MATHEMATICS

\displaystyle \text{Time Allowed : 3 hours} \hspace{5.0cm} \text{Maximum Marks : 100 }  

\displaystyle \textbf{General Instructions:}
\displaystyle \text{(i) All questions are compulsory.}
\displaystyle \text{(ii) Please check that this Question Paper contains 26 questions.}
\displaystyle \text{(iii) Marks for each question are indicated against it.}
\displaystyle \text{(iv) Question 1 to 6 in Section-A are Very Short Answer Type Questions carrying }
\displaystyle \text{one mark each.}
\displaystyle \text{(v) Question 7 to 19 in Section-B are Long Answer I Type Questions carrying }
\displaystyle \text{4 marks each.}
\displaystyle \text{(vi) Question 20 to 26 in Section-C are Long Answer II Type Questions carrying }
\displaystyle \text{6 marks each.}
\displaystyle \text{(vii) Please write down the serial number of the Question before attempting it.}
\displaystyle \text{}

\displaystyle \textbf{SECTION - A}
\displaystyle \textbf{Question numbers 1 to 6 carry 1 Mark each}

\displaystyle \text{1. The two vectors } \widehat{i} + \widehat{k} \text{ and } 3\widehat{i} - \widehat{j} + 4\widehat{k} \text{ represent the two sides AB and AC, }
\displaystyle \text{respectively of a } \triangle ABC.   \text{ Find the length of the median through A.}
\displaystyle \text{Answer:}


\displaystyle \text{In the given } \triangle ABC, \text{ we have}
\displaystyle \overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB} \text{ (Triangle Law of Vector Addition)}
\displaystyle = (3\widehat{i} - \widehat{j} + 4\widehat{k}) - (\widehat{j} + \widehat{k}) = 3\widehat{i} - 2\widehat{j} + 3\widehat{k}
\displaystyle \text{Here, AD is the median.}
\displaystyle \therefore \overrightarrow{BD} = \frac{1}{2}\overrightarrow{BC} = \frac{3}{2}\widehat{i} - \widehat{j} + \frac{3}{2}\widehat{k}
\displaystyle \text{In } \triangle ABD, \text{ using the triangle law of vector addition, we have } \overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD}
\displaystyle \text{So, } \overrightarrow{AD} = (\widehat{j} + \widehat{k}) + \left(\frac{3}{2}\widehat{i} - \widehat{j} + \frac{3}{2}\widehat{k}\right) = \frac{3}{2}\widehat{i} + 0\widehat{j} + \frac{5}{2}\widehat{k}
\displaystyle \therefore AD = \sqrt{\left(\frac{3}{2}\right)^{2} + 0^{2} + \left(\frac{5}{2}\right)^{2}} = \frac{1}{2}\sqrt{34}
\displaystyle \text{Hence, the length of the median through A is } \frac{1}{2}\sqrt{34} \text{ units.}

\displaystyle \text{2. Find the vector equation of a plane which is at a distance of } 5 \text{units from the }
\displaystyle \text{origin and its normal vector is } 2\widehat{i} - 3\widehat{j} + 6\widehat{k}.
\displaystyle \text{Answer:}
\displaystyle \text{Normal vector, } \overrightarrow{n} = 2\widehat{i} - 3\widehat{j} + 6\widehat{k}
\displaystyle \text{Then, } \widehat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{2\widehat{i} - 3\widehat{j} + 6\widehat{k}}{\sqrt{2^{2} + (-3)^{2} + 6^{2}}} = \frac{2\widehat{i} - 3\widehat{j} + 6\widehat{k}}{7}
\displaystyle \text{Here, } d = 5
\displaystyle \text{Vector equation of the plane is given by } \overrightarrow{r} \cdot \widehat{n} = d
\displaystyle \text{Hence, the required equation of the plane is } \overrightarrow{r} \cdot \left(\frac{2}{7}\widehat{i} - \frac{3}{7}\widehat{j} + \frac{6}{7}\widehat{k}\right) = 5

