CBSE Class 12 Mathematics Set 3- (65/2/3) Question Paper 2026 with Solution PDF

Concept:

The standard symmetrical form of a line is \( \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \), where \( (l, m, n) \) are the direction ratios.
Direction cosines \( (L, M, N) \) are related to direction ratios \( (l, m, n) \) by the formula \( L = \frac{l}{\sqrt{l^2 + m^2 + n^2}} \), \( M = \frac{m}{\sqrt{l^2 + m^2 + n^2}} \), and \( N = \frac{n}{\sqrt{l^2 + m^2 + n^2}} \).
Direction cosines are unique for a directed line but can have two sets of opposite signs for an undirected line.



Step 1: Convert the given line equation into standard form

The given equation is \( \frac{1 - x}{0} = y = z \).

We rewrite it to have positive \( x \) coefficients in the numerator:
\[ \frac{-(x - 1)}{0} = \frac{y - 0}{1} = \frac{z - 0}{1} \]

Multiplying the denominators of the first term by \( -1 \):
\[ \frac{x - 1}{0} = \frac{y - 0}{1} = \frac{z - 0}{1} \]


Step 2: Identify the direction ratios (DRs) of the line

Comparing with the standard form \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \):

The direction ratios are \( a = 0 \), \( b = 1 \), and \( c = 1 \).


Step 3: Calculate the magnitude for direction cosines

Let \( D = \sqrt{a^2 + b^2 + c^2} \):
\[ D = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} \]


Step 4: Determine the direction cosines (DCs)

The DCs are given by \( \left( \frac{a}{D}, \frac{b}{D}, \frac{c}{D} \right) \):
\[ DCs = \left( \frac{0}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \]

Since direction cosines can also be represented with reversed signs (opposite direction):
\[ DCs = \left( 0, -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right) \]

This matches option (B). Quick Tip: Always ensure the variables \( x, y, z \) have a coefficient of \( +1 \) before identifying DRs.
Recall that \( L^2 + M^2 + N^2 = 1 \) for direction cosines; use this to verify options quickly.
If denominators include zero, it indicates the line is perpendicular to that specific axis.

Concept:

Linear Programming Problem (LPP) involves optimizing a linear function subject to linear constraints.
Constraints are the system of linear inequalities or equations.
The feasible region is the set of all points that satisfy all constraints.



Step 1: Define the function being optimized

In any optimization problem, the goal is to find the maximum or minimum value of a specific function.

This specific function, usually denoted by \( Z = ax + by \), is defined as the objective function.


Step 2: Distinguish between the different LPP components

Constraints are the conditions that limit the variables (e.g., \( x + y \leq 5 \)).

The feasible function refers to the values within the valid region.

The optimal value is the actual max/min numerical result obtained at the corner points.


Step 3: Conclusion

Therefore, the function itself is called the objective function. Quick Tip: The objective function represents quantities like profit, cost, or distance.
Non-negativity restrictions (\( x, y \geq 0 \)) are almost always present in standard LPPs.

Concept:



The equation of a line passing through \( (a, 0) \) and \( (0, b) \) is \( \frac{x}{a} + \frac{y}{b} = 1 \).
If the shaded region includes the origin \( (0, 0) \), the inequality for a line \( ax + by \leq c \) (where \( c > 0 \)) is generally \( \leq \).



Step 1: Find the equation of the first boundary line


Looking at the graph, one line passes through \( (5, 0) \) and \( (0, 5) \).


Using the intercept form \( \frac{x}{5} + \frac{y}{5} = 1 \):

\[ x + y = 5 \]


Since the region is shaded below the line (towards the origin), the inequality is:

\[ x + y \leq 5 \]


Step 2: Find the equation of the second boundary line


The second line passes through \( (9, 0) \) and \( (0, 3) \).


Using the intercept form \( \frac{x}{9} + \frac{y}{3} = 1 \):


Multiplying by 9 on both sides:

\[ x + 3y = 9 \]


Since the region is shaded below this line as well, the inequality is:

\[ x + 3y \leq 9 \]


Step 3: Identify non-trivial constraints


Non-trivial constraints exclude the standard non-negativity constraints (\( x \geq 0, y \geq 0 \)).


Thus, the constraints are \( x + y \leq 5 \) and \( x + 3y \leq 9 \). Quick Tip: Use the intercept form \( \frac{x}{x-int} + \frac{y}{y-int} = 1 \) to find line equations instantly from graphs.
Check the origin \( (0, 0) \) in your derived inequality to verify the direction of the shaded region.

Concept:



Conditional Probability: \( P(X / Y) = \frac{P(X \cap Y)}{P(Y)} \).
De Morgan's Law: \( A' \cap B' = (A \cup B)' \).
Complementary Probability: \( P(E') = 1 - P(E) \).



Step 1: Apply the conditional probability formula

By definition:

\[ P(A' / B') = \frac{P(A' \cap B')}{P(B')} \]


Step 2: Simplify the numerator using De Morgan's Law


From set theory, the intersection of complements is the complement of the union:

\[ A' \cap B' = (A \cup B)' \]

Therefore, the probability is:
\[ P(A' \cap B') = P((A \cup B)') \]


Step 3: Convert to standard probability form


Using the property \( P(E') = 1 - P(E) \):

\[ P((A \cup B)') = 1 - P(A \cup B) \]


Step 4: Substitute back into the expression


Replacing the numerator in the expression from Step 1:

\[ P(A' / B') = \frac{1 - P(A \cup B)}{P(B')} \]


This matches option (D). Quick Tip: Remember \( P(A/B) + P(A'/B) = 1 \). This helps eliminate incorrect options like (A) and (B).
De Morgan's Laws are essential for probability problems involving "neither A nor B".

Concept:



Reflexive: \( (a, a) \in R \) for every \( a \in A \).
Symmetric: If \( (a, b) \in R \), then \( (b, a) \in R \).
Transitive: If \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \).



Step 1: Check for Reflexivity


The set is \( A = \{1, 2, 3\} \). For \( R \) to be reflexive, it must contain \( (1,1), (2,2), (3,3) \).


Checking the given set \( R \):

\( (1, 1) \in R \)
\( (2, 2) \in R \)
\( (3, 3) \in R \)

Since all are present, \( R \) is reflexive.


Step 2: Check for Symmetry


Checking pairs where \( a \neq b \):


For \( (1, 3) \in R \), we need \( (3, 1) \in R \). Checking \( R \): Yes, \( (3, 1) \) is present.


Since there are no other such pairs, \( R \) is symmetric.



Step 3: Check for Transitivity


We need to check all combinations \( (a, b) \) and \( (b, c) \):


1. Consider \( (1, 3) \) and \( (3, 3) \): \( (1, 3) \in R \). (Satisfied)

2. Consider \( (1, 3) \) and \( (3, 1) \): \( (1, 1) \in R \). (Satisfied)

3. Consider \( (3, 1) \) and \( (1, 3) \): \( (3, 3) \in R \). (Satisfied)

4. Consider \( (3, 1) \) and \( (1, 1) \): \( (3, 1) \in R \). (Satisfied)

Every possible chain leads to an element already in the set. Thus, \( R \) is transitive.


Step 4: Final Conclusion


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

This corresponds to option (D). Quick Tip: An equivalence relation must satisfy all three properties: reflexive, symmetric, and transitive.
If a relation is reflexive and symmetric, and has pairs like \( (x, y) \) and \( (y, x) \), it must have \( (x, x) \) and \( (y, y) \) to be transitive.

Concept:

Matrix multiplication is generally non-commutative, meaning \( AB \neq BA \) in most cases.
Algebraic identities for matrices must respect the order of multiplication.
A statement is "always true" only if it holds for all possible square matrices \( A \) and \( B \) of the same order.



Step 1: Analyze statement (i): \( (A + B)(A - B) = A^2 - B^2 \)

Let's expand the left-hand side using the distributive property:
\[ (A + B)(A - B) = A(A - B) + B(A - B) \]
\[ = A^2 - AB + BA - B^2 \]

This equals \( A^2 - B^2 \) only if \( -AB + BA = 0 \), which implies \( AB = BA \).

Since matrices are not always commutative, this is not always true.


Step 2: Analyze statement (ii): \( AB = BA \)

This is the commutative property of multiplication.

For matrices, this is a special case and does not hold for all matrices.

Therefore, it is not "always" true.


Step 3: Analyze statement (iii): \( (A + B)^2 = A^2 + AB + BA + B^2 \)

Expanding the square:
\[ (A + B)^2 = (A + B)(A + B) \]
\[ = A(A + B) + B(A + B) \]
\[ = A^2 + AB + BA + B^2 \]

This expansion is derived directly from the distributive laws of matrix addition and multiplication.

It does not assume commutativity, so it is always true.


Step 4: Analyze statement (iv): \( AB = 0 \Rightarrow A = 0 or B = 0 \)

In matrix algebra, the product of two non-zero matrices can be a zero matrix (these are called zero-divisors).

For example, if \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \), then \( AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \), but neither \( A \) nor \( B \) is zero.

So, this statement is not always true.


Step 5: Conclusion

Only statement (iii) is always true. Quick Tip: Never assume \( AB = BA \) unless the question explicitly states that the matrices commute.
Standard algebraic identities like \( (a+b)^2 = a^2 + 2ab + b^2 \) only apply to matrices if \( AB = BA \).
If you suspect a statement is false, try to think of a simple \( 2 \times 2 \) matrix counterexample involving zeros.

Concept:



A square matrix \( A \) is symmetric if \( A = A^T \), where \( A^T \) is the transpose of \( A \).
For a symmetric matrix, elements across the main diagonal are equal: \( a_{ij} = a_{ji} \).



Step 1: Apply the condition for a symmetric matrix


Given \( A = \begin{bmatrix} 1 & a & b \\ -1 & 2 & c \\ 0 & 5 & 3 \end{bmatrix} \).


The transpose is \( A^T = \begin{bmatrix} 1 & -1 & 0 \\ a & 2 & 5 \\ b & c & 3 \end{bmatrix} \).


