (\vec{x}+\vec{a}) - \vec{a} \cdot (\vec{x}+\vec{a})$

$= (\vec{x} \cdot \vec{x}) + (\vec{x} \cdot \vec{a}) - (\vec{a} \cdot \vec{x}) - (\vec{a} \cdot \vec{a})$

We know that the dot product of a vector with itself is equal to the square of its magnitude: $\vec{v} \cdot \vec{v} = |\vec{v}|^2$.

So, $\vec{x} \cdot \vec{x} = |\vec{x}|^2$ and $\vec{a} \cdot \vec{a} = |\vec{a}|^2$.

Also, the dot product is commutative, meaning $\vec{x} \cdot \vec{a} = \vec{a} \cdot \vec{x}$.

Substituting these properties into the expanded equation:

$|\vec{x}|^2 + (\vec{x} \cdot \vec{a}) - (\vec{x} \cdot \vec{a}) - |\vec{a}|^2 = 99$

The middle terms cancel out:

$|\vec{x}|^2 - |\vec{a}|^2 = 99$

We are given that $\vec{a}$ is a unit vector, which means its magnitude is 1:

Substitute the value of $|\vec{a}|$ into the equation:

$|\vec{x}|^2 - (1)^2 = 99$

$|\vec{x}|^2 - 1 = 99$

Add 1 to both sides of the equation:

$|\vec{x}|^2 = 99 + 1$

$|\vec{x}|^2 = 100$

To find $|\vec{x}|$, we take the square root of both sides. The magnitude of a vector is always non-negative.

$|\vec{x}| = \sqrt{100}$

$|\vec{x}| = 10$

Thus, the value of $|\vec{x}|$ is 10.

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Given:

A relation R defined on the set of integers Z, where $n$ is a fixed positive integer.

The relation R is defined as $a \text{ R } b$ if and only if $a - b$ is divisible by $n$, for all $a, b \in Z$.

This can be written as $a \text{ R } b \iff a - b = kn$ for some integer $k$.

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To Prove:

R is an equivalence relation on Z.

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Proof:

To show that R is an equivalence relation, we must prove that R is reflexive, symmetric, and transitive.

1. Reflexivity:

A relation R is reflexive if for every element $a \in Z$, $a \text{ R } a$.

According to the definition of R, $a \text{ R } a$ if and only if $a - a$ is divisible by $n$.

Let's consider $a - a$.

$a - a = 0$

Since $n$ is a positive integer, 0 is divisible by $n$ because $0 = 0 \cdot n$, and 0 is an integer.

Thus, $a - a$ is divisible by $n$ for all $a \in Z$.

Therefore, $a \text{ R } a$ for all $a \in Z$.

Hence, R is reflexive.

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2. Symmetry:

A relation R is symmetric if for every $a, b \in Z$, if $a \text{ R } b$, then $b \text{ R } a$.

Assume $a \text{ R } b$ for some $a, b \in Z$.

According to the definition of R, this means $a - b$ is divisible by $n$.

So, there exists an integer $k$ such that $a - b = kn$.

Now, consider $b - a$. We can write $b - a = -(a - b)$.

Substitute $a - b = kn$ into the expression for $b - a$:

$b - a = -(kn)$

$b - a = (-k)n$

Since $k$ is an integer, $-k$ is also an integer. Let $k' = -k$. Then $b - a = k'n$, where $k'$ is an integer.

This shows that $b - a$ is divisible by $n$.

According to the definition of R, $b - a$ being divisible by $n$ means $b \text{ R } a$.

Thus, if $a \text{ R } b$, then $b \text{ R } a$ for all $a, b \in Z$.

Hence, R is symmetric.

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3. Transitivity:

A relation R is transitive if for every $a, b, c \in Z$, if $a \text{ R } b$ and $b \text{ R } c$, then $a \text{ R } c$.

Assume $a \text{ R } b$ and $b \text{ R } c$ for some $a, b, c \in Z$.

According to the definition of R, $a \text{ R } b$ means $a - b$ is divisible by $n$.

So, there exists an integer $k_1$ such that $a - b = k_1 n$.

Similarly, $b \text{ R } c$ means $b - c$ is divisible by $n$.

So, there exists an integer $k_2$ such that $b - c = k_2 n$.

Now, consider $a - c$. We can write $a - c$ as the sum of $(a - b)$ and $(b - c)$:

$a - c = (a - b) + (b - c)$

Substitute the expressions for $a - b$ and $b - c$ from our assumptions:

$a - c = k_1 n + k_2 n$

Factor out $n$ from the right side:

$a - c = (k_1 + k_2)n$

Since $k_1$ and $k_2$ are integers, their sum $k_1 + k_2$ is also an integer. Let $k_3 = k_1 + k_2$.

So, $a - c = k_3 n$, where $k_3$ is an integer.

This shows that $a - c$ is divisible by $n$.

According to the definition of R, $a - c$ being divisible by $n$ means $a \text{ R } c$.

Thus, if $a \text{ R } b$ and $b \text{ R } c$, then $a \text{ R } c$ for all $a, b, c \in Z$.

Hence, R is transitive.

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Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation on Z.

We will prove the identity: $\cot^{–1} 7 + \cot^{–1} 8 + \cot^{-1} 8 = \cot^{–1} 3$.

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To Prove:

$\cot^{–1} 7 + \cot^{–1} 8 + \cot^{-1} 8 = \cot^{–1} 3$

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Proof:

We know the identity $\cot^{-1} x = \tan^{-1} \frac{1}{x}$ for $x > 0$. Since 7 and 8 are positive, we can use this identity.

Left Hand Side (LHS) = $\cot^{–1} 7 + \cot^{–1} 8 + \cot^{-1} 8$

LHS = $\tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{8} + \tan^{-1} \frac{1}{8}$

First, let's combine the first two terms using the identity $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\frac{x+y}{1-xy}\right)$, provided $xy < 1$.

For $\tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{8}$, we have $x = \frac{1}{7}$ and $y = \frac{1}{8}$.

$xy = \frac{1}{7} \times \frac{1}{8} = \frac{1}{56}$. Since $\frac{1}{56} < 1$, the identity is applicable.

$\tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{8} = \tan^{-1} \left(\frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \times \frac{1}{8}}\right)$

$= \tan^{-1} \left(\frac{\frac{8+7}{56}}{1 - \frac{1}{56}}\right)$

$= \tan^{-1} \left(\frac{\frac{15}{56}}{\frac{56-1}{56}}\right)$

$= \tan^{-1} \left(\frac{\frac{15}{56}}{\frac{55}{56}}\right)$

$= \tan^{-1} \left(\frac{15}{56} \times \frac{56}{55}\right)$

$= \tan^{-1} \left(\frac{15}{55}\right)$

$= \tan^{-1} \left(\frac{\cancel{15}^{3}}{\cancel{55}_{11}}\right)$

$= \tan^{-1} \frac{3}{11}$

Now, the LHS becomes:

LHS = $\tan^{-1} \frac{3}{11} + \tan^{-1} \frac{1}{8}$

Again, use the identity $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\frac{x+y}{1-xy}\right)$. Here, $x = \frac{3}{11}$ and $y = \frac{1}{8}$.

$xy = \frac{3}{11} \times \frac{1}{8} = \frac{3}{88}$. Since $\frac{3}{88} < 1$, the identity is applicable.

$\tan^{-1} \frac{3}{11} + \tan^{-1} \frac{1}{8} = \tan^{-1} \left(\frac{\frac{3}{11} + \frac{1}{8}}{1 - \frac{3}{11} \times \frac{1}{8}}\right)$

$= \tan^{-1} \left(\frac{\frac{3 \cdot 8 + 11 \cdot 1}{11 \cdot 8}}{1 - \frac{3}{88}}\right)$

$= \tan^{-1} \left(\frac{\frac{24+11}{88}}{\frac{88-3}{88}}\right)$

$= \tan^{-1} \left(\frac{\frac{35}{88}}{\frac{85}{88}}\right)$

$= \tan^{-1} \left(\frac{35}{88} \times \frac{88}{85}\right)$

$= \tan^{-1} \left(\frac{35}{85}\right)$

$= \tan^{-1} \left(\frac{\cancel{35}^{7}}{\cancel{85}_{17}}\right)$

$= \tan^{-1} \frac{7}{17}$

Hmm, there seems to be a typo in the question. If the identity is $\cot^{–1} 7 + \cot^{–1} 8 + \cot^{-1} 18 = \cot^{–1} 3$, let's try that.

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Proof (Assuming typo: $\cot^{–1} 7 + \cot^{–1} 8 + \cot^{-1} 18 = \cot^{–1} 3$):

LHS = $\cot^{–1} 7 + \cot^{–1} 8 + \cot^{-1} 18$

LHS = $\tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{8} + \tan^{-1} \frac{1}{18}$

From previous calculation, $\tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{8} = \tan^{-1} \frac{3}{11}$.

So, LHS = $\tan^{-1} \frac{3}{11} + \tan^{-1} \frac{1}{18}$.

Use the identity $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\frac{x+y}{1-xy}\right)$ with $x = \frac{3}{11}$ and $y = \frac{1}{18}$.

$xy = \frac{3}{11} \times \frac{1}{18} = \frac{3}{198} = \frac{1}{66}$. Since $\frac{1}{66} < 1$, the identity is applicable.

$\tan^{-1} \frac{3}{11} + \tan^{-1} \frac{1}{18} = \tan^{-1} \left(\frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{11} \times \frac{1}{18}}\right)$

$= \tan^{-1} \left(\frac{\frac{3 \cdot 18 + 11 \cdot 1}{11 \cdot 18}}{1 - \frac{3}{198}}\right)$

$= \tan^{-1} \left(\frac{\frac{54+11}{198}}{\frac{198-3}{198}}\right)$

$= \tan^{-1} \left(\frac{\frac{65}{198}}{\frac{195}{198}}\right)$

$= \tan^{-1} \left(\frac{65}{198} \times \frac{198}{195}\right)$

$= \tan^{-1} \left(\frac{65}{195}\right)$

$= \tan^{-1} \left(\frac{\cancel{65}^{1}}{\cancel{195}_{3}}\right)$

$= \tan^{-1} \frac{1}{3}$

Now, convert this back to cotangent inverse using $\tan^{-1} x = \cot^{-1} \frac{1}{x}$ for $x > 0$. Since $\frac{1}{3} > 0$, we have:

LHS = $\cot^{-1} \frac{1}{\frac{1}{3}} = \cot^{-1} 3$

This matches the Right Hand Side (RHS).

LHS = RHS

Hence, the identity $\cot^{–1} 7 + \cot^{–1} 8 + \cot^{-1} 18 = \cot^{–1} 3$ is proved.

Assuming the question had a typo and the third term was $\cot^{-1} 18$ instead of $\cot^{-1} 8$, the proof is shown above.

Let's re-examine the original question with $\cot^{-1} 8$ for the third term.

