Engineering Mathematics – GATE PI Engineering MathematicsPractice Questions & PYQs

This problem tests your understanding of the normal distribution's probability density function (PDF) and its key properties.

The given PDF is $f(x)=\frac{1}{\sqrt{8\pi}}e^{-\left \{ \frac{(x\;-\;1)^2}{8} \right \}}$. We need to calculate the definite integral $\int_{1}^{\infty }f(x)dx$.

First, let's identify the parameters of this normal distribution. The general form of a normal distribution PDF is:
$f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
Comparing the exponent terms of the given PDF and the general form:
$\frac{(x\;-\;1)^2}{8} = \frac{(x-\mu)^2}{2\sigma^2}$
From this, we can directly see that the mean $\mu = 1$.
Also, $2\sigma^2 = 8$, which implies $\sigma^2 = 4$. Therefore, the standard deviation $\sigma = \sqrt{4} = 2$.
We can verify this with the coefficient term: $\frac{1}{\sqrt{8\pi}} = \frac{1}{\sigma\sqrt{2\pi}}$. Substituting $\sigma = 2$ gives $\frac{1}{2\sqrt{2\pi}}$, confirming our $\sigma$ value.
So, X follows a normal distribution with mean $\mu = 1$ and standard deviation $\sigma = 2$.

The integral $\int_{1}^{\infty }f(x)dx$ represents the probability $P(X \ge 1)$. A crucial property of any normal distribution is its symmetry around its mean ($\mu$). The total area under the PDF curve is 1, i.e., $\int_{-\infty }^{\infty } f(x) dx = 1$. Due to symmetry, the area to the left of the mean is equal to the area to the right of the mean:
$P(X < \mu) = P(X > \mu) = \frac{1}{2}$
Since our mean $\mu = 1$, we have $P(X < 1) = P(X > 1) = \frac{1}{2}$.
For a continuous distribution, $P(X \ge 1) = P(X > 1)$.
Therefore, the integral we need to calculate is:
$\int_{1}^{\infty } f(x) dx = P(X \ge 1) = P(X > 1) = \frac{1}{2}$

The Newton-Raphson method helps us find the roots of an equation iteratively. The formula for the next approximation $x_{n+1}$ from the current approximation $x_n$ is given by:
$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
First, we express the given equation $2 \sin x = x$ as $f(x) = 0$, so $f(x) = 2 \sin x - x$.
Next, we find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(2 \sin x - x) = 2 \cos x - 1$.
Given the initial guess $x_0 = 2.0$ radians, we need to calculate $x_1$.
Step 1: Calculate $f(x_0)$.
$f(2.0) = 2 \sin(2.0) - 2.0 \approx 2 \times 0.909297 - 2.0 = 1.818594 - 2.0 = -0.181406$.
Step 2: Calculate $f'(x_0)$.
$f'(2.0) = 2 \cos(2.0) - 1 \approx 2 \times (-0.416147) - 1 = -0.832294 - 1 = -1.832294$.
Step 3: Calculate $x_1$ using the Newton-Raphson formula.
$x_1 = 2.0 - \frac{-0.181406}{-1.832294} = 2.0 - 0.099004 \approx 1.900996$.
The value of $x$ after one iteration is approximately $1.900996$, which is closest to $1.901$.

When a fair coin is tossed, the probability of getting a Head (H) is $P(H) = \frac{1}{2}$, and the probability of getting a Tail (T) is $P(T) = \frac{1}{2}$. The question asks for the probability that a head does not turn up in any of the $N$ tosses, which means every toss must result in a tail. Since each toss is an independent event, the probability of getting tails in all $N$ tosses is the product of the probabilities of getting a tail in each individual toss:
$P(\text{No Heads \in N tosses}) = P(\text{Tails on all N tosses}) = P(T) \times P(T) \times \dots \times P(T) \quad (\text{N \times })$
$P(\text{No Heads \in N tosses}) = \left ( \frac{1}{2} \right ) \times \left ( \frac{1}{2} \right ) \times \dots \times \left ( \frac{1}{2} \right ) \quad (\text{N \times })$
$P(\text{No Heads \in N tosses}) = \left ( \frac{1}{2} \right )^{N}$
This result directly matches option C.