\displaystyle \text{3. Find the maximum value of } \left| \begin{matrix} 1 & 1 & 1 \\ 1 & 1+\sin\theta & 1 \\ 1 & 1 & 1+\cos\theta \end{matrix} \right|
\displaystyle \text{Answer:}
\displaystyle \left| \begin{matrix} 1 & 1 & 1 \\ 1 & 1+\sin\theta & 1 \\ 1 & 1 & 1+\cos\theta \end{matrix} \right|
\displaystyle \text{Applying } R_{2} \rightarrow R_{2} - R_{1} \text{ and } R_{3} \rightarrow R_{3} - R_{1}
\displaystyle = \left| \begin{matrix} 1 & 1 & 1 \\ 0 & \sin\theta & 0 \\ 0 & 0 & \cos\theta \end{matrix} \right|
\displaystyle = \sin\theta \cos\theta = \frac{\sin 2\theta}{2}
\displaystyle \text{We know that, } -1 \leq \sin 2\theta \leq 1
\displaystyle \text{Therefore, required maximum value } = \frac{1}{2}

\displaystyle \text{4. If A is a square matrix such that } A^{2} = I, \text{ then find the }  \text{simplified value of }
\displaystyle  (A - I)^{3} + (A + I)^{3} - 7A.
\displaystyle \text{Answer:}
\displaystyle (A - I)^{3} + (A + I)^{3} - 7A
\displaystyle = A^{3} - I - 3A^{2}I + 3AI^{2} + A^{3} + I + 3A^{2}I + 3AI^{2} - 7A
\displaystyle = 2A^{3} + 6A^{2}I - 7A
\displaystyle = 2A \cdot A^{2} + 6A^{2} - 7A
\displaystyle = 2A \cdot I + 6I - 7A \quad (\text{Given: } A^{2} = I)
\displaystyle = 2A + 6I - 7A = A

\displaystyle \text{5. Matrix } A = \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}   \text{is given to be symmetric, find values of a and b.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: } A = \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix}
\displaystyle \Rightarrow A^{T} = \begin{bmatrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \end{bmatrix}
\displaystyle \text{A matrix is symmetric if } A = A^{T}
\displaystyle \text{Thus, } \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \end{bmatrix}
\displaystyle \text{Comparing both sides, we get } 2b = 3 \Rightarrow b = \frac{3}{2}
\displaystyle \text{And, } 3a = -2 \Rightarrow a = -\frac{2}{3}
\displaystyle \text{Therefore, } a = -\frac{2}{3} \text{ and } b = \frac{3}{2}

\displaystyle \text{6. Find the position vector of a point which divides the join of points with position }
\displaystyle \text{vectors } \overrightarrow{a} - 2\overrightarrow{b} \text{ and } 2\overrightarrow{a} + \overrightarrow{b}   \text{externally in the ratio } 2 : 1.
\displaystyle \text{Answer:}
\displaystyle \text{Let A and B be the points with position vectors } \overrightarrow{a} - 2\overrightarrow{b} \text{ and } 2\overrightarrow{a} + \overrightarrow{b} \text{ respectively.}
\displaystyle \text{Also, assume that R divides AB externally in the ratio } 2 : 1.
\displaystyle \text{Therefore, position vector of R } = \frac{2(2\overrightarrow{a} + \overrightarrow{b}) - 1(\overrightarrow{a} - 2\overrightarrow{b})}{2 - 1}
\displaystyle = \frac{4\overrightarrow{a} + 2\overrightarrow{b} - \overrightarrow{a} + 2\overrightarrow{b}}{1}
\displaystyle = 3\overrightarrow{a} + 4\overrightarrow{b}

\displaystyle \textbf{SECTION - B}
\displaystyle \textbf{Question numbers 7 to 19 carry 4 Mark each}