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Proof (Using original question: $\cot^{–1} 7 + \cot^{–1} 8 + \cot^{-1} 8 = \cot^{–1} 3$):

LHS = $\cot^{–1} 7 + \cot^{–1} 8 + \cot^{-1} 8$

LHS = $\tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{8} + \tan^{-1} \frac{1}{8}$

Combine the last two terms first, using $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\frac{x+y}{1-xy}\right)$ with $x = \frac{1}{8}$ and $y = \frac{1}{8}$.

$xy = \frac{1}{8} \times \frac{1}{8} = \frac{1}{64}$. Since $\frac{1}{64} < 1$, the identity is applicable.

$\tan^{-1} \frac{1}{8} + \tan^{-1} \frac{1}{8} = \tan^{-1} \left(\frac{\frac{1}{8} + \frac{1}{8}}{1 - \frac{1}{8} \times \frac{1}{8}}\right)$

$= \tan^{-1} \left(\frac{\frac{2}{8}}{1 - \frac{1}{64}}\right)$

$= \tan^{-1} \left(\frac{\frac{1}{4}}{\frac{64-1}{64}}\right)$

$= \tan^{-1} \left(\frac{\frac{1}{4}}{\frac{63}{64}}\right)$

$= \tan^{-1} \left(\frac{1}{4} \times \frac{64}{63}\right)$

$= \tan^{-1} \left(\frac{\cancel{64}^{16}}{4 \times 63}\right)$

$= \tan^{-1} \frac{16}{63}$

Now, the LHS becomes:

LHS = $\tan^{-1} \frac{1}{7} + \tan^{-1} \frac{16}{63}$

Use the identity $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\frac{x+y}{1-xy}\right)$ with $x = \frac{1}{7}$ and $y = \frac{16}{63}$.

$xy = \frac{1}{7} \times \frac{16}{63} = \frac{16}{441}$. Since $\frac{16}{441} < 1$, the identity is applicable.

$\tan^{-1} \frac{1}{7} + \tan^{-1} \frac{16}{63} = \tan^{-1} \left(\frac{\frac{1}{7} + \frac{16}{63}}{1 - \frac{1}{7} \times \frac{16}{63}}\right)$

$= \tan^{-1} \left(\frac{\frac{9 \cdot 1 + 16}{63}}{1 - \frac{16}{441}}\right)$

$= \tan^{-1} \left(\frac{\frac{9+16}{63}}{\frac{441-16}{441}}\right)$

$= \tan^{-1} \left(\frac{\frac{25}{63}}{\frac{425}{441}}\right)$

$= \tan^{-1} \left(\frac{25}{63} \times \frac{441}{425}\right)$

We know that $63 \times 7 = 441$ and $25 \times 17 = 425$.

$= \tan^{-1} \left(\frac{\cancel{25}^{1}}{\cancel{63}_{9}} \times \frac{\cancel{441}^{7}}{\cancel{425}_{17}}\right)$

$= \tan^{-1} \left(\frac{1}{9} \times \frac{7}{17}\right) = \tan^{-1} \frac{7}{153}$

This does not result in $\tan^{-1} \frac{1}{3}$. It seems there is indeed a typo in the question, and the third term should likely be $\cot^{-1} 18$. The proof for the corrected identity is given above.

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Solution (Solving the OR part):

We need to solve the equation $\tan^{-1} (2 + x) + \tan^{-1} (2 - x) = \tan^{-1} \frac{2}{3}$ for $- \sqrt{3} > x > \sqrt{3}$. The condition on $x$ seems unusual (perhaps it should be $x \in (-\sqrt{3}, \sqrt{3})$ or similar). Assuming the intention was a domain where the principal values are considered and the identity $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB}\right)$ is applicable. The condition $AB < 1$ must hold. Here $A = 2+x$ and $B = 2-x$.

$AB = (2+x)(2-x) = 4 - x^2$.

For the principal value identity, we need $4 - x^2 < 1$, which means $x^2 > 3$. This gives $x > \sqrt{3}$ or $x < -\sqrt{3}$. This matches the condition given in the question, which implies we might need to use a modified identity if $AB > 1$. However, the form of the equation $\tan^{-1}(\text{sum}) = \text{a value}$ suggests the standard identity might be intended.

Let's assume the standard identity is used and check the condition later.

Using $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB}\right)$:

$\tan^{-1} ((2 + x) + (2 - x))$

$\tan^{-1} \left(\frac{(2+x) + (2-x)}{1 - (2+x)(2-x)}\right) = \tan^{-1} \frac{2}{3}$

$\tan^{-1} \left(\frac{4}{1 - (4 - x^2)}\right) = \tan^{-1} \frac{2}{3}$

$\tan^{-1} \left(\frac{4}{1 - 4 + x^2}\right) = \tan^{-1} \frac{2}{3}$

$\tan^{-1} \left(\frac{4}{x^2 - 3}\right) = \tan^{-1} \frac{2}{3}$

For the principal values to be equal, the arguments must be equal:

$\frac{4}{x^2 - 3} = \frac{2}{3}$

Cross-multiply:

$4 \times 3 = 2 \times (x^2 - 3)$

$12 = 2x^2 - 6$

Add 6 to both sides:

$12 + 6 = 2x^2$

$18 = 2x^2$

Divide by 2:

$x^2 = \frac{18}{2}$

$x^2 = 9$

Take the square root of both sides:

$x = \pm \sqrt{9}$

$x = \pm 3$

Now, let's check the given condition for $x$: $- \sqrt{3} > x > \sqrt{3}$. This condition seems contradictory as it implies $x > \sqrt{3}$ and $x < -\sqrt{3}$ simultaneously, which is impossible. It should likely be $x > \sqrt{3}$ or $x < -\sqrt{3}$, or perhaps $x \notin [-\sqrt{3}, \sqrt{3}]$.

If the condition means $x > \sqrt{3}$ or $x < -\sqrt{3}$, then both $x=3$ and $x=-3$ satisfy this condition, as $\sqrt{3} \approx 1.732$.

Let's check if the application of the $\tan^{-1}$ sum formula was valid, i.e., $AB < 1$.

$AB = (2+x)(2-x) = 4 - x^2$.

If $x = 3$, $AB = 4 - 3^2 = 4 - 9 = -5$. Since $-5 < 1$, the formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB}\right)$ is valid.

If $x = -3$, $AB = 4 - (-3)^2 = 4 - 9 = -5$. Since $-5 < 1$, the formula is valid.

So, both $x=3$ and $x=-3$ are potential solutions.

Let's verify these solutions in the original equation:

For $x=3$: $\tan^{-1}(2+3) + \tan^{-1}(2-3) = \tan^{-1} 5 + \tan^{-1} (-1)$.

Using $\tan^{-1} (-y) = -\tan^{-1} y$, this is $\tan^{-1} 5 - \tan^{-1} 1$.

Using $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\frac{x-y}{1+xy}\right)$ (for $xy > -1$). $x=5, y=1, xy=5 > -1$.

$\tan^{-1} 5 - \tan^{-1} 1 = \tan^{-1} \left(\frac{5-1}{1+5 \cdot 1}\right) = \tan^{-1} \left(\frac{4}{6}\right) = \tan^{-1} \frac{2}{3}$. This matches the RHS.

For $x=-3$: $\tan^{-1}(2-3) + \tan^{-1}(2-(-3)) = \tan^{-1} (-1) + \tan^{-1} 5$. This is the same as the case for $x=3$, just with terms swapped, and the result is $\tan^{-1} \frac{2}{3}$.

Both $x=3$ and $x=-3$ are solutions, provided the domain condition is interpreted correctly as $x^2 > 3$.

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The problem statement contains a likely typo in the identity to prove (third term should be $\cot^{-1} 18$) and a potentially ambiguous or contradictory condition on $x$ in the equation to solve ($- \sqrt{3} > x > \sqrt{3}$). Assuming the intent was $x > \sqrt{3}$ or $x < -\sqrt{3}$ for the domain in the OR part, the solutions are $x = 3$ and $x = -3$. Given the structure of the problem, it is likely that the first part (proving the identity with the typo corrected) is the intended question to be answered.

Assuming the question meant to prove $\cot^{–1} 7 + \cot^{–1} 8 + \cot^{-1} 18 = \cot^{–1} 3$, the proof is provided above. If the question insists on $\cot^{–1} 7 + \cot^{–1} 8 + \cot^{-1} 8$, then it does not equal $\cot^{-1} 3$.

Final Answer based on assumed typo in the first part:

The identity $\cot^{–1} 7 + \cot^{–1} 8 + \cot^{-1} 18 = \cot^{–1} 3$ is proved above.

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Final Answer based on the OR part (assuming the condition on x means $x^2>3$):

The solutions to the equation $\tan^{-1} (2 + x) + \tan^{-1} (2 - x) = \tan^{-1} \frac{2}{3}$ are $x = 3$ and $x = -3$.

Given:

The determinant equation: $\left|\begin{matrix}x+2&x+6&x-1\\x+6&x-1&x+2\\x-1&x+2&x+6\\\end{matrix}\right|=0$.

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To Find:

The value(s) of $x$ that satisfy the equation.

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Solution:

Let the given determinant be $\Delta$.

$\Delta = \left|\begin{matrix}x+2&x+6&x-1\\x+6&x-1&x+2\\x-1&x+2&x+6\\\end{matrix}\right|$

We can simplify the determinant using elementary column operations.

Apply the operation $C_1 \to C_1 + C_2 + C_3$. This operation does not change the value of the determinant.

The sum of elements in each row is $(x+2) + (x+6) + (x-1) = 3x+7$.

So, the elements in the first column after the operation become:

$C_{11} = (x+2) + (x+6) + (x-1) = 3x+7$

$C_{21} = (x+6) + (x-1) + (x+2) = 3x+7$

$C_{31} = (x-1) + (x+2) + (x+6) = 3x+7$

The determinant becomes:

$\Delta = \left|\begin{matrix}3x+7&x+6&x-1\\3x+7&x-1&x+2\\3x+7&x+2&x+6\\\end{matrix}\right|$

Factor out the common factor $(3x+7)$ from the first column:

$\Delta = (3x+7) \left|\begin{matrix}1&x+6&x-1\\1&x-1&x+2\\1&x+2&x+6\\\end{matrix}\right|$

Now, apply elementary row operations to create zeros in the first column below the first element. These operations do not change the value of the determinant.

Apply $R_2 \to R_2 - R_1$:

New Row 2 elements:

$(1) - (1) = 0$

$(x-1) - (x+6) = x - 1 - x - 6 = -7$

$(x+2) - (x-1) = x + 2 - x + 1 = 3$

Apply $R_3 \to R_3 - R_1$:

New Row 3 elements:

$(1) - (1) = 0$

$(x+2) - (x+6) = x + 2 - x - 6 = -4$

$(x+6) - (x-1) = x + 6 - x + 1 = 7$

The determinant simplifies to:

$\Delta = (3x+7) \left|\begin{matrix}1&x+6&x-1\\0&-7&3\\0&-4&7\\\end{matrix}\right|$

Now, expand the determinant along the first column, as it contains two zeros.