To find the probability of getting a correct sequence, we need to divide the number of favorable outcomes (correct sequences) by the total number of possible outcomes (all distinct sequences).

Given:

• Total number of distinct components = 5
• Number of correct sequential assembly ways = 2

The total number of possible sequences for 5 distinct components is given by $5!$:
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
The number of favorable sequences (correct assembly ways) is 2.

Using the probability formula, $P(\text{Event}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$:
$P(\text{Correct Sequence}) = \frac{2}{5!}$

Work is done when a force causes an object to move in the direction of the force. When a constant force $\vec{f}$ displaces an object by a distance $\vec{d}$, and the angle between the force and displacement is $\theta$, the work done ($W$) is given by the dot product of the force and displacement vectors. This is expressed as $W = \vec{f}.\vec{d} = fd\cos\theta$. Therefore, the work done by a force is the scalar product of the force and the displacement. The SI unit for work is Joule (J).

Understanding how matrices transform vectors is crucial for competitive exams like GATE. A linear transformation matrix, when multiplied by a vector

, shows us how the vector's components change, which in turn reveals its geometric interpretation.

Let's break down each matrix:

For P:
The matrix is

. When we apply this to a vector

:

.
Notice that the $y$-component becomes $0$, while the $x$-component remains $x$. This means any point $(x, y)$ is mapped to $(x, 0)$, which is a projection onto the x-axis. So, P-3.

For Q:
The matrix is

. Applying it to

:

.
Here, the $x$-component becomes $0$, and the $y$-component remains $y$. This maps $(x, y)$ to $(0, y)$, representing a projection onto the y-axis. So, Q-4.

For R:
The matrix is

. Let's see its effect on

:

.
The $x$-component stays the same ($x$), but the $y$-component is multiplied by $3$ ($3y$). This is a stretch in the y-axis by a factor of $3$. So, R-1.

For S:
The matrix is

. Applying it to

:

.
Both the $x$-component and the $y$-component are scaled by the same factor of $4$. This is a uniform stretch in x and y-axes. So, S-2.

Combining these results, we get P-3, Q-4, R-1, S-2.

To find the curve represented by the solution to the differential equation $6yy' - 25x = 0$, we first rewrite $y'$ as $\frac{dy}{dx}$:
$6y \frac{dy}{dx} - 25x = 0$
Next, we separate the variables by moving the $x$ term to the right side and arranging the differentials:
$6y \frac{dy}{dx} = 25x$
$6y \, dy = 25x \, dx$
Now, we integrate both sides:
$\int 6y \, dy = \int 25x \, dx$
Performing the integration, we get:
$6 \cdot \frac{y^2}{2} = 25 \cdot \frac{x^2}{2} + C_1$
Simplifying the equation yields:
$3y^2 = \frac{25}{2}x^2 + C_1$
Rearranging the terms to the standard form of conic sections, we have:
$3y^2 - \frac{25}{2}x^2 = C_1$
This equation is in the form $Ay^2 + Bx^2 = C$. Here, the coefficient of $y^2$ is $A=3$ (positive), and the coefficient of $x^2$ is $B=-\frac{25}{2}$ (negative). Since the coefficients of $x^2$ and $y^2$ have opposite signs, the equation represents a hyperbola. Depending on the value of $C_1$, this is either a hyperbola opening along the y-axis (if $C_1 > 0$), a hyperbola opening along the x-axis (if $C_1 < 0$), or a pair of intersecting lines (if $C_1 = 0$, which is a degenerate hyperbola). Therefore, the solution represents a family of hyperbolas.

This problem involves a special type of differential equation called a Cauchy-Euler equation. These equations have the form $ax^2y'' + bxy' + cy = 0$. For such equations, we assume a solution of the form $y = x^r$.

