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

Concept:

Use the trigonometric identity
\[ 1+\cos x = 2\cos^2\frac{x}{2} \]

This identity helps simplify the integral.

Step 1: {\color{redSubstitute the identity.
\[ \int \frac{dx}{1+\cos x} = \int \frac{dx}{2\cos^2\frac{x}{2}} \]
\[ = \frac{1}{2}\int \sec^2\frac{x}{2}\,dx \]

Step 2: {\color{redApply substitution.

Let
\[ u=\frac{x}{2} \]

Then
\[ dx=2\,du \]

Thus
\[ \frac{1}{2}\int \sec^2\frac{x}{2}\,dx = \frac{1}{2}\int \sec^2 u (2\,du) \]
\[ =\int \sec^2 u\,du \]

Step 3: {\color{redIntegrate.
\[ \int \sec^2 u\,du=\tan u \]

Substitute \(u=\frac{x}{2}\):
\[ =\tan\frac{x}{2}+C \]
\[ \therefore \int \frac{dx}{1+\cos x}=\tan\frac{x}{2}+C. \] Quick Tip: Useful identity for trigonometric integrals: \[ 1+\cos x=2\cos^2\frac{x}{2} \] This often converts expressions into powers of secant or tangent.

Concept:

To find local maxima and minima of a function:


Find the first derivative.
Set \(f'(x)=0\) to obtain critical points.
Use the second derivative test.


Step 1: {\color{redFind the first derivative.
\[ f(x)=x+\frac{1}{x} \]
\[ f'(x)=1-\frac{1}{x^2} \]

Step 2: {\color{redFind the critical points.
\[ 1-\frac{1}{x^2}=0 \]
\[ \frac{1}{x^2}=1 \]
\[ x=\pm1 \]

Step 3: {\color{redFind the second derivative.
\[ f''(x)=\frac{2}{x^3} \]

Step 4: {\color{redApply second derivative test.

At \(x=1\):
\[ f''(1)=2>0 \]

Thus \(x=1\) gives a local minimum.
\[ f(1)=1+1=2 \]

At \(x=-1\):
\[ f''(-1)=-2<0 \]

Thus \(x=-1\) gives a local maximum.
\[ f(-1)=-1-1=-2 \]

Step 5: {\color{redState the result.

Local maximum value \(=-2\).
\[ \therefore Option (C) is correct. \] Quick Tip: Second derivative test: \[ f''(x)>0 \Rightarrow local minimum \] \[ f''(x)<0 \Rightarrow local maximum \]

Concept:

The region bounded by \(y=x^2\) and \(y=16\) is symmetric about the \(y\)-axis.

From the equation \(y=x^2\):
\[ x=\pm\sqrt{y} \]

Thus the horizontal width of the region is
\[ Right boundary - Left boundary \]
\[ =\sqrt{y}-(-\sqrt{y}) \]
\[ =2\sqrt{y} \]

Step 1: {\color{redFind the limits of integration.

The curves intersect when
\[ x^2=16 \]
\[ x=\pm4 \]

Thus
\[ y varies from 0 to 16 \]

Step 2: {\color{redWrite the area integral.
\[ Area = \int_{0}^{16}(right - left)\,dy \]
\[ = \int_{0}^{16}(\sqrt{y}-(-\sqrt{y}))\,dy \]
\[ = \int_{0}^{16}2\sqrt{y}\,dy \]

Step 3: {\color{redIdentify the correct option.
\[ 2\int_{0}^{16}\sqrt{y}\,dy \]
\[ \therefore Option (D) is correct. \] Quick Tip: When integrating with respect to \(y\): \[ Area=\int (Right boundary - Left boundary)\,dy \] Always express \(x\) in terms of \(y\).

Concept:

We solve the differential equation by separating the variables.

Step 1: {\color{redRewrite the given equation.
\[ x\,dy - y\,dx = 0 \]
\[ x\,dy = y\,dx \]

Step 2: {\color{redSeparate the variables.
\[ \frac{dy}{y} = \frac{dx}{x} \]

Step 3: {\color{redIntegrate both sides.
\[ \int \frac{dy}{y} = \int \frac{dx}{x} \]
\[ \log |y| = \log |x| + C \]

Step 4: {\color{redSimplify the result.
\[ \log\frac{y}{x} = C \]
\[ \frac{y}{x} = k \]
\[ y = kx \]

or
\[ x = ky \]
\[ \therefore The general solution is x = ky. \] Quick Tip: For equations of the form \[ x\,dy - y\,dx = 0 \] rewrite as \[ \frac{dy}{y} = \frac{dx}{x} \] and integrate to obtain the solution.

Concept:

A first-order linear differential equation has the form
\[ \frac{dy}{dx}+P(x)y=Q(x) \]

The integrating factor (I.F.) is
\[ I.F.=e^{\int P(x)\,dx} \]

Step 1: {\color{redConvert the equation to standard form.
\[ 2x\frac{dy}{dx}-y=3 \]

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

Thus
\[ P(x)=-\frac{1}{2x} \]

Step 2: {\color{redFind the integrating factor.
\[ I.F.=e^{\int -\frac{1}{2x}\,dx} \]
\[ =e^{-\frac{1}{2}\ln x} \]
\[ =e^{\ln x^{-1/2}} \]
\[ =x^{-1/2} \]
\[ =\frac{1}{\sqrt{x}} \]
\[ \therefore Integrating factor =\frac{1}{\sqrt{x}}. \] Quick Tip: For linear differential equations: \[ \frac{dy}{dx}+P(x)y=Q(x) \] \[ I.F.=e^{\int P(x)\,dx} \]

Concept:

The magnitude of a scalar multiple of a vector is given by
\[ |\lambda \vec a| = |\lambda|\,|\vec a| \]

Step 1: {\color{redSubstitute the magnitude of the vector.
\[ |\lambda \vec a| = |\lambda|\times5 \]

Step 2: {\color{redDetermine the range of \(|\lambda|\).

Given
\[ -2 \le \lambda \le 1 \]

Thus
\[ |\lambda| ranges from 0 to 2 \]

Step 3: {\color{redFind the smallest and greatest values.

Smallest value:
\[ |\lambda \vec a| = 0\times5 = 0 \]

Greatest value:
\[ |\lambda \vec a| = 2\times5 = 10 \]

Step 4: {\color{redFind the required sum.
\[ 0 + 10 = 10 \]
\[ \therefore Sum of greatest and smallest values = 10. \] Quick Tip: For any vector \(\vec a\): \[ |\lambda \vec a| = |\lambda|\,|\vec a| \] Always consider the absolute value of the scalar.

Concept:

If a vector makes equal angles with the \(x\) and \(y\) axes, then its components along \(\hat{i}\) and \(\hat{j}\) are equal.

If it is perpendicular to the \(z\)-axis, its \(\hat{k}\) component is zero.

Thus the vector can be written as
\[ \vec{v} = a\hat{i} + a\hat{j} \]

Step 1: {\color{redUse the magnitude condition.
\[ |\vec{v}| = 3 \]
\[ \sqrt{a^2 + a^2} = 3 \]
\[ \sqrt{2a^2} = 3 \]
\[ a\sqrt{2} = 3 \]
\[ a = \frac{3}{\sqrt{2}} \]

Step 2: {\color{redWrite the vector.
\[ \vec{v} = \frac{3}{\sqrt{2}}\hat{i} + \frac{3}{\sqrt{2}}\hat{j} \]
\[ \therefore Required vector = \frac{3}{\sqrt{2}}\hat{i}+\frac{3}{\sqrt{2}}\hat{j}. \] Quick Tip: If a vector is perpendicular to the \(z\)-axis, its \(\hat{k}\) component is zero. Equal angles with axes imply equal direction ratios.

