Step 1: Understanding the Question:

The question asks about the nature of the optimal values (maximum and minimum) of an objective function in a Linear Programming Problem (LPP) when the feasible region is bounded.


Step 2: Key Formula or Approach:

This question is based on a fundamental theorem of linear programming. The theorem states that if the feasible region for an LPP is bounded, then the objective function \(Z\) has both a maximum and a minimum value on this region, and each of these occurs at a corner point (vertex) of the region.


Step 3: Detailed Explanation:

A bounded feasible region is a region that can be enclosed within a circle. It is a closed and bounded set of points.

According to the corner point theorem in LPP, the optimal solution (either maximum or minimum) must occur at one of the corner points of the feasible region.

Since the region is bounded, it has a finite number of corner points.

By evaluating the objective function at each of these corner points, we can find the maximum and minimum values.

A bounded region guarantees the existence of both a maximum and a minimum value for a continuous linear function like \(Z = ax + by\).


Step 4: Final Answer:

Therefore, if the feasible region is bounded, the objective function will have both a maximum and a minimum value.

This corresponds to option (C).
Quick Tip: Remember the key LPP theorems:
1. \textbf{Bounded Feasible Region:} Both maximum and minimum values exist.
2. \textbf{Unbounded Feasible Region:} An optimal value may or may not exist. If it exists, it will be at a corner point. A maximum might not exist if the region extends infinitely in the direction of increasing Z, and a minimum might not exist if it extends infinitely in the direction of decreasing Z.

Step 1: Understanding the Question:

We need to find a unit vector that is perpendicular to two given vectors, \( \vec{a} = \hat{i} - \hat{j} \) and \( \vec{b} = \hat{i} + \hat{j} \).


Step 2: Key Formula or Approach:

A vector perpendicular to two vectors \( \vec{a} \) and \( \vec{b} \) is given by their cross product, \( \vec{a} \times \vec{b} \).

A unit vector in the direction of a vector \( \vec{v} \) is given by \( \hat{v} = \dfrac{\vec{v}}{|\vec{v}|} \).


Step 3: Detailed Explanation:

Let the given vectors be \( \vec{a} = \hat{i} - \hat{j} + 0\hat{k} \) and \( \vec{b} = \hat{i} + \hat{j} + 0\hat{k} \).

First, we find the cross product \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -1 & 0
1 & 1 & 0 \end{vmatrix} \] \[ = \hat{i}((-1)(0) - (1)(0)) - \hat{j}((1)(0) - (1)(0)) + \hat{k}((1)(1) - (1)(-1)) \] \[ = \hat{i}(0) - \hat{j}(0) + \hat{k}(1 - (-1)) \] \[ = 2\hat{k} \]
Let \( \vec{c} = 2\hat{k} \). This vector is perpendicular to both \( \vec{a} \) and \( \vec{b} \).

Next, we find the unit vector in the direction of \( \vec{c} \). We need the magnitude of \( \vec{c} \):
\[ |\vec{c}| = |2\hat{k}| = \sqrt{0^2 + 0^2 + 2^2} = \sqrt{4} = 2 \]
The unit vector \( \hat{c} \) is:
\[ \hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{2\hat{k}}{2} = \hat{k} \]

Step 4: Final Answer:

The unit vector perpendicular to the given vectors is \( \hat{k} \).

This corresponds to option (A).
Quick Tip: The cross product \( \vec{a} \times \vec{b} \) gives a vector perpendicular to the plane containing \( \vec{a} \) and \( \vec{b} \). Remember that \( -(\vec{a} \times \vec{b}) \) is also perpendicular. In this case, \( -\hat{k} \) is also a correct answer, but only \( \hat{k} \) is listed as an option among (A) and (B).

Step 1: Understanding the Question:

We are given the value of a definite integral, \( \alpha \), and asked to find the value of a related definite integral. This suggests using techniques like integration by parts to connect the two integrals.


Step 2: Key Formula or Approach:

We will use integration by parts on the given integral \( \alpha = \displaystyle \int_{0}^{1} \frac{e^x}{1+x}\,dx \).

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \).


Step 3: Detailed Explanation:

Let's apply integration by parts to the given integral \( \alpha \).

Let \( u = \dfrac{1}{1+x} \) and \( dv = e^x \, dx \).

Then, \( du = -\dfrac{1}{(1+x)^2} \, dx \) and \( v = \int e^x \, dx = e^x \).

Using the formula for definite integration by parts \( \int_{a}^{b} u \, dv = [uv]_{a}^{b} - \int_{a}^{b} v \, du \):
\[ \alpha = \int_{0}^{1} \frac{e^x}{1+x}\,dx = \left[ \frac{1}{1+x} \cdot e^x \right]_{0}^{1} - \int_{0}^{1} e^x \left( -\frac{1}{(1+x)^2} \right) \,dx \]
First, evaluate the term in the brackets:
\[ \left[ \frac{e^x}{1+x} \right]_{0}^{1} = \left( \frac{e^1}{1+1} \right) - \left( \frac{e^0}{1+0} \right) = \frac{e}{2} - \frac{1}{1} = \frac{e}{2} - 1 \]
Now substitute this back into the equation for \( \alpha \):
\[ \alpha = \left( \frac{e}{2} - 1 \right) - \int_{0}^{1} -\frac{e^x}{(1+x)^2} \,dx \] \[ \alpha = \frac{e}{2} - 1 + \int_{0}^{1} \frac{e^x}{(1+x)^2} \,dx \]
We need to find the value of \( \displaystyle \int_{0}^{1} \frac{e^x}{(1+x)^2}\,dx \). Let's rearrange the equation to solve for it:
\[ \int_{0}^{1} \frac{e^x}{(1+x)^2} \,dx = \alpha - \left( \frac{e}{2} - 1 \right) = \alpha - \frac{e}{2} + 1 \]

Step 4: Final Answer:

The value of the integral is \( \alpha + 1 - \dfrac{e}{2} \).

This corresponds to option (B).
Quick Tip: Look for the pattern \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \). Here, if \( f(x) = \frac{1}{1+x} \), then \( f'(x) = -\frac{1}{(1+x)^2} \). The problem involves these two terms. Applying integration by parts on \( \int e^x f(x) \, dx \) often leads to an expression involving \( \int e^x f'(x) \, dx \).

Step 1: Understanding the Question:

We are given an indefinite integral and its result in terms of a constant \(k\). We need to evaluate the integral to find the value of \(k\).


Step 2: Key Formula or Approach:

This integral can be solved using the method of substitution. We will also need the standard integral formula \( \int a^u \, du = \dfrac{a^u}{\ln a} + C \).


Step 3: Detailed Explanation:

Let the integral be \( I = \displaystyle \int \frac{2^{\frac{1}{x}}}{x^2}\,dx \).

Let's use substitution. Let \( t = \dfrac{1}{x} \).

Differentiating both sides with respect to \(x\):
\[ \frac{dt}{dx} = -\frac{1}{x^2} \]
Rearranging, we get:
\[ dt = -\frac{1}{x^2} \, dx \quad or \quad -dt = \frac{1}{x^2} \, dx \]
Now substitute \(t\) and \(-dt\) into the integral \(I\):
\[ I = \int 2^t (-dt) = - \int 2^t \, dt \]
Using the formula \( \int a^t \, dt = \dfrac{a^t}{\ln a} + C \), with \( a=2 \):
\[ I = - \left( \frac{2^t}{\ln 2} \right) + C \]
In calculus, \( \log \) is often used to denote the natural logarithm \( \ln \). So, \( \ln 2 = \log 2 \).
\[ I = - \frac{2^t}{\log 2} + C \]
Now, substitute back \( t = \dfrac{1}{x} \):
\[ I = - \frac{2^{\frac{1}{x}}}{\log 2} + C = \left( -\frac{1}{\log 2} \right) 2^{\frac{1}{x}} + C \]
We are given that \( I = k\,2^{\frac{1}{x}} + C \).

Comparing the two expressions, we can see that:
\[ k = -\frac{1}{\log 2} \]

Step 4: Final Answer:

The value of \(k\) is \( -\dfrac{1}{\log 2} \).

This corresponds to option (A).
Quick Tip: When you see a complex function in an integral like \(2^{1/x}\), check if its derivative is also present. The derivative of \(1/x\) is \(-1/x^2\), and a factor of \(1/x^2\) is present in the integrand. This is a strong indicator that substitution is the correct method.

Step 1: Understanding the Question:

The question asks for the inverse of a given 3x3 matrix \(A\).


Step 2: Key Formula or Approach:

The given matrix \(A\) is a diagonal matrix. The inverse of a diagonal matrix is another diagonal matrix where each diagonal element is the reciprocal of the corresponding element in the original matrix.
Alternatively, we can check the property \( A \cdot A^{-1} = I \), where \(I\) is the identity matrix.


Step 3: Detailed Explanation:

Method 1: Using the property of diagonal matrices

The matrix \(A\) is a diagonal matrix: \[ A = \begin{bmatrix} -1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} \]
The inverse \(A^{-1}\) is found by taking the reciprocal of each diagonal element: \[ A^{-1} = \begin{bmatrix} (-1)^{-1} & 0 & 0
0 & (1)^{-1} & 0
0 & 0 & (1)^{-1} \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} \]
We observe that \( A^{-1} = A \).


Method 2: Verifying \( A \cdot A = I \)

Since option (D) is the matrix \(A\) itself, let's check if \( A \cdot A = I \). If it is, then \(A\) is its own inverse.
\[ A \cdot A = \begin{bmatrix} -1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} (-1)(-1) + 0 + 0 & 0+0+0 & 0+0+0
0+0+0 & 0(0)+1(1)+0(0) & 0+0+0
0+0+0 & 0+0+0 & 0(0)+0(0)+1(1) \end{bmatrix} \] \[ = \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} = I \]
Since \( A \cdot A = I \), it follows that \( A^{-1} = A \).


Step 4: Final Answer:

The inverse of the matrix \(A\) is \(A\) itself, which is \( \begin{bmatrix} -1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} \).

This corresponds to option (D).
Quick Tip: This matrix has special properties that lead to a quick solution.
1. \textbf{Diagonal Matrix Rule:} The inverse of a diagonal matrix is found by taking the reciprocal of each diagonal element. The reciprocals of -1, 1, and 1 are -1, 1, and 1, respectively, giving the original matrix back.
2. \textbf{Involutory Matrix Rule:} A matrix is called involutory if \(A^2 = I\). For such matrices, \(A^{-1} = A\). You can quickly verify here that \(A \cdot A = I\).
Recognizing either of these properties instantly solves the problem.

Step 1: Understanding the Question:

We are given an equation involving two equal 2x2 matrices. We need to find the values of \(x\) and \(y\) to calculate the value of the expression \(x-y\).


Step 2: Key Formula or Approach:

The principle of equality of matrices states that if two matrices are equal, their corresponding elements must be equal.


Step 3: Detailed Explanation:

By equating the corresponding elements of the two matrices, we get a system of four linear equations:

1) \( x+y = 9 \)

2) \( 3y = 4x+y \)

3) \( 3x = x+6 \)

4) \( x+3 = y \)


Let's solve these equations. Equation (3) is the simplest to start with:

From (3):
\[ 3x = x+6 \] \[ 2x = 6 \] \[ x = 3 \]
Now we can use this value of \(x\) to find \(y\). Let's use equation (4):

From (4):
\[ y = x+3 \] \[ y = 3+3 \] \[ y = 6 \]
We have found \(x=3\) and \(y=6\). We should verify these values using the other two equations to ensure consistency.

Check with (1): \(x+y = 3+6 = 9\). This is correct.

Check with (2): Left side is \(3y = 3(6) = 18\). Right side is \(4x+y = 4(3)+6 = 12+6 = 18\). This is also correct.

The values are consistent. Now we can calculate the required expression \(x-y\):
\[ x-y = 3-6 = -3 \]

Step 4: Final Answer:

The value of \(x-y\) is \(-3\).

This corresponds to option (B).
Quick Tip: When solving a system of equations derived from matrix equality, always start with the simplest equation (one with only one variable, if possible). After finding the values, it's a good practice to quickly check them against the other equations to avoid errors.

Step 1: Understanding the Question:

We are given the probabilities of two events, \(M\) and \(N\), and their intersection. We need to find the conditional probability of event \(M\) given that event \(N\) has occurred, denoted by \(P(M|N)\).


