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

Given matrix equation:

$\begin{bmatrix}x\ +\ y\\x\ -\ y\\\end{bmatrix}=\left[\begin{matrix}2&1\\4&3\\\end{matrix}\right]\left[\begin{matrix}1\\-2\\\end{matrix}\right]$

First, perform the matrix multiplication on the right side:

$\left[\begin{matrix}2&1\\4&3\\\end{matrix}\right]\left[\begin{matrix}1\\-2\\\end{matrix}\right] = \begin{bmatrix} (2)(1) + (1)(-2) \\ (4)(1) + (3)(-2) \end{bmatrix}$

$= \begin{bmatrix} 2 - 2 \\ 4 - 6 \end{bmatrix}$

$= \begin{bmatrix} 0 \\ -2 \end{bmatrix}$

Now, equate the left and right matrices:

$\begin{bmatrix}x\ +\ y\\x\ -\ y\\\end{bmatrix} = \begin{bmatrix} 0 \\ -2 \end{bmatrix}$

Equating the corresponding elements gives a system of linear equations:

Add equation (1) and equation (2):

$(x + y) + (x - y) = 0 + (-2)$

$2x = -2$

$x = \frac{-2}{2}$

$x = -1$

Substitute the value of $x = -1$ into equation (1):

$(-1) + y = 0$

$y = 0 - (-1)$

$y = 1$

Thus, the solution is $(x, y) = (-1, 1)$.

The correct option is (C) (-1, 1).

Given:

Vertices of the triangle: $A = (-2, 4)$, $B = (2, k)$, $C = (5, 4)$.

Area of the triangle = 35 sq. units.

To Find:

The value of $k$.

Solution:

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substitute the coordinates of the given vertices $A(-2, 4)$, $B(2, k)$, and $C(5, 4)$ into the formula:

$(x_1, y_1) = (-2, 4)$

$(x_2, y_2) = (2, k)$

$(x_3, y_3) = (5, 4)$

Area $= \frac{1}{2} |(-2)(k - 4) + (2)(4 - 4) + (5)(4 - k)|$

Area $= \frac{1}{2} |(-2k + 8) + (2)(0) + (20 - 5k)|$

Area $= \frac{1}{2} |-2k + 8 + 0 + 20 - 5k|$

Area $= \frac{1}{2} |-7k + 28|$

We are given that the area is 35 sq. units. Therefore,

$\frac{1}{2} |-7k + 28| = 35$

$|-7k + 28| = 35 \times 2$

$|-7k + 28| = 70$

Since the absolute value is 70, the expression inside can be either 70 or -70. This gives two possible cases:

Case 1:

$-7k + 28 = 70$

$-7k = 70 - 28$

$-7k = 42$

$k = \frac{42}{-7}$

$k = -6$

Case 2:

$-7k + 28 = -70$

$-7k = -70 - 28$

$-7k = -98$

$k = \frac{-98}{-7}$

$k = 14$

The possible values for $k$ are $-6$ and $14$.

Checking the given options, we see that $-6$ is one of the options.

The value of $k$ is -6.

The correct option is (D) – 6.

Given:

Equation of the curve: $y^2 = 4x$

Equation of the line: $y = x + 1$

To Find:

The point where the line is tangent to the curve.

Solution:

If the line $y = x + 1$ is tangent to the curve $y^2 = 4x$, then there should be exactly one point of intersection between the line and the curve.

We can find the points of intersection by substituting the equation of the line into the equation of the curve.

Substitute $y = x + 1$ into $y^2 = 4x$:

$(x + 1)^2 = 4x$

Expand the left side:

$x^2 + 2x + 1 = 4x$

Rearrange the equation to form a quadratic equation in $x$:

$x^2 + 2x - 4x + 1 = 0$

$x^2 - 2x + 1 = 0$

Factor the quadratic equation:

$(x - 1)^2 = 0$

This equation has one solution for $x$:

From equation (i), we get:

$x = 1$

Now, substitute the value of $x$ back into the equation of the line $y = x + 1$ to find the corresponding $y$ coordinate:

$y = 1 + 1$

$y = 2$

The point of intersection is $(1, 2)$. Since there is only one point of intersection, the line is tangent to the curve at this point.

The point of tangency is $(1, 2)$.

Comparing with the given options, the correct option is (A).

The correct option is (A) (1, 2).

Alternate Solution (Using Differentiation):

The equation of the curve is $y^2 = 4x$.

Differentiate both sides with respect to $x$ to find the slope of the tangent:

$\frac{d}{dx}(y^2) = \frac{d}{dx}(4x)$

$2y \frac{dy}{dx} = 4$

$\frac{dy}{dx} = \frac{4}{2y}$

$\frac{dy}{dx} = \frac{2}{y}$

This is the slope of the tangent to the curve at any point $(x, y)$ on the curve.

The equation of the given line is $y = x + 1$. The slope of this line is the coefficient of $x$, which is 1.

If the line is tangent to the curve at a point $(x, y)$, then the slope of the tangent to the curve at that point must be equal to the slope of the line.

$\frac{dy}{dx} = 1$

So, $\frac{2}{y} = 1$

$y = 2$

Now, substitute this value of $y$ back into the equation of the curve $y^2 = 4x$ to find the corresponding $x$ coordinate:

$(2)^2 = 4x$

$4 = 4x$

$x = \frac{4}{4}$

$x = 1$

The point of tangency is $(1, 2)$.

Given:

The size of the matrix is 2 × 2.

The elements $a_{ij}$ are given by the formula:

$a_{ij} = \begin{cases} \frac{-3i+j}{2} & , & \text{if } i \neq j \\ (i + j)^2 & , & \text{if } i = j \end{cases}$

To Construct:

A 2 × 2 matrix $\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$.

Solution:

We need to calculate each element of the 2 × 2 matrix using the given formula.

For the element $a_{11}$: $i=1$, $j=1$. Since $i=j$, we use the second case of the formula.

$a_{11} = (i + j)^2 = (1 + 1)^2 = (2)^2 = 4$

For the element $a_{12}$: $i=1$, $j=2$. Since $i \neq j$, we use the first case of the formula.

$a_{12} = \frac{-3i+j}{2} = \frac{-3(1)+2}{2} = \frac{-3+2}{2} = \frac{-1}{2}$

For the element $a_{21}$: $i=2$, $j=1$. Since $i \neq j$, we use the first case of the formula.

$a_{21} = \frac{-3i+j}{2} = \frac{-3(2)+1}{2} = \frac{-6+1}{2} = \frac{-5}{2}$

For the element $a_{22}$: $i=2$, $j=2$. Since $i=j$, we use the second case of the formula.

$a_{22} = (i + j)^2 = (2 + 2)^2 = (4)^2 = 16$

Now, we assemble the matrix using the calculated elements:

The 2 × 2 matrix is $\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} 4 & -\frac{1}{2} \\ -\frac{5}{2} & 16 \end{bmatrix}$.

Given:

Function $y = \tan^{-1}(e^x)$.

To Find:

The value of the derivative $\frac{dy}{dx}$ at $x=0$.

Solution:

We need to find the derivative of $y = \tan^{-1}(e^x)$ with respect to $x$. We will use the chain rule.

The derivative of $\tan^{-1}(u)$ with respect to $u$ is $\frac{1}{1+u^2}$.

Let $u = e^x$. Then $\frac{du}{dx} = \frac{d}{dx}(e^x) = e^x$.

Applying the chain rule, $\frac{dy}{dx} = \frac{d}{du}(\tan^{-1}(u)) \cdot \frac{du}{dx}$:

$\frac{dy}{dx} = \frac{1}{1+u^2} \cdot e^x$

Substitute $u = e^x$ back into the expression:

$\frac{dy}{dx} = \frac{1}{1+(e^x)^2} \cdot e^x$

$\frac{dy}{dx} = \frac{e^x}{1+e^{2x}}$

Now, we need to find the value of the derivative at $x=0$. Substitute $x=0$ into the derivative expression:

$\left(\frac{dy}{dx}\right)_{x=0} = \frac{e^0}{1+e^{2(0)}}$

We know that $e^0 = 1$ and $e^{2(0)} = e^0 = 1$.

$\left(\frac{dy}{dx}\right)_{x=0} = \frac{1}{1+1}$

$\left(\frac{dy}{dx}\right)_{x=0} = \frac{1}{2}$

The value of the derivative of $\tan^{-1}(e^x)$ w.r.t. $x$ at the point $x=0$ is $\frac{1}{2}$.

Given:

The Cartesian equations of a line are $\frac{x\ -\ 3}{2}=\frac{y\ +\ 2}{-5}=\frac{z\ -\ 6}{3}$.

To Find:

The vector equation of the line.

Solution:

The general form of the Cartesian equation of a line passing through the point $(x_1, y_1, z_1)$ with direction ratios $a, b, c$ is:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

Comparing the given equation with the general form, we can identify the point through which the line passes and its direction ratios.

The given equation is $\frac{x\ -\ 3}{2}=\frac{y\ -\ (-2)}{-5}=\frac{z\ -\ 6}{3}$.

The line passes through the point $(x_1, y_1, z_1) = (3, -2, 6)$.

The direction ratios of the line are $(a, b, c) = (2, -5, 3)$.

The position vector of the point $(3, -2, 6)$ is $\vec{a} = 3\hat{i} - 2\hat{j} + 6\hat{k}$.

The direction vector parallel to the line with direction ratios $(2, -5, 3)$ is $\vec{b} = 2\hat{i} - 5\hat{j} + 3\hat{k}$.

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

$\vec{r} = \vec{a} + \lambda \vec{b}$

where $\lambda$ is a scalar parameter.

Substituting the values of $\vec{a}$ and $\vec{b}$, the vector equation of the given line is:

$\vec{r} = (3\hat{i} - 2\hat{j} + 6\hat{k}) + \lambda (2\hat{i} - 5\hat{j} + 3\hat{k})$

Given:

The integral to evaluate is $\int\limits_{-\pi}^{\pi}\left({\sin}^{83}x+x^{123}\right)dx$.

To Evaluate:

The definite integral.

Solution:

We can split the integral into two parts based on the sum property of integrals:

$\int\limits_{-\pi}^{\pi}\left({\sin}^{83}x+x^{123}\right)dx = \int\limits_{-\pi}^{\pi}{\sin}^{83}x \, dx + \int\limits_{-\pi}^{\pi}x^{123} \, dx$

Let's consider the first integral $\int\limits_{-\pi}^{\pi}{\sin}^{83}x \, dx$.

Let $f(x) = \sin^{83}x$. We need to determine if $f(x)$ is an even or odd function.

$f(-x) = \sin^{83}(-x) = (\sin(-x))^{83}$

Since $\sin(-x) = -\sin x$, we have:

$f(-x) = (-\sin x)^{83}$

Since the power 83 is odd, $(-\sin x)^{83} = -\sin^{83}x$.

So, $f(-x) = -\sin^{83}x = -f(x)$.

This means that $f(x) = \sin^{83}x$ is an odd function.

For a definite integral of an odd function over a symmetric interval $[-a, a]$, the value of the integral is 0.

$\int\limits_{-a}^{a} f(x) \, dx = 0$, if $f(x)$ is an odd function.

In this case, $a = \pi$, and $f(x) = \sin^{83}x$ is odd.

Therefore, $\int\limits_{-\pi}^{\pi}{\sin}^{83}x \, dx = 0$.

Now, let's consider the second integral $\int\limits_{-\pi}^{\pi}x^{123} \, dx$.

Let $g(x) = x^{123}$. We need to determine if $g(x)$ is an even or odd function.

$g(-x) = (-x)^{123}$

Since the power 123 is odd, $(-x)^{123} = -x^{123}$.

So, $g(-x) = -x^{123} = -g(x)$.

This means that $g(x) = x^{123}$ is an odd function.

Similar to the first integral, for a definite integral of an odd function over a symmetric interval $[-a, a]$, the value is 0.

In this case, $a = \pi$, and $g(x) = x^{123}$ is odd.

Therefore, $\int\limits_{-\pi}^{\pi}x^{123} \, dx = 0$.

Now, we sum the results of the two integrals:

$\int\limits_{-\pi}^{\pi}\left({\sin}^{83}x+x^{123}\right)dx = \int\limits_{-\pi}^{\pi}{\sin}^{83}x \, dx + \int\limits_{-\pi}^{\pi}x^{123} \, dx = 0 + 0 = 0$.

The value of the integral is 0.

Given:

The integral to evaluate is $\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \ dx$.

To Evaluate:

The indefinite integral.

Solution:

We can simplify the expression inside the square root in the denominator.

We know that $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$.

So, $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$.

Now, the denominator becomes $\sqrt{1 + \sin 2x} = \sqrt{(\sin x + \cos x)^2}$.

The square root of a square is the absolute value, i.e., $\sqrt{a^2} = |a|$.

So, $\sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|$.

The integral becomes $\int \frac{\sin x + \cos x}{|\sin x + \cos x|} \ dx$.

The value of $\frac{\sin x + \cos x}{|\sin x + \cos x|}$ depends on the sign of $\sin x + \cos x$.

If $\sin x + \cos x > 0$, then $|\sin x + \cos x| = \sin x + \cos x$, and the integrand is $\frac{\sin x + \cos x}{\sin x + \cos x} = 1$.

If $\sin x + \cos x < 0$, then $|\sin x + \cos x| = -(\sin x + \cos x)$, and the integrand is $\frac{\sin x + \cos x}{-(\sin x + \cos x)} = -1$.

Assuming the context implies the interval where $\sin x + \cos x > 0$, the integral simplifies to:

$\int 1 \ dx$

Evaluating the integral:

$\int 1 \ dx = x + C$

where $C$ is the constant of integration.

If the context implied the interval where $\sin x + \cos x < 0$, the integral would be $\int -1 \ dx = -x + C$. However, without a specified interval, the most common simplification for such problems often assumes the form where the numerator matches the simplified denominator.

Thus, the value of the integral is $x + C$.

Given:

Vector $\vec{a}=\hat{i}+3\hat{j}+\hat{k}$

Vector $\vec{b}=3\hat{i}-2\hat{j}+\lambda\hat{k}$

The vectors $\vec{a}$ and $\vec{b}$ are perpendicular to each other.

To Find:

The value of $\lambda$.

Solution:

Two vectors $\vec{a}$ and $\vec{b}$ are perpendicular if and only if their dot product $\vec{a} \cdot \vec{b}$ is equal to zero.