Since \( A = A^T \), we equate corresponding elements:

\( a_{12} = a_{21} \implies a = -1 \)
\( a_{13} = a_{31} \implies b = 0 \)
\( a_{23} = a_{32} \implies c = 5 \)



Step 2: Calculate the required expression


We need to find the value of \( 3a + b + c \).


Substitute the values found in Step 1:
\[ 3a + b + c = 3(-1) + 0 + 5 \]
\[ = -3 + 5 \]
\[ = 2 \] Quick Tip: For symmetry, just "reflect" the elements across the diagonal to find unknowns.
Remember that the diagonal elements themselves don't provide any information about symmetry constraints.

Concept:

The transpose \( A' \) is obtained by interchanging rows and columns.
\( I \) is the identity matrix of the same order.
Matrix addition is performed element-wise.



Step 1: Find the transpose \( A' \)

Given \( A = \begin{bmatrix} \tan x & \cot x \\ -\cot x & \tan x \end{bmatrix} \).

Then \( A' = \begin{bmatrix} \tan x & -\cot x \\ \cot x & \tan x \end{bmatrix} \).


Step 2: Compute \( A + A' \)
\[ A + A' = \begin{bmatrix} \tan x + \tan x & \cot x + (-\cot x) \\ -\cot x + \cot x & \tan x + \tan x \end{bmatrix} \]
\[ A + A' = \begin{bmatrix} 2 \tan x & 0 \\ 0 & 2 \tan x \end{bmatrix} \]


Step 3: Equate with \( 2I \)

Given \( A + A' = 2I \):
\[ \begin{bmatrix} 2 \tan x & 0 \\ 0 & 2 \tan x \end{bmatrix} = 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
\[ \begin{bmatrix} 2 \tan x & 0 \\ 0 & 2 \tan x \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \]

Equating corresponding elements:
\[ 2 \tan x = 2 \implies \tan x = 1 \]


Step 4: Solve for \( x \) in the given interval

We are looking for \( x \in [0, \pi/2] \) such that \( \tan x = 1 \).

The only solution in this interval is \( x = \pi/4 \). Quick Tip: For questions involving \( A + A' \), off-diagonal elements \( a_{ij} + a_{ji} \) must equal the corresponding elements in the target matrix.
Check your interval carefully; if the interval was larger, there might be more solutions like \( 5\pi/4 \).

Concept:



Inverse of a scalar multiple of a matrix: If \( k \) is a non-zero scalar and \( A \) is an invertible matrix, then \( (kA)^{-1} = \frac{1}{k} A^{-1} \).
This property arises because \( (kA) \cdot (\frac{1}{k} A^{-1}) = (k \cdot \frac{1}{k}) (A \cdot A^{-1}) = 1 \cdot I = I \).



Step 1: Identify the scalar and apply the property


In the expression \( (3A)^{-1} \), the scalar \( k \) is \( 3 \).


Applying the property \( (kA)^{-1} = \frac{1}{k} A^{-1} \):

\[ (3A)^{-1} = \frac{1}{3} A^{-1} \]


Step 2: Verify with the definition of inverse


Let \( B = \frac{1}{3} A^{-1} \).


Check if \( (3A)B = I \):

\[ (3A) \left( \frac{1}{3} A^{-1} \right) = \left( 3 \cdot \frac{1}{3} \right) (A \cdot A^{-1}) \]

\[ = 1 \cdot I = I \]



Since the product is the identity matrix, \( \frac{1}{3} A^{-1} \) is indeed the inverse of \( 3A \). Quick Tip: Don't confuse \( (kA)^{-1} = \frac{1}{k} A^{-1} \) with the determinant property \( |kA| = k^n |A| \).
Inverses work by reciprocating the scalar, while determinants scale by the power of the matrix order.

Concept:

Expansion of a \( 3 \times 3 \) determinant using minors and cofactors.
Vieta's formulas for a quadratic equation \( ax^2 + bx + c = 0 \): Sum of roots is \( -b/a \).



Step 1: Expand the determinant along Row 1
\[ \Delta = -1 \begin{vmatrix} a & -1 \\ 4 & 2a \end{vmatrix} - (-2) \begin{vmatrix} -2 & -1 \\ 0 & 2a \end{vmatrix} + 5 \begin{vmatrix} -2 & a \\ 0 & 4 \end{vmatrix} \]

Calculate the \( 2 \times 2 \) determinants:
\[ \Delta = -1(a(2a) - (-1)(4)) + 2((-2)(2a) - (-1)(0)) + 5((-2)(4) - a(0)) \]
\[ \Delta = -1(2a^2 + 4) + 2(-4a) + 5(-8) \]
\[ \Delta = -2a^2 - 4 - 8a - 40 \]
\[ \Delta = -2a^2 - 8a - 44 \]


Step 2: Set the expression equal to -86 and solve the quadratic
\[ -2a^2 - 8a - 44 = -86 \]

Add 86 to both sides:
\[ -2a^2 - 8a + 42 = 0 \]

Divide the entire equation by -2 to simplify:
\[ a^2 + 4a - 21 = 0 \]


Step 3: Find the sum of values of \( a \)

For a quadratic equation \( x^2 + px + q = 0 \), the sum of roots is \( -p \).

Here, the equation is in the variable \( a \) with coefficient \( p = 4 \).

Sum of possible values of \( a = -4 \).


Step 4: Verification (Optional)

Factoring the quadratic: \( a^2 + 7a - 3a - 21 = 0 \implies (a+7)(a-3) = 0 \).

The roots are \( a = -7 \) and \( a = 3 \).

Sum \( = -7 + 3 = -4 \). The result is consistent. Quick Tip: When asked for the sum of roots, save time by using Vieta's formulas instead of fully solving for each root.
Always look for the row or column with the most zeros to expand the determinant more efficiently.

Concept:

Implicit differentiation: Differentiating both sides of an equation with respect to \( x \) where \( y \) is treated as a function of \( x \).
Product rule of differentiation: \( \frac{d}{dx}[u(x)v(x)] = u \frac{dv}{dx} + v \frac{du}{dx} \).



Step 1: Differentiate both sides of the equation with respect to \( x \)

The given equation is \( x + y = xy \).

Differentiating with respect to \( x \):
\[ \frac{d}{dx}(x) + \frac{d}{dx}(y) = \frac{d}{dx}(xy) \]

Applying the power rule and product rule:
\[ 1 + \frac{dy}{dx} = x \frac{dy}{dx} + y(1) \]


Step 2: Isolate the terms containing \( \frac{dy}{dx} \)

Rearranging the equation to group all \( \frac{dy}{dx} \) terms on one side:
\[ 1 - y = x \frac{dy}{dx} - \frac{dy}{dx} \]


Step 3: Solve for \( \frac{dy}{dx} \)

Factor out \( \frac{dy}{dx} \) on the right-hand side:
\[ 1 - y = \frac{dy}{dx}(x - 1) \]
\[ \frac{dy}{dx} = \frac{1 - y}{x - 1} \]

This matches option (D). Quick Tip: When an equation involves products of \( x \) and \( y \), always apply the product rule carefully.
Alternatively, you can solve for \( y \) first: \( y(x - 1) = x \implies y = \frac{x}{x - 1} \), then use the quotient rule.

Concept:

Integration of trigonometric functions.
Substitution method: Replacing part of the integrand to simplify the integration process.
Trigonometric identity: \( \sec x + \tan x = \frac{1 + \sin x}{\cos x} \).



Step 1: Simplify the integrand using basic trigonometric identities

Convert the secant and tangent functions into sine and cosine:
\[ \sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x} \]

The integral becomes:
\[ I = \int \frac{1}{\frac{1 + \sin x}{\cos x}} dx = \int \frac{\cos x}{1 + \sin x} dx \]


Step 2: Perform a substitution

Let \( u = 1 + \sin x \).

Then, differentiating with respect to \( x \):
\[ du = \cos x \, dx \]


Step 3: Evaluate the integral

Substitute \( u \) and \( du \) into the integral:
\[ I = \int \frac{1}{u} du \]
\[ I = \log |u| + C \]

Substituting back for \( u \):
\[ I = \log |1 + \sin x| + C \]

This matches option (D). Quick Tip: Always look for "derivative-function" pairs in trigonometric integrals; here \( \cos x \) is the derivative of \( \sin x \).
Rationalizing the denominator by multiplying by \( (\sec x - \tan x) \) is another valid method.

Concept:

Critical points occur where the first derivative \( f'(x) = 0 \).
Second derivative test: If \( f'(c) = 0 \), then \( f''(c) > 0 \) indicates a local minimum and \( f''(c) < 0 \) indicates a local maximum.



Step 1: Find the first derivative and critical points

Given \( f(x) = x + x^{-1} \).

Differentiating:
\[ f'(x) = 1 - \frac{1}{x^2} \]

Set \( f'(x) = 0 \):
\[ 1 - \frac{1}{x^2} = 0 \implies x^2 = 1 \implies x = 1, -1 \]


Step 2: Find the second derivative

Differentiating \( f'(x) \):
\[ f''(x) = \frac{d}{dx}(1 - x^{-2}) = 2x^{-3} = \frac{2}{x^3} \]


Step 3: Test each critical point

For \( x = 1 \):
\[ f''(1) = \frac{2}{1^3} = 2 > 0 \]

Thus, \( x = 1 \) is a point of local minimum. Local minimum value \( f(1) = 1 + \frac{1}{1} = 2 \).


For \( x = -1 \):
\[ f''(-1) = \frac{2}{(-1)^3} = -2 < 0 \]

Thus, \( x = -1 \) is a point of local maximum. Local maximum value \( f(-1) = -1 + \frac{1}{-1} = -2 \).


Step 4: Compare results with options

Option (A): Local max is 2 (False, it is -2).

Option (B): Local min is -2 (False, it is 2).

Option (C): Local max is -2 (True).