Step 2: Key Formula or Approach:

The formula for conditional probability is: \[ P(M|N) = \frac{P(M \cap N)}{P(N)} \]

Step 3: Detailed Explanation:

We are given the following values: \( P(M) = 0.6 \)
\( P(N) = 0.2 \)
\( P(M \cap N) = 0.15 \)

Substituting these values into the formula for \(P(M|N)\): \[ P(M|N) = \frac{0.15}{0.2} \]
To simplify the fraction, we can write it as: \[ P(M|N) = \frac{15/100}{2/10} = \frac{15}{100} \times \frac{10}{2} = \frac{15}{20} = \frac{3}{4} \]
The calculated value is \(3/4 = 0.75\). However, this is not among the given options:
(A) \(7/8 = 0.875\)

(B) \(2/5 = 0.4\)

(C) \(1/2 = 0.5\)

(D) \(2/3 \approx 0.667\)

There seems to be a typo in the question's data. In exam situations, such errors can occur. A common typo is a single digit error. Let's consider a plausible typo. If \( P(M \cap N) \) was \(0.10\) instead of \(0.15\), the calculation would be: \[ P(M|N) = \frac{0.10}{0.2} = \frac{1}{2} \]
This matches option (C). This is a very likely intended question. Let's proceed with this assumption.


Step 4: Final Answer:

Assuming a typo in the question where \(P(M \cap N) = 0.10\), the conditional probability is \( \dfrac{1}{2} \).

This corresponds to option (C).
Quick Tip: If your calculation result for a multiple-choice question does not match any of the options, double-check your arithmetic first. If it's still different, consider the possibility of a typo in the question's data. Try to see if a small change in one of the given numbers (e.g., 0.15 to 0.10) leads to one of the options. This can help you select the most probable intended answer.

Step 1: Understanding the Question:

The question asks to identify which of the given functions is not a homogeneous function.


Step 2: Key Formula or Approach:

A function \( F(x, y) \) is defined as homogeneous of degree \(n\) if, for any non-zero constant \( \lambda \), the following condition holds: \[ F(\lambda x, \lambda y) = \lambda^n F(x, y) \]
We will test each option against this definition.


Step 3: Detailed Explanation:

Let's check each option:

(A) \( F(x,y) = y^2-xy \)
\( F(\lambda x, \lambda y) = (\lambda y)^2 - (\lambda x)(\lambda y) = \lambda^2 y^2 - \lambda^2 xy = \lambda^2 (y^2 - xy) = \lambda^2 F(x,y) \).
This is a homogeneous function of degree 2.


(B) \( F(x,y) = 7x-3y \)
\( F(\lambda x, \lambda y) = 7(\lambda x) - 3(\lambda y) = \lambda(7x-3y) = \lambda^1 F(x,y) \).
This is a homogeneous function of degree 1.


(C) \( F(x,y) = \sin^2\left(\frac{y}{x}\right)+\frac{y}{x} \)
\( F(\lambda x, \lambda y) = \sin^2\left(\frac{\lambda y}{\lambda x}\right)+\frac{\lambda y}{\lambda x} = \sin^2\left(\frac{y}{x}\right)+\frac{y}{x} = \lambda^0 F(x,y) \).
This is a homogeneous function of degree 0.


(D) \( F(x,y) = \tan x-\sec y \)
\( F(\lambda x, \lambda y) = \tan(\lambda x) - \sec(\lambda y) \).
This expression cannot be written in the form \( \lambda^n (\tan x - \sec y) \). Therefore, it is not a homogeneous function.


Step 4: Final Answer:

The function \( \tan x - \sec y \) is not a homogeneous function.

This corresponds to option (D).
Quick Tip: A quick way to check for homogeneity: For polynomial functions, check if all terms have the same total degree. For example, in \(y^2-xy\), \(y^2\) has degree 2 and \(xy\) has degree \(1+1=2\). So it's homogeneous. Functions that can be expressed entirely in terms of \(y/x\) or \(x/y\) are homogeneous of degree 0.

Step 1: Understanding the Question:

We are given a relationship between three vectors and their magnitudes. We need to find the angle between two of these vectors, \( \vec{b} \) and \( \vec{c} \).


Step 2: Key Formula or Approach:

The angle \( \theta \) between two vectors \( \vec{b} \) and \( \vec{c} \) is related to their dot product by the formula: \[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta \]
We will use the given vector sum \( \vec a + \vec b + \vec c = \vec 0 \) to find the value of \( \vec{b} \cdot \vec{c} \).


Step 3: Detailed Explanation:

From the given relation, \( \vec a + \vec b + \vec c = \vec 0 \), we can isolate the vectors whose angle we need to find. \[ \vec b + \vec c = -\vec a \]
Now, we take the dot product of each side with itself, which is equivalent to squaring the magnitude: \[ |\vec b + \vec c|^2 = |-\vec a|^2 \] \[ (\vec b + \vec c) \cdot (\vec b + \vec c) = |\vec a|^2 \]
Expand the left side: \[ \vec b \cdot \vec b + 2(\vec b \cdot \vec c) + \vec c \cdot \vec c = |\vec a|^2 \]
Using the property \( \vec{v} \cdot \vec{v} = |\vec{v}|^2 \): \[ |\vec b|^2 + 2(\vec b \cdot \vec c) + |\vec c|^2 = |\vec a|^2 \]
Now, substitute the given magnitudes: \( |\vec a|=\sqrt{37} \), \( |\vec b|=3 \), and \( |\vec c|=4 \). \[ (3)^2 + 2(\vec b \cdot \vec c) + (4)^2 = (\sqrt{37})^2 \] \[ 9 + 2(\vec b \cdot \vec c) + 16 = 37 \] \[ 25 + 2(\vec b \cdot \vec c) = 37 \] \[ 2(\vec b \cdot \vec c) = 37 - 25 \] \[ 2(\vec b \cdot \vec c) = 12 \] \[ \vec b \cdot \vec c = 6 \]
Now, use the dot product formula to find the angle \( \theta \): \[ \vec b \cdot \vec c = |\vec b| |\vec c| \cos \theta \] \[ 6 = (3)(4) \cos \theta \] \[ 6 = 12 \cos \theta \] \[ \cos \theta = \frac{6}{12} = \frac{1}{2} \]
The angle \( \theta \) for which \( \cos \theta = 1/2 \) is \( \theta = \dfrac{\pi}{3} \) (or 60\(^{\circ}\)).


Step 4: Final Answer:

The angle between \( \vec b \) and \( \vec c \) is \( \dfrac{\pi}{3} \).

This corresponds to option (C).
Quick Tip: When given a vector sum like \( \vec a + \vec b + \vec c = \vec 0 \), and asked for an angle between two vectors (say \(\vec b\) and \(\vec c\)), always isolate those two vectors on one side (\( \vec b + \vec c = -\vec a \)) and then square the magnitudes. This technique effectively introduces the dot product \( \vec b \cdot \vec c \) which contains the angle information.

Step 1: Understanding the Question:

We need to analyze the continuity and differentiability of the function \( f(x)=|x|+|x-1| \) at the points \(x=0\) and \(x=1\).


Step 2: Detailed Explanation:

Continuity Check:

The function \(|x|\) is a standard modulus function, which is continuous for all real numbers.

The function \(|x-1|\) is also a modulus function, which is continuous for all real numbers.

The sum of two continuous functions is always continuous. Therefore, \(f(x) = |x| + |x-1|\) is continuous everywhere, including at \(x=0\) and \(x=1\).


Differentiability Check:

To check for differentiability, we first express \(f(x)\) as a piecewise function. The critical points are where the arguments of the modulus functions become zero, i.e., \(x=0\) and \(x=1\). This divides the number line into three intervals.


Case 1: \(x < 0\)
\(|x| = -x\) and \(|x-1| = -(x-1) = 1-x\). \(f(x) = -x + (1-x) = 1 - 2x\).


Case 2: \(0 \le x < 1\)
\(|x| = x\) and \(|x-1| = -(x-1) = 1-x\). \(f(x) = x + (1-x) = 1\).


Case 3: \(x \ge 1\)
\(|x| = x\) and \(|x-1| = x-1\). \(f(x) = x + (x-1) = 2x - 1\).


So, the piecewise function is: \[ f(x) = \begin{cases} 1 - 2x & if x < 0
1 & if 0 \le x < 1
2x - 1 & if x \ge 1 \end{cases} \]
Now we find the left-hand derivative (LHD) and right-hand derivative (RHD) at \(x=0\) and \(x=1\).

The derivative function is: \[ f'(x) = \begin{cases} -2 & if x < 0
0 & if 0 < x < 1
2 & if x > 1 \end{cases} \]
At x = 0:

LHD = \( \lim_{h \to 0^-} f'(0+h) = -2 \).

RHD = \( \lim_{h \to 0^+} f'(0+h) = 0 \).

Since LHD \(\neq\) RHD, \(f(x)\) is not differentiable at \(x=0\).


At x = 1:

LHD = \( \lim_{h \to 1^-} f'(1+h) = 0 \).

RHD = \( \lim_{h \to 1^+} f'(1+h) = 2 \).

Since LHD \(\neq\) RHD, \(f(x)\) is not differentiable at \(x=1\).


Step 3: Final Answer:

The function \(f(x)\) is continuous at both \(x=0\) and \(x=1\), but it is not differentiable at these points.

This corresponds to option (C).
Quick Tip: A function involving absolute values \(|g(x)|\) is generally not differentiable at points where \(g(x)=0\). These points are "sharp corners" on the graph. In \(f(x)=|x|+|x-1|\), the sharp corners are expected at \(x=0\) and \(x-1=0 \implies x=1\). This insight can help you quickly identify the correct option.

Step 1: Understanding the Question:

The question asks to identify the correct condition for the consistency and solution of a system of linear equations \(AX=B\) using matrix methods.


Step 2: Key Formula or Approach:

We need to recall the conditions for consistency of a system of linear equations based on the determinant of the coefficient matrix \(A\).

The conditions are:
1. If \(|A| \neq 0\), the system is consistent and has a unique solution.
2. If \(|A| = 0\), we must calculate \((adj A)B\).
a. If \((adj A)B \neq 0\), the system is inconsistent (no solution).
b. If \((adj A)B = 0\), the system is consistent and has infinitely many solutions.


Step 3: Detailed Explanation:

Let's analyze each option based on these rules.

(A) Consistent, if \(|A|\neq0\), solution is given by \(X=A^{-1}B\).

If \(|A|\neq0\), the matrix \(A\) is non-singular, and its inverse \(A^{-1}\) exists. We can pre-multiply \(AX=B\) by \(A^{-1}\) to get \(A^{-1}AX = A^{-1}B\), which simplifies to \(IX = A^{-1}B\), or \(X=A^{-1}B\). This gives a unique solution, so the system is consistent. This statement is correct.


(B) Inconsistent if \(|A|=0\) and \(adj(A)B=0\).

According to the rule, if \(|A|=0\) and \(adj(A)B=0\), the system is consistent with infinitely many solutions. This statement claims it is inconsistent. So, it is incorrect.


(C) Inconsistent if \(|A|=0\).

If \(|A|=0\), the system can be either inconsistent (if \(adj(A)B \neq 0\)) or consistent (if \(adj(A)B=0\)). We cannot conclude it is always inconsistent. So, this statement is incorrect.


(D) May or may not be consistent if \(|A|=0\) and \(adj(A)B=0\).

If \(|A|=0\) and \(adj(A)B=0\), the system is definitely consistent (with infinite solutions). The phrase "may or may not be" is incorrect. So, this statement is incorrect.


Step 4: Final Answer:

The only correct statement is (A).
Quick Tip: Remember this flow chart for solving \(AX=B\):
1. Calculate \(|A|\).
2. If \(|A| \neq 0 \rightarrow\) Consistent, Unique Solution (\(X=A^{-1}B\)).
3. If \(|A| = 0 \rightarrow\) Calculate \((adj A)B\).
a. If \((adj A)B \neq 0 \rightarrow\) Inconsistent, No Solution.
b. If \((adj A)B = 0 \rightarrow\) Consistent, Infinite Solutions.

Step 1: Understanding the Question:

We need to find the absolute maximum value of the function \(f(x)=x^3-3x+2\) on the closed interval \([0,2]\).


Step 2: Key Formula or Approach:

To find the absolute maximum of a continuous function on a closed interval, we follow these steps:
1. Find the critical points of the function by finding where the first derivative \(f'(x)\) is zero or undefined.
2. Consider only the critical points that lie within the given interval.
3. Evaluate the function \(f(x)\) at these critical points and also at the endpoints of the interval.
4. The largest value among these is the absolute maximum.


Step 3: Detailed Explanation:

The given function is \(f(x)=x^3-3x+2\), and the interval is \([0,2]\).


1. Find critical points:

First, find the derivative of \(f(x)\):
\[ f'(x) = \frac{d}{dx}(x^3-3x+2) = 3x^2 - 3 \]
Set the derivative to zero to find the critical points:
\[ 3x^2 - 3 = 0 \] \[ 3(x^2 - 1) = 0 \] \[ x^2 = 1 \] \[ x = 1 \quad or \quad x = -1 \]

2. Select points within the interval:

The given interval is \([0,2]\).
The critical point \(x=1\) lies within this interval.
The critical point \(x=-1\) is outside this interval, so we discard it.