Given the differential equation: $x^2y'' + xy' - y = 0$.

1. Assume a solution form: Let $y = x^r$.
2. Find derivatives:
   $y' = \frac{dy}{dx} = rx^{r-1}$
   $y'' = \frac{d^2y}{dx^2} = r(r-1)x^{r-2}$
3. Substitute into the differential equation:
   $x^2(r(r-1)x^{r-2}) + x(rx^{r-1}) - x^r = 0$
4. Simplify and form the characteristic equation:
   $r(r-1)x^r + rx^r - x^r = 0$
   Factor out $x^r$ (assuming $x \neq 0$):
   $x^r [r(r-1) + r - 1] = 0$
   Since $x^r \neq 0$, the expression in the brackets must be zero:
   $r(r-1) + r - 1 = 0$
5. Solve the characteristic equation for 'r':
   $r^2 - r + r - 1 = 0$
   $r^2 - 1 = 0$
   $(r - 1)(r + 1) = 0$
   This gives us two distinct real roots: $r_1 = 1$ and $r_2 = -1$.
6. Form the general solution: For distinct real roots $r_1$ and $r_2$, the general solution to a Cauchy-Euler equation is $y = c_1 x^{r_1} + c_2 x^{r_2}$.
   Substituting our roots:
   $y = c_1 x^1 + c_2 x^{-1}$
   This can be written as:
   $y = c_1x + \frac{c_2}{x}$

Comparing this derived solution with the given options, it matches option C.

The function is defined as $f(x) = x + x + x + \dots$ ($x$ times). This means $x$ is added to itself $x$ times, which can be written as the summation: $f(x) = \sum_{i=1}^{x} x$ This simplifies to $f(x) = x × x = x^2$. However, the crucial part is the phrase "($x$ \times )". In mathematics, performing an operation a certain "number of \times " implies that this number must be a positive integer. We can add something 3 \times or 5 \times , but not $\sqrt{2}$ \times or $-4$ \times \in this context. Therefore, for the expression "$x + x + x + \dots$ ($x$ \times )" to be meaningful, $x$ must be a positive integer ($x \in \{1, 2, 3, \dots\}$).

• Option A (All real values of $x$): Incorrect. While $x^2$ is defined for all real numbers, the operation "added $x$ times" requires $x$ to be a positive integer.
• Option B (Only at positive integer values of $x$): Correct. This aligns with the requirement that the number of times an operation is performed must be a positive integer.
• Option C (Only at negative integer values of $x$): Incorrect. Adding something '$-3$ times' is not a standard interpretation for such a definition.
• Option D (Only at rational values of $x$): Incorrect. Rational numbers include fractions and negative numbers, which are not allowed for the "($x$ times)" constraint.

Simpson's rule is a numerical integration technique used to approximate definite integrals. Unlike simpler methods, Simpson's rule approximates the function within each pair of subintervals using a parabola, which is a quadratic curve defined by a second-degree polynomial. By fitting a parabola through three points (the two endpoints and the midpoint of the interval), it integrates the area under this quadratic approximation. This approach generally yields more accurate results than methods like the Trapezoidal rule (which uses straight lines) or the Rectangle rule (which uses constants).

We need to find the value of the limit: $\text{lim}_{(x,y) \to (0,0)} \frac{x^2 - xy}{\sqrt{x} - \sqrt{y}}$ As $(x,y)$ approaches $(0,0)$, the expression becomes the indeterminate form $\frac{0}{0}$. The square roots imply $x ≥ 0$ and $y ≥ 0$, and the denominator requires $\sqrt{x} - \sqrt{y} \neq 0$, so $x \neq y$.