Option (D): Local min (2) \( < \) Local max (-2) (False). Quick Tip: Note that for this function, the local minimum value (2) is actually greater than the local maximum value (-2). This is common in functions with asymptotes.
Recall the AM-GM inequality: for \( x > 0 \), \( x + \frac{1}{x} \geq 2 \). This confirms the minimum at \( x = 1 \).

Concept:



Area between curves: Area can be calculated by integrating along the \( x \)-axis or \( y \)-axis.
Symmetry: The parabola \( y = x^2 \) is symmetric about the \( y \)-axis.
Integration along \( y \)-axis: \( Area = \int_{c}^{d} (x_{right} - x_{left}) dy \).



Step 1: Identify the geometry and intersection points


The curve is \( y = x^2 \) and the line is \( y = 16 \).


The curves intersect when \( x^2 = 16 \implies x = \pm 4 \).


The region is bounded between the parabola and the horizontal line \( y = 16 \).



Step 2: Set up the area integral along the \( y \)-axis


Integrating along the \( y \)-axis is often easier for this shape.


Limits for \( y \) are from 0 to 16.


For a given \( y \), the \( x \)-values are \( x = \pm \sqrt{y} \).


The total width at height \( y \) is \( \sqrt{y} - (-\sqrt{y}) = 2\sqrt{y} \).



Step 3: Formulate the final integral


Total Area \( = \int_0^{16} (x_{right} - x_{left}) dy \)

\[ Area = \int_0^{16} (\sqrt{y} - (-\sqrt{y})) dy \]

\[ Area = \int_0^{16} 2\sqrt{y} dy = 2 \int_0^{16} \sqrt{y} dy \]

This matches option (D). Quick Tip: Always sketch the region first to determine if integration along \( x \) or \( y \) is more straightforward.
Integration along the \( x \)-axis would be \( \int_{-4}^{4} (16 - x^2) dx \), which is not listed as a simplified option.

Concept:

Variable Separable Method: A technique to solve differential equations by moving all terms involving \( y \) to one side and all terms involving \( x \) to the other.
Integration of power functions: \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \).



Step 1: Separate the variables

The given equation is \( x^2 dy + y^2 dx = 0 \).

Rearranging:
\[ x^2 dy = -y^2 dx \]

Divide both sides by \( x^2 y^2 \) (assuming \( x, y \neq 0 \)):
\[ \frac{dy}{y^2} = -\frac{dx}{x^2} \]
\[ y^{-2} dy = -x^{-2} dx \]


Step 2: Integrate both sides
\[ \int y^{-2} dy = \int -x^{-2} dx \]

Applying the integration formula:
\[ \frac{y^{-1}}{-1} = -\left( \frac{x^{-1}}{-1} \right) + C \]
\[ -\frac{1}{y} = \frac{1}{x} + C \]


Step 3: Simplify to the standard form

Multiply the equation by \( -1 \):
\[ \frac{1}{y} = -\frac{1}{x} - C \]
\[ \frac{1}{y} + \frac{1}{x} = -C \]

Let \( k = -C \) be the arbitrary constant:
\[ \frac{1}{y} + \frac{1}{x} = k \]

This matches option (C). Quick Tip: When separating variables, always move the differentials \( dx \) and \( dy \) to the numerators first.
Constant terms can be rewritten (e.g., \( -C \) as \( k \)) to match the format of given options.

Concept:

A linear differential equation of the first order has the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \).
The Integrating Factor (IF) is calculated using the formula: \( IF = e^{\int P(x) dx} \).



Step 1: Convert the given equation to the standard form

The given equation is \( 2x \frac{dy}{dx} - y = 3 \).

Divide the entire equation by \( 2x \) to make the coefficient of \( \frac{dy}{dx} \) equal to 1:
\[ \frac{dy}{dx} - \frac{y}{2x} = \frac{3}{2x} \]
\[ \frac{dy}{dx} + \left( -\frac{1}{2x} \right)y = \frac{3}{2x} \]


Step 2: Identify \( P(x) \)

Comparing with the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \):
\[ P(x) = -\frac{1}{2x} \]


Step 3: Calculate the Integrating Factor (IF)
\[ IF = e^{\int P(x) dx} = e^{\int -\frac{1}{2x} dx} \]

Take the constant out of the integral:
\[ IF = e^{-\frac{1}{2} \int \frac{1}{x} dx} \]

Using \( \int \frac{1}{x} dx = \log x \):
\[ IF = e^{-\frac{1}{2} \log x} = e^{\log x^{-1/2}} \]


Step 4: Simplify the expression

Using the property \( e^{\log f(x)} = f(x) \):
\[ IF = x^{-1/2} = \frac{1}{x^{1/2}} = \frac{1}{\sqrt{x}} \]

This matches option (B). Quick Tip: Always ensure the coefficient of \( \frac{dy}{dx} \) is 1 before identifying \( P(x) \).
Remember logarithmic properties: \( n \log a = \log a^n \) and \( e^{\log m} = m \).

Concept:



The magnitude of a scalar multiple of a vector is \( |k\vec{a}| = |k| \cdot |\vec{a}| \).
Absolute value \( |\lambda| \) represents the distance from zero on the number line.



Step 1: Express the magnitude in terms of \( \lambda \) and \( |\vec{a}| \)


Given \( |\vec{a}| = 5 \).


The expression to evaluate is \( f(\lambda) = |\lambda \vec{a}| = |\lambda| \cdot |\vec{a}| = 5|\lambda| \).



Step 2: Determine the range of \( |\lambda| \)


Given the interval \( -2 \leq \lambda \leq 1 \).


The absolute value \( |\lambda| \) will range from its minimum value at zero to its maximum value at the endpoint furthest from zero.


Minimum \( |\lambda| = 0 \) (at \( \lambda = 0 \)).


Maximum \( |\lambda| = |-2| = 2 \).



Step 3: Find the greatest and smallest values of \( 5|\lambda| \)


Smallest value \( = 5 \times 0 = 0 \).


Greatest value \( = 5 \times 2 = 10 \).



Step 4: Calculate the sum


Sum \( = Greatest Value + Smallest Value \)


Sum \( = 10 + 0 = 10 \).

This matches option (C). Quick Tip: Magnitude is always a non-negative quantity.
When dealing with intervals for absolute values, check if zero lies within the interval; if it does, the minimum absolute value is 0.

Concept:

A vector perpendicular to the \( z \)-axis lies in the \( xy \)-plane, meaning its \( z \)-component is zero.
Direction cosines \( (L, M, N) \) satisfy \( L^2 + M^2 + N^2 = 1 \).
A vector of magnitude \( r \) is given by \( \vec{v} = r(L\hat{i} + M\hat{j} + N\hat{k}) \).



Step 1: Determine the direction cosines

Let the vector make angles \( \alpha, \beta, \gamma \) with the axes.

Given \( \alpha = \beta \) (equal angles with \( x \) and \( y \)).

Since it is perpendicular to the \( z \)-axis, \( \gamma = 90^\circ \).

Then \( L = \cos \alpha \), \( M = \cos \alpha \), and \( N = \cos 90^\circ = 0 \).


Step 2: Solve for the direction cosines

Using the identity \( L^2 + M^2 + N^2 = 1 \):
\[ \cos^2 \alpha + \cos^2 \alpha + 0^2 = 1 \]
\[ 2\cos^2 \alpha = 1 \implies \cos^2 \alpha = \frac{1}{2} \]
\[ \cos \alpha = \pm \frac{1}{\sqrt{2}} \]


Step 3: Construct the vector

Magnitude \( r = 3 \).
\[ \vec{v} = 3(L\hat{i} + M\hat{j} + N\hat{k}) \]

Assuming positive components for the matching option:
\[ \vec{v} = 3\left( \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k} \right) \]
\[ \vec{v} = \frac{3}{\sqrt{2}}\hat{i} + \frac{3}{\sqrt{2}}\hat{j} \]


Step 4: Rationalize the coefficients
\[ \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \]

So, \( \vec{v} = \frac{3\sqrt{2}}{2}\hat{i} + \frac{3\sqrt{2}}{2}\hat{j} \).

This matches option (C). Quick Tip: "Perpendicular to \( z \)-axis" immediately tells you there is no \( \hat{k} \) component.
For any vector in 2D space making equal angles with axes, the direction cosines are \( (1/\sqrt{2}, 1/\sqrt{2}) \).

Concept:

Dot product: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \).
Cross product magnitude: \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \).
Identity: \( \sin^2 \theta + \cos^2 \theta = 1 \).



Step 1: Evaluate Assertion (A)

Let \( |\vec{a}| = a \) and \( |\vec{b}| = b \).

LHS \( = (ab \sin \theta)^2 + (ab \cos \theta)^2 \)

LHS \( = a^2b^2 \sin^2 \theta + a^2b^2 \cos^2 \theta \)

LHS \( = a^2b^2 (\sin^2 \theta + \cos^2 \theta) = a^2b^2 \cdot 1 = a^2b^2 \)

RHS \( = a^2b^2 \).

Since LHS \( = \) RHS, Assertion (A) is true. (This is Lagrange's Identity).


Step 2: Evaluate Reason (R)

Starting from the RHS of Reason (R):
\( (\vec{a} \cdot \vec{b}) \tan \theta = (ab \cos \theta) \left( \frac{\sin \theta}{\cos \theta} \right) \)
\( = ab \sin \theta = |\vec{a} \times \vec{b}| \).

Reason (R) is true for \( \theta \neq \pi/2 \) (where tan is defined).


Step 3: Determine if (R) explains (A)

If we take Reason (R) and square it:
\( |\vec{a} \times \vec{b}|^2 = (\vec{a} \cdot \vec{b})^2 \tan^2 \theta \)

Adding \( (\vec{a} \cdot \vec{b})^2 \) to both sides:
\( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = (\vec{a} \cdot \vec{b})^2 (\tan^2 \theta + 1) \)
\( = (\vec{a} \cdot \vec{b})^2 \sec^2 \theta = (ab \cos \theta)^2 \cdot \frac{1}{\cos^2 \theta} = a^2b^2 \).