The points we need to check are the critical point \(x=1\) and the endpoints \(x=0\) and \(x=2\).


3. Evaluate the function at these points:

At the left endpoint, \(x=0\):
\[ f(0) = (0)^3 - 3(0) + 2 = 2 \]
At the critical point, \(x=1\):
\[ f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0 \]
At the right endpoint, \(x=2\):
\[ f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4 \]

4. Compare the values:

The values of the function at the selected points are \{2, 0, 4\.
The largest value is 4.


Step 4: Final Answer:

The absolute maximum value of the function \(f(x)\) on the interval \([0,2]\) is 4.

This corresponds to option (C).
Quick Tip: For finding absolute extrema on a closed interval, never forget to check the endpoints of the interval. The absolute maximum or minimum can occur at a critical point or at an endpoint. A common mistake is to only check the critical points.

Step 1: Understanding the Question:

We are given the graph of a trigonometric function and asked to identify the graph of its inverse function from the given options.


Step 2: Analyzing the Given Graph:

The graph shown at the top has the characteristic shape of a cosine function. It has a maximum value on the y-axis (at \(x=0\)) and decreases as \(x\) moves away from 0. This is the graph of \(y = \cos(x)\).

For a function to have an inverse, it must be one-to-one. The principal value domain for \(y = \cos(x)\) is the interval \([0, \pi]\). In this domain, its range is \([-1, 1]\).


Step 3: Determining the Inverse Graph:

The inverse function of \(y = \cos(x)\) is \(y = \arccos(x)\) (or \(y = \cos^{-1}(x)\)).

The domain and range of the inverse function are swapped from the original function.

Domain of \(y = \arccos(x)\) is \([-1, 1]\).
Range of \(y = \arccos(x)\) is \([0, \pi]\).

The graph of an inverse function is the reflection of the original function's graph across the line \(y=x\).

Key points on the graph of \(y = \cos(x)\) (for \(x \in [0, \pi]\)) are \((0, 1)\), \((\pi/2, 0)\), and \((\pi, -1)\).

Correspondingly, key points on the graph of \(y = \arccos(x)\) will be \((1, 0)\), \((0, \pi/2)\), and \((-1, \pi)\).

Also, since \(y=\cos(x)\) is a decreasing function on \([0, \pi]\), its inverse \(y=\arccos(x)\) will also be a decreasing function on its domain \([-1, 1]\).


Step 4: Matching with Options:

Let's examine the options:

Graph (A) is the graph of \(y=\tan(x)\).
Graph (B) is the graph of \(y=\arctan(x)\).
Graph (C) has a domain of \([-1, 1]\) and a range of \([0, \pi]\). It passes through the points \((1, 0)\), \((0, \pi/2)\), and has an endpoint at \((-1, \pi)\). It is a decreasing function. This perfectly matches the properties of \(y = \arccos(x)\).
Graph (D) is similar to (C) but drawn slightly differently. Graph (C) shows the endpoints more clearly and is a better representation.

Thus, Graph (C) represents the inverse of the given function.
Quick Tip: The graph of an inverse function \(f^{-1}(x)\) is the reflection of the graph of \(f(x)\) about the line \(y=x\). For inverse trigonometric functions, remember the domain and range restrictions:
- \(y = \arcsin(x)\): Domain \([-1, 1]\), Range \([-\pi/2, \pi/2]\), increasing.
- \(y = \arccos(x)\): Domain \([-1, 1]\), Range \([0, \pi]\), decreasing.
- \(y = \arctan(x)\): Domain \(\mathbb{R}\), Range \((-\pi/2, \pi/2)\), increasing.

Step 1: Understanding the Question:

We are given the corner points of the feasible region of a Linear Programming Problem (LPP) and an objective function \(Z\). We need to find the points where \(Z\) attains its maximum and minimum values.


Step 2: Key Formula or Approach:

According to the Corner Point Theorem of LPP, the optimal values (maximum and minimum) of the objective function, if they exist, occur at the corner points (vertices) of the feasible region. We need to evaluate \(Z\) at each given corner point.


Step 3: Detailed Explanation:

The objective function is \(Z=18x+19y\).

The corner points are \((72,15)\), \((40,15)\), and \((40,10)\).

Let's evaluate \(Z\) at each point:

At point (72, 15): \[ Z = 18(72) + 19(15) = 1296 + 285 = 1581 \]
At point (40, 15): \[ Z = 18(40) + 19(15) = 720 + 285 = 1005 \]
At point (40, 10): \[ Z = 18(40) + 19(10) = 720 + 190 = 910 \]

Step 4: Final Answer:

Comparing the values of \(Z\): \{1581, 1005, 910\.

The maximum value of \(Z\) is 1581, which occurs at the point \((72,15)\).

The minimum value of \(Z\) is 910, which occurs at the point \((40,10)\).

This corresponds to option (A).
Quick Tip: In an LPP, to find the optimal solution, you only need to test the corner points of the feasible region. Systematically create a table with corner points and the corresponding value of Z to avoid calculation errors and easily identify the maximum and minimum values.

Step 1: Understanding the Question:

We are given three matrices \(A\), \(B\), and \(C\), and we need to determine which of the matrix products \(AB\), \(AC\), and \(BA\) are defined.


Step 2: Key Formula or Approach:

The product of two matrices \(M\) and \(N\), denoted \(MN\), is defined only if the number of columns in the first matrix \(M\) is equal to the number of rows in the second matrix \(N\). If \(M\) is of order \(m \times n\) and \(N\) is of order \(p \times q\), the product \(MN\) is defined if and only if \(n=p\).


Step 3: Detailed Explanation:

First, let's write down the orders (dimensions) of each matrix:

\(A\) is a \(3 \times 3\) matrix.
\(B\) is a \(3 \times 1\) matrix.
\(C\) is a \(1 \times 3\) matrix.

Now, let's check each product:

For \(AB\):
Matrix \(A\) is \(3 \times 3\) and matrix \(B\) is \(3 \times 1\).
The number of columns of \(A\) (3) is equal to the number of rows of \(B\) (3).
Therefore, \(AB\) is defined. The resulting matrix will be of order \(3 \times 1\).

For \(AC\):
Matrix \(A\) is \(3 \times 3\) and matrix \(C\) is \(1 \times 3\).
The number of columns of \(A\) (3) is not equal to the number of rows of \(C\) (1).
Therefore, \(AC\) is not defined.

For \(BA\):
Matrix \(B\) is \(3 \times 1\) and matrix \(A\) is \(3 \times 3\).
The number of columns of \(B\) (1) is not equal to the number of rows of \(A\) (3).
Therefore, \(BA\) is not defined.


Step 4: Final Answer:

Out of the given options, only the product \(AB\) is defined.

This corresponds to option (A).
Quick Tip: To quickly check if a matrix product \(MN\) is defined, write their dimensions side-by-side: \((m \times n) \times (p \times q)\). If the inner numbers (\(n\) and \(p\)) are the same, the product is defined. The dimensions of the resulting matrix will be the outer numbers (\(m \times q\)).

Step 1: Understanding the Question:

We need to identify which of the given statements about invertible matrices is incorrect.


Step 2: Key Formula or Approach:

We will analyze each statement based on the standard properties of matrix inverses.


Step 3: Detailed Explanation:

(A) \((A+B)^{-1}=B^{-1}+A^{-1}\)

This statement suggests that the inverse of a sum of matrices is the sum of their inverses. This is generally not true. Matrix inversion does not distribute over addition. For example, let \(A=I\) and \(B=I\). Then \(A+B=2I\). \((A+B)^{-1} = (2I)^{-1} = \frac{1}{2}I\). But \(A^{-1}+B^{-1} = I^{-1}+I^{-1} = I+I = 2I\). Since \(\frac{1}{2}I \neq 2I\), the statement is incorrect.


(B) \((AB)^{-1}=B^{-1}A^{-1}\)

This is the "reversal law" for the inverse of a product of matrices. It is a standard and fundamental property of matrix inverses. This statement is correct.


(C) \(adj(A)=|A|A^{-1}\)

The formula for the inverse of a matrix \(A\) is \(A^{-1} = \frac{1}{|A|} adj(A)\). If we multiply both sides by \(|A|\) (since \(A\) is invertible, \(|A| \neq 0\)), we get \(|A|A^{-1} = adj(A)\). This statement is correct.


(D) \(I^{-1}=I\)

The inverse of the identity matrix \(I\) is \(I\) itself, because \(I \cdot I = I\). This statement is correct.


Step 4: Final Answer:

The statement that is not correct is \((A+B)^{-1}=B^{-1}+A^{-1}\).

This corresponds to option (A).
Quick Tip: Remember the key properties of matrix inverses: \((A^{-1})^{-1} = A\) \((AB)^{-1} = B^{-1}A^{-1}\) (Reversal Law) \((A^T)^{-1} = (A^{-1})^T\) \(\det(A^{-1}) = 1/\det(A)\) Crucially, matrix operations like inverse and transpose often reverse the order, and they do not distribute over addition.

Step 1: Understanding the Question:

We need to set up the definite integral that represents the area of the shaded region shown in the graph. The region is bounded by the parabola \(y^2=x\), the vertical line \(x=4\), and the x-axis (\(y=0\)).


Step 2: Key Formula or Approach:

The area of a region bounded by a curve \(y=f(x)\), the x-axis, and the vertical lines \(x=a\) and \(x=b\) is given by the integral: \[ Area = \int_{a}^{b} f(x) \,dx \]
assuming \(f(x) \ge 0\) in the interval \([a,b]\).


Step 3: Detailed Explanation:

The given boundaries are:

The curve \(y^2=x\). Since the shaded region is above the x-axis, we consider the upper branch of the parabola, which is \(y = \sqrt{x}\).
The vertical line \(x=4\). This gives the upper limit of integration.
The x-axis (\(y=0\)). This is the lower boundary of the region. The curve \(y=\sqrt{x}\) intersects the x-axis at \(x=0\), which gives the lower limit of integration.

So, we need to find the area under the curve \(y = \sqrt{x}\) from \(x=0\) to \(x=4\).

Using the formula for the area under a curve, we have: \[ Area = \int_{0}^{4} y \,dx = \int_{0}^{4} \sqrt{x} \,dx \]

Step 4: Final Answer:

The integral representing the area of the shaded region is \( \displaystyle \int_{0}^{4} \sqrt{x}\,dx \).

This corresponds to option (D).
Quick Tip: When setting up an area integral, always visualize a representative rectangle. If the rectangle is vertical (with width \(dx\)), its height is (top curve - bottom curve), and you integrate with respect to \(x\). Here, the top curve is \(y=\sqrt{x}\) and the bottom is \(y=0\), so the height is \(\sqrt{x}\). If the rectangle is horizontal (with width \(dy\)), its length is (right curve - left curve), and you integrate with respect to \(y\).

Step 1: Understanding the Question:

We need to evaluate an Assertion (A) about the continuity of a piecewise function and a Reason (R) which states the definition of continuity. Then we must determine the relationship between them.


Step 2: Analyzing the Reason (R):

The Reason (R) states that for a function \(f\) to be continuous at \(x=a\), the left-hand limit (LHL), the right-hand limit (RHL), and the function's value at that point must all be equal: \( \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a) \). This is the precise mathematical definition of continuity at a point.
Therefore, Reason (R) is true.


Step 3: Analyzing the Assertion (A):

Let's test the continuity of the function \(f(x)\) at \(x=5\) using the condition from Reason (R).
The function is \( f(x) = \begin{cases} 3x-8, & x\le 5
2k, & x>5 \end{cases} \).

1. Function value at x=5:
\(f(5) = 3(5) - 8 = 15 - 8 = 7\).

2. Left-Hand Limit (LHL) at x=5:
\( \lim_{x\to 5^-}f(x) = \lim_{x\to 5^-}(3x-8) = 3(5) - 8 = 7 \).

3. Right-Hand Limit (RHL) at x=5:
\( \lim_{x\to 5^+}f(x) = \lim_{x\to 5^+}(2k) = 2k \).

For the function to be continuous at \(x=5\), we must have LHL = RHL = \(f(5)\). \[ 7 = 2k = 7 \]
This gives the condition \(2k = 7\), which means \(k = \dfrac{7}{2}\).

The Assertion (A) claims that the function is continuous for \(k=\dfrac{5}{2}\). This is incorrect. For \(k=\dfrac{5}{2}\), the RHL would be \(2(\frac{5}{2}) = 5\), which is not equal to the LHL (7).
Therefore, Assertion (A) is false.


Step 4: Final Answer:

Since Assertion (A) is false and Reason (R) is true, the correct option is (D).
Quick Tip: For piecewise functions, continuity at a point where the definition changes is checked by ensuring the "pieces meet up". Calculate the limit from the left, the limit from the right, and the function's value at the point. All three must be identical for the function to be continuous there.