$\vec{a} \cdot \vec{b} = 0$

Calculate the dot product of $\vec{a}$ and $\vec{b}$:

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

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

$\vec{a} \cdot \vec{b} = 3 - 6 + \lambda$

$\vec{a} \cdot \vec{b} = -3 + \lambda$

Set the dot product equal to zero:

$-3 + \lambda = 0$

Solve for $\lambda$:

$\lambda = 3$

Thus, the value of $\lambda$ for which the vectors are perpendicular is 3.

The answer is 3.

Given:

Vector $\vec{a}=\hat{i}+3\hat{j}+\hat{k}$

Vector $\vec{b}=2\hat{i}-3\hat{j}+6\hat{k}$

To Find:

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

Solution:

The projection of vector $\vec{a}$ along vector $\vec{b}$ is a scalar value given by the formula:

Projection of $\vec{a}$ along $\vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$

First, calculate the dot product $\vec{a} \cdot \vec{b}$:

$\vec{a} \cdot \vec{b} = (\hat{i}+3\hat{j}+\hat{k}) \cdot (2\hat{i}-3\hat{j}+6\hat{k})$

$\vec{a} \cdot \vec{b} = (1)(2) + (3)(-3) + (1)(6)$

$\vec{a} \cdot \vec{b} = 2 - 9 + 6$

$\vec{a} \cdot \vec{b} = 8 - 9$

$\vec{a} \cdot \vec{b} = -1$

Next, calculate the magnitude of vector $\vec{b}$, $|\vec{b}|$:

$|\vec{b}| = \sqrt{(2)^2 + (-3)^2 + (6)^2}$

$|\vec{b}| = \sqrt{4 + 9 + 36}$

$|\vec{b}| = \sqrt{49}$

$|\vec{b}| = 7$

Now, calculate the projection using the formula:

Projection of $\vec{a}$ along $\vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{-1}{7}$

The projection of $\vec{a}$ along $\vec{b}$ is $-\frac{1}{7}$.

The answer is $-\frac{1}{7}$.

Given:

Expression to simplify: $\cot^{-1} \left\{\frac{\sqrt{1\ +\ sin\ x}\ +\ \sqrt{1\ -\ sin\ x}}{\sqrt{1\ +\ sin\ x}\ -\ \sqrt{1\ -\ sin\ x}}\right\}$

Domain: $0 < x < \frac{\pi}{2}$

To Prove:

$\cot^{-1} \left\{\frac{\sqrt{1\ +\ sin\ x}\ +\ \sqrt{1\ -\ sin\ x}}{\sqrt{1\ +\ sin\ x}\ -\ \sqrt{1\ -\ sin\ x}}\right\}=\frac{x}{2}$

Proof:

Consider the expression inside the $\cot^{-1}$ function. Let the expression be $E$.

$E = \frac{\sqrt{1\ +\ sin\ x}\ +\ \sqrt{1\ -\ sin\ x}}{\sqrt{1\ +\ sin\ x}\ -\ \sqrt{1\ -\ sin\ x}}$

We use the half-angle identities for $1 \pm \sin x$. Recall that $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}$.

So, $1 + \sin x = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\sin \frac{x}{2} + \cos \frac{x}{2})^2$.

And $1 - \sin x = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\sin \frac{x}{2} - \cos \frac{x}{2})^2$.

For the given domain $0 < x < \frac{\pi}{2}$, we have $0 < \frac{x}{2} < \frac{\pi}{4}$.

In this interval, $\cos \frac{x}{2} > \sin \frac{x}{2} > 0$.

Therefore, $\sqrt{1 + \sin x} = \sqrt{(\sin \frac{x}{2} + \cos \frac{x}{2})^2} = |\sin \frac{x}{2} + \cos \frac{x}{2}| = \sin \frac{x}{2} + \cos \frac{x}{2}$ (since $\sin \frac{x}{2} + \cos \frac{x}{2} > 0$ in this interval).

And $\sqrt{1 - \sin x} = \sqrt{(\sin \frac{x}{2} - \cos \frac{x}{2})^2} = |\sin \frac{x}{2} - \cos \frac{x}{2}| = \cos \frac{x}{2} - \sin \frac{x}{2}$ (since $\cos \frac{x}{2} - \sin \frac{x}{2} > 0$ in this interval).

Substitute these simplified terms into the expression $E$:

$E = \frac{(\sin \frac{x}{2} + \cos \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\sin \frac{x}{2} + \cos \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}$

Simplify the numerator and the denominator:

Numerator $= \sin \frac{x}{2} + \cos \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2} = 2 \cos \frac{x}{2}$

Denominator $= \sin \frac{x}{2} + \cos \frac{x}{2} - \cos \frac{x}{2} + \sin \frac{x}{2} = 2 \sin \frac{x}{2}$

So, $E = \frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} = \cot \frac{x}{2}$.

Now, the left-hand side of the identity is $\cot^{-1}(E) = \cot^{-1}(\cot \frac{x}{2})$.

We know that $\cot^{-1}(\cot \theta) = \theta$ if $0 < \theta < \pi$.

For the given domain $0 < x < \frac{\pi}{2}$, we have $0 < \frac{x}{2} < \frac{\pi}{4}$.

Since $0 < \frac{x}{2} < \frac{\pi}{4}$, this value is within the interval $(0, \pi)$.

Therefore, $\cot^{-1}(\cot \frac{x}{2}) = \frac{x}{2}$.

Thus, $\cot^{-1} \left\{\frac{\sqrt{1\ +\ sin\ x}\ +\ \sqrt{1\ -\ sin\ x}}{\sqrt{1\ +\ sin\ x}\ -\ \sqrt{1\ -\ sin\ x}}\right\}=\frac{x}{2}$.

Hence Proved.

Or

Given:

The equation $\sin^{–1} x + \sin^{–1} 2x = \frac{\pi}{3}$, with $x > 0$.

To Solve:

Find the value of $x$.

Solution:

Let $\sin^{-1} x = \theta$. Since $x > 0$, $\theta$ is in the interval $(0, \frac{\pi}{2}]$.

The given equation becomes $\theta + \sin^{-1}(2x) = \frac{\pi}{3}$.

So, $\sin^{-1}(2x) = \frac{\pi}{3} - \theta$.

For $\sin^{-1}(2x)$ to be defined, we must have $|2x| \le 1$, which means $|x| \le \frac{1}{2}$. Since $x > 0$, we have $0 < x \le \frac{1}{2}$.

Also, $x = \sin \theta$. So $0 < \sin \theta \le \frac{1}{2}$.

Since $\theta \in (0, \frac{\pi}{2}]$, this implies $0 < \theta \le \frac{\pi}{6}$.

For the equation $\sin^{-1}(2x) = \frac{\pi}{3} - \theta$ to hold, the value $\frac{\pi}{3} - \theta$ must be in the principal value range of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

$-\frac{\pi}{2} \le \frac{\pi}{3} - \theta \le \frac{\pi}{2}$

Subtracting $\frac{\pi}{3}$ from all parts:

$-\frac{\pi}{2} - \frac{\pi}{3} \le -\theta \le \frac{\pi}{2} - \frac{\pi}{3}$

$-\frac{3\pi + 2\pi}{6} \le -\theta \le \frac{3\pi - 2\pi}{6}$

$-\frac{5\pi}{6} \le -\theta \le \frac{\pi}{6}$

Multiplying by -1 and reversing the inequalities:

$-\frac{\pi}{6} \le \theta \le \frac{5\pi}{6}$.

Combining the conditions on $\theta$: $\theta \in (0, \frac{\pi}{6}] \cap [-\frac{\pi}{6}, \frac{5\pi}{6}] = (0, \frac{\pi}{6}]$.

Now, take the sine of both sides of $\sin^{-1}(2x) = \frac{\pi}{3} - \theta$. Since $\frac{\pi}{3} - \theta$ is in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, $\sin(\sin^{-1}(2x)) = 2x$ is valid.

$2x = \sin(\frac{\pi}{3} - \theta)$

Substitute $x = \sin \theta$:

$2\sin \theta = \sin(\frac{\pi}{3} - \theta)$

Using the sine subtraction formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$:

$2\sin \theta = \sin \frac{\pi}{3} \cos \theta - \cos \frac{\pi}{3} \sin \theta$

$2\sin \theta = \frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta$

Multiply the equation by 2:

$4\sin \theta = \sqrt{3} \cos \theta - \sin \theta$

Rearrange the terms:

$4\sin \theta + \sin \theta = \sqrt{3} \cos \theta$

$5\sin \theta = \sqrt{3} \cos \theta$

Since $\theta \in (0, \frac{\pi}{6}]$, $\cos \theta > 0$. We can divide by $\cos \theta$:

$\frac{\sin \theta}{\cos \theta} = \frac{\sqrt{3}}{5}$

$\tan \theta = \frac{\sqrt{3}}{5}$

We need to find $x = \sin \theta$. Since $\theta \in (0, \frac{\pi}{6}]$, $\theta$ is in the first quadrant. We can construct a right triangle with opposite side $\sqrt{3}$ and adjacent side 5.

The hypotenuse is $\sqrt{(\sqrt{3})^2 + 5^2} = \sqrt{3 + 25} = \sqrt{28} = 2\sqrt{7}$.

So, $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\sqrt{3}}{2\sqrt{7}}$.

Therefore, $x = \sin \theta = \frac{\sqrt{3}}{2\sqrt{7}}$.

To rationalize the denominator:

$x = \frac{\sqrt{3}}{2\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{3 \times 7}}{2 \times 7} = \frac{\sqrt{21}}{14}$.

We must check if this value of $x$ satisfies the condition $0 < x \le \frac{1}{2}$.

The value $x = \frac{\sqrt{21}}{14}$ is positive since $\sqrt{21} > 0$.

To check $x \le \frac{1}{2}$, we can check $x^2 \le (\frac{1}{2})^2 = \frac{1}{4}$.

$x^2 = \left(\frac{\sqrt{21}}{14}\right)^2 = \frac{21}{196}$.

Is $\frac{21}{196} \le \frac{1}{4}$?

$21 \times 4 \le 1 \times 196$

$84 \le 196$. This inequality is true.

Thus, the value $x = \frac{\sqrt{21}}{14}$ satisfies the condition $0 < x \le \frac{1}{2}$.

The value of $x$ is $\frac{\sqrt{21}}{14}$.

Given:

The identity to prove is $\left|\begin{matrix}b+c&c+a&a+b\\q+r&r+p&p+q\\y+z&z+x&x+y\\\end{matrix}\right|=2\left|\begin{matrix}a&b&c\\p&q&r\\x&y&z\\\end{matrix}\right|$.

To Prove:

The given identity using properties of determinants.

Proof:

Let LHS denote the left-hand side of the identity.

LHS $= \left|\begin{matrix}b+c&c+a&a+b\\q+r&r+p&p+q\\y+z&z+x&x+y\\\end{matrix}\right|$

Apply the column operation $C_1 \to C_1 + C_2 + C_3$:

LHS $= \left|\begin{matrix}(b+c)+(c+a)+(a+b)&c+a&a+b\\(q+r)+(r+p)+(p+q)&r+p&p+q\\(y+z)+(z+x)+(x+y)&z+x&x+y\\\end{matrix}\right|$

LHS $= \left|\begin{matrix}2a+2b+2c&c+a&a+b\\2p+2q+2r&r+p&p+q\\2x+2y+2z&z+x&x+y\\\end{matrix}\right|$

Take out the common factor 2 from the first column ($C_1$):

LHS $= 2 \left|\begin{matrix}a+b+c&c+a&a+b\\p+q+r&r+p&p+q\\x+y+z&z+x&x+y\\\end{matrix}\right|$

Apply the column operation $C_2 \to C_2 - C_1$:

LHS $= 2 \left|\begin{matrix}(c+a)-(a+b+c)&c+a&a+b\\(r+p)-(p+q+r)&r+p&p+q\\(z+x)-(x+y+z)&z+x&x+y\\\end{matrix}\right|$

LHS $= 2 \left|\begin{matrix}-b&c+a&a+b\\-q&r+p&p+q\\-y&z+x&x+y\\\end{matrix}\right|$

Apply the column operation $C_3 \to C_3 - C_1$:

LHS $= 2 \left|\begin{matrix}-b&(c+a)-(a+b+c)&a+b\\-q&(r+p)-(p+q+r)&p+q\\-y&(z+x)-(x+y+z)&x+y\\\end{matrix}\right|$

LHS $= 2 \left|\begin{matrix}-b&-b&a+b\\-q&-q&p+q\\-y&-y&x+y\\\end{matrix}\right|$

Apply the column operation $C_1 \to C_1 + C_2$:

LHS $= 2 \left|\begin{matrix}(-b)+(-b)&-b&a+b\\(-q)+(-q)&-q&p+q\\(-y)+(-y)&-y&x+y\\\end{matrix}\right|$

LHS $= 2 \left|\begin{matrix}-2b&-b&a+b\\-2q&-q&p+q\\-2y&-y&x+y\\\end{matrix}\right|$

Take out the common factor -2 from the first column ($C_1$):

LHS $= 2 \times (-2) \left|\begin{matrix}b&-b&a+b\\q&-q&p+q\\y&-y&x+y\\\end{matrix}\right|$

LHS $= -4 \left|\begin{matrix}b&-b&a+b\\q&-q&p+q\\y&-y&x+y\\\end{matrix}\right|$

Apply the column operation $C_2 \to C_2 + C_1$:

LHS $= -4 \left|\begin{matrix}b&(-b)+b&a+b\\q&(-q)+q&p+q\\y&(-y)+y&x+y\\\end{matrix}\right|$

LHS $= -4 \left|\begin{matrix}b&0&a+b\\q&0&p+q\\y&0&x+y\\\end{matrix}\right|$

The determinant has a column of zeros ($C_2$). The value of such a determinant is 0.

LHS $= -4 \times 0 = 0$.

This does not match the RHS $2\left|\begin{matrix}a&b&c\\p&q&r\\x&y&z\\\end{matrix}\right|$ which is generally non-zero. My column operations were incorrect or in a bad order.

Let's restart with the C1->C1+C2+C3 step where I got the factor 2 correctly.