\displaystyle \text{7. Find the general solution of the following differential equation :}
\displaystyle (1 + y^{2}) + (x - e^{\tan^{-1}y}) \frac{dy}{dx} = 0
\displaystyle \text{Answer:}
\displaystyle \text{7. Given: } (1 + y^{2}) + (x - e^{\tan^{-1}y})\frac{dy}{dx} = 0
\displaystyle \text{Let } \tan^{-1}y = t \Rightarrow y = \tan t
\displaystyle \Rightarrow \frac{dy}{dx} = \sec^{2}t \frac{dt}{dx}
\displaystyle \text{Therefore, the equation becomes}
\displaystyle (1 + \tan^{2}t) + (x - e^{t})\sec^{2}t \frac{dt}{dx} = 0
\displaystyle \Rightarrow \sec^{2}t + (x - e^{t})\sec^{2}t \frac{dt}{dx} = 0
\displaystyle \Rightarrow \sec^{2}t \left[1 + (x - e^{t})\frac{dt}{dx}\right] = 0
\displaystyle \Rightarrow 1 + (x - e^{t})\frac{dt}{dx} = 0
\displaystyle \Rightarrow (x - e^{t})\frac{dt}{dx} = -1
\displaystyle \Rightarrow x - e^{t} = -\frac{dx}{dt}
\displaystyle \Rightarrow \frac{dx}{dt} + x = e^{t}
\displaystyle \text{The above equation is a linear first order differential equation}
\displaystyle \text{of the form } \frac{dx}{dt} + Px = Q
\displaystyle \text{Integrating factor, I.F. } = e^{\int 1 dt} = e^{t}
\displaystyle \text{Solution is given by } x \cdot \text{I.F.} = \int Q \cdot \text{I.F.} \, dt + C
\displaystyle \Rightarrow x e^{t} = \int e^{t} \cdot e^{t} dt + C = \int e^{2t} dt + C
\displaystyle \Rightarrow x e^{t} = \frac{1}{2} e^{2t} + C
\displaystyle \text{Substituting } t = \tan^{-1}y
\displaystyle \Rightarrow x e^{\tan^{-1}y} = \frac{1}{2} e^{2\tan^{-1}y} + C
\displaystyle \Rightarrow e^{2\tan^{-1}y} = 2x e^{\tan^{-1}y} + C
\displaystyle \text{It is the required general solution.}

\displaystyle \text{8. Show that the vectors } \overrightarrow{a}, \overrightarrow{b} \text{ and } \overrightarrow{c} \text{ are coplanar if } \overrightarrow{a} + \overrightarrow{b}, \overrightarrow{b} + \overrightarrow{c} \text{ and }
\displaystyle \overrightarrow{c} + \overrightarrow{a} \text{ are coplanar.}
\displaystyle \text{Answer:}
\displaystyle \text{8. It is given that } \overrightarrow{a} + \overrightarrow{b}, \overrightarrow{b} + \overrightarrow{c} \text{ and } \overrightarrow{c} + \overrightarrow{a} \text{ are coplanar.}
\displaystyle \text{Scalar triple product } = 0
\displaystyle [\overrightarrow{a} + \overrightarrow{b}, \overrightarrow{b} + \overrightarrow{c}, \overrightarrow{c} + \overrightarrow{a}] = 0
\displaystyle \Rightarrow [\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}] + [\overrightarrow{a}, \overrightarrow{c}, \overrightarrow{a}] + [\overrightarrow{b}, \overrightarrow{b}, \overrightarrow{c}] + [\overrightarrow{b}, \overrightarrow{c}, \overrightarrow{a}] = 0
\displaystyle \Rightarrow [\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}] + 0 + 0 + [\overrightarrow{b}, \overrightarrow{c}, \overrightarrow{a}] = 0
\displaystyle \Rightarrow [\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}] + [\overrightarrow{b}, \overrightarrow{c}, \overrightarrow{a}] = 0
\displaystyle \Rightarrow 2[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}] = 0
\displaystyle \Rightarrow [\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}] = 0
\displaystyle \text{Therefore, the vectors } \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \text{ are coplanar.}