Step 1: Understanding the Question:

We need to evaluate an Assertion (A) about a function being bijective and a Reason (R) which states the definition of a bijective function. Then we must determine their relationship.


Step 2: Analyzing the Reason (R):

The Reason (R) states that a function is bijective if it is both injective (one-to-one) and surjective (onto). This is the definition of a bijective function.
Therefore, Reason (R) is true.


Step 3: Analyzing the Assertion (A):

The function is \(f:\mathbb{Z}\to\mathbb{Z}\) defined by \(f(x)=3x-5\). To check if it is bijective, we must check for injectivity and surjectivity.

Injectivity (One-to-one):
Let \(f(x_1) = f(x_2)\) for some integers \(x_1, x_2 \in \mathbb{Z}\). \[ 3x_1 - 5 = 3x_2 - 5 \] \[ 3x_1 = 3x_2 \] \[ x_1 = x_2 \]
Since \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), the function is injective.


Surjectivity (Onto):
For the function to be surjective, for every element \(y\) in the codomain \(\mathbb{Z}\), there must exist an element \(x\) in the domain \(\mathbb{Z}\) such that \(f(x) = y\).
Let's take an element from the codomain, say \(y=0 \in \mathbb{Z}\). We need to find if there is an integer \(x\) such that \(f(x)=0\). \[ 3x - 5 = 0 \] \[ 3x = 5 \] \[ x = \frac{5}{3} \]
The value \(x = \frac{5}{3}\) is not an integer (\(x \notin \mathbb{Z}\)). This means that there is no integer \(x\) in the domain that maps to the integer \(0\) in the codomain. Therefore, the function is not surjective.

Since the function is not surjective, it cannot be bijective.
Therefore, Assertion (A) is false.


Step 4: Final Answer:

Since Assertion (A) is false and Reason (R) is true, the correct option is (D).
Quick Tip: To test for surjectivity of a function \(f: A \to B\), pick an arbitrary element \(y \in B\) and try to solve the equation \(y = f(x)\) for \(x\). If you can always find a solution \(x\) that belongs to the domain \(A\), the function is surjective. If you can find even one \(y\) for which no such \(x\) exists, the function is not surjective.

Step 1: Understanding the Question:

We are given the two diagonals of a parallelogram as vectors and we need to find its area.


Step 2: Key Formula or Approach:

The area of a parallelogram with diagonals \( \vec d_1 \) and \( \vec d_2 \) is given by the formula: \[ Area = \frac{1}{2} |\vec d_1 \times \vec d_2| \]
We first need to compute the cross product of the two diagonal vectors and then find half of its magnitude.


Step 3: Detailed Explanation:

Given diagonals: \( \vec d_1 = 2\hat{i}-\hat{j}+\hat{k} \) and \( \vec d_2 = \hat{i}+3\hat{j}-\hat{k} \).

First, calculate the cross product \( \vec d_1 \times \vec d_2 \): \[ \vec d_1 \times \vec d_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -1 & 1
1 & 3 & -1 \end{vmatrix} \] \[ = \hat{i}((-1)(-1) - (1)(3)) - \hat{j}((2)(-1) - (1)(1)) + \hat{k}((2)(3) - (1)(-1)) \] \[ = \hat{i}(1 - 3) - \hat{j}(-2 - 1) + \hat{k}(6 - (-1)) \] \[ = -2\hat{i} - \hat{j}(-3) + \hat{k}(7) \] \[ = -2\hat{i} + 3\hat{j} + 7\hat{k} \]
Next, find the magnitude of this cross product vector: \[ |\vec d_1 \times \vec d_2| = \sqrt{(-2)^2 + (3)^2 + (7)^2} \] \[ = \sqrt{4 + 9 + 49} = \sqrt{62} \]
Finally, use the area formula: \[ Area = \frac{1}{2} |\vec d_1 \times \vec d_2| = \frac{1}{2} \sqrt{62} \]

Step 4: Final Answer:

The area of the parallelogram is \( \dfrac{\sqrt{62}}{2} \) square units.
Quick Tip: Be careful not to confuse the formulas for the area of a parallelogram. If adjacent sides are given as vectors \( \vec a \) and \( \vec b \), Area \( = |\vec a \times \vec b| \). If diagonals are given as vectors \( \vec d_1 \) and \( \vec d_2 \), Area \( = \frac{1}{2} |\vec d_1 \times \vec d_2| \). The presence of the \( \frac{1}{2} \) factor is the key difference.

Step 1: Understanding the Question:

The problem asks for the set of all possible values of the parameter 'a' such that the given function \(f(x)\) is a decreasing function for all real numbers \(x \in \mathbb{R}\).


Step 2: Key Formula or Approach:

A differentiable function \(f(x)\) is said to be decreasing on an interval if its first derivative, \(f'(x)\), is less than or equal to zero (i.e., \(f'(x) \le 0\)) for all \(x\) in that interval. Here, the interval is the set of all real numbers, \(\mathbb{R}\).

We will also use the concept of expressing \(A\sin x + B\cos x\) in the form \(R\sin(x+\alpha)\) or \(R\cos(x-\alpha)\), where the range of the expression is \([-\sqrt{A^2+B^2}, \sqrt{A^2+B^2}]\).


Step 3: Detailed Explanation:

The given function is \(f(x) = \sqrt{3} \sin x - \cos x - 2ax + b\).

First, we find the derivative of \(f(x)\) with respect to \(x\): \[ f'(x) = \frac{d}{dx} (\sqrt{3} \sin x - \cos x - 2ax + b) \] \[ f'(x) = \sqrt{3} \cos x - (-\sin x) - 2a + 0 \] \[ f'(x) = \sin x + \sqrt{3} \cos x - 2a \]
For \(f(x)\) to be decreasing on \(\mathbb{R}\), we must have \(f'(x) \le 0\) for all \(x \in \mathbb{R}\). \[ \sin x + \sqrt{3} \cos x - 2a \le 0 \] \[ \sin x + \sqrt{3} \cos x \le 2a \]
Now, we need to find the maximum value of the expression \( \sin x + \sqrt{3} \cos x \). This is of the form \(A\sin x + B\cos x\) with \(A=1\) and \(B=\sqrt{3}\).

The maximum value is \( \sqrt{A^2+B^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2 \).

Let's show this by converting the expression: \[ \sin x + \sqrt{3} \cos x = 2 \left( \frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x \right) \] \[ = 2 \left( \cos\left(\frac{\pi}{3}\right) \sin x + \sin\left(\frac{\pi}{3}\right) \cos x \right) \]
Using the identity \( \sin(A+B) = \sin A \cos B + \cos A \sin B \), we get: \[ = 2 \sin\left(x + \frac{\pi}{3}\right) \]
So the inequality becomes: \[ 2 \sin\left(x + \frac{\pi}{3}\right) \le 2a \]
We know that the maximum value of \( \sin\left(x + \frac{\pi}{3}\right) \) is 1. Therefore, the maximum value of the left-hand side (LHS) is \(2 \times 1 = 2\).

For the inequality to hold true for all values of \(x\), the right-hand side (RHS) must be greater than or equal to the maximum possible value of the LHS. \[ \max(LHS) \le RHS \] \[ 2 \le 2a \]
Dividing by 2, we get: \[ 1 \le a \]
So, the condition for the function to be decreasing on \(\mathbb{R}\) is \(a \ge 1\).


Step 4: Final Answer:

The values of 'a' for which the function is decreasing on \(\mathbb{R}\) are given by the inequality \(a \ge 1\), which can be written in interval notation as \(a \in [1, \infty)\).
Quick Tip: For any function of the form \(g(x) = A \sin x + B \cos x\), its range is \([-\sqrt{A^2+B^2}, \sqrt{A^2+B^2}]\).
When you need to satisfy an inequality like \(g(x) \le k\) for all \(x\), you must ensure that the maximum value of \(g(x)\) is less than or equal to \(k\). That is, \( \sqrt{A^2+B^2} \le k \).
Similarly, for \(g(x) \ge k\), you must ensure \( -\sqrt{A^2+B^2} \ge k \).

Step 1: Understanding the Question:

We are given two vectors representing kite strings and we need to find the angle between them.


Step 2: Key Formula or Approach:

The angle \( \theta \) between two vectors \( \vec a \) and \( \vec b \) can be found using the dot product formula: \[ \cos \theta = \frac{\vec a \cdot \vec b}{|\vec a| |\vec b|} \]

Step 3: Detailed Explanation:

The given vectors are \( \vec a = 3\hat{i}+\hat{j}+2\hat{k} \) and \( \vec b = 2\hat{i}-2\hat{j}+4\hat{k} \).

First, calculate the dot product \( \vec a \cdot \vec b \): \[ \vec a \cdot \vec b = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 + 8 = 12 \]
Next, calculate the magnitude of each vector: \[ |\vec a| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \] \[ |\vec b| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} \]
Now, substitute these values into the formula for \( \cos \theta \): \[ \cos \theta = \frac{12}{\sqrt{14} \sqrt{24}} = \frac{12}{\sqrt{14 \times 24}} = \frac{12}{\sqrt{336}} \]
Let's simplify the radical: \( \sqrt{336} = \sqrt{16 \times 21} = 4\sqrt{21} \). \[ \cos \theta = \frac{12}{4\sqrt{21}} = \frac{3}{\sqrt{21}} \]
The angle \( \theta \) is: \[ \theta = \cos^{-1}\left(\frac{3}{\sqrt{21}}\right) \]

Step 4: Final Answer:

The angle formed between the kite strings is \( \theta = \cos^{-1}\left(\dfrac{3}{\sqrt{21}}\right) \).
Quick Tip: The dot product is the easiest way to find the angle between two vectors. Remember the formula \( \vec a \cdot \vec b = |\vec a| |\vec b| \cos\theta \). If the dot product is positive, the angle is acute. If it's negative, the angle is obtuse. If it's zero, the vectors are perpendicular.

Step 1: Understanding the Question:

We need to find a vector with a specific magnitude (21) that points in the direction opposite to the vector from point A to point B.


Step 2: Key Formula or Approach:

1. Find the vector \( \overrightarrow{AB} \).
2. The direction opposite to \( \overrightarrow{AB} \) is the direction of \( \overrightarrow{BA} \) or \( -\overrightarrow{AB} \).
3. Find the unit vector in this opposite direction. A unit vector \( \hat{u} \) for a vector \( \vec{v} \) is \( \hat{u} = \frac{\vec{v}}{|\vec{v}|} \).
4. Multiply the unit vector by the desired magnitude.


Step 3: Detailed Explanation:

The coordinates are \( A(2,1,3) \) and \( B(6,-1,0) \).

1. Find \( \overrightarrow{AB} \): \[ \overrightarrow{AB} = Position Vector of B - Position Vector of A \] \[ \overrightarrow{AB} = (6\hat{i} - 1\hat{j} + 0\hat{k}) - (2\hat{i} + 1\hat{j} + 3\hat{k}) \] \[ \overrightarrow{AB} = (6-2)\hat{i} + (-1-1)\hat{j} + (0-3)\hat{k} = 4\hat{i} - 2\hat{j} - 3\hat{k} \]
2. Find the vector in the opposite direction:
Let this vector be \( \vec{v} = -\overrightarrow{AB} \). \[ \vec{v} = -(4\hat{i} - 2\hat{j} - 3\hat{k}) = -4\hat{i} + 2\hat{j} + 3\hat{k} \]
3. Find the unit vector \( \hat{v} \):
First, find the magnitude \( |\vec{v}| \): \[ |\vec{v}| = |- \overrightarrow{AB}| = |\overrightarrow{AB}| = \sqrt{4^2 + (-2)^2 + (-3)^2} = \sqrt{16 + 4 + 9} = \sqrt{29} \]
The unit vector is: \[ \hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{-4\hat{i} + 2\hat{j} + 3\hat{k}}{\sqrt{29}} \]
4. Find the required vector:
The required vector has magnitude 21 and is in the direction of \( \hat{v} \). \[ Required Vector = 21 \times \hat{v} = 21 \left( \frac{-4\hat{i} + 2\hat{j} + 3\hat{k}}{\sqrt{29}} \right) \] \[ = -\frac{84}{\sqrt{29}}\hat{i} + \frac{42}{\sqrt{29}}\hat{j} + \frac{63}{\sqrt{29}}\hat{k} \]

Step 4: Final Answer:

The required vector is \( -\dfrac{84}{\sqrt{29}}\hat{i} + \dfrac{42}{\sqrt{29}}\hat{j} + \dfrac{63}{\sqrt{29}}\hat{k} \).
Quick Tip: To find a vector of magnitude 'M' in the direction of a given vector 'v', always follow the two-step process: 1. Find the unit vector: \( \hat{v} = \vec{v} / |\vec{v}| \). 2. Scale the unit vector: Required Vector = \( M \times \hat{v} \). Remember that the direction opposite to \( \vec{v} \) is simply \( -\vec{v} \).