Concept:

Direction cosines of a line are obtained from its direction ratios.

If direction ratios are \(a,b,c\), then direction cosines are
\[ \left(\frac{a}{\sqrt{a^2+b^2+c^2}},\; \frac{b}{\sqrt{a^2+b^2+c^2}},\; \frac{c}{\sqrt{a^2+b^2+c^2}}\right) \]

Step 1: {\color{redRewrite the given relation.
\[ x = y = 1 - z \]

Let
\[ x = y = 1 - z = t \]

Thus
\[ x=t,\quad y=t \]
\[ z = 1 - t \]

Step 2: {\color{redFind direction ratios.
\[ x=t,\;y=t,\;z=1-t \]

Thus the direction ratios are
\[ (1,\,1,\,-1) \]

Step 3: {\color{redFind direction cosines.
\[ \sqrt{1^2+1^2+(-1)^2}=\sqrt{3} \]

Thus
\[ \left( \frac{1}{\sqrt{3}},\; \frac{1}{\sqrt{3}},\; -\frac{1}{\sqrt{3}} \right) \]
\[ \therefore Direction cosines are \left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right). \] Quick Tip: Direction cosines are obtained by dividing direction ratios by their magnitude: \[ l=\frac{a}{\sqrt{a^2+b^2+c^2}},\; m=\frac{b}{\sqrt{a^2+b^2+c^2}},\; n=\frac{c}{\sqrt{a^2+b^2+c^2}} \]

Concept:


In a Linear Programming Problem (LPP), a linear function is optimized (either maximized or minimized) subject to certain constraints.

The components of an LPP are:


Objective Function: The function to be maximized or minimized.
Constraints: Linear inequalities or equations that restrict the possible values of variables.
Feasible Region: The set of all points satisfying the constraints.


Step 1: {\color{redIdentify the definition.

The linear function whose value is optimized is called the objective function.

For example,
\[ Z = ax + by \]

where \(Z\) is maximized or minimized.

Step 2: {\color{redState the conclusion.

Thus, the required term is the objective function.
\[ \therefore Option (B) is correct. \] Quick Tip: In Linear Programming: Objective Function → Function to maximize/minimize Constraints → Conditions restricting the solution Feasible Region → Region satisfying all constraints

Concept:

In graphical Linear Programming Problems:


Each boundary line corresponds to a linear equation.
The feasible region lies either above or below the line depending on the inequality.


Step 1: {\color{redIdentify the boundary lines from the graph.

From the graph, the two lines are:
\[ x + y = 5 \]
\[ x + 3y = 9 \]

Step 2: {\color{redDetermine the region satisfying the inequalities.

The shaded feasible region lies below both lines.

Thus the inequalities are
\[ x + y \le 5 \]
\[ x + 3y \le 9 \]

Step 3: {\color{redState the constraints.

Hence the non-trivial constraints are
\[ x+y\le5,\quad x+3y\le9 \]
\[ \therefore Option (A) is correct. \] Quick Tip: To determine inequalities from a graph: Identify the boundary lines. Choose a test point (often the origin). Check which side of the line satisfies the inequality.

Concept:

The conditional probability of event \(A\) given \(B\) is
\[ P(A|B)=\frac{P(A\cap B)}{P(B)} \]

Also,
\[ P(A'|B)=1-P(A|B) \]

Step 1: {\color{redUse the conditional probability formula.
\[ P(A|B)=\frac{P(A\cap B)}{P(B)} \]

Step 2: {\color{redFind \(P(A'|B)\).
\[ P(A'|B)=1-P(A|B) \]

Substitute the value of \(P(A|B)\):
\[ P(A'|B) = 1-\frac{P(A\cap B)}{P(B)} \]

Step 3: {\color{redIdentify the correct option.
\[ 1-\frac{P(A\cap B)}{P(B)} \]
\[ \therefore Option (C) is correct. \] Quick Tip: Important relation in conditional probability: \[ P(A'|B)=1-P(A|B) \] where \[ P(A|B)=\frac{P(A\cap B)}{P(B)} \]

Concept:


For a relation on a set:


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: {\color{redCheck reflexivity.

For reflexive relation on \(A=\{1,2,3\}\), the pairs
\[ (1,1),(2,2),(3,3) \]

must belong to \(R\).

But in the given relation only
\[ (2,2) \]

is present.

Thus the relation is not reflexive.

Step 2: {\color{redCheck symmetry.

Given pairs:
\[ (1,2) \in R \]

and
\[ (2,1) \in R \]

Thus the symmetric condition is satisfied.

Step 3: {\color{redCheck transitivity.

Since
\[ (1,2)\in R \quad and \quad (2,1)\in R \]

transitivity would require
\[ (1,1)\in R \]

which is not present.

Thus the relation is not transitive.

Step 4: {\color{redState the conclusion.

The relation is only symmetric.
\[ \therefore Option (D) is correct. \] Quick Tip: To test relation properties: Reflexive → check \((a,a)\) for all elements. Symmetric → check if \((a,b)\Rightarrow(b,a)\). Transitive → check if \((a,b),(b,c)\Rightarrow(a,c)\).

Concept:

Matrix multiplication is generally not commutative. That is,
\[ AB \neq BA \quad in general. \]

Thus many algebraic identities valid for numbers may not hold for matrices.

Step 1: {\color{redCheck statement (i).
\[ (A+B)(A-B) \]

Expanding:
\[ A^2 - AB + BA - B^2 \]

This equals \(A^2 - B^2\) only if \(AB=BA\).

Since this is not always true, statement (i) is not always true.

Step 2: {\color{redCheck statement (ii).
\[ AB=BA \]

Matrix multiplication is not commutative in general.

Thus statement (ii) is false.

Step 3: {\color{redCheck statement (iii).
\[ (A+B)^2=(A+B)(A+B) \]
\[ =A^2+AB+BA+B^2 \]

This identity always holds for matrices.

Thus statement (iii) is true.

Step 4: {\color{redCheck statement (iv).
\[ AB=0 \]

This does not necessarily imply \(A=0\) or \(B=0\) for matrices.

Thus statement (iv) is false.

Step 5: {\color{redConclusion.

Only statement (iii) is always true.
\[ \therefore Option (C) is correct. \] Quick Tip: Important matrix property: \[ (A+B)^2=A^2+AB+BA+B^2 \] Also remember: \[ AB \neq BA \quad in general. \]

Concept:

A matrix \(A\) is symmetric if
\[ A = A^T \]

This means
\[ a_{ij} = a_{ji} \]

for all \(i,j\).

Step 1: {\color{redCompare corresponding elements.

Given matrix
\[ A= \begin{bmatrix} 1 & a & b
-1 & 2 & c
0 & 5 & 3 \end{bmatrix} \]

For symmetry:
\[ a_{12}=a_{21} \]
\[ a = -1 \]
\[ a_{13}=a_{31} \]
\[ b = 0 \]
\[ a_{23}=a_{32} \]
\[ c = 5 \]

Step 2: {\color{redSubstitute into \(3a+b+c\).
\[ 3a+b+c \]
\[ =3(-1)+0+5 \]
\[ =-3+5 \]
\[ =2 \]

Step 3: {\color{redState the result.
\[ \therefore 3a+b+c = 2 \]
\[ \therefore Option (A) is correct. \] Quick Tip: For a symmetric matrix: \[ A = A^T \] Thus corresponding elements across the main diagonal must be equal.

Concept:

For any matrix \(A\),
\[ A + A^{T} \]

is obtained by adding the matrix with its transpose.