Since Reason (R) provides a direct relationship that leads to the truth of Assertion (A), it is the correct explanation. Quick Tip: Identity in (A) is known as Lagrange's Identity.
Reason (R) is simply a different way to express the relationship between sine and cosine components.

Concept:



Direction Cosines \( (L, M, N) \) must satisfy the fundamental constraint: \( L^2 + M^2 + N^2 = 1 \).
Direction cosines are the values \( \cos \alpha, \cos \beta, \cos \gamma \), where \( \alpha, \beta, \gamma \) are the angles with the axes.



Step 1: Evaluate Assertion (A)


For a line to have direction cosines \( < 1, 1, 1 > \), the sum of their squares must equal 1.


Calculation:

\( L^2 + M^2 + N^2 = 1^2 + 1^2 + 1^2 = 1 + 1 + 1 = 3 \).


Since \( 3 \neq 1 \), a line cannot have these direction cosines.


Assertion (A) is false.



Step 2: Evaluate Reason (R)


The statement is \( \cos \theta = 1 \) is possible for \( \theta = 0 \).


By trigonometric table values, \( \cos 0^\circ = 1 \).


So, Reason (R) is true.


Step 3: Conclusion


Assertion (A) is false and Reason (R) is true. Quick Tip: Direction cosines always lie between -1 and 1, inclusive, but their squared sum must be exactly 1.
If components are \( <1, 1, 1> \), they are direction ratios (DRs), and the corresponding DCs would be \( <1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}> \).

Concept:

The symmetrical form of a line in 3D is \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \), where \( (x_1, y_1, z_1) \) is a point on the line and \( (a, b, c) \) are the direction ratios.
A general point on this line can be represented by equating the expression to a constant parameter \( \lambda \).
By solving for \( x, y, \) and \( z \) in terms of \( \lambda \), we can apply any given conditions to find the specific point.



Step 1: Express the coordinates of a general point on the line in terms of a parameter

Let the given equation of the line be set equal to a parameter \( \lambda \):
\[ x = \frac{y-1}{2} = \frac{z-2}{3} = \lambda \]


From this, we can write the individual coordinates as:
\[ x = \lambda \]
\[ \frac{y-1}{2} = \lambda \implies y = 2\lambda + 1 \]
\[ \frac{z-2}{3} = \lambda \implies z = 3\lambda + 2 \]


So, any general point \( P \) on the line has coordinates \( (\lambda, 2\lambda + 1, 3\lambda + 2) \).


Step 2: Apply the given condition to find the value of the parameter

The problem states that the \( y \) coordinate is 3 times the \( x \) coordinate.

Mathematically, this means:
\[ y = 3x \]


Substitute the expressions for \( x \) and \( y \) from Step 1:
\[ 2\lambda + 1 = 3(\lambda) \]
\[ 2\lambda + 1 = 3\lambda \]


Subtract \( 2\lambda \) from both sides:
\[ 1 = \lambda \]
\[ \lambda = 1 \]


Step 3: Determine the coordinates of the specific point

Now, substitute \( \lambda = 1 \) back into the expressions for the coordinates:
\[ x = \lambda = 1 \]
\[ y = 2(1) + 1 = 3 \]
\[ z = 3(1) + 2 = 5 \]


Therefore, the required coordinates are \( (1, 3, 5) \). Quick Tip: Using a parameter \( \lambda \) is the standard way to find specific points on a line given geometric constraints.
Always verify your final answer by checking if the point satisfies both the line equation and the given condition.
\( \frac{1}{1} = \frac{3-1}{2} = \frac{5-2}{3} \implies 1 = 1 = 1 \), and \( 3 = 3(1) \). Correct.

Concept:

A function \( f: A \rightarrow B \) is said to be onto (surjective) if every element in the codomain \( B \) has at least one pre-image in the domain \( A \).
To check for onto, we let \( y \in B \) and try to find \( x \in A \) such that \( f(x) = y \).
If there is any value in the codomain for which no pre-image exists, the function is not onto.



Step 1: Set the function equal to \( y \) and solve for \( x \)

Let \( y = f(x) \), where \( y \in R \) (the codomain).
\[ y = \frac{x-2}{x-3} \]


Multiply both sides by \( (x-3) \):
\[ y(x-3) = x - 2 \]
\[ xy - 3y = x - 2 \]


Rearrange the terms to group \( x \) on one side:
\[ xy - x = 3y - 2 \]
\[ x(y-1) = 3y - 2 \]


Divide by \( (y-1) \):
\[ x = \frac{3y-2}{y-1} \]


Step 2: Check if every value of \( y \) in the codomain has a pre-image

From the expression \( x = \frac{3y-2}{y-1} \), we can see that \( x \) is defined for all real values of \( y \) except when the denominator is zero.

The denominator is zero when:
\[ y - 1 = 0 \implies y = 1 \]


This means that for the value \( y = 1 \) in the codomain \( R \), there is no corresponding real value of \( x \) in the domain.

Even if we check the domain restriction, \( x = 3 \) would require \( 3y-2 = 3y-3 \), which is impossible, so \( y=1 \) truly has no pre-image.


Step 3: Conclusion

Since there exists an element \( y = 1 \) in the codomain \( R \) which does not have a pre-image in the domain \( R - \{3\} \), the function \( f \) is not onto. Quick Tip: For rational functions \( \frac{ax+b}{cx+d} \), the value \( y = a/c \) is usually excluded from the range.
If the codomain were changed to \( R - \{1\} \), then the function would be onto.

Concept:

A function \( f: A \rightarrow B \) is injective (one-to-one) if \( f(a_1) = f(a_2) \) implies that \( a_1 = a_2 \) for all \( a_1, a_2 \in A \).
For a mapping of ordered pairs, we assume \( f(x_1, y_1) = f(x_2, y_2) \) and check if it forces \( x_1 = x_2 \) and \( y_1 = y_2 \).



Step 1: Assume two inputs yield the same output

Let \( (x_1, y_1) \) and \( (x_2, y_2) \) be two elements in the domain \( Z \times Z \) such that:
\[ f(x_1, y_1) = f(x_2, y_2) \]


Step 2: Equate the corresponding components of the output pairs

Using the definition \( f(x, y) = (2y, 3x) \):
\[ (2y_1, 3x_1) = (2y_2, 3x_2) \]


This vector equality implies two scalar equations:

1) \( 2y_1 = 2y_2 \)

2) \( 3x_1 = 3x_2 \)


Step 3: Solve the resulting equations

From the first equation:
\[ 2y_1 = 2y_2 \implies y_1 = y_2 \]


From the second equation:
\[ 3x_1 = 3x_2 \implies x_1 = x_2 \]


Step 4: Verify the injective property

Since \( x_1 = x_2 \) and \( y_1 = y_2 \), it follows that the ordered pairs are identical:
\[ (x_1, y_1) = (x_2, y_2) \]


As \( f(x_1, y_1) = f(x_2, y_2) \) leads directly to \( (x_1, y_1) = (x_2, y_2) \), the function \( f \) is injective. Quick Tip: Injective checks focus on ensuring different inputs never map to the same output.
Linear transformations like \( (ay, bx) \) where \( a, b \neq 0 \) are always injective over numerical sets.

Concept:

Derivatives of parametric functions: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).
Chain Rule: \( \frac{d}{dx}[e^{f(x)}] = e^{f(x)} \cdot f'(x) \).
Standard Derivatives: \( \frac{d}{dt}(\sin^{-1} t) = \frac{1}{\sqrt{1-t^2}} \) and \( \frac{d}{dt}(\cos^{-1} t) = \frac{-1}{\sqrt{1-t^2}} \).
Trigonometric Identity: \( \sin^{-1} t + \cos^{-1} t = \frac{\pi}{2} \).



Step 1: Calculate the derivative of \( x \) with respect to \( t \)

Given \( x = e^{\sin^{-1} t} \).

Differentiating with respect to \( t \) using the chain rule:
\[ \frac{dx}{dt} = e^{\sin^{-1} t} \cdot \frac{d}{dt}(\sin^{-1} t) \]
\[ \frac{dx}{dt} = e^{\sin^{-1} t} \cdot \frac{1}{\sqrt{1-t^2}} = \frac{x}{\sqrt{1-t^2}} \]


Step 2: Calculate the derivative of \( y \) with respect to \( t \)

Given \( y = e^{\cos^{-1} t} \).

Differentiating with respect to \( t \) using the chain rule:
\[ \frac{dy}{dt} = e^{\cos^{-1} t} \cdot \frac{d}{dt}(\cos^{-1} t) \]
\[ \frac{dy}{dt} = e^{\cos^{-1} t} \cdot \left( \frac{-1}{\sqrt{1-t^2}} \right) = \frac{-y}{\sqrt{1-t^2}} \]


Step 3: Find the expression for \( \frac{dy}{dx} \)

Using the parametric derivative formula:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{-y}{\sqrt{1-t^2}}}{\frac{x}{\sqrt{1-t^2}}} \]
\[ \frac{dy}{dx} = -\frac{y}{x} \]


Step 4: Evaluate at the given point \( t = 1/\sqrt{2} \)

At \( t = \frac{1}{\sqrt{2}} \):
\[ \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \implies x = e^{\pi/4} \]
\[ \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \implies y = e^{\pi/4} \]


Substituting these into the derivative expression:
\[ \left( \frac{dy}{dx} \right)_{t = 1/\sqrt{2}} = -\frac{e^{\pi/4}}{e^{\pi/4}} = -1 \] Quick Tip: Alternative Shortcut: Notice \( xy = e^{\sin^{-1} t} \cdot e^{\cos^{-1} t} = e^{\sin^{-1} t + \cos^{-1} t} = e^{\pi/2} \).
Differentiating \( xy = e^{\pi/2} \) implicitly gives \( x \frac{dy}{dx} + y = 0 \), so \( \frac{dy}{dx} = -y/x \). This is much faster!