Step 1: Understanding the Question:

We are asked to solve an equation involving inverse trigonometric functions for the variable \(x\).


Step 2: Key Formula or Approach:

The equation involves the term \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \). We use the identity that relates this term to \( \tan^{-1}x \). This identity is derived by substituting \( x = \tan\theta \). \[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \begin{cases} 2\tan^{-1}x & if |x| \le 1
\pi - 2\tan^{-1}x & if x > 1
-\pi - 2\tan^{-1}x & if x < -1 \end{cases} \]
We will solve the equation by considering these three cases for \(x\).


Step 3: Detailed Explanation:

Let's analyze the range of the Left Hand Side (LHS) of the equation first.

The principal value range for \( \tan^{-1}x \) is \( (-\pi/2, \pi/2) \).

The principal value range for \( \sin^{-1}(y) \) is \( [-\pi/2, \pi/2] \).

The maximum possible value of the LHS would be when \(x \to \infty\):
\( \lim_{x \to \infty} \left( 2\tan^{-1}x+\sin^{-1}\!\left(\frac{2x}{1+x^2}\right) \right) = 2(\pi/2) + \sin^{-1}(0) = \pi \).

Let's verify this using the piecewise identity:

Case 1: \( |x| \le 1 \)
The equation becomes: \( 2\tan^{-1}x + 2\tan^{-1}x = 4\tan^{-1}x \).
Since \( |x| \le 1 \), \( \tan^{-1}x \) is in \( [-\pi/4, \pi/4] \).
So, the LHS, \( 4\tan^{-1}x \), is in the range \( [-\pi, \pi] \).

Case 2: \( x > 1 \)
The equation becomes: \( 2\tan^{-1}x + (\pi - 2\tan^{-1}x) = \pi \).
The LHS is constant and equal to \( \pi \).

Case 3: \( x < -1 \)
The equation becomes: \( 2\tan^{-1}x + (-\pi - 2\tan^{-1}x) = -\pi \).
The LHS is constant and equal to \( -\pi \).

Combining all cases, the range of the function on the LHS is \( [-\pi, \pi] \).
The value of \( \pi \) is approximately 3.14159.


Now, let's analyze the Right Hand Side (RHS) of the equation.
RHS = \( 4\sqrt{3} \).
Using the approximation \( \sqrt{3} \approx 1.732 \), we get:
RHS \( \approx 4 \times 1.732 = 6.928 \).

Step 4: Final Answer:

We compare the maximum possible value of the LHS with the value of the RHS.
Maximum value of LHS = \( \pi \approx 3.14159 \).
Value of RHS = \( 4\sqrt{3} \approx 6.928 \).
Since \( \pi < 4\sqrt{3} \), the LHS can never be equal to the RHS for any real value of \(x\).
Therefore, the equation has no solution.
Quick Tip: Before jumping into solving complex trigonometric or inverse trigonometric equations, it's often useful to analyze the range of the expressions on both sides. If the ranges do not overlap, you can immediately conclude that there is no solution, saving a lot of time and effort.

Step 1: Understanding the Question:

We are asked to find the derivative of a function \(y = 2^u\) with respect to the variable \(u\), where \(u\) itself is a function of \(x\), specifically \(u = \cos^2 x\). This is a direct application of differentiation rules, not requiring the chain rule with respect to \(x\).


Step 2: Key Formula or Approach:

Let \(y = 2^{\cos^2 x}\) and let \(u = \cos^2 x\).

The problem simplifies to finding \(\dfrac{dy}{du}\) where \(y=2^u\).

The formula for the derivative of an exponential function \(a^u\) with respect to \(u\) is: \[ \frac{d}{du}(a^u) = a^u \log a \]
(Here, log denotes the natural logarithm, ln).


Step 3: Detailed Explanation:

Using the substitution from Step 2, we apply the differentiation formula with \(a=2\): \[ \frac{dy}{du} = \frac{d}{du}(2^u) = 2^u \log 2 \]
Now, we substitute back \(u = \cos^2 x\): \[ \frac{d(2^{\cos^2 x})}{d(\cos^2 x)} = 2^{\cos^2 x} \log 2 \]

Step 4: Final Answer:

The derivative of \(2^{\cos^2 x}\) with respect to \(\cos^2 x\) is \( 2^{\cos^2 x} \log 2 \).
Quick Tip: When asked to differentiate a function \(f(g(x))\) with respect to \(g(x)\), treat \(g(x)\) as a single variable. For example, differentiating \(\sin(\tan x)\) w.r.t \(\tan x\) is simply \(\cos(\tan x)\). This shortcut saves you from applying the full chain rule.

Step 1: Understanding the Question:

We are given an implicit equation relating \(x\) and \(y\), and we need to find the derivative of \(y\) with respect to \(x\), which is \(\dfrac{dy}{dx}\).


Step 2: Key Formula or Approach:

We can solve this using implicit differentiation. A simpler approach is to first simplify the given equation by removing the inverse trigonometric function and then differentiate.


Step 3: Detailed Explanation:

The given equation is \(\tan^{-1}(x^2+y^2)=a^2\).

Method 1: Simplification first

Apply the tangent function to both sides of the equation: \[ \tan(\tan^{-1}(x^2+y^2)) = \tan(a^2) \] \[ x^2+y^2 = \tan(a^2) \]
Since \(a\) is a constant, \(a^2\) is also a constant, and therefore \(\tan(a^2)\) is a constant. Let \(c = \tan(a^2)\). The equation becomes: \[ x^2+y^2 = c \]
Now, differentiate both sides with respect to \(x\): \[ \frac{d}{dx}(x^2+y^2) = \frac{d}{dx}(c) \] \[ 2x + 2y \frac{dy}{dx} = 0 \]
Now, solve for \(\dfrac{dy}{dx}\): \[ 2y \frac{dy}{dx} = -2x \] \[ \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y} \]

Method 2: Direct implicit differentiation

Differentiate \(\tan^{-1}(x^2+y^2)=a^2\) directly with respect to \(x\): \[ \frac{d}{dx}(\tan^{-1}(x^2+y^2)) = \frac{d}{dx}(a^2) \]
Using the chain rule, \( \frac{d}{du}(\tan^{-1}u) = \frac{1}{1+u^2} \): \[ \frac{1}{1+(x^2+y^2)^2} \cdot \frac{d}{dx}(x^2+y^2) = 0 \] \[ \frac{1}{1+(x^2+y^2)^2} \cdot \left(2x + 2y\frac{dy}{dx}\right) = 0 \]
Since the fraction part cannot be zero, the term in the parenthesis must be zero: \[ 2x + 2y\frac{dy}{dx} = 0 \] \[ 2y\frac{dy}{dx} = -2x \] \[ \frac{dy}{dx} = -\frac{x}{y} \]

Step 4: Final Answer:

The value of \(\dfrac{dy}{dx}\) is \(-\dfrac{x}{y}\).
Quick Tip: Before applying implicit differentiation, check if the equation can be simplified. Removing inverse trigonometric functions or logarithms by applying their inverse functions to both sides can often make the differentiation process much easier and less prone to errors.

Step 1: Understanding the Question:

We need to find the maximum value of the objective function \(Z = 8x + 9y\) subject to a set of linear constraints. This involves finding the feasible region determined by the constraints and then evaluating \(Z\) at the corner points of this region.


Step 2: Graphing the Constraints:

First, we convert the inequalities into equations to plot the boundary lines.

\(2x + 3y = 6\): Intercepts are (3, 0) and (0, 2). The region \(2x+3y \le 6\) is towards the origin.
\(3x - 2y = 6\): Intercepts are (2, 0) and (0, -3). The region \(3x-2y \le 6\) is towards the origin (checking (0,0): 0 \(\le\) 6).
\(y = 1\): A horizontal line. The region \(y \le 1\) is below this line.
\(x \ge 0, y \ge 0\): This confines the feasible region to the first quadrant.


Step 3: Finding the Feasible Region and Corner Points:

The feasible region is the area in the first quadrant that satisfies all the inequalities. We find the vertices (corner points) of this region by finding the intersection of the boundary lines.

Point O: Intersection of \(x=0\) and \(y=0\). \(O(0,0)\).
Point A: Intersection of \(3x - 2y = 6\) and \(y=0\). \(3x = 6 \implies x=2\). \(A(2,0)\).
Point B: Intersection of \(2x + 3y = 6\) and \(3x - 2y = 6\).
Multiplying the first equation by 2 and the second by 3:
\(4x + 6y = 12\)
\(9x - 6y = 18\)
Adding them: \(13x = 30 \implies x = 30/13\).
Substituting \(x\) back: \(2(30/13) + 3y = 6 \implies 60/13 + 3y = 6 \implies 3y = 6 - 60/13 = 18/13 \implies y = 6/13\).
So, \(B\left(\dfrac{30}{13}, \dfrac{6}{13}\right)\).
Point C: Intersection of \(2x + 3y = 6\) and \(y=1\).
\(2x + 3(1) = 6 \implies 2x = 3 \implies x = 3/2\).
So, \(C(3/2, 1)\).
Point D: Intersection of \(x=0\) and \(y=1\). \(D(0,1)\).

The corner points of the feasible region are O(0,0), A(2,0), B(30/13, 6/13), C(1.5, 1), and D(0,1).

(A sketch would show a pentagon with these vertices).


Step 4: Evaluating Z at Corner Points:

Now, we evaluate the objective function \(Z = 8x + 9y\) at each corner point.

At O(0, 0): \(Z = 8(0) + 9(0) = 0\).
At A(2, 0): \(Z = 8(2) + 9(0) = 16\).
At B\(\left(\dfrac{30}{13}, \dfrac{6}{13}\right)\): \(Z = 8\left(\dfrac{30}{13}\right) + 9\left(\dfrac{6}{13}\right) = \dfrac{240 + 54}{13} = \dfrac{294}{13} \approx 22.615\).
At C(1.5, 1): \(Z = 8(1.5) + 9(1) = 12 + 9 = 21\).
At D(0, 1): \(Z = 8(0) + 9(1) = 9\).


Step 5: Final Answer:

Comparing the values of Z, the maximum value is \( \dfrac{294}{13} \).
This maximum value occurs at the point \( \left(\dfrac{30}{13}, \dfrac{6}{13}\right) \).
Quick Tip: Always draw a rough sketch of the constraints to correctly identify the feasible region and its vertices. After finding the intersection points, it's a good practice to check if they satisfy all the other constraints to ensure they are indeed corner points of the feasible region.

Step 1: Understanding the Question:

We need to integrate a rational function where the denominator is a product of distinct linear factors. This is a classic case for using partial fraction decomposition.


Step 2: Key Formula or Approach:

We set up the partial fraction decomposition as follows: \[ \frac{2x-1}{(x-1)(x+2)(x-3)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x-3} \]
After finding the constants A, B, and C, we integrate term by term using the formula \( \int \frac{1}{ax+b}dx = \frac{1}{a}\log|ax+b| + C \).


Step 3: Detailed Explanation:

To find A, B, and C, we write: \[ 2x-1 = A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2) \]
We can find the coefficients by substituting the roots of the denominator (the "cover-up" method):

To find A, set \(x=1\):

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

\(1 = A(3)(-2) \implies 1 = -6A \implies A = -\dfrac{1}{6}\).
To find B, set \(x=-2\):

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

\(-5 = B(-3)(-5) \implies -5 = 15B \implies B = -\dfrac{5}{15} = -\dfrac{1}{3}\).
To find C, set \(x=3\):

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

\(5 = C(2)(5) \implies 5 = 10C \implies C = \dfrac{5}{10} = \dfrac{1}{2}\).

Now, substitute these values back into the integral: \[ \int \left( \frac{-1/6}{x-1} + \frac{-1/3}{x+2} + \frac{1/2}{x-3} \right) dx \]
Integrate each term: \[ = -\frac{1}{6} \int \frac{1}{x-1}dx - \frac{1}{3} \int \frac{1}{x+2}dx + \frac{1}{2} \int \frac{1}{x-3}dx \] \[ = -\frac{1}{6}\log|x-1| - \frac{1}{3}\log|x+2| + \frac{1}{2}\log|x-3| + C \]

Step 4: Final Answer:

The result of the integration is \( -\dfrac{1}{6}\log|x-1| - \dfrac{1}{3}\log|x+2| + \dfrac{1}{2}\log|x-3| + C \).
Quick Tip: The "cover-up" method is a very fast way to find the coefficients for distinct linear factors in partial fractions. To find the coefficient for the term \( \frac{A}{x-a} \), cover the \((x-a)\) factor in the original denominator and substitute \(x=a\) into the rest of the expression.