To simplify, first factor the numerator: $x^2 - xy = x(x-y)$. The expression becomes:
$\frac{x(x-y)}{\sqrt{x} - \sqrt{y}}$
Next, rationalize the denominator by multiplying the numerator and denominator by its conjugate, $\sqrt{x} + \sqrt{y}$:
$\frac{x(x-y)}{\sqrt{x} - \sqrt{y}} \times \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x} + \sqrt{y}}$
The denominator simplifies to $(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) = (\sqrt{x})^2 - (\sqrt{y})^2 = x - y$.
So the expression becomes:
$\frac{x(x-y)(\sqrt{x} + \sqrt{y})}{x - y}$
Since we are taking the limit as $(x,y) \to (0,0)$ but $(x,y) \neq (0,0)$, we can assume $x \neq y$ and cancel the $(x-y)$ term:
$x(\sqrt{x} + \sqrt{y})$
Now, substitute $x=0$ and $y=0$ into the simplified expression to evaluate the limit:
$\text{lim}_{(x,y) \to (0,0)} x(\sqrt{x} + \sqrt{y}) = 0(\sqrt{0} + \sqrt{0}) = 0(0 + 0) = 0$

To determine the concavity of the curve $y = x^4$, we need to find its second derivative. First, calculate the first derivative, $y'$, which describes the slope: $y' = \frac{d}{dx}(x^4) = 4x^{4-1} = 4x^3$ Next, calculate the second derivative, $y''$, which indicates the concavity: $y'' = \frac{d}{dx}(4x^3) = 4 \times 3x^{3-1} = 12x^2$ Now, we apply the second derivative test: if $y'' > 0$, the curve is concave up; if $y'' < 0$, it's concave down. For $y'' = 12x^2$:

• If $x > 0$, $x^2$ is positive, so $12x^2 > 0$.
• If $x < 0$, $x^2$ is positive, so $12x^2 > 0$.
• If $x = 0$, $y'' = 12(0)^2 = 0$.
  Since $x^2 \ge 0$ for all real $x$, it follows that $y'' = 12x^2 \ge 0$ for all values of $x$. As the second derivative is always non-negative, the curve $y = x^4$ is concave up for all values of $x$.

The Simpson's 1/3rd rule is used to approximate definite integrals by fitting quadratic curves through sets of three points. The general formula for Simpson's 1/3rd rule is:
\int_{x_0}^{x_n} ydx = \frac{h}{3}\left\{ {\left( {{y_0} + {y_n} \right) + 4\left( {{y_1} + {y_3} + {y_5} + \ldots } \right) + 2\left( {{y_2} + {y_4} + {y_6} + \ldots } \right)} \right\}
where $h = \frac{{b - a}}{n}$. For a polynomial function of degree less than 3, this rule gives the exact integral value.

Given the data:
$x$: 1, 3, 5
$y$: 2, 6, 4
Here, $a = 1$, $b = 5$, and $n = 2$.
First, calculate $h$: $h = \frac{{b - a}}{n} = \frac{{5 - 1}}{2} = 2$.
From the given data, we have $y_0 = 2$, $y_1 = 6$, and $y_2 = 4$.
Applying Simpson's 1/3rd rule for three points ($n=2$), the formula simplifies to:
$\int ydx = \frac{h}{3}\left[ {{y_0} + 4{y_1} + {y_2}} \right]$
Substitute the values:
$\int ydx = \frac{{2}}{3}\left[ {2+ 4 (6)+4} \right] = 20$

To find the direction where the directional derivative of a scalar function $\phi$ is maximum, we need to calculate its gradient vector, $\nabla\phi$, at the given point. The gradient vector points in the direction of the maximum rate of increase of the function.

Given the function $\phi = 2xz - y^2$, its gradient $\nabla\phi$ is found by taking partial derivatives:
$\nabla\phi = \frac{\partial\phi}{\partial x}\mathbf{i} + \frac{\partial\phi}{\partial y}\mathbf{j} + \frac{\partial\phi}{\partial z}\mathbf{k}$
Calculating the partial derivatives:
$\frac{\partial\phi}{\partial x} = 2z$
$\frac{\partial\phi}{\partial y} = -2y$
$\frac{\partial\phi}{\partial z} = 2x$
So, the gradient vector is $\nabla\phi = (2z)\mathbf{i} - (2y)\mathbf{j} + (2x)\mathbf{k}$.