Concept:

To find the absolute maximum of a function on a closed interval, we evaluate the function at its critical points (where \( f'(x) = 0 \)) and at the endpoints of the interval.
Trigonometric identity: \( \sin^2 x = 1 - \cos^2 x \).



Step 1: Simplify the function expression

Rewrite the function in terms of a single trigonometric ratio (cosine):
\[ f(x) = \cos x + (1 - \cos^2 x) \]
\[ f(x) = 1 + \cos x - \cos^2 x \]


Step 2: Find critical points by differentiating

Differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = 0 - \sin x - 2\cos x (-\sin x) \]
\[ f'(x) = -\sin x + 2\sin x \cos x \]
\[ f'(x) = \sin x (2\cos x - 1) \]


Set \( f'(x) = 0 \) to find stationary points:

1) \( \sin x = 0 \implies x = 0, \pi \) (These are the endpoints of our interval).

2) \( 2\cos x - 1 = 0 \implies \cos x = 1/2 \implies x = \pi/3 \).


Step 3: Evaluate the function at critical points and endpoints

The candidate points are \( x = 0, \pi/3, \pi \).


At \( x = 0 \):
\[ f(0) = 1 + \cos 0 - \cos^2 0 = 1 + 1 - 1^2 = 1 \]


At \( x = \pi/3 \):
\[ f(\pi/3) = 1 + \cos(\pi/3) - \cos^2(\pi/3) = 1 + 1/2 - (1/2)^2 \]
\[ f(\pi/3) = 1 + 0.5 - 0.25 = 1.25 = 5/4 \]


At \( x = \pi \):
\[ f(\pi) = 1 + \cos\pi - \cos^2\pi = 1 + (-1) - (-1)^2 \]
\[ f(\pi) = 1 - 1 - 1 = -1 \]


Step 4: Identify the absolute maximum

Comparing the values: \( f(0) = 1 \), \( f(\pi/3) = 1.25 \), and \( f(\pi) = -1 \).

The absolute maximum value is \( 5/4 \) or \( 1.25 \). Quick Tip: You can substitute \( u = \cos x \) to treat this as a quadratic optimization problem \( g(u) = 1 + u - u^2 \) on \( u \in [-1, 1] \).
The vertex of a parabola \( -u^2 + u + 1 \) occurs at \( u = -b/2a = -1/(-2) = 1/2 \).

Concept:

Volume of a solid hemisphere: \( V = \frac{2}{3}\pi r^3 \).
Total Surface Area of a solid hemisphere: \( S = 3\pi r^2 \) (Curved surface area \( 2\pi r^2 \) + base area \( \pi r^2 \)).
Uniform rate of increase means \( \frac{dV}{dt} = k \), where \( k \) is a constant.
Chain Rule for related rates: \( \frac{dy}{dt} = \frac{dy}{dr} \cdot \frac{dr}{dt} \).



Step 1: Relate the rate of change of volume to the rate of change of radius

Given \( V = \frac{2}{3}\pi r^3 \).

Differentiating with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dr}\left(\frac{2}{3}\pi r^3\right) \cdot \frac{dr}{dt} \]
\[ \frac{dV}{dt} = \left(\frac{2}{3}\pi \cdot 3r^2\right) \cdot \frac{dr}{dt} = 2\pi r^2 \frac{dr}{dt} \]


Since the volume increases at a uniform rate \( k \):
\[ k = 2\pi r^2 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{k}{2\pi r^2} \]


Step 2: Find the rate of change of surface area

Given \( S = 3\pi r^2 \).

Differentiating with respect to time \( t \):
\[ \frac{dS}{dt} = \frac{d}{dr}(3\pi r^2) \cdot \frac{dr}{dt} \]
\[ \frac{dS}{dt} = 6\pi r \frac{dr}{dt} \]


Step 3: Substitute the expression for \( \frac{dr}{dt} \) into the surface area rate equation

Substitute \( \frac{dr}{dt} = \frac{k}{2\pi r^2} \) into the equation from Step 2:
\[ \frac{dS}{dt} = 6\pi r \cdot \left( \frac{k}{2\pi r^2} \right) \]
\[ \frac{dS}{dt} = \frac{3k}{r} \]


Step 4: Conclusion

Since \( 3k \) is a constant, we have:
\[ \frac{dS}{dt} \propto \frac{1}{r} \]

This proves that the surface area varies inversely as its radius. Quick Tip: Always clarify if "surface area" refers to the curved surface area (\( 2\pi r^2 \)) or the total surface area (\( 3\pi r^2 \)) for a solid.
In this case, the proportionality holds for both (\( 2k/r \) vs \( 3k/r \)), so the proof remains valid regardless of the specific definition used.

Concept:

Parallelogram law of vector addition: The sum of two vectors representing adjacent sides of a parallelogram is the diagonal.
Midpoint property: In \( \Delta ABC \), if \( D \) is the midpoint of \( BC \), then \( \vec{AD} \) is half the vector sum of the sides \( AB \) and \( AC \).
Magnitude of a vector \( x\hat{i} + y\hat{j} + z\hat{k} = \sqrt{x^2 + y^2 + z^2} \).



Step 1: Prove the vector relation for the median

Let the position vectors of vertices \( A, B, \) and \( C \) be \( \vec{a}, \vec{b}, \) and \( \vec{c} \) respectively.

Then \( \vec{AB} = \vec{b} - \vec{a} \) and \( \vec{AC} = \vec{c} - \vec{a} \).


Since \( D \) is the midpoint of \( BC \), its position vector \( \vec{d} \) is given by:
\[ \vec{d} = \frac{\vec{b} + \vec{c}}{2} \]


The median vector \( \vec{AD} \) is given by \( \vec{d} - \vec{a} \):
\[ \vec{AD} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2} \]
\[ \vec{AD} = \frac{(\vec{b} - \vec{a}) + (\vec{c} - \vec{a})}{2} \]
\[ \vec{AD} = \frac{\vec{AB} + \vec{AC}}{2} \] (Proved).


Step 2: Calculate the median vector \( \vec{AD} \)

Substitute the given vectors \( \vec{AB} = \hat{j} + \hat{k} \) and \( \vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k} \):
\[ \vec{AD} = \frac{(\hat{j} + \hat{k}) + (3\hat{i} - \hat{j} + 4\hat{k})}{2} \]
\[ \vec{AD} = \frac{3\hat{i} + 0\hat{j} + 5\hat{k}}{2} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k} \]


Step 3: Find the length of the median

The length of the median is the magnitude of vector \( \vec{AD} \):
\[ |\vec{AD}| = \sqrt{\left(\frac{3}{2}\right)^2 + 0^2 + \left(\frac{5}{2}\right)^2} \]
\[ |\vec{AD}| = \sqrt{\frac{9}{4} + \frac{25}{4}} = \sqrt{\frac{34}{4}} \]
\[ |\vec{AD}| = \frac{\sqrt{34}}{2} units \] Quick Tip: For any triangle, the median vector is the arithmetic mean of the vectors forming the adjacent sides originating from the same vertex.
You can simplify \( \frac{\sqrt{34}}{2} \) to \( \sqrt{\frac{34}{4}} = \sqrt{8.5} \), but keeping the fraction is standard.

Concept:



Total probability of an event and its complement: \( P(E) + P(E') = 1 \).
Bernoulli trials for independent events: The outcome of one shot does not affect the other.
Complementary events: \( P(at least one) = 1 - P(none) \).



Step 1: Calculate individual probabilities for hitting and missing


Let \( p \) be the probability of hitting the target and \( q \) be the probability of missing it.


Given that \( p = 3q \).



We know that \( p + q = 1 \).


Substituting \( p \):

\[ 3q + q = 1 \implies 4q = 1 \implies q = 1/4 \]


Then, \( p = 3(1/4) = 3/4 \).



Step 2: Calculate the probability that the target is hit


The target is considered "hit" if at least one of the two shots hits the target.


It is easier to calculate the complement: both shots miss.


\( P(both miss) = q \times q = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \).

\( P(target hit) = 1 - P(both miss) = 1 - \frac{1}{16} = \frac{15}{16} \). Quick Tip: For "at least" problems, always consider calculating the "none" scenario and subtracting it from 1.
Ensure that probabilities of exhaustive events sum to 1 before proceeding.

Concept:

Total probability of an event and its complement: \( P(E) + P(E') = 1 \).
Bernoulli trials for independent events: The outcome of one shot does not affect the other.
Complementary events: \( P(at least one) = 1 - P(none) \).



Step 1: Calculate individual probabilities for hitting and missing

Let \( p \) be the probability of hitting the target and \( q \) be the probability of missing it.

Given that \( p = 3q \).


We know that \( p + q = 1 \).

Substituting \( p \):
\[ 3q + q = 1 \implies 4q = 1 \implies q = 1/4 \]

Then, \( p = 3(1/4) = 3/4 \).


Step 2: Calculate the probability that at least one shot misses the target

At least one miss means the outcome is NOT "both hits".

\( P(both hit) = p \times p = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \).

\( P(at least one miss) = 1 - P(both hit) = 1 - \frac{9}{16} = \frac{7}{16} \). Quick Tip: For "at least" problems, always consider calculating the complementary event.
Ensure that probabilities of exhaustive events sum to 1 before proceeding.

Concept:

Sample space \( S \) for \( n \) objects is \( n! \).
Conditional Probability: \( P(E/F) = \frac{n(E \cap F)}{n(F)} \).



Step 1: List the sample space and identify event F

Let M = Mother, F = Father, S = Son.

Total arrangements = \( 3! = 6 \).
\( S = \{ MFS, MSF, FMS, FSM, SMF, SFM \} \).


Event F (Father in the middle):
\( F = \{ MFS, SFM \} \).

So, \( n(F) = 2 \).


Step 2: Identify event E and the intersection

Event E (Son on one end):
\( E = \{ SMF, SFM, MFS, FMS \} \).


The intersection \( E \cap F \) includes outcomes where Father is in the middle AND Son is on an end:
\( E \cap F = \{ MFS, SFM \} \).