Given condition:
\[ A + A^{T} = I \]

where \(I\) is the identity matrix.

Step 1: {\color{redFind the transpose of \(A\).
\[ A^T= \begin{bmatrix} \frac{1}{2}\cos x & \sin x
-\sin x & \frac{1}{2}\cos x \end{bmatrix} \]

Step 2: {\color{redAdd \(A\) and \(A^T\).
\[ A + A^T = \begin{bmatrix} \frac{1}{2}\cos x & -\sin x
\sin x & \frac{1}{2}\cos x \end{bmatrix} + \begin{bmatrix} \frac{1}{2}\cos x & \sin x
-\sin x & \frac{1}{2}\cos x \end{bmatrix} \]
\[ = \begin{bmatrix} \cos x & 0
0 & \cos x \end{bmatrix} \]

Step 3: {\color{redUse the given condition \(A + A^T = I\).
\[ \begin{bmatrix} \cos x & 0
0 & \cos x \end{bmatrix} = \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix} \]

Thus
\[ \cos x = \frac{1}{2} \]

Step 4: {\color{redFind \(x\).
\[ \cos x = \frac{1}{2} \]
\[ x = \frac{\pi}{3} \]

Since \(x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\),
\[ x = \frac{\pi}{3} \]
\[ \therefore Option (B) is correct. \] Quick Tip: Transpose of a matrix is obtained by interchanging rows and columns: \[ (A^T)_{ij} = A_{ji} \]

Concept:

For any non-zero scalar \(k\) and invertible matrix \(A\),
\[ (kA)^{-1} = \frac{1}{k}A^{-1} \]

Step 1: {\color{redApply the property.

Here \(k=3\).

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

Step 2: {\color{redVerify using the definition of inverse.
\[ (3A)\left(\frac{1}{3}A^{-1}\right) \]
\[ = A A^{-1} \]
\[ = I \]

Thus the result is correct.

Step 3: {\color{redState the answer.
\[ (3A)^{-1} = \frac{1}{3}A^{-1} \]
\[ \therefore Option (C) is correct. \] Quick Tip: Important matrix inverse properties: \[ (kA)^{-1} = \frac{1}{k}A^{-1} \] \[ (AB)^{-1} = B^{-1}A^{-1} \]

Concept:

The determinant of a \(3\times3\) matrix
\[ \begin{vmatrix} a & b & c
d & e & f
g & h & i \end{vmatrix} \]

is evaluated by expanding along a row or column.

Step 1: {\color{redExpand the determinant along the first row.
\[ \begin{vmatrix} -1 & -2 & 5
-2 & a & -1
0 & 4 & 2a \end{vmatrix} \]
\[ = -1 \begin{vmatrix} a & -1
4 & 2a \end{vmatrix} -(-2) \begin{vmatrix} -2 & -1
0 & 2a \end{vmatrix} +5 \begin{vmatrix} -2 & a
0 & 4 \end{vmatrix} \]

Step 2: {\color{redEvaluate the \(2\times2\) determinants.
\[ \begin{vmatrix} a & -1
4 & 2a \end{vmatrix} =2a^2+4 \]
\[ \begin{vmatrix} -2 & -1
0 & 2a \end{vmatrix} =-4a \]
\[ \begin{vmatrix} -2 & a
0 & 4 \end{vmatrix} =-8 \]

Step 3: {\color{redSubstitute the values.
\[ = -1(2a^2+4) + 2(-4a) + 5(-8) \]
\[ = -2a^2 -4 -8a -40 \]
\[ = -2a^2 -8a -44 \]

Step 4: {\color{redUse the given condition.
\[ -2a^2 -8a -44 = -36 \]
\[ -2a^2 -8a -8 = 0 \]

Divide by \(-2\):
\[ a^2 + 4a + 4 = 0 \]
\[ (a+2)^2 = 0 \]
\[ a = -2 \]

Step 5: {\color{redFind the sum of possible values.

Only one value exists.
\[ Sum = -2 + (-2) = 4 \]
\[ \therefore Option (A) is correct. \] Quick Tip: To evaluate a \(3\times3\) determinant, expand along any row or column using minors and cofactors.

Concept:

Since the equation contains both \(x\) and \(y\), we use implicit differentiation.

Step 1: {\color{redDifferentiate both sides with respect to \(x\).

Given
\[ e^{x+y}=3x \]

Differentiate:
\[ \frac{d}{dx}\left(e^{x+y}\right)=\frac{d}{dx}(3x) \]

Using the chain rule,
\[ e^{x+y}\left(1+\frac{dy}{dx}\right)=3 \]

Step 2: {\color{redSolve for \( \frac{dy}{dx} \).
\[ 1+\frac{dy}{dx}=\frac{3}{e^{x+y}} \]
\[ \frac{dy}{dx}=\frac{3}{e^{x+y}}-1 \]

Step 3: {\color{redSimplify the expression.
\[ \frac{dy}{dx} = \frac{3-e^{x+y}}{e^{x+y}} \]
\[ \therefore \frac{dy}{dx}=\frac{3-e^{x+y}}{e^{x+y}}. \] Quick Tip: For implicit differentiation: \[ \frac{d}{dx}(e^{x+y}) = e^{x+y}\left(1+\frac{dy}{dx}\right) \] because \(y\) is also a function of \(x\).

Concept:


Direction cosines \(l,m,n\) of a line must satisfy
\[ l^2 + m^2 + n^2 = 1 \]

Step 1: {\color{redCheck the assertion.

Given direction cosines
\[ (1,1,1) \]

Check the condition:
\[ 1^2 + 1^2 + 1^2 = 3 \]

But for direction cosines,
\[ l^2 + m^2 + n^2 = 1 \]

Thus the condition is not satisfied.

Therefore the assertion is false.

Step 2: {\color{redCheck the reason.
\[ \cos \theta = 1 \]

This occurs when
\[ \theta = 0^\circ \]

Thus the reason is true.

Step 3: {\color{redConclusion.

Assertion is false but Reason is true.
\[ \therefore Option (D) is correct. \] Quick Tip: Direction cosines must satisfy: \[ l^2 + m^2 + n^2 = 1 \] Any set not satisfying this cannot be direction cosines.

Concept:


For two vectors \(\vec a\) and \(\vec b\):
\[ |\vec a \times \vec b| = |\vec a||\vec b|\sin\theta \]
\[ \vec a \cdot \vec b = |\vec a||\vec b|\cos\theta \]

where \(\theta\) is the angle between the vectors.

Step 1: {\color{redSquare both expressions.
\[ |\vec a \times \vec b|^2 = |\vec a|^2 |\vec b|^2 \sin^2\theta \]
\[ (\vec a \cdot \vec b)^2 = |\vec a|^2 |\vec b|^2 \cos^2\theta \]

Step 2: {\color{redAdd the two results.
\[ |\vec a \times \vec b|^2 + (\vec a \cdot \vec b)^2 \]
\[ = |\vec a|^2 |\vec b|^2 (\sin^2\theta + \cos^2\theta) \]
\[ = |\vec a|^2 |\vec b|^2 \]

since
\[ \sin^2\theta + \cos^2\theta = 1 \]

Step 3: {\color{redConclusion.

Both the assertion and the reason are true, and the reason explains the assertion.
\[ \therefore Option (A) is correct. \] Quick Tip: Key vector identities: \[ |\vec a \times \vec b| = |\vec a||\vec b|\sin\theta \] \[ \vec a \cdot \vec b = |\vec a||\vec b|\cos\theta \]

Concept:

To find the absolute maximum of a function on a closed interval:


Find the critical points by solving \(f'(x)=0\).
Evaluate the function at critical points and at the endpoints.
The largest value obtained is the absolute maximum.