Restarting Proof:

LHS $= \left|\begin{matrix}b+c&c+a&a+b\\q+r&r+p&p+q\\y+z&z+x&x+y\\\end{matrix}\right|$

Apply $C_1 \to C_1 + C_2 + C_3$:

LHS $= \left|\begin{matrix}2(a+b+c)&c+a&a+b\\2(p+q+r)&r+p&p+q\\2(x+y+z)&z+x&x+y\\\end{matrix}\right|$

Take 2 common from $C_1$:

LHS $= 2 \left|\begin{matrix}a+b+c&c+a&a+b\\p+q+r&r+p&p+q\\x+y+z&z+x&x+y\\\end{matrix}\right|$

Apply $C_2 \to C_2 - C_1$:

LHS $= 2 \left|\begin{matrix}-b&c+a&a+b\\-q&r+p&p+q\\-y&z+x&x+y\\\end{matrix}\right|$

The elements in the first column should be $a+b+c$, $p+q+r$, $x+y+z$ after taking 2 out. So $C_2 \to C_2 - C_1$ should use the current $C_1$.

LHS $= 2 \left|\begin{matrix}a+b+c&(c+a)-(a+b+c)&a+b\\p+q+r&(r+p)-(p+q+r)&p+q\\x+y+z&(z+x)-(x+y+z)&x+y\\\end{matrix}\right|$

LHS $= 2 \left|\begin{matrix}a+b+c&-b&a+b\\p+q+r&-q&p+q\\x+y+z&-y&x+y\\\end{matrix}\right|$

Apply $C_3 \to C_3 - C_1$:

LHS $= 2 \left|\begin{matrix}a+b+c&-b&(a+b)-(a+b+c)\\p+q+r&-q&(p+q)-(p+q+r)\\x+y+z&-y&(x+y)-(x+y+z)\\\end{matrix}\right|$

LHS $= 2 \left|\begin{matrix}a+b+c&-b&-c\\p+q+r&-q&-r\\x+y+z&-y&-z\\\end{matrix}\right|$

Apply $C_1 \to C_1 + C_2 + C_3$:

LHS $= 2 \left|\begin{matrix}(a+b+c)+(-b)+(-c)&-b&-c\\(p+q+r)+(-q)+(-r)&-q&-r\\(x+y+z)+(-y)+(-z)&-y&-z\\\end{matrix}\right|$

LHS $= 2 \left|\begin{matrix}a&-b&-c\\p&-q&-r\\x&-y&-z\\\end{matrix}\right|$

Take out the common factor -1 from the second column ($C_2$) and the third column ($C_3$). This introduces a factor of $(-1) \times (-1) = 1$.

LHS $= 2 \times (-1) \times (-1) \left|\begin{matrix}a&b&c\\p&q&r\\x&y&z\\\end{matrix}\right|$

LHS $= 2 \left|\begin{matrix}a&b&c\\p&q&r\\x&y&z\\\end{matrix}\right|$

This is equal to the RHS of the given identity.

LHS = RHS

Hence Proved.

Given:

The function $f(x) = |x + 1| + |x + 2|$.

To Discuss:

The continuity of $f(x)$ at $x = -1$ and $x = -2$.

Solution:

The function involves absolute values. We can rewrite the function in a piecewise form by considering the critical points where the expressions inside the absolute value become zero, which are $x = -1$ and $x = -2$.

We analyze the function in the intervals $(-\infty, -2)$, $[-2, -1)$, and $[-1, \infty)$.

If $x < -2$, then $x+1 < 0$ and $x+2 < 0$. So $|x+1| = -(x+1)$ and $|x+2| = -(x+2)$.

$f(x) = -(x+1) - (x+2) = -x - 1 - x - 2 = -2x - 3$ for $x < -2$.

If $-2 \le x < -1$, then $x+1 < 0$ and $x+2 \ge 0$. So $|x+1| = -(x+1)$ and $|x+2| = x+2$.

$f(x) = -(x+1) + (x+2) = -x - 1 + x + 2 = 1$ for $-2 \le x < -1$.

If $x \ge -1$, then $x+1 \ge 0$ and $x+2 \ge 0$. So $|x+1| = x+1$ and $|x+2| = x+2$.

$f(x) = (x+1) + (x+2) = 2x + 3$ for $x \ge -1$.

Thus, the function can be written as:

$f(x) = \begin{cases} -2x - 3 & , & x < -2 \\ 1 & , & -2 \le x < -1 \\ 2x + 3 & , & x \ge -1 \end{cases}$

Continuity at $x = -1$:

For the function to be continuous at $x = -1$, we need to check if $\lim\limits_{x \to -1} f(x) = f(-1)$.

1. Find $f(-1)$: Using the definition for $x \ge -1$,

$f(-1) = 2(-1) + 3 = -2 + 3 = 1$.

2. Find the limits as $x \to -1$:

Left-hand limit (LHL): $\lim\limits_{x \to -1^-} f(x)$. For $x < -1$ but close to -1 (specifically $-2 \le x < -1$), $f(x) = 1$.

$\lim\limits_{x \to -1^-} f(x) = \lim\limits_{x \to -1^-} 1 = 1$.

Right-hand limit (RHL): $\lim\limits_{x \to -1^+} f(x)$. For $x \ge -1$ but close to -1, $f(x) = 2x + 3$.

$\lim\limits_{x \to -1^+} f(x) = \lim\limits_{x \to -1^+} (2x + 3) = 2(-1) + 3 = -2 + 3 = 1$.

Since the LHL and RHL are equal, the limit exists:

$\lim\limits_{x \to -1} f(x) = 1$.

3. Compare the limit and the function value:

$\lim\limits_{x \to -1} f(x) = 1$ and $f(-1) = 1$.

Since $\lim\limits_{x \to -1} f(x) = f(-1)$, the function $f(x)$ is continuous at $x = -1$.

Continuity at $x = -2$:

For the function to be continuous at $x = -2$, we need to check if $\lim\limits_{x \to -2} f(x) = f(-2)$.

1. Find $f(-2)$: Using the definition for $-2 \le x < -1$,

$f(-2) = 1$.

2. Find the limits as $x \to -2$:

Left-hand limit (LHL): $\lim\limits_{x \to -2^-} f(x)$. For $x < -2$, $f(x) = -2x - 3$.

$\lim\limits_{x \to -2^-} f(x) = \lim\limits_{x \to -2^-} (-2x - 3) = -2(-2) - 3 = 4 - 3 = 1$.

Right-hand limit (RHL): $\lim\limits_{x \to -2^+} f(x)$. For $x > -2$ but close to -2 (specifically $-2 \le x < -1$), $f(x) = 1$.

$\lim\limits_{x \to -2^+} f(x) = \lim\limits_{x \to -2^+} 1 = 1$.

Since the LHL and RHL are equal, the limit exists:

$\lim\limits_{x \to -2} f(x) = 1$.

3. Compare the limit and the function value:

$\lim\limits_{x \to -2} f(x) = 1$ and $f(-2) = 1$.

Since $\lim\limits_{x \to -2} f(x) = f(-2)$, the function $f(x)$ is continuous at $x = -2$.

Conclusion:

The function $f(x) = |x + 1| + |x + 2|$ is continuous at both $x = -1$ and $x = -2$.

Note that the function is the sum of two absolute value functions, which are known to be continuous everywhere. The sum of continuous functions is also continuous everywhere. Hence, $f(x)$ is continuous for all real numbers $x$, including at $x=-1$ and $x=-2$.

Given:

Parametric equations:

$x = 2 \cos \theta – \cos 2\theta$

$y = 2 \sin \theta – \sin 2\theta$

To Find:

The value of $\frac{d^2y}{{dx}^2}$ at $\theta=\frac{\pi}{2}$.

Solution:

First, find $\frac{dx}{d\theta}$ by differentiating $x$ with respect to $\theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta}(2 \cos \theta – \cos 2\theta)$

$\frac{dx}{d\theta} = 2(-\sin \theta) - (-\sin 2\theta)(2)$

$\frac{dx}{d\theta} = -2\sin \theta + 2\sin 2\theta$

Next, find $\frac{dy}{d\theta}$ by differentiating $y$ with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta}(2 \sin \theta – \sin 2\theta)$

$\frac{dy}{d\theta} = 2(\cos \theta) - (\cos 2\theta)(2)$

$\frac{dy}{d\theta} = 2\cos \theta - 2\cos 2\theta$

Now, find $\frac{dy}{dx}$ using the formula $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$:

$\frac{dy}{dx} = \frac{2\cos \theta - 2\cos 2\theta}{-2\sin \theta + 2\sin 2\theta}$

$\frac{dy}{dx} = \frac{2(\cos \theta - \cos 2\theta)}{2(\sin 2\theta - \sin \theta)}$

$\frac{dy}{dx} = \frac{\cos \theta - \cos 2\theta}{\sin 2\theta - \sin \theta}$

To find $\frac{d^2y}{dx^2}$, we use the formula $\frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta}\left(\frac{dy}{dx}\right)}{\frac{dx}{d\theta}}$.

Let $Y = \frac{dy}{dx} = \frac{\cos \theta - \cos 2\theta}{\sin 2\theta - \sin \theta}$. We need to find $\frac{dY}{d\theta}$ using the quotient rule.

Let $u = \cos \theta - \cos 2\theta$, so $\frac{du}{d\theta} = -\sin \theta - (-\sin 2\theta)(2) = -\sin \theta + 2\sin 2\theta$.

Let $v = \sin 2\theta - \sin \theta$, so $\frac{dv}{d\theta} = (\cos 2\theta)(2) - \cos \theta = 2\cos 2\theta - \cos \theta$.

$\frac{dY}{d\theta} = \frac{v \frac{du}{d\theta} - u \frac{dv}{d\theta}}{v^2}$

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

Now, evaluate $\frac{d^2y}{dx^2} = \frac{dY/d\theta}{dx/d\theta}$ at $\theta=\frac{\pi}{2}$.

First, evaluate the necessary trigonometric functions at $\theta = \frac{\pi}{2}$:

$\sin \frac{\pi}{2} = 1$

$\cos \frac{\pi}{2} = 0$

$\sin (2 \cdot \frac{\pi}{2}) = \sin \pi = 0$

$\cos (2 \cdot \frac{\pi}{2}) = \cos \pi = -1$

Evaluate $\frac{dx}{d\theta}$ at $\theta=\frac{\pi}{2}$:

$\left(\frac{dx}{d\theta}\right)_{\theta=\pi/2} = -2\sin \frac{\pi}{2} + 2\sin \pi = -2(1) + 2(0) = -2$

Evaluate the numerator of $\frac{dY}{d\theta}$ at $\theta=\frac{\pi}{2}$:

$(\sin \pi - \sin \frac{\pi}{2})(-\sin \frac{\pi}{2} + 2\sin \pi) - (\cos \frac{\pi}{2} - \cos \pi)(2\cos \pi - \cos \frac{\pi}{2})$

$= (0 - 1)(-1 + 2(0)) - (0 - (-1))(2(-1) - 0)$

$= (-1)(-1) - (1)(-2)$

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

Evaluate the denominator of $\frac{dY}{d\theta}$ at $\theta=\frac{\pi}{2}$:

$(\sin \pi - \sin \frac{\pi}{2})^2 = (0 - 1)^2 = (-1)^2 = 1$

So, $\left(\frac{dY}{d\theta}\right)_{\theta=\pi/2} = \frac{3}{1} = 3$.

Finally, evaluate $\frac{d^2y}{dx^2}$ at $\theta=\frac{\pi}{2}$:

$\left(\frac{d^2y}{dx^2}\right)_{\theta=\pi/2} = \frac{\left(\frac{dY}{d\theta}\right)_{\theta=\pi/2}}{\left(\frac{dx}{d\theta}\right)_{\theta=\pi/2}} = \frac{3}{-2} = -\frac{3}{2}$

The value of $\frac{d^2y}{{dx}^2}$ at $\theta=\frac{\pi}{2}$ is $-\frac{3}{2}$.

or

Given:

The equation $x\sqrt{1\ +\ y}+y\sqrt{1+x}=0$, where $-1 < x < 1$.

To Prove:

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

Proof:

The given equation is $x\sqrt{1\ +\ y}+y\sqrt{1+x}=0$.

Rearrange the terms to isolate one of the square root terms:

$x\sqrt{1+y} = -y\sqrt{1+x}$

Square both sides of the equation to eliminate the square roots:

$(x\sqrt{1+y})^2 = (-y\sqrt{1+x})^2$

$x^2(1+y) = y^2(1+x)$

$x^2 + x^2y = y^2 + y^2x$

Rearrange the terms to group $x^2$ and $y^2$ and terms with $xy$:

$x^2 - y^2 = y^2x - x^2y$

Factor both sides:

$(x-y)(x+y) = xy(y-x)$

$(x-y)(x+y) = -xy(x-y)$

Move all terms to one side:

$(x-y)(x+y) + xy(x-y) = 0$

Factor out the common term $(x-y)$:

$(x-y)(x+y+xy) = 0$

This equation holds if either $x-y = 0$ or $x+y+xy = 0$.

Case 1: $x-y = 0 \implies y = x$.

Substitute $y=x$ into the original equation: $x\sqrt{1+x} + x\sqrt{1+x} = 0 \implies 2x\sqrt{1+x} = 0$.

This implies $x=0$ or $\sqrt{1+x}=0 \implies x=-1$.

Given the domain $-1 < x < 1$, the only point where $x=y$ satisfies the original equation is $(0, 0)$.

Case 2: $x+y+xy = 0$.

Rearrange this equation to solve for $y$ in terms of $x$:

$y + xy = -x$

$y(1+x) = -x$

Since $-1 < x < 1$, $1+x \ne 0$. We can divide by $1+x$:

$y = \frac{-x}{1+x}$

We need to confirm that all points satisfying $y = \frac{-x}{1+x}$ also satisfy the original equation $x\sqrt{1\ +\ y}+y\sqrt{1+x}=0$ for $-1 < x < 1$. We have already verified this in the thought process; the substitution yields 0.

The relation $y = \frac{-x}{1+x}$ represents the curve defined by the original equation for $-1 < x < 1$.

Now, we differentiate $y = \frac{-x}{1+x}$ with respect to $x$ using the quotient rule.

Let $u = -x$ and $v = 1+x$. Then $\frac{du}{dx} = -1$ and $\frac{dv}{dx} = 1$.

$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

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

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

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

Thus, we have proved that $\frac{dy}{dx}=\frac{-1}{\left(1+x\right)^2}$.

Hence Proved.

Given:

Diameter of cone = 10 cm, so radius $R = \frac{10}{2} = 5$ cm.