So, \( n(E \cap F) = 2 \).


Step 3: Calculate conditional probability
\[ P(E/F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{2} = 1 \] Quick Tip: When Father is in the middle (\( MFS \) or \( SFM \)), the Son is automatically at one of the ends.
This explains why the conditional probability is 1 (a certain event under the given condition).

Concept:

Integrals of the form \( \int \frac{px + q}{\sqrt{ax^2 + bx + c}} dx \).
Substitute \( px + q = A \frac{d}{dx}(ax^2 + bx + c) + B \).
Standard Integral: \( \int \frac{1}{\sqrt{x^2 - a^2}} dx = \log |x + \sqrt{x^2 - a^2}| + C \).



Step 1: Express numerator in terms of the derivative of the quadratic

Let \( 3x - 1 = A(2x - 4) + B \).

Comparing coefficients of \( x \): \( 2A = 3 \implies A = 3/2 \).

Comparing constants: \( -4A + B = -1 \implies -4(3/2) + B = -1 \implies -6 + B = -1 \implies B = 5 \).


Step 2: Split the integral
\[ I = \frac{3}{2} \int \frac{2x - 4}{\sqrt{x^2 - 4x}} dx + 5 \int \frac{1}{\sqrt{x^2 - 4x}} dx \]

Let \( I = I_1 + I_2 \).


Step 3: Evaluate the first integral \( I_1 \)

For \( I_1 \), put \( x^2 - 4x = t \), then \( (2x - 4)dx = dt \).
\[ I_1 = \frac{3}{2} \int t^{-1/2} dt = \frac{3}{2} [2\sqrt{t}] = 3\sqrt{x^2 - 4x} \]


Step 4: Evaluate the second integral \( I_2 \)

Complete the square: \( x^2 - 4x = (x - 2)^2 - 4 \).
\[ I_2 = 5 \int \frac{dx}{\sqrt{(x-2)^2 - 2^2}} = 5 \log |(x - 2) + \sqrt{x^2 - 4x}| \]


Step 5: Combine the results
\[ I = 3\sqrt{x^2 - 4x} + 5 \log |x - 2 + \sqrt{x^2 - 4x}| + C \] Quick Tip: Splitting the numerator allows you to solve the radical part using simple substitution.
Completing the square is the standard way to handle the remaining constant numerator part.

Concept:

Definite integral property: \( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \).



Step 1: Simplify the integrand
\[ I = \int_{\pi/12}^{5\pi/12} \frac{dx}{1 + \frac{\sqrt{\cos x}}{\sqrt{\sin x}}} = \int_{\pi/12}^{5\pi/12} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \quad \dots(1) \]


Step 2: Apply the integral property

Here \( a + b = \frac{\pi}{12} + \frac{5\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2} \).

Replacing \( x \) by \( \pi/2 - x \):
\[ I = \int_{\pi/12}^{5\pi/12} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} dx \]
\[ I = \int_{\pi/12}^{5\pi/12} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx \quad \dots(2) \]


Step 3: Add equations (1) and (2)
\[ 2I = \int_{\pi/12}^{5\pi/12} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \int_{\pi/12}^{5\pi/12} 1 \, dx \]
\[ 2I = [x]_{\pi/12}^{5\pi/12} = \frac{5\pi}{12} - \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3} \]


Step 4: Final value
\[ I = \frac{\pi}{6} \] Quick Tip: When limits add up to \( \pi/2 \) and the integrand involves \( \sin/ \cos \) or \( \tan/ \cot \), the property usually leads to a constant integrand.

Concept:

Definition of absolute value: \( |x| = x \) if \( x \geq 0 \) and \( |x| = -x \) if \( x < 0 \).
Definite integral property: \( \int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx \).



Step 1: Split the integral at zero
\[ I = \int_{-\pi/6}^{0} (\sin(-x) + \cos(-x)) dx + \int_{0}^{\pi/2} (\sin x + \cos x) dx \]
\[ I = \int_{-\pi/6}^{0} (-\sin x + \cos x) dx + \int_{0}^{\pi/2} (\sin x + \cos x) dx \]


Step 2: Integrate the two parts

For the first part:
\[ [\cos x + \sin x]_{-\pi/6}^{0} = (1 + 0) - \left( \frac{\sqrt{3}}{2} - \frac{1}{2} \right) = \frac{3 - \sqrt{3}}{2} \]


For the second part:
\[ [-\cos x + \sin x]_{0}^{\pi/2} = (0 + 1) - (-1 + 0) = 2 \]


Step 3: Sum the results
\[ I = \frac{3 - \sqrt{3}}{2} + 2 = \frac{3 - \sqrt{3} + 4}{2} = \frac{7 - \sqrt{3}}{2} \] Quick Tip: Always handle modulus functions by splitting the domain where the inner function changes sign.
Recall that \( \cos(-x) = \cos x \) and \( \sin(-x) = -\sin x \).

Concept:

Finding a function from its derivative involves integration.
\( F(x) = \int F'(x) dx + C \).
Use substitution for logarithmic integrals.



Step 1: Set up the integral for F(x)
\[ F(x) = \int \frac{1}{e^x + 1} dx \]

Multiply numerator and denominator by \( e^{-x} \):
\[ F(x) = \int \frac{e^{-x}}{1 + e^{-x}} dx \]


Step 2: Perform substitution

Let \( 1 + e^{-x} = u \), then \( -e^{-x} dx = du \implies e^{-x} dx = -du \).
\[ F(x) = \int -\frac{1}{u} du = -\log |u| + C \]
\[ F(x) = -\log(1 + e^{-x}) + C \]


Step 3: Use the initial condition to find C

Given \( F(0) = \log(1/2) \).
\[ -\log(1 + e^0) + C = \log(1/2) \implies -\log 2 + C = \log(1/2) \]
\[ -\log 2 + C = -\log 2 \implies C = 0 \]


Step 4: Write the final function
\[ F(x) = -\log(1 + e^{-x}) = \log\left(\frac{1}{1 + 1/e^x}\right) = \log\left(\frac{e^x}{e^x + 1}\right) \] Quick Tip: The trick of multiplying by \( e^{-x} \) turns the denominator into a sum whose derivative is in the numerator.
Alternatively, solve using \( 1 = (e^x + 1) - e^x \) in the numerator.

Concept:

Homogeneous differential equation: Substitutions \( y = vx \).
Variable separable form.



Step 1: Convert to homogeneous form and substitute

Divide by \( x \): \( \frac{dy}{dx} = \frac{y}{x} - \sin^2\left(\frac{y}{x}\right) \).

Let \( y = vx \), then \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
\[ v + x \frac{dv}{dx} = v - \sin^2 v \implies x \frac{dv}{dx} = -\sin^2 v \]


Step 2: Separate variables and integrate
\[ \frac{dv}{\sin^2 v} = -\frac{dx}{x} \implies \int \csc^2 v dv = -\int \frac{1}{x} dx \]
\[ -\cot v = -\log |x| + C \implies \cot(y/x) = \log |x| + C \]


Step 3: Solve for C

Using \( y(1) = \pi/6 \):
\[ \cot(\pi/6) = \log(1) + C \implies \sqrt{3} = 0 + C \implies C = \sqrt{3} \]


Step 4: General solution
\[ \cot(y/x) = \log |x| + \sqrt{3} \] Quick Tip: The presence of \( y/x \) inside a function is a clear hint to use homogeneous substitution.

Concept:

Linear differential equation in \( x \): \( \frac{dx}{dy} + P(y)x = Q(y) \).
Integrating Factor \( IF = e^{\int P(y) dy} \).



Step 1: Identify standard form

Divide by \( y \log y \):
\[ \frac{dx}{dy} + \frac{1}{y \log y} x = \frac{2}{y^2 \log y} \]
\( P(y) = \frac{1}{y \log y} \) and \( Q(y) = \frac{2}{y^2 \log y} \).


Step 2: Calculate IF
\[ \int P(y) dy = \int \frac{1/y}{\log y} dy = \log(\log y) \]
\[ IF = e^{\log(\log y)} = \log y \]


Step 3: Solve the equation
\[ x \cdot IF = \int Q(y) \cdot IF \, dy + C \]
\[ x \log y = \int \frac{2}{y^2 \log y} \cdot \log y \, dy = \int \frac{2}{y^2} dy \]
\[ x \log y = -\frac{2}{y} + C \] Quick Tip: Notice that the log term in \( Q(y) \) cancels out with the \( IF \), simplifying the final integration.

Concept:

Linear Programming involves identifying a feasible region bounded by linear inequalities.
Corner Point Theorem states that the optimal value of the objective function occurs at one of the vertices of the feasible region.



Step 1: Convert inequalities to equations and find intercepts

For \( L_1 \): \( x + y = 250 \). Intercepts are \( (250, 0) \) and \( (0, 250) \).

For \( L_2 \): \( 25x + 40y = 7000 \implies 5x + 8y = 1400 \).

At \( x = 0 \), \( 8y = 1400 \implies y = 175 \). Point: \( (0, 175) \).

At \( y = 0 \), \( 5x = 1400 \implies x = 280 \). Point: \( (280, 0) \).


Step 2: Determine the point of intersection

Solving \( x + y = 250 \) and \( 5x + 8y = 1400 \):

From the first equation, \( x = 250 - y \).

Substitute into the second: \( 5(250 - y) + 8y = 1400 \)
\[ 1250 - 5y + 8y = 1400 \implies 3y = 150 \implies y = 50 \]

Then \( x = 250 - 50 = 200 \). Intersection point is \( (200, 50) \).


Step 3: Identify the corner points of the feasible region

Since both constraints are \( \leq \), the region is towards the origin.

The vertices of the feasible region are: \( O(0,0), A(250, 0), B(200, 50), C(0, 175) \).