$\Delta = (3x+7) \cdot \left[ 1 \cdot \left|\begin{matrix}-7&3\\-4&7\\\end{matrix}\right| - 0 \cdot \left|\begin{matrix}x+6&x-1\\-4&7\\\end{matrix}\right| + 0 \cdot \left|\begin{matrix}x+6&x-1\\-7&3\\\end{matrix}\right| \right]$

$\Delta = (3x+7) \cdot [ 1 \cdot ((-7)(7) - (3)(-4)) - 0 + 0 ]$

$\Delta = (3x+7) \cdot [(-49) - (-12)]$

$\Delta = (3x+7) \cdot [-49 + 12]$

$\Delta = (3x+7) \cdot (-37)$

The given equation is $\Delta = 0$.

So, $(3x+7)(-37) = 0$

Since $-37 \neq 0$, for the product to be zero, the other factor must be zero:

$3x+7 = 0$

Solve this linear equation for $x$:

$3x = -7$

$x = -\frac{7}{3}$

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The value of $x$ that satisfies the equation is $-\frac{7}{3}$.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases} \frac{k \cos 2x}{\pi - 4x} & , & x \neq \frac{\pi}{4} \\ 5 & , & x = \frac{\pi}{4} \end{cases}$

We are given that the function is continuous at $x = \frac{\pi}{4}$.

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To Determine:

The value of the constant $k$.

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Solution:

For a function $f(x)$ to be continuous at a point $x=a$, the following condition must be satisfied:

$\lim\limits_{x \to a} f(x) = f(a)$

In this problem, the point is $a = \frac{\pi}{4}$. So, for $f(x)$ to be continuous at $x = \frac{\pi}{4}$, we must have:

$\lim\limits_{x \to \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right)$

From the definition of the function, we are given that $f\left(\frac{\pi}{4}\right) = 5$.

Now, we need to evaluate the limit of $f(x)$ as $x$ approaches $\frac{\pi}{4}$:

$\lim\limits_{x \to \frac{\pi}{4}} f(x) = \lim\limits_{x \to \frac{\pi}{4}} \frac{k \cos 2x}{\pi - 4x}$

Let's check the form of the limit by substituting $x = \frac{\pi}{4}$.

Numerator: $k \cos\left(2 \times \frac{\pi}{4}\right) = k \cos\left(\frac{\pi}{2}\right) = k \times 0 = 0$

Denominator: $\pi - 4\left(\frac{\pi}{4}\right) = \pi - \pi = 0$

The limit is in the indeterminate form $\frac{0}{0}$. We can evaluate this limit using substitution.

Let $h = x - \frac{\pi}{4}$. As $x \to \frac{\pi}{4}$, $h \to 0$.

From the substitution, $x = \frac{\pi}{4} + h$.

Substitute this into the limit expression:

$\lim\limits_{h \to 0} \frac{k \cos\left(2\left(\frac{\pi}{4} + h\right)\right)}{\pi - 4\left(\frac{\pi}{4} + h\right)}$

$= \lim\limits_{h \to 0} \frac{k \cos\left(\frac{\pi}{2} + 2h\right)}{\pi - \pi - 4h}$

$= \lim\limits_{h \to 0} \frac{k \cos\left(\frac{\pi}{2} + 2h\right)}{-4h}$

Using the trigonometric identity $\cos\left(\frac{\pi}{2} + \theta\right) = -\sin\theta$, we have $\cos\left(\frac{\pi}{2} + 2h\right) = -\sin(2h)$.

So, the limit becomes:

$= \lim\limits_{h \to 0} \frac{k (-\sin(2h))}{-4h}$

$= \lim\limits_{h \to 0} \frac{-k \sin(2h)}{-4h}$

$= \lim\limits_{h \to 0} \frac{k \sin(2h)}{4h}$

We can rewrite this limit to use the standard result $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

$= \lim\limits_{h \to 0} \frac{k}{4} \frac{\sin(2h)}{h}$

Multiply the numerator and denominator of the fraction $\frac{\sin(2h)}{h}$ by 2:

$= \lim\limits_{h \to 0} \frac{k}{4} \frac{\sin(2h)}{h} \times \frac{2}{2}$

$= \lim\limits_{h \to 0} \frac{k}{2} \frac{\sin(2h)}{2h}$

Now, let $u = 2h$. As $h \to 0$, $u \to 0$. The limit becomes:

$= \frac{k}{2} \lim\limits_{u \to 0} \frac{\sin u}{u}$

Using the standard limit $\lim\limits_{u \to 0} \frac{\sin u}{u} = 1$:

$= \frac{k}{2} \times 1$

$= \frac{k}{2}$

So, we have $\lim\limits_{x \to \frac{\pi}{4}} f(x) = \frac{k}{2}$.

For continuity at $x = \frac{\pi}{4}$, we equate the limit to the function value:

$\frac{k}{2} = f\left(\frac{\pi}{4}\right)$

$\frac{k}{2} = 5$

Multiply both sides by 2:

$k = 5 \times 2$

$k = 10$

Thus, the value of $k$ for which the function is continuous at $x = \frac{\pi}{4}$ is 10.

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Given:

The function $y = e^{a\cos^{-1}x}$.

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To Show:

$(1 – x^2) \frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0$.

(Note: The question in the prompt appears to have a typo, and the standard result for this function is $(1 – x^2) \frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0$. The proof will be provided for the standard form.)

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Proof:

We are given the function $y = e^{a\cos^{-1}x}$.

Differentiate $y$ with respect to $x$ using the chain rule:

$\frac{dy}{dx} = \frac{d}{dx}(e^{a\cos^{-1}x})$

Let $u = a\cos^{-1}x$. Then $\frac{du}{dx} = a \frac{d}{dx}(\cos^{-1}x) = a \left(\frac{-1}{\sqrt{1-x^2}}\right) = \frac{-a}{\sqrt{1-x^2}}$.

So, $\frac{dy}{dx} = e^{a\cos^{-1}x} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = e^{a\cos^{-1}x} \left(\frac{-a}{\sqrt{1-x^2}}\right)$

Substitute $y = e^{a\cos^{-1}x}$ back into the equation:

$\frac{dy}{dx} = \frac{-ay}{\sqrt{1-x^2}}$

Rearrange the equation to eliminate the square root:

$\sqrt{1-x^2} \frac{dy}{dx} = -ay$

Now, differentiate this equation with respect to $x$. Use the product rule on the left side: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

Let $u = \sqrt{1-x^2}$ and $v = \frac{dy}{dx}$.

$\frac{du}{dx} = \frac{d}{dx}((1-x^2)^{1/2}) = \frac{1}{2}(1-x^2)^{-1/2} \cdot \frac{d}{dx}(1-x^2) = \frac{1}{2\sqrt{1-x^2}} (-2x) = \frac{-x}{\sqrt{1-x^2}}$.

$\frac{dv}{dx} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2}$.

Applying the product rule on the left side of $\sqrt{1-x^2} \frac{dy}{dx} = -ay$:

$\sqrt{1-x^2} \frac{d}{dx}\left(\frac{dy}{dx}\right) + \frac{dy}{dx} \frac{d}{dx}(\sqrt{1-x^2}) = \frac{d}{dx}(-ay)$

$\sqrt{1-x^2} \frac{d^2y}{dx^2} + \frac{dy}{dx} \left(\frac{-x}{\sqrt{1-x^2}}\right) = -a \frac{dy}{dx}$

$\sqrt{1-x^2} \frac{d^2y}{dx^2} - \frac{x}{\sqrt{1-x^2}} \frac{dy}{dx} = -a \frac{dy}{dx}$

To eliminate the square root in the denominator, multiply the entire equation by $\sqrt{1-x^2}$:

$\sqrt{1-x^2} \left(\sqrt{1-x^2} \frac{d^2y}{dx^2}\right) - \sqrt{1-x^2} \left(\frac{x}{\sqrt{1-x^2}} \frac{dy}{dx}\right) = \sqrt{1-x^2} \left(-a \frac{dy}{dx}\right)$

$(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = -a \sqrt{1-x^2} \frac{dy}{dx}$

From the equation $\sqrt{1-x^2} \frac{dy}{dx} = -ay$, we can substitute $-ay$ for $\sqrt{1-x^2} \frac{dy}{dx}$ on the right side:

$(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = a (-ay)$

$(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = -a^2 y$

Move the term $-a^2 y$ to the left side of the equation:

$(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + a^2 y = 0$

Wait, the target is $(1 – x^2) \frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0$. Let's check the step where $-a \sqrt{1-x^2} \frac{dy}{dx}$ became $a(-ay)$.

We had $\sqrt{1-x^2} \frac{dy}{dx} = -ay$.

So, $-a \left(\sqrt{1-x^2} \frac{dy}{dx}\right) = -a (-ay) = a^2 y$. This is incorrect. I should substitute $-ay$ for $\sqrt{1-x^2} \frac{dy}{dx}$ on the RHS of the equation *before* multiplying by $\sqrt{1-x^2}$.

Let's restart from $\sqrt{1-x^2} \frac{d^2y}{dx^2} - \frac{x}{\sqrt{1-x^2}} \frac{dy}{dx} = -a \frac{dy}{dx}$.

Multiply the second term on the LHS by $\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}$ to get a common denominator if needed, but multiplying the whole equation by $\sqrt{1-x^2}$ is simpler.

Multiply the equation $\sqrt{1-x^2} \frac{d^2y}{dx^2} - \frac{x}{\sqrt{1-x^2}} \frac{dy}{dx} = -a \frac{dy}{dx}$ by $\sqrt{1-x^2}$:

$(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = -a \sqrt{1-x^2} \frac{dy}{dx}$

From the equation $\sqrt{1-x^2} \frac{dy}{dx} = -ay$, substitute this into the RHS:

$(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = -a (-ay)$

$(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = a^2 y$ [This step was correct earlier, but the target equation has a minus sign]

Move the term $a^2 y$ to the left side:

$(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0$

This matches the standard form of the required differential equation.

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The final answer is $(1 – x^2) \frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0$.

We will solve the first part of the question: finding the equation of the tangent to the curve $x = \sin^3 t$, $y = \cos^2 t$ at $t = \frac{\pi}{4}$.

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Given:

The parametric equations of the curve are $x = \sin^3 t$ and $y = \cos^2 t$.

The point of tangency corresponds to $t = \frac{\pi}{4}$.

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To Find:

The equation of the tangent line to the curve at $t = \frac{\pi}{4}$.

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Solution:

First, we find the coordinates $(x_0, y_0)$ of the point on the curve at $t = \frac{\pi}{4}$.

$x_0 = \sin^3 \left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^3 = \frac{1}{(\sqrt{2})^3} = \frac{1}{2\sqrt{2}}$

$y_0 = \cos^2 \left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$

The point of tangency is $\left(\frac{1}{2\sqrt{2}}, \frac{1}{2}\right)$.