Depth of cone = 10 cm, so height $H = 10$ cm.

Rate of water poured into the cone $\frac{dV}{dt} = 4 \text{ cm}^3/\text{min}$.

The current depth of water is $h = 6$ cm.

To Find:

The rate at which the water level is rising ($\frac{dh}{dt}$) when $h = 6$ cm.

Solution:

Let $r$ be the radius and $h$ be the height of the water cone at any instant $t$. The volume of water in the cone is given by $V = \frac{1}{3}\pi r^2 h$.

The full cone and the water cone are similar. By similar triangles, the ratio of the radius to the height is constant:

$\frac{r}{h} = \frac{R}{H}$

From (i), we get the relationship between $r$ and $h$ for the water cone:

$r = \frac{5}{10} h = \frac{1}{2} h$

Substitute this relationship into the volume formula to express $V$ solely in terms of $h$:

$V = \frac{1}{3}\pi \left(\frac{1}{2}h\right)^2 h$

$V = \frac{1}{3}\pi \left(\frac{h^2}{4}\right) h$

Now, differentiate both sides of equation (ii) with respect to time $t$ to relate the rates of change:

$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{\pi}{12} h^3\right)$

Using the chain rule, $\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt}$

We are given $\frac{dV}{dt} = 4 \text{ cm}^3/\text{min}$ and we want to find $\frac{dh}{dt}$ when $h = 6$ cm.

Substitute these values into equation (iii):

$4 = \frac{\pi}{4} (6)^2 \frac{dh}{dt}$

$4 = \frac{\pi}{4} (36) \frac{dh}{dt}$

$4 = 9\pi \frac{dh}{dt}$

Solve for $\frac{dh}{dt}$:

$\frac{dh}{dt} = \frac{4}{9\pi}$

The rate at which the water level is rising at the instant when the depth is 6 cm is $\frac{4}{9\pi}$ cm/min.

OR

Given:

The function $f(x) = x^3 + \frac{1}{x^3}$, $x \neq 0$.

To Find:

The intervals in which $f(x)$ is (i) increasing and (ii) decreasing.

Solution:

To determine the intervals of increasing and decreasing, we need to find the first derivative of the function, $f'(x)$, and analyze its sign.

$f(x) = x^3 + x^{-3}$

Differentiate $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(x^{-3})$

$f'(x) = 3x^{3-1} + (-3)x^{-3-1}$

$f'(x) = 3x^2 - 3x^{-4}$

$f'(x) = 3x^2 - \frac{3}{x^4}$

To find the critical points, we set $f'(x) = 0$ or find where $f'(x)$ is undefined. The function is undefined at $x=0$, but the domain is $x \neq 0$.

Set $f'(x) = 0$:

$3x^2 - \frac{3}{x^4} = 0$

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

Divide by 3 (since $3 \neq 0$):

$x^2 = \frac{1}{x^4}$

Multiply by $x^4$ (since $x \neq 0$):

$x^2 \cdot x^4 = 1$

$x^6 = 1$

Taking the sixth root, we get $x = \pm 1$.

The critical points are $x = -1$ and $x = 1$. These points, along with the point $x=0$ (where the function is undefined), divide the number line into the intervals $(-\infty, -1)$, $(-1, 0)$, $(0, 1)$, and $(1, \infty)$.

We test the sign of $f'(x) = 3x^2 - \frac{3}{x^4} = 3\left(\frac{x^6 - 1}{x^4}\right)$ in each interval.

Since $x^4 > 0$ for $x \neq 0$, the sign of $f'(x)$ is determined by the sign of $(x^6 - 1)$.

Interval $(-\infty, -1)$: Choose a test value, e.g., $x = -2$.

$x^6 - 1 = (-2)^6 - 1 = 64 - 1 = 63$.

Since $63 > 0$, $x^6 - 1 > 0$. Thus, $f'(x) > 0$.

Interval $(-1, 0)$: Choose a test value, e.g., $x = -0.5$.

$x^6 - 1 = (-0.5)^6 - 1 = \left(\frac{1}{64}\right) - 1 = -\frac{63}{64}$.

Since $-\frac{63}{64} < 0$, $x^6 - 1 < 0$. Thus, $f'(x) < 0$.

Interval $(0, 1)$: Choose a test value, e.g., $x = 0.5$.

$x^6 - 1 = (0.5)^6 - 1 = \left(\frac{1}{64}\right) - 1 = -\frac{63}{64}$.

Since $-\frac{63}{64} < 0$, $x^6 - 1 < 0$. Thus, $f'(x) < 0$.

Interval $(1, \infty)$: Choose a test value, e.g., $x = 2$.

$x^6 - 1 = (2)^6 - 1 = 64 - 1 = 63$.

Since $63 > 0$, $x^6 - 1 > 0$. Thus, $f'(x) > 0$.

The function $f(x)$ is increasing where $f'(x) > 0$ and decreasing where $f'(x) < 0$.

(i) The function is increasing on the intervals $(-\infty, -1)$ and $(1, \infty)$.

(ii) The function is decreasing on the intervals $(-1, 0)$ and $(0, 1)$.

Given:

The integral $\int{\frac{3x\ -\ 2}{\left(x+2\right)\left(x+1\right)^2}dx}$

To Evaluate:

The indefinite integral.

Solution:

We will use the method of partial fraction decomposition to evaluate the integral.

Let $\frac{3x - 2}{(x+2)(x+1)^2} = \frac{A}{x+2} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$.

Multiply both sides by the denominator $(x+2)(x+1)^2$:

$3x - 2 = A(x+1)^2 + B(x+2)(x+1) + C(x+2)$

Expand the right side:

$3x - 2 = A(x^2 + 2x + 1) + B(x^2 + 3x + 2) + C(x+2)$

$3x - 2 = Ax^2 + 2Ax + A + Bx^2 + 3Bx + 2B + Cx + 2C$

Group terms by powers of $x$:

$3x - 2 = (A+B)x^2 + (2A+3B+C)x + (A+2B+2C)$

Equate the coefficients of the corresponding powers of $x$ on both sides:

Coefficient of $x^2$: $A+B = 0$

Coefficient of $x$: $2A+3B+C = 3$

Constant term: $A+2B+2C = -2$

Alternatively, we can substitute specific values of $x$ into the equation $3x - 2 = A(x+1)^2 + B(x+2)(x+1) + C(x+2)$.

Set $x = -1$:

$3(-1) - 2 = A(-1+1)^2 + B(-1+2)(-1+1) + C(-1+2)$

$-3 - 2 = A(0)^2 + B(1)(0) + C(1)$

$-5 = C$

Set $x = -2$:

$3(-2) - 2 = A(-2+1)^2 + B(-2+2)(-2+1) + C(-2+2)$

$-6 - 2 = A(-1)^2 + B(0)(-1) + C(0)$

$-8 = A(1) + 0 + 0$

$-8 = A$

Now substitute the values of $A$ and $C$ into the equation $A+B=0$:

$-8 + B = 0$

$B = 8$

So the partial fraction decomposition is:

$\frac{3x - 2}{(x+2)(x+1)^2} = \frac{-8}{x+2} + \frac{8}{x+1} + \frac{-5}{(x+1)^2}$

Now integrate each term:

$\int \frac{3x - 2}{(x+2)(x+1)^2} dx = \int \frac{-8}{x+2} dx + \int \frac{8}{x+1} dx + \int \frac{-5}{(x+1)^2} dx$

$\int \frac{-8}{x+2} dx = -8 \int \frac{1}{x+2} dx = -8 \log|x+2|$

$\int \frac{8}{x+1} dx = 8 \int \frac{1}{x+1} dx = 8 \log|x+1|$

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

Combining the results and adding the constant of integration $C$:

$\int \frac{3x - 2}{(x+2)(x+1)^2} dx = -8 \log|x+2| + 8 \log|x+1| + \frac{5}{x+1} + C$

Using logarithm properties, we can write $8 \log|x+1| - 8 \log|x+2| = 8 (\log|x+1| - \log|x+2|) = 8 \log\left|\frac{x+1}{x+2}\right|$.

The final result is $8 \log\left|\frac{x+1}{x+2}\right| + \frac{5}{x+1} + C$.

OR

Given:

The integral $\int \left( \log(\log x) + \frac{1}{(\log x)^2} \right) \ dx$

To Evaluate:

The indefinite integral.

Solution:

Consider the derivative of the function $F(x) = x \log(\log x) - \frac{x}{\log x}$. We will evaluate the derivative using the product rule and the quotient rule.

First term: $\frac{d}{dx}(x \log(\log x))$. Using the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$ with $u=x$ and $v=\log(\log x)$.

$\frac{du}{dx} = \frac{d}{dx}(x) = 1$

$\frac{dv}{dx} = \frac{d}{dx}(\log(\log x))$. Using the chain rule, $\frac{d}{dx}(\log(\log x)) = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) = \frac{1}{\log x} \cdot \frac{1}{x}$.

So, $\frac{d}{dx}(x \log(\log x)) = x \cdot \frac{1}{x \log x} + \log(\log x) \cdot 1 = \frac{1}{\log x} + \log(\log x)$.

Second term: $\frac{d}{dx}\left(\frac{x}{\log x}\right)$. Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$ with $u=x$ and $v=\log x$.

$\frac{du}{dx} = \frac{d}{dx}(x) = 1$

$\frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$.

So, $\frac{d}{dx}\left(\frac{x}{\log x}\right) = \frac{(\log x)(1) - x(\frac{1}{x})}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2} = \frac{\log x}{(\log x)^2} - \frac{1}{(\log x)^2} = \frac{1}{\log x} - \frac{1}{(\log x)^2}$.

Now, subtract the second derivative from the first:

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

$= \log(\log x) + \frac{1}{\log x} - \frac{1}{\log x} + \frac{1}{(\log x)^2}$

$= \log(\log x) + \frac{1}{(\log x)^2}$.

We see that the integrand is the derivative of $x \log(\log x) - \frac{x}{\log x}$.

Therefore, the integral is equal to this function plus the constant of integration.

$\int \left( \log(\log x) + \frac{1}{(\log x)^2} \right) \ dx = x \log(\log x) - \frac{x}{\log x} + C$.

Given:

The integral to evaluate is $\int\limits_{0}^{\pi}{\frac{x\ sin\ x}{1\ +\ {cos}^2\ x}dx}$.

To Evaluate:

The definite integral.

Solution:

Let the given integral be $I$.

$I = \int\limits_{0}^{\pi}{\frac{x\ sin\ x}{1\ +\ {cos}^2\ x}dx}$

We use the property of definite integrals: $\int\limits_{0}^{a} f(x) dx = \int\limits_{0}^{a} f(a-x) dx$.

Here, $a = \pi$. So, replace $x$ with $\pi - x$ in the integral:

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

We know that $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$, which means $\cos^2(\pi - x) = (-\cos x)^2 = \cos^2 x$.

Substitute these into the integral:

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

Split the integrand into two parts:

$I = \int\limits_{0}^{\pi}{\left(\frac{\pi \sin x}{1\ +\ {cos}^2 x} - \frac{x \sin x}{1\ +\ {cos}^2 x}\right)dx}$

$I = \int\limits_{0}^{\pi}{\frac{\pi \sin x}{1\ +\ {cos}^2 x}dx} - \int\limits_{0}^{\pi}{\frac{x \sin x}{1\ +\ {cos}^2 x}dx}$

The second integral on the right side is the original integral $I$.

$I = \int\limits_{0}^{\pi}{\frac{\pi \sin x}{1\ +\ {cos}^2 x}dx} - I$

Add $I$ to both sides of the equation:

$2I = \int\limits_{0}^{\pi}{\frac{\pi \sin x}{1\ +\ {cos}^2 x}dx}$

$2I = \pi \int\limits_{0}^{\pi}{\frac{\sin x}{1\ +\ {cos}^2 x}dx}$

Now, let's evaluate the integral $J = \int\limits_{0}^{\pi}{\frac{\sin x}{1\ +\ {cos}^2 x}dx}$.

Let $u = \cos x$. Then, the differential $du = \frac{d}{dx}(\cos x) dx = -\sin x \ dx$. So, $\sin x \ dx = -du$.

Change the limits of integration according to the substitution:

When $x = 0$, $u = \cos(0) = 1$.

When $x = \pi$, $u = \cos(\pi) = -1$.

Substitute $u$ and $du$ into the integral $J$, and change the limits:

$J = \int\limits_{1}^{-1}{\frac{-du}{1\ +\ u^2}}$

$J = - \int\limits_{1}^{-1}{\frac{1}{1\ +\ u^2}du}$

Use the property $\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx$ to reverse the limits:

$J = \int\limits_{-1}^{1}{\frac{1}{1\ +\ u^2}du}$

The integral of $\frac{1}{1+u^2}$ is $\tan^{-1} u$.

$J = [\tan^{-1} u]_{-1}^{1}$

$J = \tan^{-1}(1) - \tan^{-1}(-1)$

Using the principal values of the inverse tangent function:

$J = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right)$

$J = \frac{\pi}{4} + \frac{\pi}{4}$

$J = \frac{2\pi}{4} = \frac{\pi}{2}$

Now substitute the value of $J$ back into the equation for $2I$:

$2I = \pi \cdot J$

$2I = \pi \cdot \frac{\pi}{2}$

$2I = \frac{\pi^2}{2}$

Solve for $I$:

$I = \frac{\pi^2}{2} \cdot \frac{1}{2}$

$I = \frac{\pi^2}{4}$

The value of the integral is $\frac{\pi^2}{4}$.

Given:

A family of circles which pass through the origin and whose centres lie on the x-axis.

To Find:

The differential equation of this family of circles.

Solution:

Let the centre of a circle in this family be $(a, 0)$, since the centre lies on the x-axis.

Since the circle passes through the origin $(0, 0)$, the distance from the centre $(a, 0)$ to the origin $(0, 0)$ is the radius of the circle.

Radius $r = \sqrt{(a-0)^2 + (0-0)^2} = \sqrt{a^2} = |a|$.

The equation of a circle with centre $(h, k)$ and radius $r$ is given by $(x-h)^2 + (y-k)^2 = r^2$.

Substituting the centre $(a, 0)$ and radius $r = |a|$, the equation of any circle in the given family is:

$(x - a)^2 + (y - 0)^2 = a^2$

$(x - a)^2 + y^2 = a^2$

Expand the equation:

$x^2 - 2ax + a^2 + y^2 = a^2$

Subtract $a^2$ from both sides:

$x^2 - 2ax + y^2 = 0$

This is the equation of the family of circles. The parameter $a$ is the arbitrary constant. Since there is one arbitrary constant, the order of the differential equation will be 1.