\displaystyle \text{9. Find the vector and Cartesian equations of the line through } (1, 2, -4)
\displaystyle \text{ and perpendicular to the two lines}
\displaystyle \overrightarrow{r} = (8\widehat{i} - 19\widehat{j} + 10\widehat{k}) + \lambda(3\widehat{i} - 16\widehat{j} + 7\widehat{k})
\displaystyle \overrightarrow{r} = (15\widehat{i} + 29\widehat{j} + 5\widehat{k}) + \mu(3\widehat{i} + 8\widehat{j} - 5\widehat{k})
\displaystyle \text{Answer:}
\displaystyle \text{9. The equations of the given lines are}
\displaystyle \overrightarrow{r} = (8\widehat{i} - 19\widehat{j} + 10\widehat{k}) + \lambda(3\widehat{i} - 16\widehat{j} + 7\widehat{k})
\displaystyle \overrightarrow{r} = (15\widehat{i} + 29\widehat{j} + 5\widehat{k}) + \mu(3\widehat{i} + 8\widehat{j} - 5\widehat{k})
\displaystyle \text{Vector parallel to first line } \overrightarrow{b_{1}} = 3\widehat{i} - 16\widehat{j} + 7\widehat{k}
\displaystyle \text{Vector parallel to second line } \overrightarrow{b_{2}} = 3\widehat{i} + 8\widehat{j} - 5\widehat{k}
\displaystyle \text{Required line is perpendicular to both, so direction } \overrightarrow{b} = \overrightarrow{b_{1}} \times \overrightarrow{b_{2}}
\displaystyle \Rightarrow \overrightarrow{b} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}
\displaystyle = 24\widehat{i} + 36\widehat{j} + 72\widehat{k}
\displaystyle \text{Since line passes through } (1,2,-4), \overrightarrow{a} = \widehat{i} + 2\widehat{j} - 4\widehat{k}
\displaystyle \text{Vector equation is } \overrightarrow{r} = \overrightarrow{a} + k\overrightarrow{b}
\displaystyle \Rightarrow \overrightarrow{r} = (\widehat{i} + 2\widehat{j} - 4\widehat{k}) + k(24\widehat{i} + 36\widehat{j} + 72\widehat{k})
\displaystyle \text{Cartesian equation is } \frac{x - 1}{24} = \frac{y - 2}{36} = \frac{z + 4}{72}