Next, we find the slope of the tangent line, $\frac{dy}{dx}$, using the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

Calculate $\frac{dx}{dt}$:

$\frac{dx}{dt} = \frac{d}{dt}(\sin^3 t) = 3\sin^2 t \cdot \frac{d}{dt}(\sin t) = 3\sin^2 t \cos t$

Calculate $\frac{dy}{dt}$:

$\frac{dy}{dt} = \frac{d}{dt}(\cos^2 t) = 2\cos t \cdot \frac{d}{dt}(\cos t) = 2\cos t (-\sin t) = -2\sin t \cos t$

Now, find $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-2\sin t \cos t}{3\sin^2 t \cos t}$

Assuming $\sin t \neq 0$ and $\cos t \neq 0$, we can cancel common terms:

$\frac{dy}{dx} = \frac{-2}{3\sin t}$

Evaluate the slope $m$ at $t = \frac{\pi}{4}$:

$m = \left(\frac{dy}{dx}\right)_{t=\frac{\pi}{4}} = \frac{-2}{3\sin \left(\frac{\pi}{4}\right)} = \frac{-2}{3 \cdot \frac{1}{\sqrt{2}}} = \frac{-2\sqrt{2}}{3}$

The equation of the tangent line in point-slope form is $y - y_0 = m(x - x_0)$.

Substitute the point $\left(\frac{1}{2\sqrt{2}}, \frac{1}{2}\right)$ and the slope $m = \frac{-2\sqrt{2}}{3}$:

$y - \frac{1}{2} = \frac{-2\sqrt{2}}{3} \left(x - \frac{1}{2\sqrt{2}}\right)$

Multiply both sides by 6 to clear denominators:

$6\left(y - \frac{1}{2}\right) = 6 \left(\frac{-2\sqrt{2}}{3} \left(x - \frac{1}{2\sqrt{2}}\right)\right)$

$6y - 3 = -4\sqrt{2} \left(x - \frac{1}{2\sqrt{2}}\right)$

$6y - 3 = -4\sqrt{2}x + 4\sqrt{2} \cdot \frac{1}{2\sqrt{2}}$

$6y - 3 = -4\sqrt{2}x + 2$

Rearrange the terms to get the equation in the form $Ax + By + C = 0$:

$4\sqrt{2}x + 6y - 3 - 2 = 0$

$4\sqrt{2}x + 6y - 5 = 0$

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The equation of the tangent to the curve at $t = \frac{\pi}{4}$ is $4\sqrt{2}x + 6y - 5 = 0$.

Given:

The definite integral $\int\limits_{0}^{\frac{\pi}{6}}{{\sin}^4x\ {\cos}^3x\ x\ dx}$.

Note: The integrand contains 'x' multiplying the trigonometric terms, and also 'dx'. Based on the typical complexity of problems at this level and the structure of the trigonometric part of the integrand ($\sin^m x \cos^n x$), it is highly probable that the 'x' multiplying the terms is a typographical error, and the intended integral is $\int\limits_{0}^{\frac{\pi}{6}}{{\sin}^4x\ {\cos}^3x\ dx}$. The solution will proceed with this assumption.

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To Evaluate:

$\int\limits_{0}^{\frac{\pi}{6}}{{\sin}^4x\ {\cos}^3x\ dx}$ (assuming the 'x' in the integrand was a typo)

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Solution:

We need to evaluate the integral $I = \int\limits_{0}^{\frac{\pi}{6}} \sin^4 x \cos^3 x \, dx$.

The power of $\cos x$ is odd (3). We can separate one factor of $\cos x$ and convert the remaining even power of $\cos x$ into terms of $\sin x$ using the identity $\cos^2 x = 1 - \sin^2 x$.

$I = \int\limits_{0}^{\frac{\pi}{6}} \sin^4 x \cos^2 x \cos x \, dx$

$I = \int\limits_{0}^{\frac{\pi}{6}} \sin^4 x (1 - \sin^2 x) \cos x \, dx$

Now, we use the substitution method.

Let $u = \sin x$.

Differentiating both sides with respect to $x$, we get $\frac{du}{dx} = \cos x$, which means $du = \cos x \, dx$.

We also need to change the limits of integration according to the substitution:

When $x = 0$, $u = \sin(0) = 0$.

When $x = \frac{\pi}{6}$, $u = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

Substituting $u = \sin x$ and $du = \cos x \, dx$ into the integral, and changing the limits, we get:

$I = \int\limits_{0}^{1/2} u^4 (1 - u^2) \, du$

Expand the integrand:

$I = \int\limits_{0}^{1/2} (u^4 - u^6) \, du$

Now, integrate term by term with respect to $u$:

$I = \left[ \frac{u^{4+1}}{4+1} - \frac{u^{6+1}}{6+1} \right]_{0}^{1/2}$

$I = \left[ \frac{u^5}{5} - \frac{u^7}{7} \right]_{0}^{1/2}$

Evaluate the expression at the upper and lower limits:

$I = \left( \frac{(1/2)^5}{5} - \frac{(1/2)^7}{7} \right) - \left( \frac{(0)^5}{5} - \frac{(0)^7}{7} \right)$

$I = \left( \frac{1/32}{5} - \frac{1/128}{7} \right) - (0 - 0)$

$I = \frac{1}{32 \times 5} - \frac{1}{128 \times 7}$

$I = \frac{1}{160} - \frac{1}{896}$

To combine these fractions, find a common denominator. The least common multiple of 160 and 896 is 4480.

$160 \times 28 = 4480$

$896 \times 5 = 4480$

$I = \frac{1 \times 28}{160 \times 28} - \frac{1 \times 5}{896 \times 5}$

$I = \frac{28}{4480} - \frac{5}{4480}$

$I = \frac{28 - 5}{4480}$

$I = \frac{23}{4480}$

---

Assuming the integrand was $\sin^4 x \cos^3 x$, the value of the integral is $\frac{23}{4480}$.

If the integrand was indeed $x \sin^4 x \cos^3 x$, the evaluation would require integration by parts and would be significantly more complex, involving terms with $\pi$ and $\sqrt{3}$. Given the likely level of the question, the typo assumption is reasonable.

Given:

The indefinite integral $\int \frac{3x + 1}{2x^2 - 2x + 3} dx$.

---

To Evaluate:

The given indefinite integral.

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Solution:

The integral is of the form $\int \frac{Px+Q}{Ax^2+Bx+C} dx$. We express the numerator $3x+1$ in terms of the derivative of the denominator $2x^2 - 2x + 3$, which is $\frac{d}{dx}(2x^2 - 2x + 3) = 4x - 2$.

We write $3x+1 = L(4x-2) + M$ for some constants $L$ and $M$.

Expanding the right side, we get $3x+1 = 4Lx - 2L + M$.

Comparing the coefficients of $x$ on both sides:

$3 = 4L$

$L = \frac{3}{4}$

Comparing the constant terms:

$1 = -2L + M$

Substitute the value of $L$:

$1 = -2\left(\frac{3}{4}\right) + M$

$1 = -\frac{3}{2} + M$}

$M = 1 + \frac{3}{2} = \frac{2}{2} + \frac{3}{2} = \frac{5}{2}$

So, we can rewrite the numerator as $3x+1 = \frac{3}{4}(4x-2) + \frac{5}{2}$.

The integral becomes:

$\int \frac{\frac{3}{4}(4x-2) + \frac{5}{2}}{2x^2 - 2x + 3} dx$}

We can split this into two separate integrals:

$I = \int \frac{\frac{3}{4}(4x-2)}{2x^2 - 2x + 3} dx + \int \frac{\frac{5}{2}}{2x^2 - 2x + 3} dx$

$I = \frac{3}{4} \int \frac{4x-2}{2x^2 - 2x + 3} dx + \frac{5}{2} \int \frac{1}{2x^2 - 2x + 3} dx$

For the first integral, let $u = 2x^2 - 2x + 3$. Then $du = (4x-2) dx$.

$\int \frac{4x-2}{2x^2 - 2x + 3} dx = \int \frac{du}{u} = \log|u| + C_1 = \log|2x^2 - 2x + 3| + C_1$

Since the discriminant of $2x^2 - 2x + 3$ is $(-2)^2 - 4(2)(3) = 4 - 24 = -20 < 0$ and the coefficient of $x^2$ (2) is positive, the quadratic $2x^2 - 2x + 3$ is always positive for all real $x$. Thus, we can write $\log|2x^2 - 2x + 3|$ as $\log(2x^2 - 2x + 3)$.

So, the first part of the integral is $\frac{3}{4} \log(2x^2 - 2x + 3)$.

For the second integral $\int \frac{1}{2x^2 - 2x + 3} dx$, we complete the square in the denominator:

$2x^2 - 2x + 3 = 2(x^2 - x + \frac{3}{2})$

$= 2\left(x^2 - x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + \frac{3}{2}\right)$

$= 2\left(\left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + \frac{6}{4}\right)$

$= 2\left(\left(x - \frac{1}{2}\right)^2 + \frac{5}{4}\right)$

$= 2\left(\left(x - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{5}}{2}\right)^2\right)$

The second integral becomes:

$\frac{5}{2} \int \frac{1}{2\left(\left(x - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{5}}{2}\right)^2\right)} dx$

$= \frac{5}{2} \cdot \frac{1}{2} \int \frac{1}{\left(x - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{5}}{2}\right)^2} dx$

$= \frac{5}{4} \int \frac{1}{\left(x - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{5}}{2}\right)^2} dx$

This integral is of the form $\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C_2$.

Here $u = x - \frac{1}{2}$ (so $du = dx$) and $a = \frac{\sqrt{5}}{2}$.

$= \frac{5}{4} \cdot \frac{1}{\frac{\sqrt{5}}{2}} \tan^{-1}\left(\frac{x - \frac{1}{2}}{\frac{\sqrt{5}}{2}}\right) + C_2$

$= \frac{5}{4} \cdot \frac{2}{\sqrt{5}} \tan^{-1}\left(\frac{\frac{2x - 1}{2}}{\frac{\sqrt{5}}{2}}\right) + C_2$

$= \frac{5}{2\sqrt{5}} \tan^{-1}\left(\frac{2x - 1}{\sqrt{5}}\right) + C_2$

$= \frac{\cancel{5}^{\sqrt{5}}}{2\cancel{\sqrt{5}}} \tan^{-1}\left(\frac{2x - 1}{\sqrt{5}}\right) + C_2$

$= \frac{\sqrt{5}}{2} \tan^{-1}\left(\frac{2x - 1}{\sqrt{5}}\right) + C_2$

Combining the results of the two integrals:

$I = \frac{3}{4} \log(2x^2 - 2x + 3) + \frac{\sqrt{5}}{2} \tan^{-1}\left(\frac{2x - 1}{\sqrt{5}}\right) + C$

where $C = C_1 + C_2$ is the constant of integration.

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The value of the integral is $\frac{3}{4} \log(2x^2 - 2x + 3) + \frac{\sqrt{5}}{2} \tan^{-1}\left(\frac{2x - 1}{\sqrt{5}}\right) + C$.

Given:

The differential equation is $2y e^\frac{x}{y} dx + \left(y-2x e^\frac{x}{y}\right) dy = 0$.

The initial condition is $x = 0$ when $y = 1$.

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To Find:

A particular solution of the given differential equation satisfying the initial condition.

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Solution:

The given differential equation is $2y e^\frac{x}{y} dx + \left(y-2x e^\frac{x}{y}\right) dy = 0$.

We can rewrite this equation in the form $\frac{dx}{dy}$.