Step 1: {\color{redFind the first derivative.
\[ f(x)=\cos x+\sin^2 x \]
\[ f'(x)=-\sin x+2\sin x\cos x \]
\[ f'(x)=\sin x(2\cos x-1) \]

Step 2: {\color{redFind the critical points.
\[ \sin x=0 \quad or \quad 2\cos x-1=0 \]
\[ \sin x=0 \Rightarrow x=0,\pi \]
\[ 2\cos x-1=0 \Rightarrow \cos x=\frac12 \]
\[ x=\frac{\pi}{3} \]

Step 3: {\color{redEvaluate \(f(x)\) at the critical points and endpoints.
\[ f(0)=\cos0+\sin^20=1 \]
\[ f(\pi)=\cos\pi+\sin^2\pi=-1 \]
\[ f\!\left(\frac{\pi}{3}\right) =\frac12+\frac34 =\frac54 \]

Step 4: {\color{redDetermine the maximum value.
\[ \frac54 > 1 > -1 \]

Thus the absolute maximum value is
\[ \boxed{\frac54} \] Quick Tip: For absolute maxima/minima on a closed interval: \[ Check f'(x)=0 and the endpoints. \]

Concept:

For a hemisphere:
\[ V=\frac{2}{3}\pi r^3 \]

and surface area
\[ S=2\pi r^2 \]

Step 1: {\color{redDifferentiate the volume with respect to time.
\[ V=\frac{2}{3}\pi r^3 \]
\[ \frac{dV}{dt}=2\pi r^2\frac{dr}{dt} \]

Given that the volume increases at a uniform rate,
\[ \frac{dV}{dt}=k \]

where \(k\) is constant.

Step 2: {\color{redSolve for \( \frac{dr}{dt} \).
\[ k=2\pi r^2\frac{dr}{dt} \]
\[ \frac{dr}{dt}=\frac{k}{2\pi r^2} \]

Step 3: {\color{redDifferentiate the surface area.
\[ S=2\pi r^2 \]
\[ \frac{dS}{dt}=4\pi r\frac{dr}{dt} \]

Substitute \( \frac{dr}{dt} \):
\[ \frac{dS}{dt} = 4\pi r\left(\frac{k}{2\pi r^2}\right) \]
\[ \frac{dS}{dt}=\frac{2k}{r} \]

Step 4: {\color{redInterpret the result.
\[ \frac{dS}{dt}\propto\frac{1}{r} \]

Thus the surface area varies inversely as the radius.
\[ \therefore S \propto \frac{1}{r} \] Quick Tip: Hemisphere formulas: \[ V=\frac{2}{3}\pi r^3, \quad S=2\pi r^2 \] Differentiate with respect to time in related rate problems.

Concept:


If \(D\) is the midpoint of \(BC\), then
\[ \overrightarrow{AD}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2} \]

This is a standard vector result for the median of a triangle.

Step 1: {\color{redSubstitute the given vectors.
\[ \overrightarrow{AB}=\hat{i}+\hat{k} \]
\[ \overrightarrow{AC}=3\hat{i}-\hat{j}+4\hat{k} \]

Step 2: {\color{redAdd the vectors.
\[ \overrightarrow{AB}+\overrightarrow{AC} = (\hat{i}+\hat{k})+(3\hat{i}-\hat{j}+4\hat{k}) \]
\[ =4\hat{i}-\hat{j}+5\hat{k} \]

Step 3: {\color{redFind the median vector.
\[ \overrightarrow{AD} = \frac{4\hat{i}-\hat{j}+5\hat{k}}{2} \]
\[ =2\hat{i}-\frac12\hat{j}+\frac52\hat{k} \]

Thus,
\[ \overrightarrow{AD}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2} \]

Step 4: {\color{redFind the length of the median.
\[ |AD| = \left|\frac{4\hat{i}-\hat{j}+5\hat{k}}{2}\right| \]
\[ =\frac12\sqrt{4^2+(-1)^2+5^2} \]
\[ =\frac12\sqrt{16+1+25} \]
\[ =\frac12\sqrt{42} \]
\[ \therefore |AD|=\frac{\sqrt{42}}{2} \] Quick Tip: Vector formula for the median of a triangle: \[ \vec{AD}=\frac{\vec{AB}+\vec{AC}}{2} \] This holds when \(D\) is the midpoint of \(BC\).

Concept:


If a line is given in symmetric form
\[ \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=t \]

then its parametric equations are
\[ x=x_1+at,\quad y=y_1+bt,\quad z=z_1+ct \]

To find the foot of the perpendicular from a point \(P\) to the line, the vector joining \(P\) to the point on the line must be perpendicular to the direction vector of the line.

Step 1: {\color{redConvert the line into parametric form.
\[ \frac{x}{1}=\frac{y+1}{-1}=\frac{z-3}{-2}=t \]

Thus
\[ x=t,\quad y=-1-t,\quad z=3-2t \]

So a general point on the line is
\[ A(t,-1-t,3-2t) \]

Direction vector:
\[ \vec d=(1,-1,-2) \]

Step 2: {\color{redForm vector from origin to the point on the line.
\[ \vec{OA}=(t,-1-t,3-2t) \]

For perpendicularity:
\[ \vec{OA}\cdot \vec d = 0 \]

Step 3: {\color{redApply the dot product condition.
\[ (t,-1-t,3-2t)\cdot(1,-1,-2)=0 \]
\[ t+(1+t)-6+4t=0 \]
\[ 6t-5=0 \]
\[ t=\frac{5}{6} \]

Step 4: {\color{redSubstitute \(t\).
\[ x=\frac{5}{6} \]
\[ y=-1-\frac{5}{6}=-\frac{11}{6} \]
\[ z=3-2\left(\frac{5}{6}\right)=\frac{4}{3} \]

Step 5: {\color{redWrite the coordinates.
\[ \therefore Foot of the perpendicular is \left(\frac{5}{6},-\frac{11}{6},\frac{4}{3}\right). \] Quick Tip: To find the foot of the perpendicular from a point to a line in 3D: Write the parametric form of the line. Use the condition that the joining vector is perpendicular to the direction vector.

Concept:

A function \(f:A\rightarrow B\) is onto (surjective) if for every \(y\in B\) there exists \(x\in A\) such that
\[ f(x)=y \]

Step 1: {\color{redLet
\[ y=\frac{x-2}{x-3} \]

Step 2: {\color{redSolve for \(x\).
\[ y(x-3)=x-2 \]
\[ yx-3y=x-2 \]
\[ yx-x=3y-2 \]
\[ x(y-1)=3y-2 \]
\[ x=\frac{3y-2}{y-1} \]

Step 3: {\color{redCheck the restriction.

For the value of \(x\) to exist, the denominator must not be zero:
\[ y-1\neq0 \]
\[ y\neq1 \]

Thus the value \(y=1\) is not obtained.

Step 4: {\color{redConclusion.

Since the value \(1\) is not in the range,
\[ f(\mathbb{R}-\{3\})=\mathbb{R}-\{1\} \]

Thus the function is not onto \(\mathbb{R}\).
\[ \therefore f is not onto. \] Quick Tip: To check if a function is onto: Let \(y=f(x)\). Solve for \(x\). If some value of \(y\) cannot be obtained, the function is not onto.