To find the differential equation, we differentiate the equation $x^2 - 2ax + y^2 = 0$ with respect to $x$ implicitly:

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

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

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

$2x - 2a + 2y \frac{dy}{dx} = 0$

Divide the entire equation by 2:

$x - a + y \frac{dy}{dx} = 0$

Now, we need to eliminate the arbitrary constant $a$. From the equation $x - a + y \frac{dy}{dx} = 0$, we can express $a$ in terms of $x$, $y$, and $\frac{dy}{dx}$:

$a = x + y \frac{dy}{dx}$

Substitute this expression for $a$ back into the equation of the family of circles $x^2 - 2ax + y^2 = 0$:

$x^2 - 2 \left(x + y \frac{dy}{dx}\right) x + y^2 = 0$

$x^2 - 2x^2 - 2x \left(y \frac{dy}{dx}\right) + y^2 = 0$

$x^2 - 2x^2 - 2xy \frac{dy}{dx} + y^2 = 0$

Combine the $x^2$ terms:

$-x^2 - 2xy \frac{dy}{dx} + y^2 = 0$

Rearrange the terms, usually writing the highest derivative term or positive leading coefficient first:

$y^2 - x^2 - 2xy \frac{dy}{dx} = 0$

Or, multiplying by -1:

$x^2 - y^2 + 2xy \frac{dy}{dx} = 0$

The differential equation of all circles passing through the origin and whose centres lie on the x-axis is $y^2 - x^2 - 2xy \frac{dy}{dx} = 0$ or $x^2 - y^2 + 2xy \frac{dy}{dx} = 0$.

Given:

The differential equation $x^2y \, dx – (x^3 + y^3) \, dy = 0$.

To Solve:

Find the general solution of the differential equation.

Solution:

Rearrange the given differential equation to the form $\frac{dy}{dx}$:

$(x^3 + y^3) \, dy = x^2y \, dx$

$\frac{dy}{dx} = \frac{x^2y}{x^3 + y^3}$

This is a homogeneous differential equation because the numerator $x^2y$ has degree 3 ($2+1=3$) and the denominator $x^3 + y^3$ has degree 3. Let $f(x, y) = \frac{x^2y}{x^3 + y^3}$.

$f(\lambda x, \lambda y) = \frac{(\lambda x)^2 (\lambda y)}{(\lambda x)^3 + (\lambda y)^3} = \frac{\lambda^3 x^2 y}{\lambda^3 x^3 + \lambda^3 y^3} = \frac{\lambda^3 x^2 y}{\lambda^3 (x^3 + y^3)} = \lambda^0 \frac{x^2 y}{x^3 + y^3} = \lambda^0 f(x, y)$.

To solve a homogeneous differential equation, we use the substitution $y = vx$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x\frac{dv}{dx}$.

Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ into the differential equation:

$v + x\frac{dv}{dx} = \frac{x^2(vx)}{x^3 + (vx)^3}$

$v + x\frac{dv}{dx} = \frac{vx^3}{x^3 + v^3x^3}$

$v + x\frac{dv}{dx} = \frac{vx^3}{x^3(1 + v^3)}$

$v + x\frac{dv}{dx} = \frac{v}{1 + v^3}$

Separate the variables $v$ and $x$:

$x\frac{dv}{dx} = \frac{v}{1 + v^3} - v$

$x\frac{dv}{dx} = \frac{v - v(1 + v^3)}{1 + v^3}$

$x\frac{dv}{dx} = \frac{v - v - v^4}{1 + v^3}$

$x\frac{dv}{dx} = \frac{-v^4}{1 + v^3}$

$\frac{1 + v^3}{v^4} dv = -\frac{1}{x} dx$

$\left(\frac{1}{v^4} + \frac{v^3}{v^4}\right) dv = -\frac{1}{x} dx$

$\left(v^{-4} + \frac{1}{v}\right) dv = -\frac{1}{x} dx$

Integrate both sides:

$\int \left(v^{-4} + \frac{1}{v}\right) dv = \int -\frac{1}{x} dx$

$\frac{v^{-3}}{-3} + \log|v| = -\log|x| + C$

$-\frac{1}{3v^3} + \log|v| = -\log|x| + C$

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

$-\frac{1}{3(y/x)^3} + \log\left|\frac{y}{x}\right| = -\log|x| + C$

$-\frac{x^3}{3y^3} + \log|y| - \log|x| = -\log|x| + C$

Add $\log|x|$ to both sides:

$-\frac{x^3}{3y^3} + \log|y| = C$

This is the general solution. We can rearrange it slightly if desired, e.g., multiply by -3 and redefine the constant:

$\frac{x^3}{y^3} - 3\log|y| = -3C$

Let $C_1 = -3C$.

$\left(\frac{x}{y}\right)^3 - 3\log|y| = C_1$

The general solution is $\boxed{-\frac{x^3}{3y^3} + \log|y| = C}$ (or an equivalent form).

Given:

$\vec{a}\times\vec{b}=\vec{a}\times\vec{c}$

$\vec{a}\neq\vec{0}$

$\vec{b}\neq\vec{c}$

To Show:

$\vec{b} = \vec{c} + \lambda\vec{a}$ for some scalar $\lambda$.

Proof:

We are given the equation $\vec{a}\times\vec{b}=\vec{a}\times\vec{c}$.

Subtract $\vec{a}\times\vec{c}$ from both sides of the equation:

$\vec{a}\times\vec{b} - \vec{a}\times\vec{c} = \vec{0}$

Using the distributive property of the cross product, $\vec{x} \times \vec{y} - \vec{x} \times \vec{z} = \vec{x} \times (\vec{y} - \vec{z})$, we can write:

$\vec{a} \times (\vec{b} - \vec{c}) = \vec{0}$

The cross product of two vectors is the zero vector if and only if the two vectors are parallel (collinear).

So, the vector $\vec{a}$ must be parallel to the vector $(\vec{b} - \vec{c})$.

If two non-zero vectors are parallel, one can be expressed as a scalar multiple of the other.

We are given that $\vec{a} \neq \vec{0}$.

Since $\vec{a}$ is parallel to $(\vec{b} - \vec{c})$, there exists some scalar $\lambda$ such that:

$\vec{b} - \vec{c} = \lambda \vec{a}$

Add $\vec{c}$ to both sides of the equation:

$\vec{b} = \vec{c} + \lambda \vec{a}$

This shows that $\vec{b}$ can be expressed in the form $\vec{c} + \lambda \vec{a}$ for some scalar $\lambda$.

The condition $\vec{b}\neq\vec{c}$ given in the problem implies that $\vec{b} - \vec{c}$ is a non-zero vector. Since $\vec{a} \times (\vec{b} - \vec{c}) = \vec{0}$ and $\vec{a} \neq \vec{0}$, it confirms that $(\vec{b} - \vec{c})$ must be parallel to $\vec{a}$. Because $\vec{b} - \vec{c} \neq \vec{0}$ and $\vec{a} \neq \vec{0}$, the scalar $\lambda$ must be non-zero in the relationship $\vec{b} - \vec{c} = \lambda \vec{a}$.

Thus, we have shown that if $\vec{a}\times\vec{b}=\vec{a}\times\vec{c}$, $\vec{a}\neq\vec{0}$ and $\vec{b}\neq\vec{c}$, then $\vec{b} = \vec{c} + \lambda\vec{a}$ for some scalar $\lambda$ (which must be non-zero in this case).

Hence Shown.

Given:

Equation of the first line: $\vec{r}=\left(\lambda-1\right)\hat{i}+\left(\lambda+1\right)\hat{j}-\left(1+\lambda\right)\hat{k}$

Equation of the second line: $\vec{r}=\left(1-\mu\right)\hat{i}+\left(2\mu-1\right)\hat{j}+\left(\mu+2\right)\hat{k}$

To Find:

The shortest distance between the two lines.

Solution:

The vector equation of a line is given by $\vec{r} = \vec{a} + t\vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is a vector parallel to the line.

Rewrite the first line's equation:

$\vec{r} = (-1 + \lambda)\hat{i} + (1 + \lambda)\hat{j} + (-1 - \lambda)\hat{k}$

$\vec{r} = (-\hat{i} + \hat{j} - \hat{k}) + \lambda (\hat{i} + \hat{j} - \hat{k})$

Comparing this with $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$, we get:

$\vec{a}_1 = -\hat{i} + \hat{j} - \hat{k}$

$\vec{b}_1 = \hat{i} + \hat{j} - \hat{k}$

Rewrite the second line's equation:

$\vec{r} = (1 - \mu)\hat{i} + (-1 + 2\mu)\hat{j} + (2 + \mu)\hat{k}$

$\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu (-\hat{i} + 2\hat{j} + \hat{k})$

Comparing this with $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$, we get:

$\vec{a}_2 = \hat{i} - \hat{j} + 2\hat{k}$

$\vec{b}_2 = -\hat{i} + 2\hat{j} + \hat{k}$

The formula for the shortest distance between two skew lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is:

$d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$

Calculate $\vec{a}_2 - \vec{a}_1$:

$\vec{a}_2 - \vec{a}_1 = (\hat{i} - \hat{j} + 2\hat{k}) - (-\hat{i} + \hat{j} - \hat{k})$

$\vec{a}_2 - \vec{a}_1 = \hat{i} - \hat{j} + 2\hat{k} + \hat{i} - \hat{j} + \hat{k}$

$\vec{a}_2 - \vec{a}_1 = 2\hat{i} - 2\hat{j} + 3\hat{k}$

Calculate $\vec{b}_1 \times \vec{b}_2$:

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ -1 & 2 & 1 \end{vmatrix}$

$= \hat{i}((1)(1) - (-1)(2)) - \hat{j}((1)(1) - (-1)(-1)) + \hat{k}((1)(2) - (1)(-1))$

$= \hat{i}(1 + 2) - \hat{j}(1 - 1) + \hat{k}(2 + 1)$

$= 3\hat{i} - 0\hat{j} + 3\hat{k}$

$\vec{b}_1 \times \vec{b}_2 = 3\hat{i} + 3\hat{k}$

Calculate $|\vec{b}_1 \times \vec{b}_2|$:

$|\vec{b}_1 \times \vec{b}_2| = |3\hat{i} + 0\hat{j} + 3\hat{k}|$

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{3^2 + 0^2 + 3^2} = \sqrt{9 + 0 + 9} = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$

Calculate $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (2\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} + 0\hat{j} + 3\hat{k})$

$= (2)(3) + (-2)(0) + (3)(3)$

$= 6 + 0 + 9 = 15$

Substitute the calculated values into the shortest distance formula:

$d = \left| \frac{15}{3\sqrt{2}} \right|$

$d = \frac{15}{3\sqrt{2}}$

$d = \frac{5}{\sqrt{2}}$

Rationalize the denominator:

$d = \frac{5}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$

The shortest distance between the two lines is $\frac{5\sqrt{2}}{2}$ units.

Given:

Total number of cards in a pack = 52.

Number of cards of each suit (Hearts, Diamonds, Clubs, Spades) = 13.

One card is lost from the pack.

Two cards are drawn from the remaining 51 cards, and both are found to be hearts.

To Find:

The probability that the missing card is a heart.

Solution:

Let H be the event that the lost card is a heart.

Let NH be the event that the lost card is not a heart.

The prior probability of the lost card being a heart is:

$P(H) = \frac{\text{Number of hearts}}{\text{Total number of cards}} = \frac{13}{52} = \frac{1}{4}$

The prior probability of the lost card not being a heart is:

$P(NH) = 1 - P(H) = 1 - \frac{1}{4} = \frac{3}{4}$

Alternatively, there are 3 suits other than hearts, each with 13 cards. So, the number of non-hearts is $3 \times 13 = 39$.

$P(NH) = \frac{\text{Number of non-hearts}}{\text{Total number of cards}} = \frac{39}{52} = \frac{3}{4}$

Let E be the event that two cards drawn from the remaining 51 cards are hearts.

We need to find the probability of event E occurring given the lost card was a heart, i.e., $P(E|H)$.

If the lost card was a heart, the remaining pack contains 51 cards, with 12 hearts and 39 non-hearts.

The number of ways to draw 2 cards from the remaining 51 is $\binom{51}{2}$.

$\binom{51}{2} = \frac{51 \times 50}{2 \times 1} = 51 \times 25 = 1275$

The number of ways to draw 2 hearts from the remaining 12 hearts is $\binom{12}{2}$.

$\binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 6 \times 11 = 66$

$P(E|H) = \frac{\text{Number of ways to draw 2 hearts from 12}}{\text{Number of ways to draw 2 cards from 51}} = \frac{66}{1275}$

We need to find the probability of event E occurring given the lost card was not a heart, i.e., $P(E|NH)$.

If the lost card was not a heart, the remaining pack contains 51 cards, with 13 hearts and 38 non-hearts.

The number of ways to draw 2 cards from the remaining 51 is $\binom{51}{2} = 1275$.

The number of ways to draw 2 hearts from the remaining 13 hearts is $\binom{13}{2}$.

$\binom{13}{2} = \frac{13 \times 12}{2 \times 1} = 13 \times 6 = 78$

$P(E|NH) = \frac{\text{Number of ways to draw 2 hearts from 13}}{\text{Number of ways to draw 2 cards from 51}} = \frac{78}{1275}$

We want to find the probability that the lost card was a heart given that two hearts were drawn from the remaining pack, i.e., $P(H|E)$.

Using Bayes' Theorem:

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

Using the Law of Total Probability, the probability of event E is:

$P(E) = P(E|H) P(H) + P(E|NH) P(NH)$

$P(E) = \left(\frac{66}{1275}\right) \left(\frac{1}{4}\right) + \left(\frac{78}{1275}\right) \left(\frac{3}{4}\right)$

$P(E) = \frac{66}{1275 \times 4} + \frac{78 \times 3}{1275 \times 4}$

$P(E) = \frac{66}{5100} + \frac{234}{5100}$

$P(E) = \frac{66 + 234}{5100}$

$P(E) = \frac{300}{5100}$

$P(E) = \frac{30}{510} = \frac{3}{51} = \frac{1}{17}$

Now, substitute the values into Bayes' Theorem:

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

$P(H|E) = \frac{\left(\frac{66}{1275}\right) \left(\frac{1}{4}\right)}{\frac{300}{5100}}$

$P(H|E) = \frac{\frac{66}{5100}}{\frac{300}{5100}}$

$P(H|E) = \frac{66}{300}$

Simplify the fraction:

$P(H|E) = \frac{\cancel{6}^{11}}{\cancel{300}^{50}}$

$P(H|E) = \frac{11}{50}$

The probability of the missing card being a heart is $\frac{11}{50}$.