$2y e^\frac{x}{y} dx = -(y-2x e^\frac{x}{y}) dy$

$2y e^\frac{x}{y} dx = (2x e^\frac{x}{y} - y) dy$

$\frac{dx}{dy} = \frac{2x e^\frac{x}{y} - y}{2y e^\frac{x}{y}}$

Divide the numerator and the denominator by $y$:

$\frac{dx}{dy} = \frac{\frac{2x}{y} e^\frac{x}{y} - \frac{y}{y}}{\frac{2y}{y} e^\frac{x}{y}} = \frac{2\left(\frac{x}{y}\right) e^\frac{x}{y} - 1}{2 e^\frac{x}{y}}$

This is a homogeneous differential equation, as $\frac{dx}{dy}$ can be expressed as a function of $\frac{x}{y}$.

We use the substitution $x = vy$.

Differentiating $x = vy$ with respect to $y$ using the product rule, we get:

$\frac{dx}{dy} = v \cdot \frac{dy}{dy} + y \cdot \frac{dv}{dy} = v + y \frac{dv}{dy}$

Substitute $x = vy$ (which means $v = x/y$) and $\frac{dx}{dy} = v + y \frac{dv}{dy}$ into the differential equation:

$v + y \frac{dv}{dy} = \frac{2v e^v - 1}{2 e^v}$

Separate the variables by moving $v$ to the right side:

$y \frac{dv}{dy} = \frac{2v e^v - 1}{2 e^v} - v$

$y \frac{dv}{dy} = \frac{2v e^v - 1 - v (2 e^v)}{2 e^v}$

$y \frac{dv}{dy} = \frac{2v e^v - 1 - 2v e^v}{2 e^v}$

$y \frac{dv}{dy} = \frac{-1}{2 e^v}$

Now, separate the variables $v$ and $y$:

$2 e^v dv = -\frac{1}{y} dy$

Integrate both sides:

$\int 2 e^v dv = \int -\frac{1}{y} dy$

$2 \int e^v dv = - \int \frac{1}{y} dy$

$2 e^v = -\log|y| + C$

Substitute back $v = \frac{x}{y}$:

$2 e^{x/y} = -\log|y| + C$

This is the general solution of the differential equation.

Now, we use the initial condition $x = 0$ when $y = 1$ to find the value of the constant $C$.

Substitute $x=0$ and $y=1$ into the general solution:

$2 e^{0/1} = -\log|1| + C$

$2 e^0 = -\log(1) + C$

We know that $e^0 = 1$ and $\log(1) = 0$ (natural logarithm).

$2 \cdot 1 = -0 + C$

$2 = C$

Substitute the value of $C = 2$ back into the general solution to obtain the particular solution:

$2 e^{x/y} = -\log|y| + 2$}

Since the initial condition is at $y=1$, which is positive, the solution near this point is for $y > 0$. Thus, $|y| = y$, and $\log|y| = \log y$.

The particular solution is $2 e^{x/y} = -\log y + 2$.

We can rearrange the terms if desired:

$2 e^{x/y} + \log y = 2$

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The particular solution of the differential equation is $2 e^{x/y} + \log y = 2$.

Given:

The vectors $\vec{a} = 2\hat{i}-2\hat{j}+\hat{k}$, $\vec{b} = \hat{i}+2\hat{j}-3\hat{k}$, and $\vec{c} = 2\hat{i}-\hat{j}+4\hat{k}$.

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To Find:

The projection of $\vec{b} + \vec{c}$ along $\vec{a}$.

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Solution:

The projection of a vector $\mathbf{v}$ along a vector $\mathbf{w}$ is given by the scalar projection formula:

$\text{proj}_{\mathbf{w}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{w}|}$

In this problem, we need to find the projection of $\vec{b} + \vec{c}$ along $\vec{a}$. So, $\mathbf{v} = \vec{b} + \vec{c}$ and $\mathbf{w} = \vec{a}$.

First, let's find the vector sum $\vec{b} + \vec{c}$:

$\vec{b} + \vec{c} = (\hat{i}+2\hat{j}-3\hat{k}) + (2\hat{i}-\hat{j}+4\hat{k})$

$\vec{b} + \vec{c} = (1+2)\hat{i} + (2-1)\hat{j} + (-3+4)\hat{k}$

$\vec{b} + \vec{c} = 3\hat{i} + \hat{j} + \hat{k}$

Next, we need to calculate the dot product of $(\vec{b} + \vec{c})$ and $\vec{a}$:

$(\vec{b} + \vec{c}) \cdot \vec{a} = (3\hat{i} + \hat{j} + \hat{k}) \cdot (2\hat{i}-2\hat{j}+\hat{k})$

Using the dot product formula for component vectors:

$(\vec{b} + \vec{c}) \cdot \vec{a} = (3)(2) + (1)(-2) + (1)(1)$

$(\vec{b} + \vec{c}) \cdot \vec{a} = 6 - 2 + 1$

$(\vec{b} + \vec{c}) \cdot \vec{a} = 5$

Now, we need to calculate the magnitude of the vector $\vec{a}$:

$|\vec{a}| = |2\hat{i}-2\hat{j}+\hat{k}|$

$|\vec{a}| = \sqrt{(2)^2 + (-2)^2 + (1)^2}$

$|\vec{a}| = \sqrt{4 + 4 + 1}$

$|\vec{a}| = \sqrt{9}$

$|\vec{a}| = 3$

Finally, we calculate the projection of $\vec{b} + \vec{c}$ along $\vec{a}$ using the formula:

Projection $= \frac{(\vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{a}|}$

Projection $= \frac{5}{3}$

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The projection of $\vec{b} + \vec{c}$ along $\vec{a}$ is $\frac{5}{3}$.

Given:

The line passes through the point with coordinates $(1, 2, -4)$. The position vector of this point is $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$.

The line is perpendicular to the two given lines:

Line 1: $\vec{r}_1 = (8\hat{i}-16\hat{j}+10\hat{k}) + \lambda (3\hat{i}-16\hat{j}+7\hat{k})$. The direction vector of this line is $\vec{b}_1 = 3\hat{i}-16\hat{j}+7\hat{k}$.

Line 2: $\vec{r}_2 = (15\hat{i}+29\hat{j}+5\hat{k}) + \mu (3\hat{i}+8\hat{j}-5\hat{k})$. The direction vector of this line is $\vec{b}_2 = 3\hat{i}+8\hat{j}-5\hat{k}$.

---

To Find:

The vector equation of the line passing through $(1, 2, -4)$ and perpendicular to the two given lines.

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Solution:

Let the required line be denoted by L. The line L passes through the point $A(1, 2, -4)$, so its position vector is $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$.

The line L is perpendicular to the lines with direction vectors $\vec{b}_1$ and $\vec{b}_2$. This means the direction vector of line L, let's call it $\vec{d}$, must be perpendicular to both $\vec{b}_1$ and $\vec{b}_2$.

A vector perpendicular to two vectors $\vec{b}_1$ and $\vec{b}_2$ is given by their cross product $\vec{b}_1 \times \vec{b}_2$.

Let's calculate $\vec{d} = \vec{b}_1 \times \vec{b}_2$:

$\vec{d} = (3\hat{i}-16\hat{j}+7\hat{k}) \times (3\hat{i}+8\hat{j}-5\hat{k})$

We can compute the cross product using a determinant:

$\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$

$\vec{d} = \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3))$

$\vec{d} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 - (-48))$

$\vec{d} = \hat{i}(24) - \hat{j}(-36) + \hat{k}(24 + 48)$

$\vec{d} = 24\hat{i} + 36\hat{j} + 72\hat{k}$

This vector $\vec{d}$ is the direction vector of the required line. Any vector parallel to $\vec{d}$ can also be used as the direction vector. We can take a common factor of 12 from the components of $\vec{d}$ to simplify it:

$\vec{d}' = \frac{1}{12}\vec{d} = \frac{1}{12}(24\hat{i} + 36\hat{j} + 72\hat{k}) = 2\hat{i} + 3\hat{j} + 6\hat{k}$

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a direction vector $\vec{d}'$ is given by $\vec{r} = \vec{a} + t\vec{d}'$, where $t$ is a scalar parameter.

Using the point $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$ and the direction vector $\vec{d}' = 2\hat{i} + 3\hat{j} + 6\hat{k}$, the vector equation of the line is:

$\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + t(2\hat{i} + 3\hat{j} + 6\hat{k})$

Note: Using $\vec{d} = 24\hat{i} + 36\hat{j} + 72\hat{k}$ as the direction vector would also be correct, resulting in the equation $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + s(24\hat{i} + 36\hat{j} + 72\hat{k})$, where the parameter $s$ would be related to $t$ by $s = t/12$. The simpler direction vector is usually preferred.

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The vector equation of the line is $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + t(2\hat{i} + 3\hat{j} + 6\hat{k})$.

Given:

There are three coins. Let the events of choosing each coin be $C_1$, $C_2$, and $C_3$.

$C_1$: Choosing the biased coin that gives tails 60% of the time.

$C_2$: Choosing the biased coin that gives heads 75% of the time.

$C_3$: Choosing the unbiased coin.

One coin is chosen at random, so the probabilities of choosing each coin are equal:

$P(C_1) = P(C_2) = P(C_3) = \frac{1}{3}$

Let $H$ be the event that the toss shows heads.

The conditional probabilities of getting a head given the choice of coin are:

For $C_1$ (60% tails): $P(H | C_1) = 1 - P(\text{Tail} | C_1) = 1 - 0.60 = 0.40$

For $C_2$ (75% heads): $P(H | C_2) = 0.75$

For $C_3$ (unbiased): $P(H | C_3) = 0.50$

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To Find:

The probability that the coin chosen was the unbiased coin, given that it showed heads. This is $P(C_3 | H)$.

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Solution:

We can use Bayes' Theorem to find $P(C_3 | H)$. Bayes' Theorem is given by:

$P(C_3 | H) = \frac{P(H | C_3) P(C_3)}{P(H)}$

First, we need to find the total probability of getting heads, $P(H)$. This can be calculated using the Law of Total Probability:

$P(H) = P(H | C_1) P(C_1) + P(H | C_2) P(C_2) + P(H | C_3) P(C_3)$

Substitute the given values:

$P(H) = (0.40) \left(\frac{1}{3}\right) + (0.75) \left(\frac{1}{3}\right) + (0.50) \left(\frac{1}{3}\right)$

$P(H) = \frac{1}{3} (0.40 + 0.75 + 0.50)$

$P(H) = \frac{1}{3} (1.65)$

$P(H) = \frac{1.65}{3}$

To work with fractions, $1.65 = \frac{165}{100}$.

$P(H) = \frac{165}{100 \times 3} = \frac{165}{300}$

Simplify the fraction by dividing numerator and denominator by 15:

$\frac{165}{300} = \frac{\cancel{165}^{11}}{\cancel{300}_{20}} = \frac{11}{20}$

So, $P(H) = \frac{11}{20}$.

Now, substitute the values into Bayes' Theorem for $P(C_3 | H)$:

$P(C_3 | H) = \frac{P(H | C_3) P(C_3)}{P(H)}$

$P(C_3 | H) = \frac{(0.50) \left(\frac{1}{3}\right)}{\frac{11}{20}}$

Convert 0.50 to a fraction: $0.50 = \frac{50}{100} = \frac{1}{2}$.