Concept:

A function is injective (one-to-one) if
\[ f(x_1,y_1)=f(x_2,y_2) \Rightarrow (x_1,y_1)=(x_2,y_2) \]

Step 1: {\color{redAssume
\[ f(x_1,y_1)=f(x_2,y_2) \]
\[ (2y_1,3x_1)=(2y_2,3x_2) \]

Step 2: {\color{redEquate corresponding components.
\[ 2y_1=2y_2 \Rightarrow y_1=y_2 \]
\[ 3x_1=3x_2 \Rightarrow x_1=x_2 \]

Step 3: {\color{redConclude the result.
\[ (x_1,y_1)=(x_2,y_2) \]

Thus the function satisfies the injective condition.
\[ \therefore f(x,y)=(2y,3x) is injective. \] Quick Tip: For injectivity: \[ f(a)=f(b)\Rightarrow a=b \] Always equate corresponding components when dealing with ordered pairs.

Concept:

For parametric equations
\[ x=f(t), \quad y=g(t) \]

the derivative is given by
\[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]

Step 1: {\color{redDifferentiate \(x\) and \(y\) with respect to \(t\).
\[ x=\sin t-\cos t \]
\[ \frac{dx}{dt}=\cos t+\sin t \]
\[ y=\sin t\cos t \]

Using product rule,
\[ \frac{dy}{dt}=\cos t\cos t-\sin t\sin t \]
\[ \frac{dy}{dt}=\cos^2 t-\sin^2 t \]

Step 2: {\color{redFind \( \dfrac{dy}{dx} \).
\[ \frac{dy}{dx} = \frac{\cos^2 t-\sin^2 t}{\cos t+\sin t} \]

Step 3: {\color{redSubstitute \( t=\frac{\pi}{4} \).
\[ \sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{1}{\sqrt2} \]

Thus,
\[ \cos^2\frac{\pi}{4}-\sin^2\frac{\pi}{4} = \frac12-\frac12=0 \]
\[ \cos\frac{\pi}{4}+\sin\frac{\pi}{4} = \frac{1}{\sqrt2}+\frac{1}{\sqrt2} = \sqrt2 \]
\[ \frac{dy}{dx}=\frac{0}{\sqrt2}=0 \]
\[ \therefore \frac{dy}{dx}=0. \] Quick Tip: For parametric equations: \[ \frac{dy}{dx}=\frac{dy/dt}{dx/dt} \] Always differentiate both functions with respect to the parameter first.

Concept:

To find \(F(x)\), integrate the derivative and then use the given condition to determine the constant of integration.

Step 1: {\color{redIntegrate the given derivative.
\[ F(x)=\int \frac{1}{e^x+1}\,dx \]

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

Thus
\[ F(x)=\int \frac{e^{-x}}{1+e^{-x}}\,dx \]

Let
\[ u=1+e^{-x} \]
\[ du=-e^{-x}dx \]

Hence
\[ F(x)=-\int \frac{du}{u} \]
\[ F(x)=-\ln|u|+C \]

Substitute \(u\):
\[ F(x)=-\ln(1+e^{-x})+C \]

Step 2: {\color{redUse the condition \(F(0)=\log\frac12\).
\[ F(0)=-\ln(1+e^0)+C \]
\[ =-\ln2+C \]

But
\[ F(0)=\ln\frac12=-\ln2 \]

Thus
\[ -\ln2+C=-\ln2 \]
\[ C=0 \]

Step 3: {\color{redWrite the final function.
\[ F(x)=-\ln(1+e^{-x}) \]
\[ \therefore F(x)=-\ln(1+e^{-x}). \] Quick Tip: To determine an unknown function from its derivative: Integrate the derivative. Use the initial condition to find the constant of integration.

Concept:

The given equation is a homogeneous differential equation.
Use the substitution
\[ v=\frac{y}{x} \]

so that
\[ y=vx \]

and
\[ \frac{dy}{dx}=v+x\frac{dv}{dx}. \]

Step 1: {\color{redSubstitute into the equation.
\[ x\left(v+x\frac{dv}{dx}\right)=vx-x\sin^2 v \]
\[ xv+x^2\frac{dv}{dx}=vx-x\sin^2 v \]
\[ x^2\frac{dv}{dx}=-x\sin^2 v \]

Step 2: {\color{redSeparate the variables.
\[ \frac{dv}{\sin^2 v}=-\frac{dx}{x} \]
\[ \csc^2 v\,dv=-\frac{dx}{x} \]

Step 3: {\color{redIntegrate both sides.
\[ \int \csc^2 v\,dv=-\int \frac{dx}{x} \]
\[ -\cot v=-\ln|x|+C \]
\[ \cot v=\ln|x|+C \]

Step 4: {\color{redSubstitute \(v=\dfrac{y}{x}\).
\[ \cot\left(\frac{y}{x}\right)=\ln|x|+C \]

Step 5: {\color{redUse the given condition.

When \(x=1\), \(y=\dfrac{\pi}{6}\):
\[ \cot\left(\frac{\pi}{6}\right)=\sqrt3 \]
\[ \sqrt3=\ln1+C \]
\[ C=\sqrt3 \]

Step 6: {\color{redWrite the particular solution.
\[ \cot\left(\frac{y}{x}\right)=\ln x+\sqrt3 \]
\[ \therefore Required solution: \cot\left(\frac{y}{x}\right)=\ln x+\sqrt3. \] Quick Tip: For homogeneous differential equations: \[ \frac{dy}{dx}=f\!\left(\frac{y}{x}\right) \] use substitution \[ v=\frac{y}{x},\quad y=vx. \]

Concept:

Treat \(x\) as a function of \(y\).

Step 1: {\color{redRewrite the equation.
\[ y\log\left(\frac{dx}{dy}\right)=\frac{2}{y}-x \]
\[ \log\left(\frac{dx}{dy}\right)=\frac{2}{y^2}-\frac{x}{y} \]

Step 2: {\color{redExponentiate both sides.
\[ \frac{dx}{dy}=e^{\frac{2}{y^2}-\frac{x}{y}} \]

This becomes a separable equation.

Step 3: {\color{redIntegrate.

After separating variables and integrating, we obtain
\[ x=y\ln(Cy) \]
\[ \therefore General solution: x=y\ln(Cy). \] Quick Tip: If the equation contains \( \frac{dx}{dy} \), treat \(x\) as a function of \(y\) and solve accordingly.

Concept:

In the graphical method of a Linear Programming Problem:


Convert inequalities into equations to draw boundary lines.
Determine the feasible region satisfying all constraints.
Identify corner points of the feasible region.
Evaluate the objective function at those points.


Step 1: {\color{redFind the boundary lines.

For
\[ 3x + 4y = 60 \]

Intercepts:
\[ x=20 \; (y=0), \qquad y=15 \; (x=0) \]

For
\[ x + 3y = 30 \]

Intercepts:
\[ x=30 \; (y=0), \qquad y=10 \; (x=0) \]

Step 2: {\color{redFind the intersection of the two lines.

Solve
\[ 3x + 4y = 60 \]
\[ x + 3y = 30 \]

Multiply the second equation by \(3\):
\[ 3x + 9y = 90 \]

Subtract:
\[ 5y = 30 \]
\[ y = 6 \]

Substitute:
\[ x + 18 = 30 \]
\[ x = 12 \]

Thus intersection point:
\[ (12,6) \]

Step 3: {\color{redDetermine feasible region vertices.

Corner points are
\[ (0,0),\quad (20,0),\quad (12,6),\quad (0,10) \]

Step 4: {\color{redEvaluate the objective function.
\[ Z = 8000x + 12000y \]
\[ Z(0,0) = 0 \]
\[ Z(20,0) = 160000 \]
\[ Z(12,6) = 8000(12) + 12000(6) \]
\[ =96000 + 72000 \]
\[ =168000 \]
\[ Z(0,10) = 120000 \]

Step 5: {\color{redFind the maximum value.
\[ Z_{\max} = 168000 \]

at
\[ (x,y) = (12,6) \]
\[ \therefore Maximum value Z = 168000 at (12,6). \] Quick Tip: In graphical LPP, the optimal value always occurs at one of the corner points of the feasible region.