Given:

Matrices $A = \left[\begin{matrix}1&-1&0\\2&3&4\\0&1&2\\\end{matrix}\right]$ and $B=\left[\begin{matrix}2&2&-4\\-4&2&-4\\2&-1&5\\\end{matrix}\right]$

System of equations:

$x - y = 3$

$2x + 3y + 4z = 17$

$y + 2z = 7$

To Verify:

$AB = BA = 6I$, where $I$ is the unit matrix of order 3.

To Solve:

The given system of equations using the verification result.

Solution:

First, let's calculate the product AB:

$AB = \left[\begin{matrix}1&-1&0\\2&3&4\\0&1&2\\\end{matrix}\right] \left[\begin{matrix}2&2&-4\\-4&2&-4\\2&-1&5\\\end{matrix}\right]$

$AB_{11} = (1)(2) + (-1)(-4) + (0)(2) = 2 + 4 + 0 = 6$

$AB_{12} = (1)(2) + (-1)(2) + (0)(-1) = 2 - 2 + 0 = 0$

$AB_{13} = (1)(-4) + (-1)(-4) + (0)(5) = -4 + 4 + 0 = 0$

$AB_{21} = (2)(2) + (3)(-4) + (4)(2) = 4 - 12 + 8 = 0$

$AB_{22} = (2)(2) + (3)(2) + (4)(-1) = 4 + 6 - 4 = 6$

$AB_{23} = (2)(-4) + (3)(-4) + (4)(5) = -8 - 12 + 20 = 0$

$AB_{31} = (0)(2) + (1)(-4) + (2)(2) = 0 - 4 + 4 = 0$

$AB_{32} = (0)(2) + (1)(2) + (2)(-1) = 0 + 2 - 2 = 0$

$AB_{33} = (0)(-4) + (1)(-4) + (2)(5) = 0 - 4 + 10 = 6$

So, $AB = \left[\begin{matrix}6&0&0\\0&6&0\\0&0&6\\\end{matrix}\right]$

$AB = 6 \left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\\\end{matrix}\right] = 6I$

Next, let's calculate the product BA:

$BA = \left[\begin{matrix}2&2&-4\\-4&2&-4\\2&-1&5\\\end{matrix}\right] \left[\begin{matrix}1&-1&0\\2&3&4\\0&1&2\\\end{matrix}\right]$

$BA_{11} = (2)(1) + (2)(2) + (-4)(0) = 2 + 4 + 0 = 6$

$BA_{12} = (2)(-1) + (2)(3) + (-4)(1) = -2 + 6 - 4 = 0$

$BA_{13} = (2)(0) + (2)(4) + (-4)(2) = 0 + 8 - 8 = 0$

$BA_{21} = (-4)(1) + (2)(2) + (-4)(0) = -4 + 4 + 0 = 0$

$BA_{22} = (-4)(-1) + (2)(3) + (-4)(1) = 4 + 6 - 4 = 6$

$BA_{23} = (-4)(0) + (2)(4) + (-4)(2) = 0 + 8 - 8 = 0$

$BA_{31} = (2)(1) + (-1)(2) + (5)(0) = 2 - 2 + 0 = 0$

$BA_{32} = (2)(-1) + (-1)(3) + (5)(1) = -2 - 3 + 5 = 0$

$BA_{33} = (2)(0) + (-1)(4) + (5)(2) = 0 - 4 + 10 = 6$

So, $BA = \left[\begin{matrix}6&0&0\\0&6&0\\0&0&6\\\end{matrix}\right]$

$BA = 6 \left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\\\end{matrix}\right] = 6I$

From the calculations, we see that $AB = 6I$ and $BA = 6I$.

Therefore, $AB = BA = 6I$.

This verifies the first part of the question.

Now, we use this result to solve the system of equations.

The given system of equations is:

$x - y + 0z = 3$

$2x + 3y + 4z = 17$

$0x + y + 2z = 7$

This system can be written in matrix form $AX = C$, where:

$A = \left[\begin{matrix}1&-1&0\\2&3&4\\0&1&2\\\end{matrix}\right]$, $X = \left[\begin{matrix}x\\y\\z\\\end{matrix}\right]$, and $C = \left[\begin{matrix}3\\17\\7\\\end{matrix}\right]$

Note that the matrix $A$ in the system of equations is the same as the matrix $A$ given in the problem.

From the verification, we have $AB = 6I$.

We know that for an invertible matrix $A$, $AA^{-1} = A^{-1}A = I$.

Comparing $AB = 6I$ with $AA^{-1} = I$, we can see a relationship between $A^{-1}$ and $B$.

Divide the equation $AB = 6I$ by 6:

$A \left(\frac{1}{6}B\right) = I$

By the definition of the inverse matrix, $A^{-1} = \frac{1}{6}B$.

To solve the matrix equation $AX = C$, we pre-multiply both sides by $A^{-1}$:

$A^{-1}(AX) = A^{-1}C$

$(A^{-1}A)X = A^{-1}C$

$IX = A^{-1}C$

$X = A^{-1}C$

Substitute $A^{-1} = \frac{1}{6}B$ into the equation for $X$:

$X = \frac{1}{6}BC$

$X = \frac{1}{6} \left[\begin{matrix}2&2&-4\\-4&2&-4\\2&-1&5\\\end{matrix}\right] \left[\begin{matrix}3\\17\\7\\\end{matrix}\right]$

Calculate the matrix product $BC$:

$BC_{11} = (2)(3) + (2)(17) + (-4)(7) = 6 + 34 - 28 = 40 - 28 = 12$

$BC_{21} = (-4)(3) + (2)(17) + (-4)(7) = -12 + 34 - 28 = 22 - 28 = -6$

$BC_{31} = (2)(3) + (-1)(17) + (5)(7) = 6 - 17 + 35 = -11 + 35 = 24$

So, $BC = \left[\begin{matrix}12\\-6\\24\\\end{matrix}\right]$

Now, calculate $X = \frac{1}{6} BC$:

$X = \frac{1}{6} \left[\begin{matrix}12\\-6\\24\\\end{matrix}\right] = \left[\begin{matrix}12/6\\-6/6\\24/6\\\end{matrix}\right] = \left[\begin{matrix}2\\-1\\4\\\end{matrix}\right]$

Since $X = \left[\begin{matrix}x\\y\\z\\\end{matrix}\right]$, we have $\left[\begin{matrix}x\\y\\z\\\end{matrix}\right] = \left[\begin{matrix}2\\-1\\4\\\end{matrix}\right]$.

Equating the elements, we get $x = 2$, $y = -1$, and $z = 4$.

Verification of the solution in the original equations:

Equation 1: $x - y = 2 - (-1) = 2 + 1 = 3$. (Matches)

Equation 2: $2x + 3y + 4z = 2(2) + 3(-1) + 4(4) = 4 - 3 + 16 = 1 + 16 = 17$. (Matches)

Equation 3: $y + 2z = -1 + 2(4) = -1 + 8 = 7$. (Matches)

The solution to the system of equations is $x = 2$, $y = -1$, and $z = 4$.

Given:

The set is $R - \{-1\}$.

The binary operation $*$ is defined by $a * b = a + b + ab$ for all $a, b \in R – \{-1\}$.

To Prove:

1. The operation $*$ is commutative on $R – \{-1\}$.

2. Find the identity element for $*$ on $R – \{-1\}$.

3. Prove that every element of $R – \{-1\}$ is invertible under $*$.

Proof of Commutativity:

For the operation $*$ to be commutative on $R – \{-1\}$, we must show that $a * b = b * a$ for all $a, b \in R – \{-1\}$.

Consider $a * b = a + b + ab$.

Consider $b * a = b + a + ba$.

Since addition and multiplication of real numbers are commutative, we have $a + b = b + a$ and $ab = ba$.

Therefore, $a + b + ab = b + a + ba$.

Thus, $a * b = b * a$ for all $a, b \in R – \{-1\}$.

Hence, the operation $*$ is commutative on $R – \{-1\}$.

Finding the Identity Element:

Let $e$ be the identity element for the operation $*$ on $R – \{-1\}$.

By definition, for every $a \in R – \{-1\}$, we must have $a * e = a$ and $e * a = a$.

Using the definition of the operation, $a * e = a + e + ae$.

So, $a + e + ae = a$.

Subtract $a$ from both sides:

$e + ae = 0$

Factor out $e$:

$e(1 + a) = 0$

Since $a \in R – \{-1\}$, we know that $a \neq -1$, so $1 + a \neq 0$.

Since $1 + a \neq 0$, we can divide by $1 + a$:

$e = \frac{0}{1+a} = 0$

We must check if this element $e=0$ is in the set $R – \{-1\}$. Since $0 \neq -1$, $e=0$ is in the set.

We should also check if $e * a = a$ gives the same result:

$e * a = e + a + ea = 0 + a + 0a = a$. This confirms $e=0$ is the right identity.

The identity element for the operation $*$ on $R – \{-1\}$ is 0.

Proving Invertibility:

For every element $a \in R – \{-1\}$, we need to find an inverse element, let's call it $b$, such that $a * b = e$ and $b * a = e$, where $e=0$ is the identity element.

Using the definition of the operation, $a * b = a + b + ab$.

We set this equal to the identity element 0:

$a + b + ab = 0$

We want to find $b$ in terms of $a$. Group the terms involving $b$:

$b + ab = -a$

Factor out $b$:

$b(1 + a) = -a$

Since $a \in R – \{-1\}$, we know that $a \neq -1$, so $1 + a \neq 0$.

We can divide by $1 + a$:

$b = \frac{-a}{1+a}$

For this element $b$ to be the inverse of $a$, it must also be in the set $R – \{-1\}$. This means $b$ cannot be equal to -1.

Assume $b = -1$ for some $a \in R – \{-1\}$.

$\frac{-a}{1+a} = -1$

$-a = -1(1+a)$

$-a = -1 - a$

$0 = -1$

This is a contradiction. Therefore, $b = \frac{-a}{1+a}$ can never be equal to -1 for any $a \in R – \{-1\}$.

Thus, for every $a \in R – \{-1\}$, the inverse element $b = \frac{-a}{1+a}$ exists and is also in $R – \{-1\}$.

Since the operation is commutative, if $a * b = 0$, then $b * a = 0$ automatically holds.

Hence, every element of $R – \{-1\}$ is invertible.

Given:

A right-angled triangle with a given hypotenuse, let's denote its length by $h$.

To Prove:

The perimeter of the triangle is maximum when the triangle is isosceles (i.e., the two legs are equal in length).

Solution:

Let the lengths of the two legs of the right-angled triangle be $x$ and $y$. By the Pythagorean theorem, we have:

$x^2 + y^2 = h^2$

Since $h$ is a given constant, $x$ and $y$ must be positive values satisfying this equation. The domain for $x$ (and $y$) is $(0, h)$.

The perimeter of the triangle is given by $P = x + y + h$.

Since $h$ is constant, maximizing the perimeter $P$ is equivalent to maximizing the sum of the lengths of the legs, $S = x + y$.

From equation (i), we can express $y$ in terms of $x$ and $h$: $y = \sqrt{h^2 - x^2}$. Since $y > 0$, we consider the positive square root.

Now, express $S$ as a function of a single variable $x$:

$S(x) = x + \sqrt{h^2 - x^2}$

The domain of this function for the problem context is $0 < x < h$.

To find the maximum value of $S(x)$, we find the critical points by differentiating $S(x)$ with respect to $x$ and setting the derivative equal to zero.

$S'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\sqrt{h^2 - x^2})$

$S'(x) = 1 + \frac{1}{2\sqrt{h^2 - x^2}} \cdot \frac{d}{dx}(h^2 - x^2)$

$S'(x) = 1 + \frac{1}{2\sqrt{h^2 - x^2}} \cdot (-2x)$

$S'(x) = 1 - \frac{x}{\sqrt{h^2 - x^2}}$

Set $S'(x) = 0$ to find the critical points:

$1 - \frac{x}{\sqrt{h^2 - x^2}} = 0$

$1 = \frac{x}{\sqrt{h^2 - x^2}}$

$\sqrt{h^2 - x^2} = x$

Square both sides (since both sides are positive in the domain $0 < x < h$):

$h^2 - x^2 = x^2$

$h^2 = 2x^2$

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

Since $x > 0$, we take the positive square root:

$x = \sqrt{\frac{h^2}{2}} = \frac{h}{\sqrt{2}}$

Now, find the corresponding value of $y$ using $y = \sqrt{h^2 - x^2}$:

$y = \sqrt{h^2 - \left(\frac{h}{\sqrt{2}}\right)^2} = \sqrt{h^2 - \frac{h^2}{2}} = \sqrt{\frac{h^2}{2}} = \frac{h}{\sqrt{2}}$

So, the critical point occurs when $x = y = \frac{h}{\sqrt{2}}$. This means the two legs of the right-angled triangle are equal in length, which is the definition of an isosceles right-angled triangle.

To confirm that this critical point corresponds to a maximum, we can use the second derivative test.

$S''(x) = \frac{d}{dx}\left(1 - x(h^2 - x^2)^{-1/2}\right)$

$S''(x) = 0 - \left[1 \cdot (h^2 - x^2)^{-1/2} + x \cdot \left(-\frac{1}{2}\right)(h^2 - x^2)^{-3/2} \cdot (-2x)\right]$

$S''(x) = -\left[\frac{1}{\sqrt{h^2 - x^2}} + \frac{x^2}{(h^2 - x^2)^{3/2}}\right]$

$S''(x) = -\frac{(h^2 - x^2) + x^2}{(h^2 - x^2)^{3/2}} = -\frac{h^2}{(h^2 - x^2)^{3/2}}$

Evaluate $S''(x)$ at the critical point $x = \frac{h}{\sqrt{2}}$:

When $x = \frac{h}{\sqrt{2}}$, $h^2 - x^2 = h^2 - \left(\frac{h}{\sqrt{2}}\right)^2 = h^2 - \frac{h^2}{2} = \frac{h^2}{2}$.