$P(C_3 | H) = \frac{\frac{1}{2} \times \frac{1}{3}}{\frac{11}{20}}$

$P(C_3 | H) = \frac{\frac{1}{6}}{\frac{11}{20}}$

$P(C_3 | H) = \frac{1}{6} \times \frac{20}{11}$

$P(C_3 | H) = \frac{1 \times 20}{6 \times 11} = \frac{20}{66}$}

Simplify the fraction by dividing numerator and denominator by 2:

$P(C_3 | H) = \frac{\cancel{20}^{10}}{\cancel{66}_{33}} = \frac{10}{33}$

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The probability that the coin was the unbiased coin, given that it showed heads, is $\frac{10}{33}$.

Given:

The matrix $A = \left(\begin{matrix}4&1&3\\2&1&1\\3&1&{-2}\\\end{matrix}\right)$.

The system of linear equations is: $4 x + 2 y + 3 z = 2$, $x + y + z = 1$, $3 x + y – 2 z = 5$.

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To Find:

The inverse of matrix A, $A^{-1}$, and use it to solve the given system of equations.

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Solution: Finding A^–1

First, we calculate the determinant of matrix A, $|A|$.

$\begin{vmatrix}4&1&3\\2&1&1\\3&1&-2\\\end{vmatrix} = 4 \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} - 1 \begin{vmatrix} 2 & 1 \\ 3 & -2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix}$

$= 4((1)(-2) - (1)(1)) - 1((2)(-2) - (1)(3)) + 3((2)(1) - (1)(3))$

$= 4(-2 - 1) - 1(-4 - 3) + 3(2 - 3)$

$= 4(-3) - 1(-7) + 3(-1)$

$= -12 + 7 - 3 = -8$

Since $|A| = -8 \neq 0$, the inverse of A exists.

Next, we find the matrix of cofactors of A. The cofactor $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$.

$C_{11} = (-1)^{1+1} \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = 1(-2 - 1) = -3$

$C_{12} = (-1)^{1+2} \begin{vmatrix} 2 & 1 \\ 3 & -2 \end{vmatrix} = -1(-4 - 3) = 7$

$C_{13} = (-1)^{1+3} \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = 1(2 - 3) = -1$

$C_{21} = (-1)^{2+1} \begin{vmatrix} 1 & 3 \\ 1 & -2 \end{vmatrix} = -1(-2 - 3) = 5$

$C_{22} = (-1)^{2+2} \begin{vmatrix} 4 & 3 \\ 3 & -2 \end{vmatrix} = 1(-8 - 9) = -17$

$C_{23} = (-1)^{2+3} \begin{vmatrix} 4 & 1 \\ 3 & 1 \end{vmatrix} = -1(4 - 3) = -1$

$C_{31} = (-1)^{3+1} \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} = 1(1 - 3) = -2$

$C_{32} = (-1)^{3+2} \begin{vmatrix} 4 & 3 \\ 2 & 1 \end{vmatrix} = -1(4 - 6) = 2$

$C_{33} = (-1)^{3+3} \begin{vmatrix} 4 & 1 \\ 2 & 1 \end{vmatrix} = 1(4 - 2) = 2$

The matrix of cofactors is $\left(\begin{matrix}-3&7&-1\\5&-17&-1\\-2&2&2\\\end{matrix}\right)$.

The adjoint of A is the transpose of the cofactor matrix.

$\text{Adj}(A) = \left(\begin{matrix}-3&5&-2\\7&-17&2\\-1&-1&2\\\end{matrix}\right)$.

The inverse of A is given by $A^{-1} = \frac{1}{|A|} \text{Adj}(A)$.

$A^{-1} = \frac{1}{-8} \left(\begin{matrix}-3&5&-2\\7&-17&2\\-1&-1&2\\\end{matrix}\right)$.

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Solution: Solving the System of Equations

The given system of linear equations is:

$4 x + 2 y + 3 z = 2$

$x + y + z = 1$

$3 x + y – 2 z = 5$

This system can be written in matrix form as $A_{sys} X = B$, where $X = \left(\begin{matrix}x\\y\\z\\\end{matrix}\right)$ and $B = \left(\begin{matrix}2\\1\\5\\\end{matrix}\right)$.

The coefficient matrix is $A_{sys} = \left(\begin{matrix}4&2&3\\1&1&1\\3&1&-2\\\end{matrix}\right)$.

Note that the coefficient matrix $A_{sys}$ is different from the matrix A given for finding the inverse. However, the phrase "Hence solve" suggests that the inverse of the given matrix A should be used. We will proceed assuming that the intended system of equations has the given matrix A as its coefficient matrix.

The assumed system is:

$4x + y + 3z = 2$

$2x + y + z = 1$

$3x + y - 2z = 5$

In matrix form, this is $A X = B$, where $A = \left(\begin{matrix}4&1&3\\2&1&1\\3&1&-2\\\end{matrix}\right)$, $X = \left(\begin{matrix}x\\y\\z\\\end{matrix}\right)$, and $B = \left(\begin{matrix}2\\1\\5\\\end{matrix}\right)$.

The solution to this system is given by $X = A^{-1}B$. We use the inverse $A^{-1}$ calculated earlier.

$X = \frac{1}{-8} \left(\begin{matrix}-3&5&-2\\7&-17&2\\-1&-1&2\\\end{matrix}\right) \left(\begin{matrix}2\\1\\5\\\end{matrix}\right)$

Perform the matrix multiplication:

$\left(\begin{matrix}-3&5&-2\\7&-17&2\\-1&-1&2\\\end{matrix}\right) \left(\begin{matrix}2\\1\\5\\\end{matrix}\right) = \left(\begin{matrix}(-3)(2) + (5)(1) + (-2)(5) \\ (7)(2) + (-17)(1) + (2)(5) \\ (-1)(2) + (-1)(1) + (2)(5) \end{matrix}\right)$

$= \left(\begin{matrix}-6 + 5 - 10 \\ 14 - 17 + 10 \\ -2 - 1 + 10 \end{matrix}\right) = \left(\begin{matrix}-11 \\ 7 \\ 7 \end{matrix}\right)$

So, $X = \frac{1}{-8} \left(\begin{matrix}-11\\7\\7\\\end{matrix}\right) = \left(\begin{matrix}\frac{-11}{-8}\\\frac{7}{-8}\\\frac{7}{-8}\\\end{matrix}\right) = \left(\begin{matrix}\frac{11}{8}\\-\frac{7}{8}\\-\frac{7}{8}\\\end{matrix}\right)$.

Equating the components, we get the values of x, y, and z:

$x = \frac{11}{8}$

$y = -\frac{7}{8}$

$z = -\frac{7}{8}$

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The inverse of matrix A is $A^{-1} = \frac{1}{-8} \left(\begin{matrix}-3&5&-2\\7&-17&2\\-1&-1&2\\\end{matrix}\right)$.

Assuming the system of equations corresponds to matrix A, the solution is $x = \frac{11}{8}$, $y = -\frac{7}{8}$, $z = -\frac{7}{8}$.

Given:

A cone with a given slant height, let's denote it by $l$.

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To Show:

The semi-vertical angle ($\alpha$) of the cone of maximum volume is $\tan^{-1} \sqrt{2}$.

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Solution:

Let $r$ be the radius of the base of the cone and $h$ be its height.

The slant height $l$ is constant.

In a right-angled triangle formed by the height, radius, and slant height, the semi-vertical angle $\alpha$ is the angle between the height and the slant height.

From trigonometry, we have the relations:

$r = l \sin \alpha$

$h = l \cos \alpha$

The volume of the cone, $V$, is given by the formula:

$V = \frac{1}{3} \pi r^2 h$

Substitute the expressions for $r$ and $h$ in terms of $l$ and $\alpha$ into the volume formula:

$V(\alpha) = \frac{1}{3} \pi (l \sin \alpha)^2 (l \cos \alpha)$

$V(\alpha) = \frac{1}{3} \pi l^2 \sin^2 \alpha \cdot l \cos \alpha$

$V(\alpha) = \frac{1}{3} \pi l^3 \sin^2 \alpha \cos \alpha$

To find the angle $\alpha$ that maximizes the volume, we differentiate $V(\alpha)$ with respect to $\alpha$ and set the derivative equal to zero. Remember that $l$ is a constant.

$\frac{dV}{d\alpha} = \frac{d}{d\alpha} \left( \frac{1}{3} \pi l^3 \sin^2 \alpha \cos \alpha \right)$

$\frac{dV}{d\alpha} = \frac{1}{3} \pi l^3 \frac{d}{d\alpha} (\sin^2 \alpha \cos \alpha)$

Using the product rule $\frac{d}{d\alpha}(uv) = u'v + uv'$, where $u = \sin^2 \alpha$ and $v = \cos \alpha$:

$u' = \frac{d}{d\alpha}(\sin^2 \alpha) = 2 \sin \alpha \cos \alpha$

$v' = \frac{d}{d\alpha}(\cos \alpha) = -\sin \alpha$

$\frac{d}{d\alpha} (\sin^2 \alpha \cos \alpha) = (2 \sin \alpha \cos \alpha) \cos \alpha + \sin^2 \alpha (-\sin \alpha)$

$= 2 \sin \alpha \cos^2 \alpha - \sin^3 \alpha$

So, $\frac{dV}{d\alpha} = \frac{1}{3} \pi l^3 (2 \sin \alpha \cos^2 \alpha - \sin^3 \alpha)$

Set $\frac{dV}{d\alpha} = 0$ for critical points:

$\frac{1}{3} \pi l^3 (2 \sin \alpha \cos^2 \alpha - \sin^3 \alpha) = 0$

Since $\frac{1}{3} \pi l^3 \neq 0$ (as $l$ is a given slant height), we must have:

$2 \sin \alpha \cos^2 \alpha - \sin^3 \alpha = 0$

Factor out $\sin \alpha$:

$\sin \alpha (2 \cos^2 \alpha - \sin^2 \alpha) = 0$

This gives two possibilities: $\sin \alpha = 0$ or $2 \cos^2 \alpha - \sin^2 \alpha = 0$.

The semi-vertical angle $\alpha$ for a cone must be in the range $0 < \alpha < \frac{\pi}{2}$. In this range, $\sin \alpha \neq 0$.

Therefore, we must have:

$2 \cos^2 \alpha - \sin^2 \alpha = 0$

$2 \cos^2 \alpha = \sin^2 \alpha$

Divide both sides by $\cos^2 \alpha$. Since $0 < \alpha < \frac{\pi}{2}$, $\cos \alpha \neq 0$, so $\cos^2 \alpha \neq 0$.

$2 = \frac{\sin^2 \alpha}{\cos^2 \alpha}$

$2 = \tan^2 \alpha$

Taking the square root of both sides:

$\tan \alpha = \pm \sqrt{2}$

Since $0 < \alpha < \frac{\pi}{2}$, the tangent of $\alpha$ must be positive.