\displaystyle \text{10. Three persons A, B and C apply for a job of Manager in a Private Company. }
\displaystyle \text{Chances of their selection (A, B and C)} \text{are in the ratio } 1 : 2 : 4. \text{The probabilities }
\displaystyle \text{that A, B and C can introduce changes to improve profits of} \text{ the company are} 
\displaystyle   0.8, 0.5 \text{ and } 0.3 \text{ respectively. If the change does not take place, find the probability }
\displaystyle \text{that it is due to the appointment of C.}
\displaystyle \text{OR}
\displaystyle \text{A and B throw a pair of dice alternately. A wins the game if he gets a total of }
\displaystyle \text{7 and B wins the game if he gets a total of 10. If A starts the game, then find the }
\displaystyle \text{probability that B wins.}
\displaystyle \text{Answer:}
\displaystyle \text{10. Let } E_{1}, E_{2} \text{ and } E_{3} \text{ denote the selection of A, B and C as managers, respectively.}
\displaystyle \therefore P(E_{1}) = \frac{x}{x + 2x + 4x} = \frac{1}{7}
\displaystyle P(E_{2}) = \frac{2x}{x + 2x + 4x} = \frac{2}{7}
\displaystyle P(E_{3}) = \frac{4x}{x + 2x + 4x} = \frac{4}{7}
\displaystyle \text{Let } A \text{ be the event denoting that the change does not take place.}
\displaystyle \therefore P(A \mid E_{1}) = \text{Probability that A does not introduce change} = 0.2
\displaystyle P(A \mid E_{2}) = \text{Probability that B does not introduce change} = 0.5
\displaystyle P(A \mid E_{3}) = \text{Probability that C does not introduce change} = 0.7
\displaystyle \therefore \text{Required probability } = P(E_{3} \mid A)
\displaystyle \text{Using Bayes' theorem, we have}
\displaystyle P(E_{3} \mid A) = \frac{P(E_{3})P(A \mid E_{3})}{P(E_{1})P(A \mid E_{1}) + P(E_{2})P(A \mid E_{2}) + P(E_{3})P(A \mid E_{3})}
\displaystyle = \frac{\frac{4}{7} \times 0.7}{\frac{1}{7} \times 0.2 + \frac{2}{7} \times 0.5 + \frac{4}{7} \times 0.7}
\displaystyle = \frac{2.8}{0.2 + 1 + 2.8} = \frac{2.8}{4} = 0.7
\displaystyle \text{OR}
\displaystyle \text{Total of 7 on the dice can be obtained in the following ways:}
\displaystyle (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)
\displaystyle \text{Probability of getting a total of 7 } = \frac{6}{36} = \frac{1}{6}
\displaystyle \text{Probability of not getting a total of 7 } = 1 - \frac{1}{6} = \frac{5}{6}
\displaystyle \text{Total of 10 on the dice can be obtained in the following ways:}
\displaystyle (4,6), (6,4), (5,5)
\displaystyle \text{Probability of getting a total of 10 } = \frac{3}{36} = \frac{1}{12}
\displaystyle \text{Probability of not getting a total of 10 } = 1 - \frac{1}{12} = \frac{11}{12}
\displaystyle \text{Let } E \text{ and } F \text{ be the two events, defined as follows:}
\displaystyle E = \text{Getting a total of 7 in a single throw of a pair of dice}
\displaystyle F = \text{Getting a total of 10 in a single throw of a pair of dice}
\displaystyle P(E) = \frac{1}{6},\; P(F) = \frac{1}{12} \Rightarrow P(\overline{E}) = \frac{5}{6},\; P(\overline{F}) = \frac{11}{12}
\displaystyle \text{A wins if he gets a total of 7 in the 1st, 3rd or 5th } \ldots \text{ throws and so on.}
\displaystyle \text{Probability of A getting a total of 7 in the 1st throw } = \frac{1}{6}
\displaystyle \text{A will get the 3rd throw if he fails in the 1st throw and B fails in the 2nd throw.}
\displaystyle \text{Probability of A getting a total of 7 in the 3rd throw}
\displaystyle = P(\overline{E})P(\overline{F})P(E) = \frac{5}{6} \times \frac{11}{12} \times \frac{1}{6}
\displaystyle \text{Similarly, probability of getting a total of 7 in the 5th throw is}
\displaystyle P(\overline{E})P(\overline{F})P(\overline{E})P(\overline{F})P(E)
\displaystyle = \frac{5}{6} \times \frac{11}{12} \times \frac{5}{6} \times \frac{11}{12} \times \frac{1}{6}
\displaystyle \text{Probability of winning of A}
\displaystyle = \frac{1}{6} + \left(\frac{5}{6} \times \frac{11}{12} \times \frac{1}{6}\right) + \left(\frac{5}{6} \times \frac{11}{12} \times \frac{5}{6} \times \frac{11}{12} \times \frac{1}{6}\right) + \ldots
\displaystyle = \frac{\frac{1}{6}}{1 - \frac{5}{6} \times \frac{11}{12}} = \frac{\frac{1}{6}}{1 - \frac{55}{72}} = \frac{\frac{1}{6}}{\frac{17}{72}} = \frac{12}{17}
\displaystyle \text{Probability of winning of B } = 1 - \text{Probability of winning of A}
\displaystyle = 1 - \frac{12}{17} = \frac{5}{17}