Concept:

Let the probability of hitting the target be \(p\).
Then the probability of missing the target is \(1-p\).

Step 1: {\color{redForm the relation.

Given
\[ p = 3(1-p) \]
\[ p = 3 - 3p \]
\[ 4p = 3 \]
\[ p = \frac{3}{4} \]

Thus
\[ P(hit)=\frac{3}{4}, \quad P(miss)=\frac{1}{4} \]

Step 2: {\color{redProbability that both shots hit the target.
\[ P(both hits) = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]

Step 3: {\color{redProbability that at least one shot misses.
\[ P(at least one miss) = 1 - P(both hits) \]
\[ = 1 - \frac{9}{16} \]
\[ = \frac{7}{16} \]

Step 4: {\color{redWrite the results.
\[ P(target hit in both shots)=\frac{9}{16} \]
\[ P(at least one miss)=\frac{7}{16} \] Quick Tip: For repeated independent trials: \[ P(both events) = P(A) \times P(B) \] For “at least one” events: \[ P(at least one) = 1 - P(none) \]

Concept:

Conditional probability formula:
\[ P(E|F)=\frac{P(E\cap F)}{P(F)} \]

Step 1: {\color{redTotal possible arrangements.

Three people can be arranged in
\[ 3! = 6 \]

ways.

Step 2: {\color{redEvent \(F\): Father in the middle.

Possible arrangements:
\[ (M, F, S), \quad (S, F, M) \]

Thus
\[ P(F)=\frac{2}{6}=\frac{1}{3} \]

Step 3: {\color{redEvent \(E\cap F\): Son on one end and Father in middle.

Both arrangements above satisfy the condition.

Thus
\[ P(E\cap F)=\frac{2}{6}=\frac{1}{3} \]

Step 4: {\color{redCompute conditional probability.
\[ P(E|F)=\frac{P(E\cap F)}{P(F)} \]
\[ =\frac{1/3}{1/3} \]
\[ =1 \]
\[ \therefore P(E|F)=1 \] Quick Tip: Conditional probability: \[ P(A|B)=\frac{P(A\cap B)}{P(B)} \] It measures the probability of \(A\) given that \(B\) has occurred.

Concept:

For integrals of the form
\[ \int \frac{f'(x)}{\sqrt{f(x)}}\,dx \]

use substitution \(u=f(x)\).

Step 1: {\color{redChoose substitution.

Let
\[ u = x^2 + 6x \]

Then
\[ \frac{du}{dx} = 2x + 6 \]
\[ du = (2x+6)dx \]

Rewrite the numerator:
\[ 2x+1 = (2x+6) -5 \]

Thus the integral becomes
\[ \int \frac{(2x+6)-5}{\sqrt{x^2+6x}}\,dx \]
\[ = \int \frac{2x+6}{\sqrt{x^2+6x}}dx - 5\int \frac{dx}{\sqrt{x^2+6x}} \]

Step 2: {\color{redEvaluate the first integral.

Using substitution \(u=x^2+6x\):
\[ \int \frac{2x+6}{\sqrt{x^2+6x}}dx = \int \frac{du}{\sqrt{u}} \]
\[ =2\sqrt{u} \]
\[ =2\sqrt{x^2+6x} \]

Step 3: {\color{redEvaluate the second integral.
\[ \int \frac{dx}{\sqrt{x^2+6x}} \]

Complete the square:
\[ x^2+6x=(x+3)^2-9 \]

Thus
\[ \int \frac{dx}{\sqrt{(x+3)^2-9}} = \ln\left|x+3+\sqrt{x^2+6x}\right| \]

Step 4: {\color{redWrite the final result.
\[ \int \frac{2x+1}{\sqrt{6x+x^2}}dx = 2\sqrt{x^2+6x} - 5\ln\left|x+3+\sqrt{x^2+6x}\right| + C \]
\[ \therefore Required integral = 2\sqrt{x^2+6x} - 5\ln\left|x+3+\sqrt{x^2+6x}\right| + C. \] Quick Tip: For integrals containing \(\sqrt{x^2+ax}\), complete the square: \[ x^2+ax=(x+\frac{a}{2})^2-\left(\frac{a}{2}\right)^2 \]

Concept:

For definite integrals, the property
\[ \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \]

is useful for simplifying expressions.

Step 1: {\color{redLet
\[ I=\int_{\frac{5\pi}{12}}^{\frac{13\pi}{12}} \frac{dx}{1+\sqrt{\cot x}} \]

Using the property
\[ x \rightarrow \frac{5\pi}{12}+\frac{13\pi}{12}-x \]
\[ = \frac{18\pi}{12}-x = \frac{3\pi}{2}-x \]

Thus
\[ I=\int_{\frac{5\pi}{12}}^{\frac{13\pi}{12}} \frac{dx}{1+\sqrt{\cot(\frac{3\pi}{2}-x)}} \]

But
\[ \cot\left(\frac{3\pi}{2}-x\right)=\tan x \]

Hence
\[ I=\int_{\frac{5\pi}{12}}^{\frac{13\pi}{12}} \frac{dx}{1+\sqrt{\tan x}} \]

Step 2: {\color{redAdd the two integrals.
\[ 2I= \int_{\frac{5\pi}{12}}^{\frac{13\pi}{12}} \left( \frac{1}{1+\sqrt{\cot x}} + \frac{1}{1+\sqrt{\tan x}} \right)dx \]

Since
\[ \sqrt{\cot x}=\frac{1}{\sqrt{\tan x}} \]

the sum simplifies to \(1\).

Thus
\[ 2I=\int_{\frac{5\pi}{12}}^{\frac{13\pi}{12}} dx \]
\[ = \frac{13\pi}{12}-\frac{5\pi}{12} \]
\[ = \frac{8\pi}{12} \]
\[ = \frac{2\pi}{3} \]

Step 3: {\color{redFind \(I\).
\[ I=\frac{1}{2}\cdot\frac{2\pi}{3} \]
\[ I=\frac{\pi}{3} \]
\[ \therefore \int_{\frac{5\pi}{12}}^{\frac{13\pi}{12}} \frac{dx}{1+\sqrt{\cot x}}=\frac{\pi}{3}. \] Quick Tip: For definite integrals: \[ \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \] This symmetry often simplifies trigonometric integrals.

Concept:

Because of the modulus function, split the interval at \(x=0\).

Step 1: {\color{redWrite the integral in two parts.
\[ I= \int_{-\frac{\pi}{6}}^{0}(\sin|x|+\cos|x|)\,dx + \int_{0}^{\frac{\pi}{6}}(\sin|x|+\cos|x|)\,dx \]

For \(x<0\):
\[ |x|=-x \]

Thus
\[ \sin|x|=\sin(-x)=-\sin x \]
\[ \cos|x|=\cos(-x)=\cos x \]

Step 2: {\color{redEvaluate the integrals.

After simplification,
\[ I=2\int_0^{\frac{\pi}{6}} (\sin x+\cos x)\,dx \]
\[ =2\left[-\cos x+\sin x\right]_0^{\frac{\pi}{6}} \]
\[ =2\left(\frac12+\frac{\sqrt3}{2}-1\right) \]
\[ =\sqrt3-1 \]
\[ \therefore \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (\sin|x|+\cos|x|)\,dx=\sqrt3-1. \] Quick Tip: For integrals involving \(|x|\), always split the interval at \(x=0\).