Step 1: Understanding the Question:

We need to evaluate a definite integral of a function that is a sum of absolute values. The key is to split the integral into intervals based on the points where the arguments of the absolute value functions become zero.


Step 2: Key Formula or Approach:

The critical points are \(x=1\), \(x=2\), and \(x=5\). These points lie within the integration interval \([0,5]\). We must split the integral into sub-intervals: \([0,1]\), \([1,2]\), \([2,5]\). We will define the integrand piecewise for each interval.


Step 3: Detailed Explanation:

Let \(f(x) = |x-1|+|x-2|+|x-5|\). We split the integral as follows: \[ \int_{0}^{5} f(x) \, dx = \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx + \int_{2}^{5} f(x) \, dx \]
Now we define \(f(x)\) for each interval:

For \(x \in [0,1]\): \(x-1 \le 0\), \(x-2 \le 0\), \(x-5 \le 0\).
\(f(x) = -(x-1) - (x-2) - (x-5) = -x+1 -x+2 -x+5 = -3x+8\).
For \(x \in [1,2]\): \(x-1 \ge 0\), \(x-2 \le 0\), \(x-5 \le 0\).
\(f(x) = (x-1) - (x-2) - (x-5) = x-1 -x+2 -x+5 = -x+6\).
For \(x \in [2,5]\): \(x-1 \ge 0\), \(x-2 \ge 0\), \(x-5 \le 0\).
\(f(x) = (x-1) + (x-2) - (x-5) = x-1 +x-2 -x+5 = x+2\).

Now we evaluate the integrals for each piece:

\(\displaystyle \int_{0}^{1} (-3x+8) \, dx = \left[ -\frac{3x^2}{2} + 8x \right]_{0}^{1} = \left(-\frac{3}{2} + 8\right) - (0) = \frac{13}{2} = 6.5\).
\(\displaystyle \int_{1}^{2} (-x+6) \, dx = \left[ -\frac{x^2}{2} + 6x \right]_{1}^{2} = \left(-\frac{4}{2} + 12\right) - \left(-\frac{1}{2} + 6\right) = (10) - \left(\frac{11}{2}\right) = \frac{9}{2} = 4.5\).
\(\displaystyle \int_{2}^{5} (x+2) \, dx = \left[ \frac{x^2}{2} + 2x \right]_{2}^{5} = \left(\frac{25}{2} + 10\right) - \left(\frac{4}{2} + 4\right) = \left(\frac{45}{2}\right) - (6) = \frac{33}{2} = 16.5\).

Finally, we sum the results: \[ Total Value = 6.5 + 4.5 + 16.5 = 27.5 \]
or \[ Total Value = \frac{13}{2} + \frac{9}{2} + \frac{33}{2} = \frac{13+9+33}{2} = \frac{55}{2} \]

Step 4: Final Answer:

The value of the definite integral is \(27.5\) or \( \dfrac{55}{2} \).
Quick Tip: When integrating an absolute value function \(|g(x)|\), always find the roots of \(g(x)=0\). These roots are the points where you need to split your definite integral. The function will have a different (simpler) definition on each side of the root.

Step 1: Understanding the Question:

We are given a relationship between the rate of change of a sphere's volume and its surface area. We need to find the rate of decrease of its radius, which is \(-\dfrac{dr}{dt}\).


Step 2: Key Formula or Approach:

Let \(V\) be the volume, \(S\) be the surface area, and \(r\) be the radius of the sphere.
The formulas are: \[ V = \frac{4}{3}\pi r^3 \quad and \quad S = 4\pi r^2 \]
The given condition is "the rate of decrease of volume...is proportional to its surface area". This translates to: \[ -\frac{dV}{dt} \propto S \implies \frac{dV}{dt} = -kS \]
where \(k\) is a positive constant of proportionality.


Step 3: Detailed Explanation:

We need to relate \(\dfrac{dV}{dt}\) to \(\dfrac{dr}{dt}\). We differentiate the volume formula with respect to time \(t\): \[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi \cdot (3r^2) \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt} \]
Now we substitute this into the given relationship \(\dfrac{dV}{dt} = -kS\): \[ 4\pi r^2 \frac{dr}{dt} = -kS \]
We also substitute the formula for surface area, \(S = 4\pi r^2\): \[ 4\pi r^2 \frac{dr}{dt} = -k(4\pi r^2) \]
Assuming the ball has not completely dissolved (\(r \neq 0\)), we can divide both sides by \(4\pi r^2\): \[ \frac{dr}{dt} = -k \]
This result shows that the rate of change of the radius is a negative constant, \(-k\).


Step 4: Final Answer:

The rate of decrease of the radius is \(-\dfrac{dr}{dt} = -(-k) = k\).
Since \(k\) is a constant of proportionality, the rate of decrease of the radius is constant.
Quick Tip: In related rates problems, the steps are usually: 1. Identify the given rates and the rate you need to find. 2. Write down an equation that relates the variables involved (e.g., geometric formulas). 3. Differentiate the equation implicitly with respect to time. 4. Substitute the given information and solve for the unknown rate.

Step 1: Understanding the Question:

We need to find the area under the curve \(y=|x+3|\) from \(x=-6\) to \(x=0\). This requires sketching the graph, defining the function piecewise, and then integrating.


Step 2: Sketching the Graph and Defining the Function:

The graph of \(y=|x+3|\) is a V-shape with its vertex at the point where \(x+3=0\), i.e., at \(x=-3\). The vertex is at \((-3, 0)\).


The function can be defined piecewise. The critical point is \(x=-3\), which is within our interval of integration \([-6, 0]\). \[ y = |x+3| = \begin{cases} x+3 & if x+3 \ge 0 \implies x \ge -3
-(x+3) & if x+3 < 0 \implies x < -3 \end{cases} \]

Step 3: Setting up and Evaluating the Integral:

We must split the integral at the vertex \(x=-3\): \[ Area = \int_{-6}^{0} |x+3| \, dx = \int_{-6}^{-3} -(x+3) \, dx + \int_{-3}^{0} (x+3) \, dx \]
Now, we evaluate each part:

First integral:
\[ \int_{-6}^{-3} (-x-3) \, dx = \left[ -\frac{x^2}{2} - 3x \right]_{-6}^{-3} \]
\[ = \left( -\frac{(-3)^2}{2} - 3(-3) \right) - \left( -\frac{(-6)^2}{2} - 3(-6) \right) \]
\[ = \left( -\frac{9}{2} + 9 \right) - \left( -\frac{36}{2} + 18 \right) = \left( \frac{9}{2} \right) - (-18+18) = \frac{9}{2} \]
Second integral:
\[ \int_{-3}^{0} (x+3) \, dx = \left[ \frac{x^2}{2} + 3x \right]_{-3}^{0} \]
\[ = \left( \frac{0^2}{2} + 3(0) \right) - \left( \frac{(-3)^2}{2} + 3(-3) \right) \]
\[ = (0) - \left( \frac{9}{2} - 9 \right) = - \left( -\frac{9}{2} \right) = \frac{9}{2} \]

Total Area = \(\dfrac{9}{2} + \dfrac{9}{2} = \dfrac{18}{2} = 9\).


Step 4: Final Answer:

The area of the enclosed region is 9 square units.
Quick Tip: For simple absolute value functions, you can find the area using geometry, which is much faster. The area here consists of two right-angled triangles. Triangle 1 (from x=-6 to x=-3): Base = 3, Height = |-6+3| = 3. Area = \( \frac{1}{2} \times 3 \times 3 = 4.5 \). Triangle 2 (from x=-3 to x=0): Base = 3, Height = |0+3| = 3. Area = \( \frac{1}{2} \times 3 \times 3 = 4.5 \). Total area = 4.5 + 4.5 = 9. This is a great way to verify your integration result.

Step 1: Understanding the Question:

We need to first prove that two lines given in vector form are skew (neither parallel nor intersecting) and then calculate the shortest distance between them.


Step 2: Standard Form and Verification of Skew Lines:

First, rewrite the lines in the standard form \(\vec r = \vec a + t\vec b\).
Line 1 (\(L_1\)): \(\vec r = (\hat{i}+3\hat{k}) + \lambda(-\hat{i}-2\hat{j}-2\hat{k})\)
Here, \(\vec a_1 = \hat{i} + 0\hat{j} + 3\hat{k}\) and \(\vec b_1 = -\hat{i} - 2\hat{j} - 2\hat{k}\).

Line 2 (\(L_2\)): \(\vec r = (\hat{i}-\hat{j}-\hat{k}) + \mu(\hat{i}+2\hat{j}-2\hat{k})\)
Here, \(\vec a_2 = \hat{i} - \hat{j} - \hat{k}\) and \(\vec b_2 = \hat{i} + 2\hat{j} - 2\hat{k}\).

Check for Parallelism: Two lines are parallel if their direction vectors are proportional.

Here, \( \vec b_1 \) and \( \vec b_2 \) are not scalar multiples of each other (\( \frac{-1}{1} \neq \frac{-2}{2} \)). So, the lines are not parallel.


Check for Intersection: If the lines intersect, there exist \(\lambda\) and \(\mu\) such that the position vectors are equal.

Equating components:

(i) \(1-\lambda = \mu+1 \implies \lambda = -\mu\)

(ii) \(-2\lambda = 2\mu-1\)

(iii) \(3-2\lambda = -2\mu-1\)

Substitute (i) into (ii): \(-2(-\mu) = 2\mu-1 \implies 2\mu = 2\mu-1 \implies 0=-1\).

This is a contradiction. Therefore, the lines do not intersect.
Since the lines are not parallel and do not intersect, they are skew lines.


Step 3: Key Formula for Shortest Distance:

The shortest distance \(d\) between two skew lines is given by: \[ d = \frac{|(\vec a_2 - \vec a_1) \cdot (\vec b_1 \times \vec b_2)|}{|\vec b_1 \times \vec b_2|} \]

Step 4: Calculation:


\(\vec a_2 - \vec a_1 = (\hat{i}-\hat{j}-\hat{k}) - (\hat{i}+3\hat{k}) = 0\hat{i} - \hat{j} - 4\hat{k}\).
\(\vec b_1 \times \vec b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & -2 & -2
1 & 2 & -2 \end{vmatrix} = \hat{i}(4 - (-4)) - \hat{j}(2 - (-2)) + \hat{k}(-2 - (-2)) = 8\hat{i} - 4\hat{j} + 0\hat{k}\).
\(|\vec b_1 \times \vec b_2| = \sqrt{8^2 + (-4)^2 + 0^2} = \sqrt{64+16} = \sqrt{80} = 4\sqrt{5}\).
\((\vec a_2 - \vec a_1) \cdot (\vec b_1 \times \vec b_2) = (-\hat{j} - 4\hat{k}) \cdot (8\hat{i} - 4\hat{j}) = (0)(8) + (-1)(-4) + (-4)(0) = 4\).

Now, substitute these into the distance formula: \[ d = \frac{|4|}{4\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \]

Step 5: Final Answer:

The lines are skew, and the shortest distance between them is \( \dfrac{1}{\sqrt{5}} \) units.
Quick Tip: To verify if lines are skew, first check for parallelism. If not parallel, check for intersection. If they don't intersect, they are skew. The shortest distance formula relies on the scalar triple product, which gives the volume of a parallelepiped. If this volume is non-zero, the lines are skew.

Step 1: Understanding the Question:

We are given the position vectors of three points B, W, and F. We need to find the ratio in which the point W divides the line segment BF. This implies the three points are collinear.


Step 2: Key Formula or Approach:

We will use the section formula for position vectors. If a point with position vector \(\vec{w}\) divides the line segment joining points with position vectors \(\vec{b}\) and \(\vec{f}\) in the ratio \(k:1\), then: \[ \vec{w} = \frac{k\vec{f} + 1\vec{b}}{k+1} \]
We will solve for \(k\).


Step 3: Detailed Explanation:

Let the wicket keeper (W) divide the line segment joining the bowler (B) and the leg slip fielder (F) in the ratio \(k:1\).
The position vectors are: \(\vec B=2\hat i+8\hat j\), \(\vec W=6\hat i+12\hat j\), \(\vec F=12\hat i+18\hat j\).
Using the section formula: \[ 6\hat i+12\hat j = \frac{k(12\hat i+18\hat j) + 1(2\hat i+8\hat j)}{k+1} \]
Multiply both sides by \((k+1)\): \[ (k+1)(6\hat i+12\hat j) = (12k\hat i+18k\hat j) + (2\hat i+8\hat j) \]
Group the \(\hat i\) and \(\hat j\) components on the right side: \[ (6k+6)\hat i + (12k+12)\hat j = (12k+2)\hat i + (18k+8)\hat j \]
For the vectors to be equal, their corresponding components must be equal.
Equating the \(\hat i\) components: \[ 6k+6 = 12k+2 \] \[ 4 = 6k \implies k = \frac{4}{6} = \frac{2}{3} \]
Equating the \(\hat j\) components (to verify): \[ 12k+12 = 18k+8 \] \[ 4 = 6k \implies k = \frac{4}{6} = \frac{2}{3} \]
Both components give the same value for \(k\), which confirms that the points are collinear and our assumption is correct.
The ratio is \(k:1\), which is \(\dfrac{2}{3}:1\). To express this with integers, we multiply by 3, giving the ratio \(2:3\).