\displaystyle \text{11. Prove that: } \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4}
\displaystyle \text{OR}
\displaystyle \text{Solve for x : } 2\tan^{-1}(\cos x) = \tan^{-1}(2\mathrm{cosec}\ x)
\displaystyle \text{Answer:}
\displaystyle \text{11. } \left(\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7}\right) + \left(\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{8}\right)
\displaystyle = \tan^{-1}\left(\frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \times \frac{1}{7}}\right) + \tan^{-1}\left(\frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \times \frac{1}{8}}\right)
\displaystyle \text{Using } \tan^{-1}X + \tan^{-1}Y = \tan^{-1}\left(\frac{X + Y}{1 - XY}\right)
\displaystyle = \tan^{-1}\frac{6}{17} + \tan^{-1}\frac{11}{23}
\displaystyle = \tan^{-1}\left(\frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \times \frac{11}{23}}\right)
\displaystyle = \tan^{-1}\left(\frac{\frac{325}{391}}{\frac{325}{391}}\right) = \tan^{-1}(1) = \frac{\pi}{4}
\displaystyle \text{Hence, proved.}
\displaystyle \text{OR}
\displaystyle 2\tan^{-1}(\cos x) = \tan^{-1}(2\mathrm{cosec}\ x)
\displaystyle \Rightarrow \tan^{-1}\left(\frac{2\cos x}{1 - \cos^{2}x}\right) = \tan^{-1}(2\mathrm{cosec}\ x)
\displaystyle \text{Using } 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1 - x^{2}}\right)
\displaystyle \Rightarrow \frac{2\cos x}{\sin^{2}x} = 2\mathrm{cosec}\ x
\displaystyle \Rightarrow \frac{\cos x}{\sin x} = 1
\displaystyle \Rightarrow \tan x = 1
\displaystyle \therefore x = \frac{\pi}{4}

\displaystyle \text{12. The monthly incomes of Aryan and Babban are in the ratio } 3:4 \text{ and their monthly}
\displaystyle \text{expenditures are in the ratio } 5:7. \text{ If each saves } \text{Rs }15000 \text{ per month, find their }
\displaystyle \text{monthly incomes using matrix method. This problem reflects which value?}
\displaystyle \text{Answer:}
\displaystyle \text{12. Let the monthly incomes of Aryan and Babban be } 3x \text{ and } 4x, \text{ respectively.}
\displaystyle \text{Suppose their monthly expenditures are } 5y \text{ and } 7y, \text{ respectively. Since each saves Rs }15000 \text{ per month,}
\displaystyle \text{Monthly savings of Aryan: } 3x - 5y = 15000
\displaystyle \text{Monthly savings of Babban: } 4x - 7y = 15000
\displaystyle \text{The above system of equations can be written in the matrix form as follows:}
\displaystyle \begin{bmatrix} 3 & -5 \\ 4 & -7 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 15000 \\ 15000 \end{bmatrix}
\displaystyle \text{or } AX = B, \text{ where}
\displaystyle A = \begin{bmatrix} 3 & -5 \\ 4 & -7 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 15000 \\ 15000 \end{bmatrix}
\displaystyle \text{Now, } |A| = \left|\begin{matrix} 3 & -5 \\ 4 & -7 \end{matrix}\right| = -21 - (-20) = -1
\displaystyle \text{Adj } A = \begin{bmatrix} -7 & 5 \\ -4 & 3 \end{bmatrix}
\displaystyle \text{So, } A^{-1} = \frac{1}{|A|}\text{Adj } A = -1 \begin{bmatrix} -7 & 5 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ 4 & -3 \end{bmatrix}
\displaystyle \therefore X = A^{-1}B
\displaystyle \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ 4 & -3 \end{bmatrix}\begin{bmatrix} 15000 \\ 15000 \end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 105000 - 75000 \\ 60000 - 45000 \end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 30000 \\ 15000 \end{bmatrix}
\displaystyle \Rightarrow x = 30000 \text{ and } y = 15000
\displaystyle \text{Therefore,}
\displaystyle \text{Monthly income of Aryan } = 3 \times \text{Rs }30000 = \text{Rs }90000
\displaystyle \text{Monthly income of Babban } = 4 \times \text{Rs }30000 = \text{Rs }120000
\displaystyle \text{Here, we are encouraged to understand the importance of savings.}
\displaystyle \text{We should save a certain part of our monthly income for the future.}