Next, we evaluate this gradient at the point (1, 3, 2) by substituting $x=1$, $y=3$, and $z=2$:
$x$-component: $2z = 2(2) = 4$
$y$-component: $-2y = -2(3) = -6$
$z$-component: $2x = 2(1) = 2$
Thus, the gradient vector at (1, 3, 2) is $\nabla\phi\bigg|_{(1,3,2)} = 4\mathbf{i} - 6\mathbf{j} + 2\mathbf{k}$.

This vector, $4\mathbf{i} - 6\mathbf{j} + 2\mathbf{k}$, represents the direction in which the directional derivative of $\phi$ is maximum at the point (1, 3, 2). Comparing this with the given options, it matches Option B.

To find $\nabla^2\varphi$ for the given scalar function $\varphi = 2x^3y^2z^4$, we need to compute the Laplacian. The Laplacian operator in Cartesian coordinates is defined as the sum of the second partial derivatives with respect to each spatial variable:
$\nabla^2\varphi = \frac{\partial^2\varphi}{\partial x^2} + \frac{\partial^2\varphi}{\partial y^2} + \frac{\partial^2\varphi}{\partial z^2}$
Let's calculate each term step-by-step.

1. Calculate $\frac{\partial^2\varphi}{\partial x^2}$:
First, we find the partial derivative with respect to $x$:
$\frac{\partial\varphi}{\partial x} = \frac{\partial}{\partial x} (2x^3y^2z^4) = 2y^2z^4 \cdot (3x^2) = 6x^2y^2z^4$
Next, we differentiate this result again with respect to $x$:
$\frac{\partial^2\varphi}{\partial x^2} = \frac{\partial}{\partial x} (6x^2y^2z^4) = 6y^2z^4 \cdot (2x) = 12xy^2z^4$

2. Calculate $\frac{\partial^2\varphi}{\partial y^2}$:
First, we find the partial derivative with respect to $y$:
$\frac{\partial\varphi}{\partial y} = \frac{\partial}{\partial y} (2x^3y^2z^4) = 2x^3z^4 \cdot (2y) = 4x^3yz^4$
Next, we differentiate this result again with respect to $y$:
$\frac{\partial^2\varphi}{\partial y^2} = \frac{\partial}{\partial y} (4x^3yz^4) = 4x^3z^4 \cdot (1) = 4x^3z^4$

3. Calculate $\frac{\partial^2\varphi}{\partial z^2}$:
First, we find the partial derivative with respect to $z$:
$\frac{\partial\varphi}{\partial z} = \frac{\partial}{\partial z} (2x^3y^2z^4) = 2x^3y^2 \cdot (4z^3) = 8x^3y^2z^3$
Next, we differentiate this result again with respect to $z$:
$\frac{\partial^2\varphi}{\partial z^2} = \frac{\partial}{\partial z} (8x^3y^2z^3) = 8x^3y^2 \cdot (3z^2) = 24x^3y^2z^2$

4. Sum the partial derivatives:
Finally, we add the three second partial derivatives to get the Laplacian:
$\nabla^2\varphi = 12xy^2z^4 + 4x^3z^4 + 24x^3y^2z^2$
Comparing this result with the given options, we find it matches Option C.

We need to find the mean, median, and mode for the given marks distribution.

1. Calculating the Mean ($\bar{x}$):
The mean is the average, calculated using the formula: $\text{Mean} (\bar{x}) = \frac{\sum fx}{\sum f}$
First, we calculate $fx$ for each mark and sum them up, along with the total frequency (\sum f$).

| Marks (x) | Students (f) | $fx$ |
| :-------- | :----------- | :--- |
| 25 | 20 | $25 \times 20 = 500$ |
| 30 | 20 | $30 \times 20 = 600$ |
| 35 | 40 | $35 \times 40 = 1400$ |
| 40 | 20 | $40 \times 20 = 800$ |
| Total | $\sum f = 100$ | $\sum fx = 3300$ |

Now, substitute these values into the mean formula: $\text{Mean} (\bar{x}) = \frac{3300}{100} = 33$
So, the mean marks are 33.