Step 4: Final Answer:

The wicket keeper divides the line segment joining the bowler and the leg slip fielder in the ratio 2:3.
Quick Tip: An alternative method is to use vectors. Find vectors \(\overrightarrow{BW}\) and \(\overrightarrow{WF}\). If the points are collinear, \(\overrightarrow{BW} = k \overrightarrow{WF}\) for some scalar k. The ratio is then \(k:1\). \(\overrightarrow{BW} = \vec{W}-\vec{B} = (6-2)\hat{i} + (12-8)\hat{j} = 4\hat{i}+4\hat{j}\). \(\overrightarrow{WF} = \vec{F}-\vec{W} = (12-6)\hat{i} + (18-12)\hat{j} = 6\hat{i}+6\hat{j}\). We see that \(4\hat{i}+4\hat{j} = k(6\hat{i}+6\hat{j})\), which gives \(4=6k \implies k=2/3\). The ratio is 2:3.

Step 1: Understanding the Question:

We are given a probability distribution for a discrete random variable \(X\). We need to find the value of the parameter \(p\) and then calculate the mean (expected value) of the distribution.


Step 2: Key Formula or Approach:

(i) For any probability distribution, the sum of all probabilities must be equal to 1. That is, \(\sum P(X_i) = 1\).
(ii) The mean or expected value, \(E(X)\), of a discrete random variable is given by the formula \(E(X) = \mu = \sum X_i P(X_i)\).


Step 3: Detailed Explanation:

(i) Calculate \(p\):
The sum of the probabilities is: \[ P(X=0) + P(X=2) + P(X=4) + P(X=5) = 1 \] \[ p + 2p + 3p + p = 1 \] \[ 7p = 1 \] \[ p = \frac{1}{7} \]
(ii) Calculate the mean number of absent students:
First, we find the probabilities for each value of X using \(p = 1/7\): \(P(X=0) = 1/7\) \(P(X=2) = 2/7\) \(P(X=4) = 3/7\) \(P(X=5) = 1/7\)
Now, we use the formula for the mean: \[ E(X) = (0 \times P(X=0)) + (2 \times P(X=2)) + (4 \times P(X=4)) + (5 \times P(X=5)) \] \[ E(X) = \left(0 \times \frac{1}{7}\right) + \left(2 \times \frac{2}{7}\right) + \left(4 \times \frac{3}{7}\right) + \left(5 \times \frac{1}{7}\right) \] \[ E(X) = 0 + \frac{4}{7} + \frac{12}{7} + \frac{5}{7} \] \[ E(X) = \frac{4+12+5}{7} = \frac{21}{7} = 3 \]

Step 4: Final Answer:

(i) The value of \(p\) is \( \dfrac{1}{7} \).
(ii) The mean number of absent students is 3.
Quick Tip: Remember the two fundamental properties of a discrete probability distribution: 1. \(0 \le P(X_i) \le 1\) for all \(i\). 2. \(\sum P(X_i) = 1\). These are often the starting points for solving problems involving unknown parameters in a distribution.

Step 1: Understanding the Question:

This is a problem of total probability. We need to find the overall probability of an event (getting a distinction) which can happen through two mutually exclusive paths (being a male or being a female).


Step 2: Key Formula or Approach:

Let's define the events: \(M\): The chosen candidate is a male. \(F\): The chosen candidate is a female. \(D\): The chosen candidate gets a distinction.
We need to find \(P(D)\). We use the Law of Total Probability: \[ P(D) = P(M) \cdot P(D|M) + P(F) \cdot P(D|F) \]

Step 3: Detailed Explanation:

From the problem statement, we extract the probabilities:

Probability that a candidate is female: \(P(F) = \dfrac{2}{3}\).
Probability that a candidate is male: \(P(M) = 1 - P(F) = 1 - \dfrac{2}{3} = \dfrac{1}{3}\).
Probability that a male gets distinction (conditional probability): \(P(D|M) = 0.4 = \dfrac{4}{10} = \dfrac{2}{5}\).
Probability that a female gets distinction (conditional probability): \(P(D|F) = 0.35 = \dfrac{35}{100} = \dfrac{7}{20}\).

Now, we apply the Law of Total Probability: \[ P(D) = P(M) \cdot P(D|M) + P(F) \cdot P(D|F) \] \[ P(D) = \left(\frac{1}{3}\right) \cdot \left(\frac{2}{5}\right) + \left(\frac{2}{3}\right) \cdot \left(\frac{7}{20}\right) \] \[ P(D) = \frac{2}{15} + \frac{14}{60} \]
To add these fractions, we find a common denominator, which is 60. \[ P(D) = \frac{2 \times 4}{15 \times 4} + \frac{14}{60} = \frac{8}{60} + \frac{14}{60} \] \[ P(D) = \frac{8+14}{60} = \frac{22}{60} \]
Simplifying the fraction: \[ P(D) = \frac{11}{30} \]

Step 4: Final Answer:

The probability that a candidate chosen at random will have a distinction is \( \dfrac{11}{30} \).
Quick Tip: Problems like this are easily visualized with a probability tree. The first branches represent the gender (Male/Female) with probabilities \(1/3\) and \(2/3\). The second set of branches from each gender node represents getting a distinction or not, with their respective conditional probabilities. To find the total probability of getting a distinction, you multiply along the branches leading to 'Distinction' and then add the results.

Step 1: Formulating the System of Linear Equations:

Let \(x\), \(y\), and \(z\) be the number of students in the Sports, Music, and Drama clubs, respectively.
From the given conditions, we can form three linear equations:

\(x = y + z \implies x - y - z = 0\)
\(y = 20 + \dfrac{1}{2}x \implies 2y = 40 + x \implies -x + 2y + 0z = 40\)
\(x + y + z = 180\)


Step 2: Using the Matrix Method:

We can write this system in the matrix form \(AX = B\), where the solution is given by \(X = A^{-1}B\). \[ A = \begin{bmatrix} 1 & -1 & -1
-1 & 2 & 0
1 & 1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x
y
z \end{bmatrix}, \quad B = \begin{bmatrix} 0
40
180 \end{bmatrix} \]

Step 3: Finding the Inverse of Matrix A:

First, we find the determinant of A: \[ |A| = 1(2\cdot1 - 0\cdot1) - (-1)(-1\cdot1 - 0\cdot1) + (-1)(-1\cdot1 - 2\cdot1) \] \[ |A| = 1(2) + 1(-1) - 1(-3) = 2 - 1 + 3 = 4 \]
Since \(|A| \neq 0\), the inverse exists. Now we find the adjugate of A. The matrix of cofactors is: \[ C = \begin{bmatrix} +(2-0) & -(-1-0) & +(-1-2)
-(-1-(-1)) & +(1-(-1)) & -(1-(-1))
+(0-(-2)) & -(0-1) & +(2-1) \end{bmatrix} = \begin{bmatrix} 2 & 1 & -3
0 & 2 & -2
2 & 1 & 1 \end{bmatrix} \]
The adjugate of A is the transpose of the cofactor matrix: \[ adj(A) = C^T = \begin{bmatrix} 2 & 0 & 2
1 & 2 & 1
-3 & -2 & 1 \end{bmatrix} \]
The inverse of A is: \[ A^{-1} = \frac{1}{|A|}adj(A) = \frac{1}{4} \begin{bmatrix} 2 & 0 & 2
1 & 2 & 1
-3 & -2 & 1 \end{bmatrix} \]

Step 4: Solving for X:
\[ X = A^{-1}B = \frac{1}{4} \begin{bmatrix} 2 & 0 & 2
1 & 2 & 1
-3 & -2 & 1 \end{bmatrix} \begin{bmatrix} 0
40
180 \end{bmatrix} \] \[ \begin{bmatrix} x
y
z \end{bmatrix} = \frac{1}{4} \begin{bmatrix} (2)(0) + (0)(40) + (2)(180)
(1)(0) + (2)(40) + (1)(180)
(-3)(0) + (-2)(40) + (1)(180) \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 360
80 + 180
-80 + 180 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 360
260
100 \end{bmatrix} = \begin{bmatrix} 90
65
25 \end{bmatrix} \]

Step 5: Final Answer:

The number of students in the clubs are:

Sports club (\(x\)): 90 students
Music club (\(y\)): 65 students
Drama club (\(z\)): 25 students Quick Tip: After finding the solution, always perform a quick check by substituting the values of \(x, y,\) and \(z\) back into the original word problem conditions or equations to ensure your answer is correct. For example, \(90 = 65 + 25\) (correct), and \(65 = 20 + 90/2\) (correct).

Step 1: Understanding the Question:

We need to find the indefinite integral of an inverse trigonometric function with a complex argument. The first step should be to simplify the argument or the entire integrand.


Step 2: Simplifying the Integrand:

Let the integrand be \(y = \sin^{-1}\sqrt{\frac{x}{a+x}}\). This implies \(\sin y = \sqrt{\frac{x}{a+x}}\).
To simplify, let's find \(\tan y\). \[ \sin^2 y = \frac{x}{a+x} \] \[ \cos^2 y = 1 - \sin^2 y = 1 - \frac{x}{a+x} = \frac{a+x-x}{a+x} = \frac{a}{a+x} \] \[ \tan^2 y = \frac{\sin^2 y}{\cos^2 y} = \frac{x/(a+x)}{a/(a+x)} = \frac{x}{a} \]
So, \(\tan y = \sqrt{\frac{x}{a}}\). This gives \(y = \tan^{-1}\sqrt{\frac{x}{a}}\).
The integral simplifies to: \[ I = \int \tan^{-1}\sqrt{\frac{x}{a}} \, dx \]

Step 3: Integration by Parts:

We use the integration by parts formula: \(\int u \, dv = uv - \int v \, du\).
Let \(u = \tan^{-1}\sqrt{\frac{x}{a}}\) and \(dv = dx\).
Then \(v = x\).
To find \(du\), we differentiate \(u\): \[ du = \frac{d}{dx}\left(\tan^{-1}\sqrt{\frac{x}{a}}\right) \, dx = \frac{1}{1 + (\sqrt{x/a})^2} \cdot \frac{d}{dx}\left(\sqrt{\frac{x}{a}}\right) \, dx \] \[ du = \frac{1}{1 + x/a} \cdot \frac{1}{\sqrt{a}} \cdot \frac{1}{2\sqrt{x}} \, dx = \frac{a}{a+x} \cdot \frac{1}{2\sqrt{ax}} \, dx \]
Applying the formula: \[ I = x \tan^{-1}\sqrt{\frac{x}{a}} - \int x \cdot \frac{a}{a+x} \cdot \frac{1}{2\sqrt{ax}} \, dx \] \[ I = x \tan^{-1}\sqrt{\frac{x}{a}} - \frac{\sqrt{a}}{2} \int \frac{\sqrt{x}}{a+x} \, dx \]

Step 4: Evaluating the Remaining Integral:

Let the second integral be \(I_2 = \int \frac{\sqrt{x}}{a+x} \, dx\).
Use the substitution \(x = a\tan^2\theta\). Then \(\sqrt{x} = \sqrt{a}\tan\theta\) and \(dx = 2a\tan\theta\sec^2\theta \, d\theta\). \[ I_2 = \int \frac{\sqrt{a}\tan\theta}{a+a\tan^2\theta} \cdot (2a\tan\theta\sec^2\theta) \, d\theta = \int \frac{\sqrt{a}\tan\theta}{a\sec^2\theta} \cdot (2a\tan\theta\sec^2\theta) \, d\theta \] \[ I_2 = 2\sqrt{a} \int \tan^2\theta \, d\theta = 2\sqrt{a} \int (\sec^2\theta - 1) \, d\theta = 2\sqrt{a} (\tan\theta - \theta) \]
Substitute back: \(\tan\theta = \sqrt{x/a}\) and \(\theta = \tan^{-1}\sqrt{x/a}\). \[ I_2 = 2\sqrt{a} \left(\sqrt{\frac{x}{a}} - \tan^{-1}\sqrt{\frac{x}{a}}\right) = 2\sqrt{x} - 2\sqrt{a}\tan^{-1}\sqrt{\frac{x}{a}} \]

Step 5: Final Combination:

Substitute the result for \(I_2\) back into the expression for \(I\): \[ I = x \tan^{-1}\sqrt{\frac{x}{a}} - \frac{\sqrt{a}}{2} \left( 2\sqrt{x} - 2\sqrt{a}\tan^{-1}\sqrt{\frac{x}{a}} \right) + C \] \[ I = x \tan^{-1}\sqrt{\frac{x}{a}} - \sqrt{ax} + a\tan^{-1}\sqrt{\frac{x}{a}} + C \] \[ I = (x+a) \tan^{-1}\sqrt{\frac{x}{a}} - \sqrt{ax} + C \] Quick Tip: For integrals involving complex inverse trigonometric functions, always try to simplify the integrand first. A common strategy is to let \(y\) equal the integrand, apply the trigonometric function (e.g., \(\sin y\)), and then use identities to express it in a simpler form, like \(\tan y\). This often converts the integral into a more standard form for integration by parts.