\displaystyle \text{13. If } x = a\sin 2t (1 + \cos 2t) \text{ and } y = b\cos 2t (1 - \cos 2t), \text{ find the values of }
\displaystyle \frac{dy}{dx} \text{ at } t = \frac{\pi}{4} \text{ and } t = \frac{\pi}{3}.
\displaystyle \text{OR}
\displaystyle \text{If } y = x^{x}, \text{ prove that } \frac{d^{2}y}{dx^{2}} - \frac{1}{y}\left(\frac{dy}{dx}\right)^{2} - \frac{y}{x} = 0.
\displaystyle \text{Answer:}
\displaystyle \text{13. Given that:}
\displaystyle x = a\sin 2t(1 + \cos 2t)
\displaystyle y = b\cos 2t(1 - \cos 2t)
\displaystyle \text{Differentiating the above equations w.r.t. } t, \text{ we get}
\displaystyle \frac{dx}{dt} = 2a\cos 2t(1 + \cos 2t) - 2a\sin^{2}2t
\displaystyle = 2a\cos 2t + 2a(\cos^{2}2t - \sin^{2}2t)
\displaystyle \frac{dx}{dt} = a(2\cos 2t + 2\cos 4t)
\displaystyle \frac{dy}{dt} = -2b\sin 2t(1 - \cos 2t) + b\cos 2t(2\sin 2t)
\displaystyle = -2b\sin 2t + 4b\sin 2t \cos 2t
\displaystyle = -2b\sin 2t + 2b\sin 4t
\displaystyle \text{and } \frac{dy}{dt} = b(-2\sin 2t + 2\sin 4t)
\displaystyle \text{Now, } \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{b(-2\sin 2t + 2\sin 4t)}{a(2\cos 2t + 2\cos 4t)}
\displaystyle \therefore \frac{dy}{dx} = \frac{b}{a}\left[\frac{-2\sin 2t + 2\sin 4t}{2\cos 2t + 2\cos 4t}\right]
\displaystyle \therefore \left.\frac{dy}{dx}\right|_{t=\frac{\pi}{4}} = \frac{b}{a}\left[\frac{-2 + 0}{0 - 2}\right] = \frac{b}{a}
\displaystyle \text{and } \left.\frac{dy}{dx}\right|_{t=\frac{\pi}{3}} = \frac{b}{a}\left[\frac{-2\sin\frac{2\pi}{3} + 2\sin\frac{4\pi}{3}}{2\cos\frac{2\pi}{3} + 2\cos\frac{4\pi}{3}}\right] = \frac{b}{a}\left[\frac{-2\sqrt{3}}{-2}\right] = \frac{\sqrt{3}b}{a}
\displaystyle \text{OR}
\displaystyle y = x^{x}
\displaystyle \text{Taking logarithm on both sides, we have}
\displaystyle \log y = x \log x
\displaystyle \frac{1}{y}\frac{dy}{dx} = \log x + x \cdot \frac{1}{x} = 1 + \log x
\displaystyle \frac{dy}{dx} = x^{x}(1 + \log x)
\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d(x^{x})}{dx}(1 + \log x) + x^{x}\left[\frac{d}{dx}(1 + \log x)\right]
\displaystyle = x^{x}(1 + \log x)(1 + \log x) + x^{x}\left(\frac{1}{x}\right)
\displaystyle = x^{x}(1 + \log x)^{2} + x^{x-1}
\displaystyle \text{Putting the values of } \frac{d^{2}y}{dx^{2}}, \frac{dy}{dx} \text{ and } y \text{ in the expression}
\displaystyle \frac{d^{2}y}{dx^{2}} - \frac{1}{y}\left(\frac{dy}{dx}\right)^{2} - \frac{y}{x}, \text{ we have}
\displaystyle x^{x}(1 + \log x)^{2} + x^{x-1} - \frac{1}{x^{x}}\left(x^{x}(1 + \log x)\right)^{2} - \frac{x^{x}}{x}
\displaystyle = x^{x}(1 + \log x)^{2} + x^{x-1} - \frac{x^{2x}(1 + \log x)^{2}}{x^{x}} - x^{x-1}
\displaystyle = x^{x}(1 + \log x)^{2} - x^{x}(1 + \log x)^{2} = 0
\displaystyle \text{Hence, proved.}