$S''\left(\frac{h}{\sqrt{2}}\right) = -\frac{h^2}{(\frac{h^2}{2})^{3/2}} = -\frac{h^2}{(\frac{h^3}{2\sqrt{2}})} = -\frac{2\sqrt{2}h^2}{h^3} = -\frac{2\sqrt{2}}{h}$

Since $h > 0$, $S''\left(\frac{h}{\sqrt{2}}\right) = -\frac{2\sqrt{2}}{h}$ is negative.

By the second derivative test, $x = \frac{h}{\sqrt{2}}$ is a point of local maximum.

The maximum value of $S(x)$ occurs at $x = \frac{h}{\sqrt{2}}$, where $y = \frac{h}{\sqrt{2}}$.

This corresponds to the case where the two legs of the right-angled triangle are equal ($x=y$), meaning the triangle is isosceles.

The maximum value of the perimeter is $P = x + y + h = \frac{h}{\sqrt{2}} + \frac{h}{\sqrt{2}} + h = \frac{2h}{\sqrt{2}} + h = \sqrt{2}h + h = (\sqrt{2} + 1)h$.

Alternate Solution (Using Trigonometry):

Let the acute angles of the right-angled triangle be $\theta$ and $\frac{\pi}{2} - \theta$, where $0 < \theta < \frac{\pi}{2}$.

Let the hypotenuse be $h$, which is given. The lengths of the legs are $x$ and $y$.

We can write $x = h \cos \theta$ and $y = h \sin \theta$.

The perimeter is $P = x + y + h = h \cos \theta + h \sin \theta + h = h(\cos \theta + \sin \theta + 1)$.

To maximize $P$, we need to maximize the function $f(\theta) = \cos \theta + \sin \theta$ for $0 < \theta < \frac{\pi}{2}$.

Find the derivative of $f(\theta)$ with respect to $\theta$:

$f'(\theta) = \frac{d}{d\theta}(\cos \theta + \sin \theta) = -\sin \theta + \cos \theta$

Set $f'(\theta) = 0$ to find critical points:

$-\sin \theta + \cos \theta = 0$

$\cos \theta = \sin \theta$

Divide by $\cos \theta$ (since $\cos \theta \neq 0$ for $0 < \theta < \frac{\pi}{2}$):

$\frac{\sin \theta}{\cos \theta} = 1$

$\tan \theta = 1$

In the interval $0 < \theta < \frac{\pi}{2}$, the only solution is $\theta = \frac{\pi}{4}$.

To confirm this is a maximum, find the second derivative:

$f''(\theta) = \frac{d}{d\theta}(-\sin \theta + \cos \theta) = -\cos \theta - \sin \theta$

Evaluate $f''(\theta)$ at $\theta = \frac{\pi}{4}$:

$f''\left(\frac{\pi}{4}\right) = -\cos \frac{\pi}{4} - \sin \frac{\pi}{4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}$

Since $f''\left(\frac{\pi}{4}\right) < 0$, the function $f(\theta)$ has a local maximum at $\theta = \frac{\pi}{4}$.

When $\theta = \frac{\pi}{4}$, the lengths of the legs are:

$x = h \cos \frac{\pi}{4} = h \frac{\sqrt{2}}{2} = \frac{h}{\sqrt{2}}$

$y = h \sin \frac{\pi}{4} = h \frac{\sqrt{2}}{2} = \frac{h}{\sqrt{2}}$

Since $x = y$, the triangle is isosceles.

The boundary values of $\theta$ (as $\theta \to 0^+$ or $\theta \to \pi/2^-$) result in degenerate triangles (where one leg is 0 and the other is $h$), yielding a perimeter of $0 + h + h = 2h$. At $\theta = \pi/4$, the perimeter is $h + \frac{h}{\sqrt{2}} + \frac{h}{\sqrt{2}} = h + \sqrt{2}h = (1+\sqrt{2})h$. Since $1+\sqrt{2} > 2$, the maximum occurs at $\theta = \pi/4$.

Both methods show that the perimeter of a right-angled triangle with a given hypotenuse is maximum when the two legs are equal in length, which means the triangle is isosceles.

Hence Proved.

Given:

The lines:

Line 1: $2x + y = 4 \implies y = 4 - 2x$

Line 2: $3x – 2y = 6 \implies 2y = 3x - 6 \implies y = \frac{3}{2}x - 3$

Line 3: $x – 3y + 5 = 0 \implies 3y = x + 5 \implies y = \frac{1}{3}x + \frac{5}{3}$

To Find:

The area of the region bounded by these lines using the method of integration.

Solution:

First, we need to find the vertices of the triangle formed by the intersection of these three lines.

Intersection of Line 1 and Line 2:

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

$7 = \frac{3}{2}x + 2x = \frac{3x + 4x}{2} = \frac{7x}{2}$

$7x = 14 \implies x = 2$

Substitute $x=2$ into $y = 4 - 2x$:

$y = 4 - 2(2) = 0$

Vertex A: (2, 0)

Intersection of Line 1 and Line 3:

$4 - 2x = \frac{1}{3}x + \frac{5}{3}$

Multiply by 3: $12 - 6x = x + 5$

$12 - 5 = x + 6x$

$7 = 7x \implies x = 1$

Substitute $x=1$ into $y = 4 - 2x$:

$y = 4 - 2(1) = 2$

Vertex B: (1, 2)

Intersection of Line 2 and Line 3:

$\frac{3}{2}x - 3 = \frac{1}{3}x + \frac{5}{3}$

Multiply by 6: $9x - 18 = 2x + 10$

$9x - 2x = 10 + 18$

$7x = 28 \implies x = 4$

Substitute $x=4$ into $y = \frac{3}{2}x - 3$:

$y = \frac{3}{2}(4) - 3 = 6 - 3 = 3$

Vertex C: (4, 3)

The vertices of the triangle are A(2, 0), B(1, 2), and C(4, 3).

To find the area using integration with respect to $x$, we integrate the difference between the upper and lower boundary curves over the relevant intervals of $x$.

The upper boundary is the line segment BC (Line 3: $y = \frac{1}{3}x + \frac{5}{3}$) from $x=1$ to $x=4$.

The lower boundary consists of two line segments:

Segment BA (Line 1: $y = 4 - 2x$) from $x=1$ to $x=2$.

Segment AC (Line 2: $y = \frac{3}{2}x - 3$) from $x=2$ to $x=4$.

The area of the triangle can be calculated as the area under the upper boundary minus the area under the lower boundaries.

Area $= \int_{1}^{4} \left(\frac{1}{3}x + \frac{5}{3}\right) dx - \left( \int_{1}^{2} (4 - 2x) dx + \int_{2}^{4} \left(\frac{3}{2}x - 3\right) dx \right)$

Evaluate the integrals:

$\int \left(\frac{1}{3}x + \frac{5}{3}\right) dx = \frac{1}{3} \cdot \frac{x^2}{2} + \frac{5}{3}x = \frac{x^2}{6} + \frac{5x}{3}$

$\int (4 - 2x) dx = 4x - \frac{2x^2}{2} = 4x - x^2$

$\int \left(\frac{3}{2}x - 3\right) dx = \frac{3}{2} \cdot \frac{x^2}{2} - 3x = \frac{3x^2}{4} - 3x$

Evaluate the definite integrals:

$\int_{1}^{4} \left(\frac{1}{3}x + \frac{5}{3}\right) dx = \left[\frac{x^2}{6} + \frac{5x}{3}\right]_{1}^{4} = \left(\frac{4^2}{6} + \frac{5(4)}{3}\right) - \left(\frac{1^2}{6} + \frac{5(1)}{3}\right) = \left(\frac{16}{6} + \frac{20}{3}\right) - \left(\frac{1}{6} + \frac{5}{3}\right)$

$= \left(\frac{8}{3} + \frac{20}{3}\right) - \left(\frac{1}{6} + \frac{10}{6}\right) = \frac{28}{3} - \frac{11}{6} = \frac{56 - 11}{6} = \frac{45}{6} = \frac{15}{2}$

$\int_{1}^{2} (4 - 2x) dx = [4x - x^2]_{1}^{2} = (4(2) - 2^2) - (4(1) - 1^2) = (8 - 4) - (4 - 1) = 4 - 3 = 1$

$\int_{2}^{4} \left(\frac{3}{2}x - 3\right) dx = \left[\frac{3x^2}{4} - 3x\right]_{2}^{4} = \left(\frac{3(4)^2}{4} - 3(4)\right) - \left(\frac{3(2)^2}{4} - 3(2)\right) = \left(\frac{48}{4} - 12\right) - \left(\frac{12}{4} - 6\right)$

$= (12 - 12) - (3 - 6) = 0 - (-3) = 3$

Now, calculate the total area:

Area $= \frac{15}{2} - (1 + 3) = \frac{15}{2} - 4 = \frac{15 - 8}{2} = \frac{7}{2}$

The area of the region bounded by the lines is $\frac{7}{2}$ square units.

OR

Given:

The integral to evaluate is $\int\limits_{1}^{4}\left({2x}^2-x\right) \ dx$.

Limits of integration are $a=1$ and $b=4$.

The function is $f(x) = 2x^2 - x$.

To Evaluate:

The definite integral as the limit of a sum.

Solution:

The definite integral $\int\limits_{a}^{b} f(x) dx$ is defined as the limit of a sum:

$\int\limits_{a}^{b} f(x) dx = \lim\limits_{n \to \infty} h \sum\limits_{i=1}^{n} f(a + ih)$, where $h = \frac{b-a}{n}$.

In this case, $a = 1$, $b = 4$, and $f(x) = 2x^2 - x$.

First, calculate $h$:

$h = \frac{b-a}{n} = \frac{4 - 1}{n} = \frac{3}{n}$.

Next, calculate $a + ih$:

$a + ih = 1 + i\left(\frac{3}{n}\right) = 1 + \frac{3i}{n}$.

Now, calculate $f(a + ih) = f\left(1 + \frac{3i}{n}\right)$:

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

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

$= 2\left(1 + \frac{6i}{n} + \frac{9i^2}{n^2}\right) - 1 - \frac{3i}{n}$

$= 2 + \frac{12i}{n} + \frac{18i^2}{n^2} - 1 - \frac{3i}{n}$

$= 1 + \left(\frac{12}{n} - \frac{3}{n}\right)i + \frac{18i^2}{n^2}$

$f\left(1 + \frac{3i}{n}\right) = 1 + \frac{9i}{n} + \frac{18i^2}{n^2}$

Now, calculate the sum $\sum\limits_{i=1}^{n} f(a + ih)$:

$\sum\limits_{i=1}^{n} \left(1 + \frac{9i}{n} + \frac{18i^2}{n^2}\right) = \sum\limits_{i=1}^{n} 1 + \frac{9}{n}\sum\limits_{i=1}^{n} i + \frac{18}{n^2}\sum\limits_{i=1}^{n} i^2$

Using the standard summation formulas $\sum_{i=1}^n 1 = n$, $\sum_{i=1}^n i = \frac{n(n+1)}{2}$, and $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$:

Sum $= n + \frac{9}{n} \cdot \frac{n(n+1)}{2} + \frac{18}{n^2} \cdot \frac{n(n+1)(2n+1)}{6}$

$= n + \frac{9(n+1)}{2} + \frac{3(n+1)(2n+1)}{n}$

Multiply the sum by $h = \frac{3}{n}$:

$h \sum\limits_{i=1}^{n} f(a + ih) = \frac{3}{n} \left[ n + \frac{9(n+1)}{2} + \frac{3(n+1)(2n+1)}{n} \right]$

$= \frac{3}{n} \cdot n + \frac{3}{n} \cdot \frac{9(n+1)}{2} + \frac{3}{n} \cdot \frac{3(n+1)(2n+1)}{n}$

$= 3 + \frac{27(n+1)}{2n} + \frac{9(n+1)(2n+1)}{n^2}$

$= 3 + \frac{27}{2}\left(\frac{n+1}{n}\right) + 9\left(\frac{n+1}{n}\right)\left(\frac{2n+1}{n}\right)$

$= 3 + \frac{27}{2}\left(1 + \frac{1}{n}\right) + 9\left(1 + \frac{1}{n}\right)\left(2 + \frac{1}{n}\right)$

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

$\int\limits_{1}^{4}\left({2x}^2-x\right) \ dx = \lim\limits_{n \to \infty} \left[ 3 + \frac{27}{2}\left(1 + \frac{1}{n}\right) + 9\left(1 + \frac{1}{n}\right)\left(2 + \frac{1}{n}\right) \right]$

As $n \to \infty$, $\frac{1}{n} \to 0$.

$= 3 + \frac{27}{2}(1 + 0) + 9(1 + 0)(2 + 0)$

$= 3 + \frac{27}{2} + 9(1)(2)$

$= 3 + \frac{27}{2} + 18$

$= 21 + \frac{27}{2}$

$= \frac{42}{2} + \frac{27}{2} = \frac{42 + 27}{2} = \frac{69}{2}$

The value of the integral evaluated as the limit of a sum is $\frac{69}{2}$.

Given:

Point P(2, 3, 7).

Plane equation: $3x - y - z = 7$.

To Find:

1. The coordinates of the foot of the perpendicular from P to the plane.

2. The length of the perpendicular.

Solution:

The equation of the plane is $3x - y - z - 7 = 0$.

The normal vector to the plane is $\vec{n} = 3\hat{i} - \hat{j} - \hat{k}$.

The line passing through the point P(2, 3, 7) and perpendicular to the plane has the direction vector parallel to the normal vector $\vec{n}$.

The equation of this line in vector form is $\vec{r} = \vec{p} + \lambda \vec{n}$, where $\vec{p}$ is the position vector of P.

$\vec{p} = 2\hat{i} + 3\hat{j} + 7\hat{k}$.

So, the equation of the line is $\vec{r} = (2\hat{i} + 3\hat{j} + 7\hat{k}) + \lambda (3\hat{i} - \hat{j} - \hat{k})$.

Any point Q on this line has the position vector $\vec{r} = (2 + 3\lambda)\hat{i} + (3 - \lambda)\hat{j} + (7 - \lambda)\hat{k}$.

Let the coordinates of Q be $(x_Q, y_Q, z_Q) = (2 + 3\lambda, 3 - \lambda, 7 - \lambda)$.

The foot of the perpendicular Q is the point where the line intersects the plane. So, the coordinates of Q must satisfy the equation of the plane $3x - y - z = 7$.