So, $\tan \alpha = \sqrt{2}$

Taking the inverse tangent of both sides:

$\alpha = \tan^{-1} \sqrt{2}$

To confirm that this angle corresponds to a maximum volume, we would ideally use the second derivative test. However, the calculation of the second derivative is lengthy. Alternatively, we can analyze the sign of $\frac{dV}{d\alpha}$ around $\alpha = \tan^{-1} \sqrt{2}$.

$\frac{dV}{d\alpha} = \frac{1}{3} \pi l^3 \sin \alpha (2 \cos^2 \alpha - \sin^2 \alpha) = \frac{1}{3} \pi l^3 \sin \alpha \cos^2 \alpha (2 - \tan^2 \alpha)$

For $0 < \alpha < \frac{\pi}{2}$, $\sin \alpha > 0$ and $\cos^2 \alpha > 0$. The sign of $\frac{dV}{d\alpha}$ is determined by $(2 - \tan^2 \alpha)$.

If $0 < \alpha < \tan^{-1} \sqrt{2}$, then $0 < \tan \alpha < \sqrt{2}$, so $\tan^2 \alpha < 2$. Thus, $(2 - \tan^2 \alpha) > 0$, and $\frac{dV}{d\alpha} > 0$ (Volume is increasing).

If $\tan^{-1} \sqrt{2} < \alpha < \frac{\pi}{2}$, then $\tan \alpha > \sqrt{2}$, so $\tan^2 \alpha > 2$. Thus, $(2 - \tan^2 \alpha) < 0$, and $\frac{dV}{d\alpha} < 0$ (Volume is decreasing).

Since the derivative changes from positive to negative at $\alpha = \tan^{-1} \sqrt{2}$, this angle corresponds to a local maximum volume. As it is the only critical point in the interval $(0, \pi/2)$, it corresponds to the maximum volume.

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Therefore, the semi-vertical angle of the cone of maximum volume and of given slant height is $\tan^{-1} \sqrt{2}$.

Given:

The definite integral $\int\limits_{1}^{3}{({3x}^2+2x+5)} \ dx$.

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To Evaluate:

The given definite integral using the method of limit of sum.

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Solution:

The formula for evaluating the definite integral $\int\limits_{a}^{b} f(x) dx$ using the method of limit of sum is:

$\int\limits_{a}^{b} f(x) dx = \lim\limits_{n \to \infty} h \sum\limits_{r=0}^{n-1} f(a+rh)$

where $h = \frac{b-a}{n}$.

In this problem, we have $a=1$, $b=3$, and $f(x) = 3x^2 + 2x + 5$.

First, calculate $h$:

$h = \frac{b-a}{n} = \frac{3-1}{n} = \frac{2}{n}$

Next, calculate $a+rh$:

$a+rh = 1 + r \left(\frac{2}{n}\right) = 1 + \frac{2r}{n}$

Now, evaluate $f(a+rh) = f\left(1 + \frac{2r}{n}\right)$:

$f\left(1 + \frac{2r}{n}\right) = 3\left(1 + \frac{2r}{n}\right)^2 + 2\left(1 + \frac{2r}{n}\right) + 5$

$= 3\left(1^2 + 2(1)\left(\frac{2r}{n}\right) + \left(\frac{2r}{n}\right)^2\right) + 2 + \frac{4r}{n} + 5$

$= 3\left(1 + \frac{4r}{n} + \frac{4r^2}{n^2}\right) + 7 + \frac{4r}{n}$

$= 3 + \frac{12r}{n} + \frac{12r^2}{n^2} + 7 + \frac{4r}{n}$

$= (3+7) + \left(\frac{12r}{n} + \frac{4r}{n}\right) + \frac{12r^2}{n^2}$

$= 10 + \frac{16r}{n} + \frac{12r^2}{n^2}$

Now, calculate the sum $\sum\limits_{r=0}^{n-1} f(a+rh)$:

$\sum\limits_{r=0}^{n-1} \left(10 + \frac{16r}{n} + \frac{12r^2}{n^2}\right)$

$= \sum\limits_{r=0}^{n-1} 10 + \sum\limits_{r=0}^{n-1} \frac{16r}{n} + \sum\limits_{r=0}^{n-1} \frac{12r^2}{n^2}$

$= 10 \sum\limits_{r=0}^{n-1} 1 + \frac{16}{n} \sum\limits_{r=0}^{n-1} r + \frac{12}{n^2} \sum\limits_{r=0}^{n-1} r^2$

Using the standard summation formulas:

$\sum\limits_{r=0}^{n-1} 1 = n$

$\sum\limits_{r=0}^{n-1} r = 0 + 1 + 2 + \dots + (n-1) = \frac{(n-1)n}{2}$

$\sum\limits_{r=0}^{n-1} r^2 = 0^2 + 1^2 + 2^2 + \dots + (n-1)^2 = \frac{(n-1)(n-1+1)(2(n-1)+1)}{6} = \frac{(n-1)n(2n-1)}{6}$

Substitute these sums back into the expression:

$\sum\limits_{r=0}^{n-1} f(a+rh) = 10(n) + \frac{16}{n} \left(\frac{(n-1)n}{2}\right) + \frac{12}{n^2} \left(\frac{(n-1)n(2n-1)}{6}\right)$

$= 10n + 8(n-1) + \frac{2}{n}(n-1)(2n-1)$

$= 10n + 8n - 8 + \frac{2}{n}(2n^2 - 3n + 1)$

$= 18n - 8 + 4n - 6 + \frac{2}{n}$

$= 22n - 14 + \frac{2}{n}$

Now, calculate $h \sum\limits_{r=0}^{n-1} f(a+rh)$:

$h \sum\limits_{r=0}^{n-1} f(a+rh) = \frac{2}{n} \left(22n - 14 + \frac{2}{n}\right)$

$= \frac{2}{n}(22n) - \frac{2}{n}(14) + \frac{2}{n}\left(\frac{2}{n}\right)$

$= 44 - \frac{28}{n} + \frac{4}{n^2}$

Finally, evaluate the limit as $n \to \infty$:

$\int\limits_{1}^{3} (3x^2 + 2x + 5) dx = \lim\limits_{n \to \infty} \left(44 - \frac{28}{n} + \frac{4}{n^2}\right)$

As $n \to \infty$, $\frac{28}{n} \to 0$ and $\frac{4}{n^2} \to 0$.

$= 44 - 0 + 0$

$= 44$

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The value of the integral is 44.

Solution:

The equation of the circle is $x^2 + y^2 = 4$. The center of the circle is $(0,0)$ and the radius is $r=2$.

The point of tangency is $(x_1, y_1) = (1, \sqrt{3})$. This point lies on the circle since $1^2 + (\sqrt{3})^2 = 1 + 3 = 4$.

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Equation of the Tangent:

The equation of the tangent to the circle $x^2 + y^2 = r^2$ at the point $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$.

Substituting $(x_1, y_1) = (1, \sqrt{3})$ and $r^2=4$, the equation of the tangent is:

$x(1) + y(\sqrt{3}) = 4$

$x + \sqrt{3}y = 4$

To find the x-intercept of the tangent, set $y=0$:

$x + \sqrt{3}(0) = 4$

$x = 4$

The tangent intersects the x-axis at point A $(4, 0)$. This point is on the positive x-axis.

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Equation of the Normal:

The normal to the circle at any point passes through the center $(0,0)$.

The normal line passes through the point of tangency $(1, \sqrt{3})$ and the center $(0,0)$.

The slope of the normal is the slope of the line connecting $(0,0)$ and $(1, \sqrt{3})$.

Slope of normal $m_n = \frac{\sqrt{3}-0}{1-0} = \sqrt{3}$.

Using the point-slope form $y - y_1 = m(x - x_1)$ with $(x_1, y_1) = (0,0)$ and $m = \sqrt{3}$:

$y - 0 = \sqrt{3}(x - 0)$

$y = \sqrt{3}x$

To find the x-intercept of the normal, set $y=0$:

$0 = \sqrt{3}x$

$x = 0$

The normal intersects the x-axis at point O $(0, 0)$. This point is the origin, which is on the positive x-axis.

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Vertices of the Triangle:

The triangle is formed by the positive x-axis, the tangent ($x + \sqrt{3}y = 4$), and the normal ($y = \sqrt{3}x$).

The vertices are the intersection points of these lines:

• Intersection of Normal and positive x-axis: O $(0,0)$.
• Intersection of Tangent and positive x-axis: A $(4,0)$.
• Intersection of Tangent and Normal: This is the point of tangency B $(1, \sqrt{3})$.

The triangle OAB has vertices O$(0,0)$, A$(4,0)$, and B$(1, \sqrt{3})$.

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Area using Integration:

The triangle OAB is bounded by the x-axis from $x=0$ to $x=4$.

The line segment OB is part of the normal $y = \sqrt{3}x$, from $x=0$ to $x=1$.

The line segment AB is part of the tangent $x + \sqrt{3}y = 4$, which can be written as $y = \frac{4-x}{\sqrt{3}}$, from $x=1$ to $x=4$.

The area of the triangle OAB can be calculated as the sum of the areas under the normal line from $x=0$ to $x=1$ and under the tangent line from $x=1$ to $x=4$.

Area = $\int\limits_{0}^{1} (\sqrt{3}x) \, dx + \int\limits_{1}^{4} \left(\frac{4-x}{\sqrt{3}}\right) \, dx$

Evaluate the first integral:

$\int\limits_{0}^{1} \sqrt{3}x \, dx = \sqrt{3} \left[\frac{x^2}{2}\right]_{0}^{1} = \sqrt{3} \left(\frac{1^2}{2} - \frac{0^2}{2}\right) = \sqrt{3} \left(\frac{1}{2} - 0\right) = \frac{\sqrt{3}}{2}$

Evaluate the second integral:

$\int\limits_{1}^{4} \frac{4-x}{\sqrt{3}} \, dx = \frac{1}{\sqrt{3}} \int\limits_{1}^{4} (4-x) \, dx = \frac{1}{\sqrt{3}} \left[4x - \frac{x^2}{2}\right]_{1}^{4}$

$= \frac{1}{\sqrt{3}} \left[\left(4(4) - \frac{4^2}{2}\right) - \left(4(1) - \frac{1^2}{2}\right)\right]$

$= \frac{1}{\sqrt{3}} \left[\left(16 - \frac{16}{2}\right) - \left(4 - \frac{1}{2}\right)\right]$

$= \frac{1}{\sqrt{3}} \left[\left(16 - 8\right) - \left(\frac{8-1}{2}\right)\right]$

$= \frac{1}{\sqrt{3}} \left[8 - \frac{7}{2}\right]$

$= \frac{1}{\sqrt{3}} \left[\frac{16-7}{2}\right] = \frac{1}{\sqrt{3}} \left[\frac{9}{2}\right] = \frac{9}{2\sqrt{3}}$

Rationalize the denominator: $\frac{9}{2\sqrt{3}} = \frac{9 \times \sqrt{3}}{2\sqrt{3} \times \sqrt{3}} = \frac{9\sqrt{3}}{2 \times 3} = \frac{9\sqrt{3}}{6} = \frac{3\sqrt{3}}{2}$.