Concept:

For the inverse sine function
\[ y=\sin^{-1}(t) \]

the argument must satisfy
\[ -1 \le t \le 1 \]

and its range is
\[ -\frac{\pi}{2} \le y \le \frac{\pi}{2}. \]

Step 1: {\color{redFind the domain of \(p(x)\).

Given
\[ p(x)=\sin^{-1}(3-2x) \]

Thus
\[ -1 \le 3-2x \le 1 \]

Solve the inequalities:
\[ -1 \le 3-2x \]
\[ -4 \le -2x \]
\[ x \le 2 \]

and
\[ 3-2x \le 1 \]
\[ -2x \le -2 \]
\[ x \ge 1 \]

Hence
\[ 1 \le x \le 2 \]

Thus the domain is
\[ [1,2]. \]

Step 2: {\color{redFind \(x\) when \(p(x)=\frac{\pi}{6}\).
\[ \sin^{-1}(3-2x)=\frac{\pi}{6} \]
\[ 3-2x=\sin\frac{\pi}{6} \]
\[ 3-2x=\frac12 \]
\[ 2x=\frac{5}{2} \]
\[ x=\frac{5}{4} \]

Step 3: {\color{redFind the range of \(2p(x)+\frac{\pi}{2}\).

Since
\[ -\frac{\pi}{2} \le p(x) \le \frac{\pi}{2} \]

Multiply by \(2\):
\[ -\pi \le 2p(x) \le \pi \]

Add \( \frac{\pi}{2} \):
\[ -\frac{\pi}{2} \le 2p(x)+\frac{\pi}{2} \le \frac{3\pi}{2} \]

Thus the range is
\[ \left[-\frac{\pi}{2},\frac{3\pi}{2}\right]. \] Quick Tip: For \(y=\sin^{-1}(x)\): \[ -1 \le x \le 1 \] and \[ -\frac{\pi}{2} \le y \le \frac{\pi}{2}. \]

Concept:

The equation of a line through two points \(A(x_1,y_1,z_1)\) and \(B(x_2,y_2,z_2)\) is
\[ \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}. \]

A line perpendicular to two given lines has direction ratios equal to the cross product of their direction vectors.

Step 1: {\color{redEquation of line \(AB\).

Direction ratios:
\[ \vec{AB}=(6-1,\,8-2,\,11-3)=(5,6,8) \]

Thus
\[ \frac{x-1}{5}=\frac{y-2}{6}=\frac{z-3}{8}=t \]

So
\[ x=1+5t,\quad y=2+6t,\quad z=3+8t \]

Step 2: {\color{redParametric form of the second line.
\[ \vec r = (4,1,0)+\lambda(6,2,1) \]

Thus
\[ x=4+6\lambda,\quad y=1+2\lambda,\quad z=\lambda \]

Step 3: {\color{redFind the point of intersection.

Equate coordinates:
\[ 1+5t=4+6\lambda \]
\[ 2+6t=1+2\lambda \]
\[ 3+8t=\lambda \]

Solving,
\[ t=-\frac13, \qquad \lambda=\frac13 \]

Substitute \(t\):
\[ x=\frac{-2}{3},\quad y=0,\quad z=\frac13 \]

Thus the intersection point is
\[ P\left(-\frac23,0,\frac13\right) \]

Step 4: {\color{redDirection ratios of required line.

Direction vectors:
\[ \vec d_1=(5,6,8) \]
\[ \vec d_2=(6,2,1) \]

Cross product:
\[ \vec d_1\times\vec d_2= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}
5&6&8
6&2&1 \end{vmatrix} \]
\[ =(-10)\hat{i}+43\hat{j}-26\hat{k} \]

Thus direction ratios:
\[ (-10,43,-26) \]

Step 5: {\color{redEquation of the required line.

Through point \(P\left(-\frac23,0,\frac13\right)\):
\[ \frac{x+\frac23}{-10} = \frac{y}{43} = \frac{z-\frac13}{-26} \]
\[ \therefore Equation of the required line obtained. \] Quick Tip: A line perpendicular to two given lines has direction ratios equal to the cross product of their direction vectors.

Concept:

A system of linear equations can be written in matrix form
\[ AX = B \]

and the solution is
\[ X = A^{-1}B \]

if \(A^{-1}\) exists.

Step 1: {\color{redWrite the coefficient matrix.
\[ A= \begin{bmatrix} 1 & -1 & 0
2 & 3 & 4
0 & 1 & 2 \end{bmatrix} \]

and
\[ X= \begin{bmatrix} x
y
z \end{bmatrix}, \qquad B= \begin{bmatrix} 8
17
7 \end{bmatrix}. \]

Step 2: {\color{redSolve the equations.

From
\[ x-y=8 \]
\[ x=y+8 \]

From
\[ y+2z=7 \]
\[ y=7-2z \]

Substitute into the second equation:
\[ 2x+3y+4z=17 \]
\[ 2(y+8)+3(7-2z)+4z=17 \]
\[ 2y+16+21-6z+4z=17 \]
\[ 2y-2z+37=17 \]
\[ 2y-2z=-20 \]
\[ y-z=-10 \]

Substitute \(y=7-2z\):
\[ 7-2z-z=-10 \]
\[ 7-3z=-10 \]
\[ z=\frac{17}{3} \]
\[ y=7-2\left(\frac{17}{3}\right)=-\frac{13}{3} \]
\[ x=y+8=\frac{11}{3} \]

Step 3: {\color{redWrite the solution.
\[ x=\frac{11}{3},\quad y=-\frac{13}{3},\quad z=\frac{17}{3} \]
\[ \therefore Solution of the system obtained. \] Quick Tip: A system of linear equations can be solved using matrices by writing it in the form \[ AX=B \] and finding \(X=A^{-1}B\).

Step 1: {\color{redExpand the determinant.
\[ A= \begin{vmatrix} 1+x & 1 & 1
1 & 1+y & 1
1 & 1 & 1+z \end{vmatrix} \]

Apply row operations
\[ R_2 \rightarrow R_2 - R_1,\quad R_3 \rightarrow R_3 - R_1 \]
\[ A= \begin{vmatrix} 1+x & 1 & 1
-x & y & 0
-x & 0 & z \end{vmatrix} \]

Expanding,
\[ A=xyz(x+y+z+1) \]

Step 2: {\color{redUse the condition \(A=0\).
\[ xyz(x+y+z+1)=0 \]

Since \(x,y,z\neq0\),
\[ x+y+z+1=0 \]

Divide by \(xyz\):
\[ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-1 \]
\[ \therefore Result proved. \] Quick Tip: Row operations simplify determinants without changing their value.

Concept:

To determine where a function is increasing or decreasing, compute its first derivative and analyze its sign.

Step 1: {\color{redDifferentiate the function.
\[ f(x)=\tan^{-1}(\sin x-\cos x) \]

Using the derivative of \(\tan^{-1}u\):
\[ \frac{d}{dx}(\tan^{-1}u)=\frac{u'}{1+u^2} \]

Let \(u=\sin x-\cos x\)
\[ u'=\cos x+\sin x \]

Thus
\[ f'(x)=\frac{\cos x+\sin x}{1+(\sin x-\cos x)^2} \]

Step 2: {\color{redDetermine the sign of the derivative.

The denominator
\[ 1+(\sin x-\cos x)^2>0 \]

always.

Thus the sign depends on
\[ \cos x+\sin x \]

Step 3: {\color{redSolve
\[ \cos x+\sin x=0 \]
\[ \tan x=-1 \]

In \((0,\pi)\),
\[ x=\frac{3\pi}{4} \]

Step 4: {\color{redDetermine intervals.

For
\[ 0 \[ \cos x+\sin x>0 \]

Thus \(f'(x)>0\) → function increasing.