Substitute the coordinates of Q into the plane equation:

$3(2 + 3\lambda) - (3 - \lambda) - (7 - \lambda) = 7$

$6 + 9\lambda - 3 + \lambda - 7 + \lambda = 7$

$(9\lambda + \lambda + \lambda) + (6 - 3 - 7) = 7$

$11\lambda - 4 = 7$

$11\lambda = 7 + 4$

$11\lambda = 11$

$\lambda = 1$

Now, substitute the value of $\lambda = 1$ back into the coordinates of Q:

$x_Q = 2 + 3(1) = 2 + 3 = 5$

$y_Q = 3 - 1 = 2$

$z_Q = 7 - 1 = 6$

The coordinates of the foot of the perpendicular Q are (5, 2, 6).

Next, find the length of the perpendicular PQ, which is the distance between the points P(2, 3, 7) and Q(5, 2, 6).

Using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

$PQ = \sqrt{(5 - 2)^2 + (2 - 3)^2 + (6 - 7)^2}$

$PQ = \sqrt{(3)^2 + (-1)^2 + (-1)^2}$

$PQ = \sqrt{9 + 1 + 1}$

$PQ = \sqrt{11}$

The length of the perpendicular from P to the plane is $\sqrt{11}$ units.

Alternate Calculation of Length of Perpendicular (using formula):

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

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

Here, the point is P(2, 3, 7), so $(x_0, y_0, z_0) = (2, 3, 7)$.

The plane equation is $3x - y - z - 7 = 0$, so $A=3$, $B=-1$, $C=-1$, $D=-7$.

Distance $= \frac{|3(2) + (-1)(3) + (-1)(7) + (-7)|}{\sqrt{3^2 + (-1)^2 + (-1)^2}}$

$= \frac{|6 - 3 - 7 - 7|}{\sqrt{9 + 1 + 1}}$

$= \frac{|3 - 14|}{\sqrt{11}}$

$= \frac{|-11|}{\sqrt{11}}$

$= \frac{11}{\sqrt{11}}$

$= \frac{11}{\sqrt{11}} \times \frac{\sqrt{11}}{\sqrt{11}} = \frac{11\sqrt{11}}{11} = \sqrt{11}$

This confirms the length of the perpendicular is $\sqrt{11}$ units.

OR

Given:

Line 1: $\vec{r}= \hat{i} + \hat{j} + \lambda ( \hat{i} + 2\hat{j} - \hat{k})$

Line 2: $\vec{r} = \hat{i} + \hat{j} + \mu (-\hat{i} + \hat{j} - 2\hat{k})$

Point (1, 1, 1).

To Find:

1. The equation of the plane containing the two lines.

2. The distance of this plane from the point (1, 1, 1).

Solution (Equation of the Plane):

From the equations of the lines, we can see that both lines pass through the point with position vector $\vec{a} = \hat{i} + \hat{j}$ (by setting $\lambda = 0$ in the first equation and $\mu = 0$ in the second equation). The coordinates of this point are (1, 1, 0).

The direction vector of the first line is $\vec{b}_1 = \hat{i} + 2\hat{j} - \hat{k}$.

The direction vector of the second line is $\vec{b}_2 = -\hat{i} + \hat{j} - 2\hat{k}$.

Since the plane contains both lines, the normal vector to the plane $\vec{n}$ is perpendicular to both $\vec{b}_1$ and $\vec{b}_2$. Thus, $\vec{n}$ is parallel to $\vec{b}_1 \times \vec{b}_2$.

Calculate the cross product $\vec{b}_1 \times \vec{b}_2$:

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{vmatrix}$

$= \hat{i}((2)(-2) - (-1)(1)) - \hat{j}((1)(-2) - (-1)(-1)) + \hat{k}((1)(1) - (2)(-1))$

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

$= -3\hat{i} + 3\hat{j} + 3\hat{k}$

We can take the normal vector $\vec{n}$ proportional to this vector, for example, $\vec{n} = -\hat{i} + \hat{j} + \hat{k}$ (by dividing by 3).

The equation of the plane passing through the point $\vec{a} = \hat{i} + \hat{j}$ with normal vector $\vec{n} = -\hat{i} + \hat{j} + \hat{k}$ is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.

$\vec{a} \cdot \vec{n} = (\hat{i} + \hat{j} + 0\hat{k}) \cdot (-\hat{i} + \hat{j} + \hat{k})$

$= (1)(-1) + (1)(1) + (0)(1) = -1 + 1 + 0 = 0$

So, the vector equation of the plane is $\vec{r} \cdot (-\hat{i} + \hat{j} + \hat{k}) = 0$.

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. The Cartesian equation is:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (-\hat{i} + \hat{j} + \hat{k}) = 0$

$-x + y + z = 0$

Or, $x - y - z = 0$.

The equation of the plane containing the given lines is $x - y - z = 0$.

Solution (Distance from the point (1, 1, 1)):

The equation of the plane is $x - y - z = 0$. Comparing this with $Ax + By + Cz + D = 0$, we have $A=1$, $B=-1$, $C=-1$, and $D=0$.

The given point is $(x_0, y_0, z_0) = (1, 1, 1)$.

The distance from the point $(x_0, y_0, z_0)$ to the plane $Ax + By + Cz + D = 0$ is:

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

Substitute the values:

$D = \frac{|(1)(1) + (-1)(1) + (-1)(1) + 0|}{\sqrt{1^2 + (-1)^2 + (-1)^2}}$

$D = \frac{|1 - 1 - 1 + 0|}{\sqrt{1 + 1 + 1}}$

$D = \frac{|-1|}{\sqrt{3}}$

$D = \frac{1}{\sqrt{3}}$

Rationalize the denominator:

$D = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

The distance of the plane from the point (1, 1, 1) is $\frac{1}{\sqrt{3}}$ units or $\frac{\sqrt{3}}{3}$ units.

Given:

Total cards in a well-shuffled pack = 52.

Number of Kings = 4.

Number of Non-Kings = 52 - 4 = 48.

Two cards are drawn successively without replacement.

To Find:

1. The probability distribution of the number of kings drawn.

2. The mean of this distribution.

3. The variance of this distribution.

Solution:

Let X be the random variable representing the number of kings drawn in two successive draws without replacement.

The possible values for X are 0, 1, or 2.

We calculate the probability for each possible value of X.

Case X = 0 (No kings drawn):

This means the first card is a non-king (NK) and the second card is a non-king (NK).

$P(X=0) = P(\text{NK on 1st draw}) \times P(\text{NK on 2nd draw | NK on 1st draw})$

$P(X=0) = \frac{48}{52} \times \frac{47}{51} = \frac{\cancel{48}^{12}}{52_{\cancel{13}}} \times \frac{47}{\cancel{51}^{17}} = \frac{12}{13} \times \frac{47}{17} = \frac{564}{221}$.

Simplifying the fraction $\frac{564}{663}$ by dividing numerator and denominator by 3: $\frac{188}{221}$.

$P(X=0) = \frac{188}{221}$.

Case X = 1 (Exactly one king drawn):

This can occur in two mutually exclusive ways: (King on 1st, Non-King on 2nd) or (Non-King on 1st, King on 2nd).

$P(\text{K on 1st, NK on 2nd}) = P(\text{K on 1st}) \times P(\text{NK on 2nd | K on 1st})$

$P(\text{K on 1st, NK on 2nd}) = \frac{4}{52} \times \frac{48}{51} = \frac{1}{13} \times \frac{48}{51} = \frac{48}{663} = \frac{16}{221}$.

$P(\text{NK on 1st, K on 2nd}) = P(\text{NK on 1st}) \times P(\text{K on 2nd | NK on 1st})$

$P(\text{NK on 1st, K on 2nd}) = \frac{48}{52} \times \frac{4}{51} = \frac{12}{13} \times \frac{4}{51} = \frac{48}{663} = \frac{16}{221}$.

$P(X=1) = P(\text{K on 1st, NK on 2nd}) + P(\text{NK on 1st, K on 2nd}) = \frac{16}{221} + \frac{16}{221} = \frac{32}{221}$.

Case X = 2 (Two kings drawn):

This means the first card is a king (K) and the second card is a king (K).

$P(X=2) = P(\text{K on 1st draw}) \times P(\text{K on 2nd draw | K on 1st draw})$

$P(X=2) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{13} \times \frac{3}{51} = \frac{3}{663} = \frac{1}{221}$.

The probability distribution of the number of kings (X) is:

Check sum of probabilities: $\frac{188}{221} + \frac{32}{221} + \frac{1}{221} = \frac{188 + 32 + 1}{221} = \frac{221}{221} = 1$. The distribution is valid.

Mean of the distribution (E[X]):

The mean of a probability distribution is given by $E[X] = \sum x_i P(X=x_i)$.

$E[X] = (0) \cdot P(X=0) + (1) \cdot P(X=1) + (2) \cdot P(X=2)$

$E[X] = (0) \cdot \frac{188}{221} + (1) \cdot \frac{32}{221} + (2) \cdot \frac{1}{221}$

$E[X] = 0 + \frac{32}{221} + \frac{2}{221} = \frac{34}{221}$.

Simplify the fraction $\frac{34}{221}$ by dividing numerator and denominator by 17:

$E[X] = \frac{\cancel{34}^{2}}{\cancel{221}^{13}} = \frac{2}{13}$.

Variance of the distribution (Var(X)):

The variance of a probability distribution is given by $Var(X) = E[X^2] - (E[X])^2$.

First, calculate $E[X^2]$:

$E[X^2] = \sum x_i^2 P(X=x_i)$

$E[X^2] = (0)^2 \cdot P(X=0) + (1)^2 \cdot P(X=1) + (2)^2 \cdot P(X=2)$

$E[X^2] = (0) \cdot \frac{188}{221} + (1) \cdot \frac{32}{221} + (4) \cdot \frac{1}{221}$

$E[X^2] = 0 + \frac{32}{221} + \frac{4}{221} = \frac{36}{221}$.

Now, calculate the variance:

$Var(X) = E[X^2] - (E[X])^2$

$Var(X) = \frac{36}{221} - \left(\frac{2}{13}\right)^2$

$Var(X) = \frac{36}{221} - \frac{4}{169}$.

Find the least common multiple of the denominators 221 and 169.

$221 = 13 \times 17$

$169 = 13^2$

$LCM(221, 169) = 13^2 \times 17 = 169 \times 17 = 2873$.

$Var(X) = \frac{36}{221} \cdot \frac{13}{13} - \frac{4}{169} \cdot \frac{17}{17}$

$Var(X) = \frac{36 \times 13}{2873} - \frac{4 \times 17}{2873}$

$Var(X) = \frac{468}{2873} - \frac{68}{2873}$

$Var(X) = \frac{468 - 68}{2873} = \frac{400}{2873}$.

The probability distribution is given in the table, the mean is $\frac{2}{13}$ and the variance is $\frac{400}{2873}$.

Formulation of the Linear Programming Problem:

Let $x$ be the quantity of Food 'I' (in kg) in the mixture.

Let $y$ be the quantity of Food 'II' (in kg) in the mixture.

The objective is to minimize the cost of the mixture. The cost of Food 'I' is $\textsf{₹} 50$/kg, and the cost of Food 'II' is $\textsf{₹} 70$/kg.

Objective function: Minimize $Z = 50x + 70y$

The constraints are based on the vitamin contents:

Vitamin A content:

Food 'I' has 2 units/kg, Food 'II' has 1 unit/kg. The mixture must have at least 8 units of Vitamin A.

$2x + y \ge 8$

Vitamin C content:

Food 'I' has 1 unit/kg, Food 'II' has 2 units/kg. The mixture must have at least 10 units of Vitamin C.

$x + 2y \ge 10$

Non-negativity constraints:

The quantities of food cannot be negative.

$x \ge 0$

$y \ge 0$

Thus, the linear programming problem is:

Minimize $Z = 50x + 70y$

Subject to:

$2x + y \ge 8$

$x + 2y \ge 10$

$x \ge 0, y \ge 0$

Graphical Solution:

We need to plot the feasible region defined by the constraints.

Consider the boundary lines:

L1: $2x + y = 8$. Points on this line: If $x=0$, $y=8$ (0, 8). If $y=0$, $2x=8$, $x=4$ (4, 0).

L2: $x + 2y = 10$. Points on this line: If $x=0$, $2y=10$, $y=5$ (0, 5). If $y=0$, $x=10$ (10, 0).

L3: $x = 0$ (y-axis)

L4: $y = 0$ (x-axis)

For $2x + y \ge 8$, the region is above or to the right of L1 (away from the origin, since (0,0) gives $0 < 8$).

For $x + 2y \ge 10$, the region is above or to the right of L2 (away from the origin, since (0,0) gives $0 < 10$).

For $x \ge 0$ and $y \ge 0$, the region is in the first quadrant.

The feasible region is the unbounded region satisfying all these conditions.

Find the corner points of the feasible region:

1. Intersection of $y = 0$ and $2x + y = 8$: $2x + 0 = 8 \implies x = 4$. Point: (4, 0).

2. Intersection of $x = 0$ and $x + 2y = 10$: $0 + 2y = 10 \implies y = 5$. Point: (0, 5).

3. Intersection of $2x + y = 8$ and $x + 2y = 10$:

From $2x + y = 8$, $y = 8 - 2x$.

Substitute into $x + 2y = 10$: $x + 2(8 - 2x) = 10$

$x + 16 - 4x = 10$

$-3x = -6 \implies x = 2$.

Substitute $x=2$ into $y = 8 - 2x$: $y = 8 - 2(2) = 8 - 4 = 4$. Point: (2, 4).

The corner points of the feasible region are (4, 0), (2, 4), and (0, 5).

Evaluate the objective function $Z = 50x + 70y$ at each corner point:

At (4, 0): $Z = 50(4) + 70(0) = 200 + 0 = 200$.

At (2, 4): $Z = 50(2) + 70(4) = 100 + 280 = 380$.

At (0, 5): $Z = 50(0) + 70(5) = 0 + 350 = 350$.

The minimum value among the corner points is 200.

Since the feasible region is unbounded and the coefficients of $x$ and $y$ in the objective function ($50 > 0$ and $70 > 0$) are positive, the minimum value found at a corner point is the minimum value over the entire feasible region.

The minimum cost is $\textsf{₹} 200$, which occurs when $x = 4$ kg of Food 'I' and $y = 0$ kg of Food 'II' are used.