Total Area = Area under Normal + Area under Tangent

Total Area = $\frac{\sqrt{3}}{2} + \frac{3\sqrt{3}}{2} = \frac{\sqrt{3} + 3\sqrt{3}}{2} = \frac{4\sqrt{3}}{2} = 2\sqrt{3}$.

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Final Answer:

The area of the triangle formed by the positive x-axis, the normal, and the tangent to the circle $x^2 + y^2 = 4$ at $(1, \sqrt{3})$ is $2\sqrt{3}$ square units.

Solution:

Let the equations of the two given planes be:

Plane 1 ($P_1$): $x + 3y + 6 = 0$

Plane 2 ($P_2$): $3x - y - 4z = 0$

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The equation of a plane passing through the intersection of the planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$, where $\lambda$ is a constant.

Substituting the equations of the planes:

$(x + 3y + 6) + \lambda(3x - y - 4z) = 0$

Rearranging the terms to get the standard form $Ax + By + Cz + D = 0$:

$x + 3y + 6 + 3\lambda x - \lambda y - 4\lambda z = 0$

$(1 + 3\lambda)x + (3 - \lambda)y + (-4\lambda)z + 6 = 0$

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We are given that the perpendicular distance of this plane from the origin $(0,0,0)$ is unity (1).

The formula for the perpendicular distance from a point $(x_0, y_0, z_0)$ to the plane $Ax + By + Cz + D = 0$ is given by:

$d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Here, $(x_0, y_0, z_0) = (0,0,0)$, $d = 1$, $A = (1 + 3\lambda)$, $B = (3 - \lambda)$, $C = (-4\lambda)$, and $D = 6$.

Substituting these values into the distance formula:

$1 = \frac{|(1 + 3\lambda)(0) + (3 - \lambda)(0) + (-4\lambda)(0) + 6|}{\sqrt{(1 + 3\lambda)^2 + (3 - \lambda)^2 + (-4\lambda)^2}}$

$1 = \frac{|6|}{\sqrt{(1 + 6\lambda + 9\lambda^2) + (9 - 6\lambda + \lambda^2) + (16\lambda^2)}}$

$1 = \frac{6}{\sqrt{9\lambda^2 + \lambda^2 + 16\lambda^2 + 6\lambda - 6\lambda + 1 + 9}}$

$1 = \frac{6}{\sqrt{26\lambda^2 + 10}}$

Cross-multiplying, we get:

$\sqrt{26\lambda^2 + 10} = 6$

Squaring both sides:

$(\sqrt{26\lambda^2 + 10})^2 = 6^2$

$26\lambda^2 + 10 = 36$

$26\lambda^2 = 36 - 10$

$26\lambda^2 = 26$

$\lambda^2 = \frac{26}{26}$

$\lambda^2 = 1$

Taking the square root of both sides, we get two possible values for $\lambda$:

$\lambda = \pm 1$

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Case 1: When $\lambda = 1$

Substitute $\lambda = 1$ into the plane equation $(1 + 3\lambda)x + (3 - \lambda)y + (-4\lambda)z + 6 = 0$:

$(1 + 3(1))x + (3 - 1)y + (-4(1))z + 6 = 0$

$(1 + 3)x + (2)y + (-4)z + 6 = 0$

$4x + 2y - 4z + 6 = 0$

Dividing the entire equation by 2:

$2x + y - 2z + 3 = 0$

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Case 2: When $\lambda = -1$

Substitute $\lambda = -1$ into the plane equation $(1 + 3\lambda)x + (3 - \lambda)y + (-4\lambda)z + 6 = 0$:

$(1 + 3(-1))x + (3 - (-1))y + (-4(-1))z + 6 = 0$

$(1 - 3)x + (3 + 1)y + (4)z + 6 = 0$

$-2x + 4y + 4z + 6 = 0$

Dividing the entire equation by -2:

$\frac{-2x}{-2} + \frac{4y}{-2} + \frac{4z}{-2} + \frac{6}{-2} = 0$

$x - 2y - 2z - 3 = 0$

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Thus, there are two such planes that satisfy the given conditions.

Final Answer:

The equations of the planes are $2x + y - 2z + 3 = 0$ and $x - 2y - 2z - 3 = 0$.

Solution:

Total number of bulbs = Number of defective bulbs + Number of good bulbs = $4 + 6 = 10$.

Number of bulbs drawn at random = 4.

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Let X be the random variable representing the number of defective bulbs drawn.

Since 4 bulbs are drawn from a lot of 10 bulbs (4 defective and 6 good), the possible values for X (number of defective bulbs drawn) are 0, 1, 2, 3, or 4.

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The total number of ways to draw 4 bulbs from 10 bulbs is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Total number of ways to draw 4 bulbs from 10 = $\binom{10}{4}$.

$\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210$.

So, the total number of possible outcomes is 210.

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Now, we find the number of ways to draw exactly $k$ defective bulbs and $4-k$ good bulbs for each possible value of X ($k=0, 1, 2, 3, 4$).

Number of ways to draw $k$ defective bulbs from 4 = $\binom{4}{k}$.

Number of ways to draw $4-k$ good bulbs from 6 = $\binom{6}{4-k}$.

The number of ways to draw $k$ defective and $4-k$ good bulbs is $\binom{4}{k} \times \binom{6}{4-k}$.

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Calculate the number of ways and probability for each value of X:

Case X = 0: (0 defective, 4 good)

Number of ways = $\binom{4}{0} \times \binom{6}{4} = 1 \times \frac{6 \times 5}{2 \times 1} = 1 \times 15 = 15$.

Probability $P(X=0) = \frac{\text{Number of ways}}{\text{Total ways}} = \frac{15}{210} = \frac{1}{14}$.

Case X = 1: (1 defective, 3 good)

Number of ways = $\binom{4}{1} \times \binom{6}{3} = 4 \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 4 \times 20 = 80$.

Probability $P(X=1) = \frac{80}{210} = \frac{8}{21}$.

Case X = 2: (2 defective, 2 good)

Number of ways = $\binom{4}{2} \times \binom{6}{2} = \frac{4 \times 3}{2 \times 1} \times \frac{6 \times 5}{2 \times 1} = 6 \times 15 = 90$.

Probability $P(X=2) = \frac{90}{210} = \frac{9}{21} = \frac{3}{7}$.

Case X = 3: (3 defective, 1 good)

Number of ways = $\binom{4}{3} \times \binom{6}{1} = 4 \times 6 = 24$.

Probability $P(X=3) = \frac{24}{210} = \frac{4}{35}$.

Case X = 4: (4 defective, 0 good)

Number of ways = $\binom{4}{4} \times \binom{6}{0} = 1 \times 1 = 1$.

Probability $P(X=4) = \frac{1}{210}$.

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The probability distribution of the number of defective bulbs (X) is:

Number of Defective Bulbs (X=k)	Probability P(X=k)
0	$\frac{15}{210} = \frac{1}{14}$
1	$\frac{80}{210} = \frac{8}{21}$
2	$\frac{90}{210} = \frac{9}{21} = \frac{3}{7}$
3	$\frac{24}{210} = \frac{4}{35}$
4	$\frac{1}{210}$

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Final Answer:

The probability distribution of the number of defective bulbs is:

X	0	1	2	3	4
P(X)	$\frac{1}{14}$	$\frac{8}{21}$	$\frac{3}{7}$	$\frac{4}{35}$	$\frac{1}{210}$

Solution:

Let $x$ be the number of chairs manufactured per week and $y$ be the number of tables manufactured per week.

We want to maximize the profit. The profit on one chair is $\textsf{₹}30$ and on one table is $\textsf{₹}60$.

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Formulation of the Linear Programming Problem:

Maximize Profit:

$Z = 30x + 60y$

Subject to the constraints:

Based on machine time requirements and availability:

Machine A: $2x + y \leq 70$

Machine B: $x + y \leq 40$

Machine C: $x + 3y \leq 90$

Also, the number of chairs and tables cannot be negative:

$x \geq 0$, $y \geq 0$

So, the LPP is:

Maximize $Z = 30x + 60y$

Subject to:

$2x + y \leq 70$

$x + y \leq 40$

$x + 3y \leq 90$

$x \geq 0$, $y \geq 0$

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Graphical Solution:

To solve graphically, we first consider the equations of the boundary lines corresponding to the inequality constraints:

L1: $2x + y = 70$

L2: $x + y = 40$

L3: $x + 3y = 90$

We find two points on each line to plot them:

For L1 ($2x + y = 70$):

If $x=0, y=70$. Point (0, 70).

If $y=0, 2x=70 \implies x=35$. Point (35, 0).

For L2 ($x + y = 40$):

If $x=0, y=40$. Point (0, 40).

If $y=0, x=40$. Point (40, 0).

For L3 ($x + 3y = 90$):

If $x=0, 3y=90 \implies y=30$. Point (0, 30).

If $y=0, x=90$. Point (90, 0).

The non-negativity constraints $x \geq 0, y \geq 0$ restrict the feasible region to the first quadrant.

We plot these lines and determine the feasible region by checking the origin (0,0) for each inequality:

$2(0) + 0 \leq 70$ (True)

$0 + 0 \leq 40$ (True)

$0 + 3(0) \leq 90$ (True)

Since the origin satisfies all inequalities, the feasible region is the area in the first quadrant below or on all three lines. The feasible region is a convex polygon.

The corner points of the feasible region are the intersection points of the boundary lines. These are:

1. Intersection of $x=0$ and $y=0$: O(0,0).

2. Intersection of $y=0$ and $2x+y=70$: $2x+0=70 \implies x=35$. Point A(35,0).

3. Intersection of $2x+y=70$ and $x+y=40$:

Subtracting $x+y=40$ from $2x+y=70$:

$(2x+y) - (x+y) = 70 - 40$

$x = 30$

Substitute $x=30$ into $x+y=40$: $30+y=40 \implies y=10$. Point B(30,10).

4. Intersection of $x+y=40$ and $x+3y=90$:

Subtracting $x+y=40$ from $x+3y=90$:

$(x+3y) - (x+y) = 90 - 40$

$2y = 50 \implies y=25$

Substitute $y=25$ into $x+y=40$: $x+25=40 \implies x=15$. Point C(15,25).

5. Intersection of $x=0$ and $x+3y=90$: $0+3y=90 \implies y=30$. Point D(0,30).

The corner points of the feasible region are O(0,0), A(35,0), B(30,10), C(15,25), and D(0,30).

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Evaluation of the Objective Function at Corner Points:

We evaluate the profit function $Z = 30x + 60y$ at each corner point:

• At O(0,0): $Z = 30(0) + 60(0) = 0$
• At A(35,0): $Z = 30(35) + 60(0) = 1050$
• At B(30,10): $Z = 30(30) + 60(10) = 900 + 600 = 1500$
• At C(15,25): $Z = 30(15) + 60(25) = 450 + 1500 = 1950$
• At D(0,30): $Z = 30(0) + 60(30) = 0 + 1800 = 1800$

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The maximum value of the profit $Z$ is 1950, which occurs at the corner point C(15,25).

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Final Answer:

To maximize profit, the firm should manufacture 15 chairs and 25 tables per week. The maximum profit obtained is $\textsf{₹}1950$.