2. Determining the Median:
The median is the middle value. For grouped data, we find the cumulative frequency. The total number of students ($N$) is 100. Since $N$ is even, the median is the average of the values at the $\left(\frac{N}{2}\right)^{th}$ and $\left(\frac{N}{2} + 1\right)^{th}$ positions.
$\frac{N}{2} = \frac{100}{2} = 50^{th}$ position.
$\left(\frac{N}{2} + 1\right) = \left(\frac{100}{2} + 1\right) = 51^{st}$ position.

Let's construct the cumulative frequency table:

| Marks (x) | Students (f) | Cumulative Frequency (cf) |
| :-------- | :----------- | :------------------------ |
| 25 | 20 | 20 |
| 30 | 20 | $20 + 20 = 40$ |
| 35 | 40 | $40 + 40 = 80$ |
| 40 | 20 | $80 + 20 = 100$ |

From the table, the 40th student has marks up to 30. Students from the 41st to the 80th position have marks 35. Therefore, both the $50^{th}$ and $51^{st}$ observations correspond to marks of 35.
$\text{Median} = \frac{50^{th} \text{ observation} + 51^{st} \text{ observation}}{2} = \frac{35 + 35}{2} = \frac{70}{2} = 35$
So, the median marks are 35.

3. Identifying the Mode:
The mode is the value with the highest frequency. We look at the 'Number of students (f)' column in the original table:

• Marks 25: Frequency 20
• Marks 30: Frequency 20
• Marks 35: Frequency 40
• Marks 40: Frequency 20

The highest frequency is 40, which corresponds to marks of 35.
Thus, the mode is 35.

Summary:
Mean = 33
Median = 35
Mode = 35

To find the standard error of the mean, we first need to understand the relationship between the number of observations, the coefficient of variation, and the allowable percentage error. The number of observations, $N$, is given by the formula $N = {\left( {\frac{{{C_V}}}{\varepsilon }} \right)^2}$, where $\varepsilon$ is the allowable percentage error of the mean and $C_V$ is the coefficient of variation. The coefficient of variation is calculated as $C_v= \frac{\sigma}{\bar{x}} \times 100$, where $\sigma$ is the standard deviation and $\bar{x}$ is the mean.

Given $\bar{x} = 80$, $\sigma = 12$, and $N = 100$:

First, calculate the coefficient of variation ($C_V$):
${C_v} = \frac{12}{80} \times 100 = 15\%$

Next, use the formula for $N$ to find $\varepsilon$:
$100 = {\left( {\frac{{15}}{\varepsilon }} \right)^2}$
$\varepsilon = 1.5\%$

Finally, the standard error of the mean is calculated by multiplying the mean by the allowable percentage error (expressed as a decimal):
$\text{Standard error of mean} = 80 \times \frac{1.5}{100} = 1.2$

To determine the nature of solutions for a system of linear equations $AX = B$, we analyze the determinant of the coefficient matrix $A$. If $\det(A) \neq 0$, the system has a unique solution. If $\det(A) = 0$, it could have no solution or infinitely many solutions.

First, let's form the coefficient matrix $A$ from the given equations:
$A = \begin{pmatrix} 1 & 2 & 4 \\ 1 & 3 & 1 \\ 3 & 2 & 3 \end{pmatrix}$
Now, we calculate the determinant of $A$:
$\det(A) = 1 \cdot \begin{vmatrix} 3 & 1 \\ 2 & 3 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & 1 \\ 3 & 3 \end{vmatrix} + 4 \cdot \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix}$
$\det(A) = 1((3 \times 3) - (1 \times 2)) - 2((1 \times 3) - (1 \times 3)) + 4((1 \times 2) - (3 \times 3))$
$\det(A) = 1(9 - 2) - 2(3 - 3) + 4(2 - 9)$
$\det(A) = 1(7) - 2(0) + 4(-7)$
$\det(A) = 7 - 0 - 28 = -21$
Since $\det(A) = -21 \neq 0$, the system of linear equations has a unique solution.