Step 4: Evaluate \( Z \) at each corner point





At \( O(0,0) \): \( Z = 4500(0) + 5000(0) = 0 \)
At \( A(250,0) \): \( Z = 4500(250) + 5000(0) = 1,125,000 \)
At \( B(200,50) \): \( Z = 4500(200) + 5000(50) = 900,000 + 250,000 = 1,150,000 \)
At \( C(0,175) \): \( Z = 4500(0) + 5000(175) = 875,000 \)



The maximum value of \( Z \) is \( 1,150,000 \) at \( (200, 50) \). Quick Tip: Always double-check intersection points by substituting them back into both original equations.
Ensure the non-negativity constraints are satisfied by keeping the region in the first quadrant.

Concept:



A function is increasing if \( f'(x) > 0 \) and decreasing if \( f'(x) < 0 \).
Derivative of \( \log u \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).



Step 1: Differentiate the function


Given \( f(x) = \log(\sin x + \cos x) \).

\[ f'(x) = \frac{1}{\sin x + \cos x} \cdot \frac{d}{dx}(\sin x + \cos x) \]

\[ f'(x) = \frac{\cos x - \sin x}{\sin x + \cos x} \]



Step 2: Determine the sign of the derivative


In the interval \( (0, \pi/2) \), \( \sin x + \cos x \) is always positive.


The sign of \( f'(x) \) depends solely on the numerator \( \cos x - \sin x \).



For increasing: \( \cos x - \sin x > 0 \implies \cos x > \sin x \implies \tan x < 1 \).


In \( (0, \pi/2) \), this happens when \( x \in (0, \pi/4) \).



For decreasing: \( \cos x - \sin x < 0 \implies \cos x < \sin x \implies \tan x > 1 \).


In \( (0, \pi/2) \), this happens when \( x \in (\pi/4, \pi/2) \). Quick Tip: Use the unit circle or tan graph to quickly identify where \( \tan x < 1 \).
\( \sin x \) and \( \cos x \) intersect at \( x = \pi/4 \) in the first quadrant.

Concept:

Volume of a cylinder is \( V = \pi r^2 h \).
If a rectangle of sides \( x, y \) is revolved about side \( x \), the radius of the resulting cylinder is \( y \) and height is \( x \).



Step 1: Express dimensions in one variable

Perimeter \( P = 2(x + y) = 30 \implies x + y = 15 \implies y = 15 - x \).


Step 2: Formulate the volume function

Revolving about side \( x \): Radius \( r = y = 15 - x \), Height \( h = x \).
\[ V(x) = \pi(15 - x)^2 x = \pi(225x - 30x^2 + x^3) \]


Step 3: Maximize the volume
\[ V'(x) = \pi(225 - 60x + 3x^2) = 3\pi(x^2 - 20x + 75) \]

Set \( V'(x) = 0 \): \( (x-5)(x-15) = 0 \).

As \( x < 15 \), we have \( x = 5 \).


Checking second derivative: \( V''(x) = 3\pi(2x - 20) \).

At \( x = 5 \), \( V''(5) = 3\pi(10 - 20) = -30\pi < 0 \). Maxima at \( x = 5 \).


Dimensions are \( x = 5 cm \) and \( y = 15 - 5 = 10 cm \). Quick Tip: To maximize volume, the side being squared (the radius) should generally be the larger side.

Concept:

Domain of \( \cos^{-1} u \) is \( -1 \leq u \leq 1 \).
Range of \( \cos^{-1} u \) is \( [0, \pi] \).



Step 1: Determine the domain
\[ -1 \leq 4x^2 - 3 \leq 1 \]

Adding 3 to all parts: \( 2 \leq 4x^2 \leq 4 \).

Dividing by 4: \( 1/2 \leq x^2 \leq 1 \).

Taking square root: \( 1/\sqrt{2} \leq |x| \leq 1 \).

Domain: \( [-1, -1/\sqrt{2}] \cup [1/\sqrt{2}, 1] \).


Step 2: Solve for \( q(x) = 0 \)
\[ \cos^{-1}(4x^2 - 3) = 0 \implies 4x^2 - 3 = \cos 0 = 1 \]
\[ 4x^2 = 4 \implies x^2 = 1 \implies x = \pm 1 \]


Step 3: Determine the range of \( 3q(x) - \pi \)

Since range of \( q(x) \) is \( [0, \pi] \):
\[ 0 \leq q(x) \leq \pi \]
\[ 0 \leq 3q(x) \leq 3\pi \]
\[ -\pi \leq 3q(x) - \pi \leq 2\pi \]

Range: \( [-\pi, 2\pi] \). Quick Tip: Always check if the specific input function \( 4x^2 - 3 \) covers the full range of the parent function \( \cos^{-1} \) to determine the range.

Concept:

Equation of a line through two points \( \vec{a} \) and \( \vec{b} \) is \( \vec{r} = \vec{a} + \mu(\vec{b} - \vec{a}) \).
Perpendicular direction is found using the cross product of direction vectors.



Step 1: Find the equation of the first line and point of intersection

Line 1: \( \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 8\hat{k}) \).

Equating components with Line 2:
\[ 1 + 4\mu = 4 + 5\lambda \]
\[ 2 + 6\mu = 1 + 2\lambda \]

Solving gives \( \mu = -1/2 \) and \( \lambda = -1 \). (Using reduced DRs \( 2, 3, 4 \) is easier).

Point of intersection: \( (-1, -1, -1) \).


Step 2: Find the perpendicular direction

DRs of Line 1 are \( (2, 3, 4) \), Line 2 are \( (5, 2, 1) \).

Direction \( = (2, 3, 4) \times (5, 2, 1) \):
\[ \vec{n} = \hat{i}(3-8) - \hat{j}(2-20) + \hat{k}(4-15) = -5\hat{i} + 18\hat{j} - 11\hat{k} \]


Step 3: Write the final line equation

Equation: \( \frac{x+1}{-5} = \frac{y+1}{18} = \frac{z+1}{-11} \). Quick Tip: Use reduced direction ratios for simpler algebra during the cross product.

Concept:



Matrix inversion property: If \( QP = kI \), then \( P^{-1} = \frac{1}{k}Q \).



Step 1: Calculate \( QP \)

\[ QP = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = 6I \]


Step 2: Solve the system


System is \( PX = B \), where \( B = [3, 17, 7]^T \).
\[ X = P^{-1}B = \frac{1}{6}QB = \frac{1}{6} \begin{bmatrix} 6 + 34 - 28 \\ -12 + 34 - 28 \\ 6 - 17 + 35 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix} \]

So \( x = 2, y = -1, z = 4 \). Quick Tip: When a question says "hence", you MUST use the computed product to find the inverse.

Concept:



Properties of determinants for simplification.



Step 1: Simplify the determinant


Take \( x, y, z \) common from \( R_1, R_2, R_3 \):

\[ \Delta = xyz \begin{vmatrix} 1/x + 1 & 1/x & 1/x \\ 1/y & 1/y + 1 & 1/y \\ 1/z & 1/z & 1/z + 1 \end{vmatrix} \]


Applying \( R_1 \to R_1 + R_2 + R_3 \):

\[ \Delta = xyz(1 + 1/x + 1/y + 1/z) \begin{vmatrix} 1 & 1 & 1 \\ 1/y & 1/y+1 & 1/y \\ 1/z & 1/z & 1/z+1 \end{vmatrix} \]


Expanding gives \( \Delta = xyz(1 + 1/x + 1/y + 1/z) \).



Step 2: Apply the condition \( \Delta = 0 \)


Since \( x, y, z \neq 0 \), we have \( 1 + 1/x + 1/y + 1/z = 0 \).

\[ x^{-1} + y^{-1} + z^{-1} = -1 \] Quick Tip: Factoring out terms to create a common row/column sum is a powerful determinant strategy.

Concept:

To find the derivative of a piecewise function in a specific interval, we identify the functional form corresponding to that interval.
Power Rule of differentiation: \( \frac{d}{dx}(x^n) = nx^{n-1} \).
Derivative of a constant is zero.



Step 1: Identify the functional branch for the given interval

The question asks for the derivative in the range \( 0 < x < 3 \).

Looking at the definition of \( f(x) \), for \( 0 \leq x < 3 \), the function is defined as:
\[ f(x) = x^4 - 4x^2 + 4 \]


Step 2: Apply the rules of differentiation

Differentiate each term of the polynomial with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(4x^2) + \frac{d}{dx}(4) \]


Applying the power rule:
\[ \frac{d}{dx}(x^4) = 4x^3 \]
\[ \frac{d}{dx}(4x^2) = 4(2x) = 8x \]
\[ \frac{d}{dx}(4) = 0 \]


Step 3: Combine the results
\[ f'(x) = 4x^3 - 8x \]

This is the derivative of the track function for the specified interval. Quick Tip: Always ensure you are differentiating the correct branch of a piecewise function by checking the interval inequalities.
Remember that at the transition points (like \( x=3 \)), the derivative might not exist; hence the interval is often given as strictly greater than or less than.

Concept:

The derivative at a specific point \( a \) is found by differentiating the functional branch that contains \( a \).
Power Rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \).



Step 1: Identify the branch containing \( x = 4 \)

The value \( 4 \) is greater than \( 3 \), so we look at the branch for \( x \geq 3 \):
\[ f(x) = x^2 + 40 \]


Step 2: Find the general derivative for this branch

Differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(40) \]
\[ f'(x) = 2x + 0 = 2x \]


Step 3: Substitute the specific value \( x = 4 \)

Substitute \( x = 4 \) into the derivative expression:
\[ f'(4) = 2(4) \]
\[ f'(4) = 8 \] Quick Tip: For piecewise functions, evaluating the derivative at a point away from the boundary is straightforward—just differentiate the relevant branch.
Always check the boundaries carefully to see if the function is differentiable at the "joint".

Concept:

A function is continuous at \( x = a \) if \( \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \).
LHL (Left Hand Limit): limit as we approach from the left.
RHL (Right Hand Limit): limit as we approach from the right.