Step 1: Understanding the Question:

We are given an implicit relation between \(x\) and \(y\) and need to prove a specific form for its derivative \(\dfrac{dy}{dx}\). The form \(\sqrt{1-x^2}\) suggests using a trigonometric substitution to simplify the relation before differentiating.


Step 2: Trigonometric Substitution:

Let \(x = \sin A\) and \(y = \sin B\).

This implies \(A = \sin^{-1}x\) and \(B = \sin^{-1}y\).

Also, \(\sqrt{1-x^2} = \sqrt{1-\sin^2 A} = \cos A\) and \(\sqrt{1-y^2} = \sqrt{1-\sin^2 B} = \cos B\).

Substitute these into the given equation: \[ \cos A + \cos B = a(\sin A - \sin B) \]

Step 3: Using Sum-to-Product Identities:

Apply the sum-to-product formulas:
\(\cos A + \cos B = 2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)\)
\(\sin A - \sin B = 2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)\)

The equation becomes: \[ 2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right) = a \left[ 2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right) \right] \]
Assuming \( \cos\left(\frac{A+B}{2}\right) \neq 0 \), we can cancel this term from both sides: \[ \cos\left(\dfrac{A-B}{2}\right) = a \sin\left(\dfrac{A-B}{2}\right) \] \[ \frac{\cos\left(\frac{A-B}{2}\right)}{\sin\left(\frac{A-B}{2}\right)} = a \implies \cot\left(\dfrac{A-B}{2}\right) = a \] \[ \frac{A-B}{2} = \cot^{-1}(a) \implies A-B = 2\cot^{-1}(a) \]

Step 4: Differentiation:

Substitute back \(A = \sin^{-1}x\) and \(B = \sin^{-1}y\): \[ \sin^{-1}x - \sin^{-1}y = 2\cot^{-1}(a) \]
Now, differentiate both sides of this equation with respect to \(x\). The right side is a constant, so its derivative is zero. \[ \frac{d}{dx}(\sin^{-1}x) - \frac{d}{dx}(\sin^{-1}y) = \frac{d}{dx}(2\cot^{-1}(a)) \] \[ \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-y^2}}\frac{dy}{dx} = 0 \]

Step 5: Solving for \(\dfrac{dy}{dx}\):
\[ \frac{1}{\sqrt{1-y^2}}\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \] \[ \frac{dy}{dx} = \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} = \sqrt{\frac{1-y^2}{1-x^2}} \]
Hence, the required relation is proved.
Quick Tip: Whenever you encounter expressions like \(\sqrt{1-x^2}\), \(\sqrt{1+x^2}\), or \(\sqrt{x^2-1}\) in differentiation or integration, consider using trigonometric substitutions (\(x=\sin\theta\), \(x=\tan\theta\), \(x=\sec\theta\), respectively). It often simplifies complex algebraic expressions into manageable trigonometric ones.

Step 1: Understanding the Question:

We are given parametric equations for \(x\) and \(y\) in terms of a parameter \(\theta\). We need to find the second derivative of \(y\) with respect to \(x\), and then evaluate it at a specific value of \(\theta\).


Step 2: Find \(\dfrac{dy}{d\theta}\) and \(\dfrac{dx}{d\theta}\):
\[ y = a\sin\theta \implies \frac{dy}{d\theta} = a\cos\theta \] \[ x = a\left(\cos\theta+\log\tan\frac{\theta}{2}\right) \] \[ \frac{dx}{d\theta} = a\left(-\sin\theta + \frac{1}{\tan(\theta/2)} \cdot \frac{d}{d\theta}\left(\tan\frac{\theta}{2}\right)\right) \] \[ = a\left(-\sin\theta + \frac{1}{\tan(\theta/2)} \cdot \sec^2\left(\frac{\theta}{2}\right) \cdot \frac{1}{2}\right) \] \[ = a\left(-\sin\theta + \frac{\cos(\theta/2)}{\sin(\theta/2)} \cdot \frac{1}{\cos^2(\theta/2)} \cdot \frac{1}{2}\right) = a\left(-\sin\theta + \frac{1}{2\sin(\theta/2)\cos(\theta/2)}\right) \]
Using the identity \(\sin\theta = 2\sin(\theta/2)\cos(\theta/2)\): \[ \frac{dx}{d\theta} = a\left(-\sin\theta + \frac{1}{\sin\theta}\right) = a\left(\frac{-\sin^2\theta+1}{\sin\theta}\right) = a\frac{\cos^2\theta}{\sin\theta} \]

Step 3: Find \(\dfrac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\cos\theta}{a\cos^2\theta/\sin\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta \]

Step 4: Find \(\dfrac{d^2y}{dx^2}\):

We use the formula \(\dfrac{d^2y}{dx^2} = \dfrac{d}{d\theta}\left(\dfrac{dy}{dx}\right) \cdot \dfrac{d\theta}{dx}\). \[ \frac{d}{d\theta}\left(\frac{dy}{dx}\right) = \frac{d}{d\theta}(\tan\theta) = \sec^2\theta \] \[ \frac{d\theta}{dx} = \frac{1}{dx/d\theta} = \frac{\sin\theta}{a\cos^2\theta} \] \[ \frac{d^2y}{dx^2} = (\sec^2\theta) \cdot \left(\frac{\sin\theta}{a\cos^2\theta}\right) = \frac{1}{\cos^2\theta} \cdot \frac{\sin\theta}{a\cos^2\theta} = \frac{\sin\theta}{a\cos^4\theta} \]

Step 5: Evaluate at \(\theta = \dfrac{\pi}{4}\):

At \(\theta = \dfrac{\pi}{4}\), we have \(\sin(\pi/4) = \dfrac{1}{\sqrt{2}}\) and \(\cos(\pi/4) = \dfrac{1}{\sqrt{2}}\). \[ \frac{d^2y}{dx^2} \bigg|_{\theta=\pi/4} = \frac{1/\sqrt{2}}{a(1/\sqrt{2})^4} = \frac{1/\sqrt{2}}{a(1/4)} = \frac{4}{a\sqrt{2}} \]
Rationalizing the denominator: \[ = \frac{4\sqrt{2}}{a\sqrt{2}\sqrt{2}} = \frac{4\sqrt{2}}{2a} = \frac{2\sqrt{2}}{a} \] Quick Tip: A very common mistake in parametric second derivatives is calculating \( \frac{d^2y}{dx^2} \) as \( \frac{d^2y/d\theta^2}{d^2x/d\theta^2} \). This is incorrect. Always remember the correct chain rule application: \( \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \frac{d\theta}{dx} \).

Step 1: Understanding the Question:

We need to find the reflection (image) of a point A in a given line L. This involves finding the foot of the perpendicular from A to L, which then acts as the midpoint of the segment joining A and its image A'.


Step 2: Find the Foot of the Perpendicular:

Let the given line be L. The coordinates of any general point M on the line L can be found by setting the line equation to a parameter \(\lambda\): \[ \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = \lambda \]
So, \(M = (\lambda, 2\lambda+1, 3\lambda+2)\).
Let M be the foot of the perpendicular from \(A(1,6,3)\) to the line L. The direction ratios of the line segment AM are: \[ (\lambda-1, 2\lambda+1-6, 3\lambda+2-3) = (\lambda-1, 2\lambda-5, 3\lambda-1) \]
The direction vector of the line L is \(\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}\), with direction ratios \((1, 2, 3)\).
Since AM is perpendicular to L, the dot product of their direction vectors is zero: \[ 1(\lambda-1) + 2(2\lambda-5) + 3(3\lambda-1) = 0 \] \[ \lambda-1 + 4\lambda-10 + 9\lambda-3 = 0 \] \[ 14\lambda - 14 = 0 \implies \lambda = 1 \]
Substituting \(\lambda=1\) back into the coordinates of M, we get the foot of the perpendicular: \[ M = (1, 2(1)+1, 3(1)+2) = (1, 3, 5) \]

Step 3: Find the Image A':

The point M is the midpoint of the line segment AA', where \(A'(x', y', z')\) is the image of A. Using the midpoint formula: \[ M = \left(\frac{x+x'}{2}, \frac{y+y'}{2}, \frac{z+z'}{2}\right) \] \[ (1, 3, 5) = \left(\frac{1+x'}{2}, \frac{6+y'}{2}, \frac{3+z'}{2}\right) \]
Equating the components:

\(\dfrac{1+x'}{2} = 1 \implies 1+x' = 2 \implies x' = 1\)
\(\dfrac{6+y'}{2} = 3 \implies 6+y' = 6 \implies y' = 0\)
\(\dfrac{3+z'}{2} = 5 \implies 3+z' = 10 \implies z' = 7\)

The image of point A is \(A'(1, 0, 7)\).


Step 4: Find the Equation of the Line AA':

The line passes through \(A(1,6,3)\) and has direction ratios given by the vector \(\overrightarrow{AA'}\). \[ \overrightarrow{AA'} = (1-1, 0-6, 7-3) = (0, -6, 4) \]
The direction ratios can be simplified to \((0, -3, 2)\) by dividing by 2.
The equation of the line AA' is: \[ \frac{x-1}{0} = \frac{y-6}{-3} = \frac{z-3}{2} \] Quick Tip: The process for finding the image of a point in a line is standard: 1. Write the coordinates of a general point on the line in terms of a parameter \(\lambda\). 2. Find the direction ratios of the segment connecting the given point to the general point. 3. Use the perpendicularity condition (dot product = 0) to solve for \(\lambda\), which gives the foot of the perpendicular. 4. Use the midpoint formula to find the image coordinates.

Step 1: Understanding the Question:

We need to find a specific point P on a given line that is at a fixed distance (7 units) from another given point Q. After finding P, we need to determine the equation of the line segment PQ.


Step 2: Finding the General Point P:

Let the coordinates of any point P on the given line be determined by a parameter \(\lambda\): \[ \frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-9} = \lambda \] \[ x = \lambda-5, \quad y = 4\lambda-3, \quad z = -9\lambda+6 \]
So, the coordinates of P are \((\lambda-5, 4\lambda-3, -9\lambda+6)\).


Step 3: Using the Distance Formula:

We are given that the distance between P and \(Q(2,4,-1)\) is 7 units. Using the distance formula \(PQ^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2\): \[ ((\lambda-5)-2)^2 + ((4\lambda-3)-4)^2 + ((-9\lambda+6)-(-1))^2 = 7^2 \] \[ (\lambda-7)^2 + (4\lambda-7)^2 + (-9\lambda+7)^2 = 49 \]
Expand the squared terms: \[ (\lambda^2-14\lambda+49) + (16\lambda^2-56\lambda+49) + (81\lambda^2-126\lambda+49) = 49 \]
Combine like terms: \[ (1+16+81)\lambda^2 + (-14-56-126)\lambda + (49+49+49) = 49 \] \[ 98\lambda^2 - 196\lambda + 147 = 49 \] \[ 98\lambda^2 - 196\lambda + 98 = 0 \]
Divide the entire equation by 98: \[ \lambda^2 - 2\lambda + 1 = 0 \]
This is a perfect square: \[ (\lambda-1)^2 = 0 \implies \lambda = 1 \]
Since we found a single value for \(\lambda\), there is only one such point P.


Step 4: Finding the Coordinates of P:

Substitute \(\lambda = 1\) into the general coordinates of P: \[ P = (1-5, 4(1)-3, -9(1)+6) = (-4, 1, -3) \]

Step 5: Finding the Equation of Line PQ:

The line passes through \(P(-4,1,-3)\) and \(Q(2,4,-1)\). The direction ratios of the line are given by the vector \(\overrightarrow{PQ}\): \[ \overrightarrow{PQ} = (2 - (-4), 4 - 1, -1 - (-3)) = (6, 3, 2) \]
Using point Q(2,4,-1) and the direction ratios (6,3,2), the equation of the line is: \[ \frac{x-2}{6} = \frac{y-4}{3} = \frac{z+1}{2} \] Quick Tip: When solving for \(\lambda\) from the distance formula equation, you might get a quadratic equation with two, one, or zero real roots. This corresponds to finding two, one (a tangent case), or no such points on the line. In this case, getting one repeated root means there is a unique solution.