This problem requires us to find the probability that it did not rain, given that the fair did not make a loss. This is a classic application of Bayes' Theorem.

First, let's define our events and list the given probabilities:

• $R$: It rains.
• $R'$: It does not rain.
• $L$: Fair makes a loss.
• $L'$: Fair does not make a loss.

Given:

• $P(R) = 70\% = 0.70$
• $P(R') = 1 - P(R) = 1 - 0.70 = 0.30$
• $P(L | R) = 80\% = 0.80$ (Loss given rain)
• $P(L | R') = 10\% = 0.10$ (Loss given no rain)

Next, we calculate the probabilities of the fair not making a loss:

• $P(L' | R) = 1 - P(L | R) = 1 - 0.80 = 0.20$ (No loss given rain)
• $P(L' | R') = 1 - P(L | R') = 1 - 0.10 = 0.90$ (No loss given no rain)

Now, we find the total probability of the fair not making a loss, $P(L')$, using the law of total probability:
$P(L') = P(L' | R) \times P(R) + P(L' | R') \times P(R')$
$P(L') = (0.20 \times 0.70) + (0.90 \times 0.30)$
$P(L') = 0.14 + 0.27$
$P(L') = 0.41$

Finally, we apply Bayes' Theorem to find $P(R' | L')$ (probability of no rain given no loss):
$P(R' | L') = \frac{P(L' | R') \times P(R')}{P(L')}$
$P(R' | L') = \frac{0.90 \times 0.30}{0.41}$
$P(R' | L') = \frac{0.27}{0.41}$
The probability is $\frac{27}{41}$.

Let's solve the equation $\sin(x) = x^2$ using the Newton-Raphson method. This method helps us find roots of an equation $f(x)=0$ iteratively. The formula for each iteration is:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

First, we need to define our function $f(x)$. From $\sin(x) = x^2$, we get $f(x) = x^2 - \sin(x)$.
Next, we find its derivative, $f'(x) = 2x - \cos(x)$.

We are given the initial guess $x_0 = 1$. Remember, all trigonometric functions operate on radians, so ensure your calculator is in radian mode or convert degrees appropriately (as noted in the "Mistake Points").

First Iteration (n = 0):

Substitute $x_0 = 1$ into $f(x)$ and $f'(x)$:
$f(x_0) = f(1) = (1)^2 - \sin(1) = 1 - \sin(1) \approx 0.159$
$f'(x_0) = f'(1) = 2(1) - \cos(1) = 2 - \cos(1) \approx 1.459$

Now, apply the Newton-Raphson formula to find $x_1$:
$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{0.159}{1.459} \approx 1 - 0.10897875 \approx 0.891$

Second Iteration (n = 1):

Now we use $x_1 = 0.891$ for the next iteration.
Substitute $x_1 = 0.891$ into $f(x)$ and $f'(x)$:
$f(x_1) = f(0.891) = (0.891)^2 - \sin(0.891) \approx 0.793881 - 0.77818 \approx 0.015701 \approx 0.0161$ (Rounding difference in original explanation)
$f'(x_1) = f'(0.891) = 2(0.891) - \cos(0.891) = 1.782 - 0.6288 \approx 1.1532 \approx 1.153$

Finally, apply the Newton-Raphson formula to find $x_2$:
$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.891 - \frac{0.0161}{1.153} \approx 0.891 - 0.01396357 \approx 0.877036 \approx 0.87$

The value of $x$ after 2 iterations is approximately $0.87$.