For
\[ \frac{3\pi}{4} \[ \cos x+\sin x<0 \]

Thus \(f'(x)<0\) → function decreasing.

Step 5: {\color{redWrite the result.

Increasing on
\[ (0,\frac{3\pi}{4}) \]

Decreasing on
\[ (\frac{3\pi}{4},\pi) \] Quick Tip: A function is increasing where \(f'(x)>0\) and decreasing where \(f'(x)<0\).

Concept:

If a rectangle is revolved about one of its sides, the resulting solid is a cylinder.
\[ V=\pi r^2 h \]

where \(r\) is the other side of the rectangle and \(h\) is the axis of rotation.

Step 1: {\color{redLet the sides of the rectangle be \(x\) and \(y\).

Given perimeter
\[ 2(x+y)=24 \]
\[ x+y=12 \]
\[ y=12-x \]

Step 2: {\color{redWrite the volume of the cylinder.

Assume the rectangle rotates about side \(x\):
\[ V=\pi y^2 x \]

Substitute \(y\):
\[ V=\pi x(12-x)^2 \]

Step 3: {\color{redMaximize the volume.

Differentiate:
\[ \frac{dV}{dx}=\pi[(12-x)^2-2x(12-x)] \]
\[ =\pi(12-x)(12-3x) \]

Set
\[ \frac{dV}{dx}=0 \]
\[ 12-3x=0 \]
\[ x=4 \]
\[ y=12-4=8 \]

Step 4: {\color{redWrite the dimensions.

Thus the rectangle dimensions are
\[ 4 cm \times 8 cm. \]
\[ \therefore Required dimensions obtained. \] Quick Tip: Optimization problems use derivatives: \[ \frac{dV}{dx}=0 \] gives maximum or minimum values.

Concept:

The given ellipse is
\[ 9x^2 + 16y^2 = 144 \]

which can be written in standard form
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]

Thus
\[ a=4, \quad b=3. \]

Step 1: {\color{redExpress \(y\) as a function of \(x\).
\[ 9x^2 + 16y^2 = 144 \]
\[ 16y^2 = 144 - 9x^2 \]
\[ y^2 = \frac{144 - 9x^2}{16} \]
\[ y = \frac{\sqrt{144-9x^2}}{4} \]

Step 2: {\color{redIntegrate the function.
\[ \int y\,dx = \int \frac{\sqrt{144-9x^2}}{4}\,dx \]

Factor \(9\):
\[ \int \frac{\sqrt{9(16-x^2)}}{4}\,dx \]
\[ = \frac{3}{4}\int \sqrt{16-x^2}\,dx \]

Using the standard integral
\[ \int \sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}+C \]

Thus
\[ \int y\,dx = \frac{3}{8}\left[x\sqrt{16-x^2}+16\sin^{-1}\frac{x}{4}\right]+C \]

Step 3: {\color{redArea enclosed by the ellipse.

Area of ellipse
\[ A=\pi ab \]
\[ =\pi(4)(3) \]
\[ =12\pi \]

Thus the area enclosed by the elliptical ground is
\[ \boxed{12\pi square units} \]


Alternative (iii): {\color{redArea of triangle \(POQ\).

Intercepts of the ellipse:
\[ x=4,\quad y=3 \]

Thus
\[ P(4,0), \quad Q(0,3) \]

Area of triangle \(POQ\):
\[ \frac{1}{2}\times4\times3 \]
\[ =6 \]
\[ \therefore Area of triangle POQ = 6. \] Quick Tip: For ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] Area \(= \pi ab\).

Concept:

To analyze piecewise functions:


Differentiate each part separately.
For continuity at \(x=a\): \( \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a) \).
For differentiability: left and right derivatives must be equal.


Step 1: {\color{redFind \(f'(x)\) for \(0 \[ f(x)=x^4-4x^2+4 \]

Differentiate:
\[ f'(x)=4x^3-8x \]
\[ \therefore f'(x)=4x^3-8x \]

for \(0


Step 2: {\color{redFind \(f'(4)\).

For \(x\ge3\)
\[ f(x)=x^2+40 \]
\[ f'(x)=2x \]

Thus
\[ f'(4)=2(4)=8 \]



Step 3: {\color{redTest continuity at \(x=3\).

Left-hand limit:
\[ \lim_{x\to3^-}f(x)=3^4-4(3^2)+4 \]
\[ =81-36+4 \]
\[ =49 \]

Right-hand limit:
\[ \lim_{x\to3^+}f(x)=3^2+40 \]
\[ =9+40 \]
\[ =49 \]

Also
\[ f(3)=3^2+40=49 \]

Thus
\[ LHL=RHL=f(3) \]
\[ \therefore f(x) is continuous at x=3. \]



Alternative (iii)(b): {\color{redTest differentiability at \(x=3\).

Left derivative:
\[ f'(x)=4x^3-8x \]
\[ f'(3^-)=4(27)-8(3) \]
\[ =108-24 \]
\[ =84 \]

Right derivative:
\[ f'(x)=2x \]
\[ f'(3^+)=2(3)=6 \]

Since
\[ 84\ne6 \]
\[ \therefore f(x) is not differentiable at x=3. \] Quick Tip: A function is differentiable at a point only if it is continuous there and its left and right derivatives are equal.

Concept:

Use basic probability and Bayes' theorem.

Let
\[ M = male, \quad F = female \]
\[ L = lung complication \]

Given:
\[ P(L|M)=\frac{170}{1000}=0.17 \]
\[ P(L|F)=\frac{120}{1000}=0.12 \]

Number of smokers = \(50\)
\[ 30 males, \quad 20 females \]

Thus
\[ P(M)=\frac{30}{50}=\frac{3}{5} \]
\[ P(F)=\frac{20}{50}=\frac{2}{5} \]



Step 1: {\color{red(i) Probability that the selected person is female.
\[ P(F)=\frac{20}{50} \]
\[ =\frac{2}{5} \]



Step 2: {\color{red(ii) Probability that a male does not suffer lung problems.
\[ P(No lung|M)=1-P(L|M) \]
\[ =1-0.17 \]
\[ =0.83 \]



Step 3: {\color{red(iii)(a) Probability that the person is female given lung complications.

Using Bayes' theorem:
\[ P(F|L)=\frac{P(F)P(L|F)}{P(M)P(L|M)+P(F)P(L|F)} \]

Substitute values:
\[ P(F|L)=\frac{\frac{2}{5}\times0.12}{\frac{3}{5}\times0.17+\frac{2}{5}\times0.12} \]
\[ =\frac{0.048}{0.102+0.048} \]
\[ =\frac{0.048}{0.15} \]
\[ =0.32 \]
\[ \therefore P(F|L)=0.32 \]



Alternative (iii)(b): {\color{redProbability that the person is male given no lung problems.
\[ P(No lung|M)=0.83 \]
\[ P(No lung|F)=1-0.12=0.88 \]

Total probability:
\[ P(No lung)=\frac{3}{5}(0.83)+\frac{2}{5}(0.88) \]
\[ =0.498+0.352 \]
\[ =0.85 \]

Thus
\[ P(M|No lung)= \frac{P(M)P(No lung|M)}{P(No lung)} \]
\[ =\frac{\frac{3}{5}\times0.83}{0.85} \]
\[ =\frac{0.498}{0.85} \]
\[ \approx0.586 \]
\[ \therefore P(M|No lung)\approx0.586 \] Quick Tip: Bayes' theorem: \[ P(A|B)=\frac{P(A)P(B|A)}{\sum P(A_i)P(B|A_i)} \] It helps update probabilities based on new evidence.