Step 1: Calculate the Left Hand Limit (LHL) at \( x = 3 \)

For \( x < 3 \), \( f(x) = x^4 - 4x^2 + 4 \).
\[ LHL = \lim_{x \to 3^-} (x^4 - 4x^2 + 4) \]
\[ LHL = (3)^4 - 4(3)^2 + 4 \]
\[ LHL = 81 - 4(9) + 4 \]
\[ LHL = 81 - 36 + 4 = 49 \]


Step 2: Calculate the Right Hand Limit (RHL) and function value \( f(3) \)

For \( x \geq 3 \), \( f(x) = x^2 + 40 \).
\[ RHL = \lim_{x \to 3^+} (x^2 + 40) \]
\[ RHL = (3)^2 + 40 = 9 + 40 = 49 \]


Also, at \( x = 3 \):
\[ f(3) = 3^2 + 40 = 49 \]


Step 3: Compare results and conclude

Since \( LHL = 49 \), \( RHL = 49 \), and \( f(3) = 49 \), we have:
\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3) \]


Therefore, the function \( f(x) \) is continuous at \( x = 3 \). Quick Tip: Continuity simply means there is no "break" or "jump" in the graph at that point.
If any of the three values (LHL, RHL, or f(a)) differ, the function is discontinuous.

Concept:

A function is differentiable at \( x = a \) if it is continuous at \( a \) and the Left Hand Derivative (LHD) equals the Right Hand Derivative (RHD).
LHD = \( \lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h} \), which can be checked using the derivative of the left branch.
RHD = \( \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h} \), which can be checked using the derivative of the right branch.



Step 1: Calculate the Left Hand Derivative (LHD) at \( x = 3 \)

From previous parts, the derivative of the left branch \( f(x) = x^4 - 4x^2 + 4 \) is \( f'(x) = 4x^3 - 8x \).

Evaluating at the limit \( x \to 3^- \):
\[ LHD = 4(3)^3 - 8(3) \]
\[ LHD = 4(27) - 24 \]
\[ LHD = 108 - 24 = 84 \]


Step 2: Calculate the Right Hand Derivative (RHD) at \( x = 3 \)

The derivative of the right branch \( f(x) = x^2 + 40 \) is \( f'(x) = 2x \).

Evaluating at the limit \( x \to 3^+ \):
\[ RHD = 2(3) = 6 \]


Step 3: Compare LHD and RHD

We observe that:
\[ LHD = 84 \]
\[ RHD = 6 \]


Since \( LHD \neq RHD \), the slopes of the tangents approaching from either side are different.


Step 4: Conclusion

Even though the function is continuous at \( x = 3 \), it is not differentiable at \( x = 3 \) because the left and right hand derivatives are not equal. Quick Tip: A "corner" or "kink" in a graph indicates a point where the function is continuous but not differentiable.
Always verify continuity first; if a function is not continuous, it cannot be differentiable.

Concept:

Probability of an event \( P(E) = \frac{Number of favorable outcomes}{Total number of outcomes} \).



Step 1: Identify the given data

Total number of smokers in the colony = 50

Number of male smokers = 30

Total number of females = Total smokers - Male smokers
\[ Number of females = 50 - 30 = 20 \]


Step 2: Calculate the probability

Let \( F \) be the event that the selected person is a female.
\[ P(F) = \frac{Number of females}{Total smokers} \]
\[ P(F) = \frac{20}{50} = \frac{2}{5} \]


In decimal form:
\[ P(F) = 0.4 \] Quick Tip: Always define your sample space clearly before calculating probabilities.
Basic probabilities are the building blocks for more complex Bayes' Theorem problems.

Concept:

The probability of the complement of an event is \( P(E') = 1 - P(E) \).



Step 1: Find the probability of suffering from lung problems

Given that 170 in 1000 males develop complications.

Let \( C \) be the event of having lung complications given the person is male.
\[ P(C|M) = \frac{170}{1000} = 0.17 \]


Step 2: Calculate the probability of not suffering from lung problems

The probability of not having complications is the complement of having them:
\[ P(C'|M) = 1 - P(C|M) \]
\[ P(C'|M) = 1 - 0.17 \]
\[ P(C'|M) = 0.83 \] Quick Tip: Use complementation to simplify calculations when "not" is mentioned in the question.

Concept:

Bayes' Theorem: \( P(A|B) = \frac{P(A) \cdot P(B|A)}{P(B)} \).



Step 1: Define the events and their probabilities

Let \( M \): Selected person is Male, \( F \): Selected person is Female.

Let \( C \): Selected person has lung complications.

\( P(M) = 30/50 = 0.6 \)
\( P(F) = 20/50 = 0.4 \)


Conditional probabilities given from the study:
\( P(C|M) = 170/1000 = 0.17 \)
\( P(C|F) = 120/1000 = 0.12 \)


Step 2: Calculate the total probability of complications \( P(C) \)

Using the theorem of total probability:
\[ P(C) = P(M)P(C|M) + P(F)P(C|F) \]
\[ P(C) = (0.6 \times 0.17) + (0.4 \times 0.12) \]
\[ P(C) = 0.102 + 0.048 = 0.150 \]


Step 3: Apply Bayes' Theorem to find \( P(F|C) \)
\[ P(F|C) = \frac{P(F) \cdot P(C|F)}{P(C)} \]
\[ P(F|C) = \frac{0.4 \times 0.12}{0.150} = \frac{0.048}{0.150} \]
\[ P(F|C) = \frac{48}{150} = \frac{8}{25} = 0.32 \] Quick Tip: Bayes' Theorem is used to find "reverse" probabilities when the outcome is known.

Concept:

Bayes' Theorem applied to the complement of the complications event.



Step 1: Define the events for 'no complications'

Let \( C' \) be the event that the person does not have lung complications.

\( P(C'|M) = 1 - 0.17 = 0.83 \)
\( P(C'|F) = 1 - 0.12 = 0.88 \)


Probabilities of selection:
\( P(M) = 0.6 \), \( P(F) = 0.4 \)


Step 2: Calculate total probability of no complications \( P(C') \)
\[ P(C') = P(M)P(C'|M) + P(F)P(C'|F) \]
\[ P(C') = (0.6 \times 0.83) + (0.4 \times 0.88) \]
\[ P(C') = 0.498 + 0.352 = 0.850 \]


Step 3: Apply Bayes' Theorem to find \( P(M|C') \)
\[ P(M|C') = \frac{P(M) \cdot P(C'|M)}{P(C')} \]
\[ P(M|C') = \frac{0.498}{0.850} = \frac{498}{850} = \frac{249}{425} \] Quick Tip: You can also calculate \( P(C') \) as \( 1 - P(C) \) from the previous part to save time.

Concept:

Transposing terms to isolate one variable in terms of the other.



Step 1: Isolate the term containing \( y \)

Given equation: \( 9x^2 + 16y^2 = 144 \)
\[ 16y^2 = 144 - 9x^2 \]


Step 2: Solve for \( y \)

Divide by 16:
\[ y^2 = \frac{144 - 9x^2}{16} = 9 - \frac{9}{16}x^2 \]
\[ y^2 = \frac{9}{16}(16 - x^2) \]


Taking square root (considering the positive branch for a function):
\[ y = \frac{3}{4} \sqrt{16 - x^2} \] Quick Tip: In area problems, we typically use the positive branch as we integrate in the upper half-plane.

Concept:

Standard Integral: \( \int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \).



Step 1: Set up the integral
\[ I = \int \frac{3}{4} \sqrt{16 - x^2} dx \]

Here \( a^2 = 16 \), so \( a = 4 \).


Step 2: Apply the standard formula
\[ I = \frac{3}{4} \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right] + C \]
\[ I = \frac{3}{4} \left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right] + C \]


Step 3: Simplify the expression
\[ I = \frac{3}{8}x\sqrt{16 - x^2} + 6\sin^{-1}\left(\frac{x}{4}\right) + C \] Quick Tip: This integral is fundamental for finding the area of circles and ellipses.

Concept:



Area of ellipse is 4 times the area in the first quadrant.
Area \( = 4 \int_{0}^{a} y dx \).



Step 1: Determine the limits


The standard form is \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \), so \( a=4 \).


The limits for the first quadrant are from 0 to 4.



Step 2: Calculate the definite integral


Using the result from (ii):

\[ Area = 4 \int_{0}^{4} \frac{3}{4} \sqrt{16 - x^2} dx \]

\[ Area = 4 \left[ \frac{3}{8}x\sqrt{16 - x^2} + 6\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4} \]


Step 3: Substitute values


Upper limit (4): \( \frac{3}{8}(4)\sqrt{0} + 6\sin^{-1}(1) = 0 + 6(\pi/2) = 3\pi \)

Lower limit (0): \( 0 + 6\sin^{-1}(0) = 0 \)


Total Area \( = 4 \times 3\pi = 12\pi sq units \). Quick Tip: Verification: Area of ellipse \( \pi ab = \pi \times 4 \times 3 = 12\pi \).

Concept:

Expanding the semi-axes by adding the track width.
Equation of a line in intercept form: \( \frac{x}{a} + \frac{y}{b} = 1 \).



Step 1: Find coordinates of P and Q

Inner semi-axes: \( a=4, b=3 \).

Width = 3 m. Thus outer semi-axes are \( a' = 4+3 = 7 \) and \( b' = 3+3 = 6 \).

Outer edge cuts x-axis at \( P(7, 0) \) and y-axis at \( Q(0, 6) \).


Step 2: Find the equation of line PQ

Using intercept form \( \frac{x}{7} + \frac{y}{6} = 1 \):
\[ y = 6 \left( 1 - \frac{x}{7} \right) = 6 - \frac{6x}{7} \]


Step 3: Find the area using integration
\[ Area = \int_{0}^{7} \left( 6 - \frac{6x}{7} \right) dx \]
\[ Area = \left[ 6x - \frac{3x^2}{7} \right]_{0}^{7} \]
\[ Area = (42 - 21) - (0) = 21 sq units \] Quick Tip: Verification: Area of right triangle \( \frac{1}{2} \times 7 \times 6 = 21 \).