\displaystyle \text{14. Find the values of p and q for which}   f(x) = \begin{cases} \frac{1 - \sin^{3}x}{3\cos^{2}x}, & x < \frac{\pi}{2} \\ p, & x = \frac{\pi}{2} \\ \frac{q(1 - \sin x)}{(\pi - 2x)^{2}}, & x > \frac{\pi}{2} \end{cases}
\displaystyle \text{is continuous at } x = \frac{\pi}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{14. } f(x) = \begin{cases} \frac{1 - \sin^{3}x}{3\cos^{2}x}, & \text{if } x < \frac{\pi}{2} \\ p, & \text{if } x = \frac{\pi}{2} \\ \frac{q(1 - \sin x)}{(\pi - 2x)^{2}}, & \text{if } x > \frac{\pi}{2} \end{cases}
\displaystyle \text{For } f(x) \text{ to be continuous at } x = \frac{\pi}{2},
\displaystyle \lim_{x \to \frac{\pi}{2}^{-}} f(x) = \lim_{x \to \frac{\pi}{2}^{+}} f(x) = f\left(\frac{\pi}{2}\right)
\displaystyle \text{L.H.L.: } \lim_{x \to \frac{\pi}{2}^{-}} f(x) = \lim_{x \to \frac{\pi}{2}^{-}} \left(\frac{1 - \sin^{3}x}{3\cos^{2}x}\right)
\displaystyle = \lim_{x \to \frac{\pi}{2}^{-}} \frac{(1 - \sin x)(1 + \sin x + \sin^{2}x)}{3(1 - \sin^{2}x)}
\displaystyle = \lim_{x \to \frac{\pi}{2}^{-}} \frac{1 + \sin x + \sin^{2}x}{3(1 + \sin x)}
\displaystyle = \frac{1 + 1 + 1}{3(1 + 1)} = \frac{1}{2}
\displaystyle \text{R.H.L.: } \lim_{x \to \frac{\pi}{2}^{+}} f(x) = \lim_{x \to \frac{\pi}{2}^{+}} \frac{q(1 - \sin x)}{(\pi - 2x)^{2}}
\displaystyle \text{Let } x = \frac{\pi}{2} + h, \text{ where } h \to 0
\displaystyle \Rightarrow \lim_{h \to 0} \frac{q\left[1 - \sin\left(\frac{\pi}{2} + h\right)\right]}{\left[\pi - 2\left(\frac{\pi}{2} + h\right)\right]^{2}}
\displaystyle = \lim_{h \to 0} \frac{q(1 - \cos h)}{4h^{2}}
\displaystyle = \lim_{h \to 0} \frac{q \cdot 2\sin^{2}\frac{h}{2}}{4h^{2}}
\displaystyle = \frac{q}{8} \lim_{h \to 0} \left(\frac{\sin\frac{h}{2}}{\frac{h}{2}}\right)^{2}
\displaystyle = \frac{q}{8}
\displaystyle \text{Now, } \lim_{x \to \frac{\pi}{2}^{-}} f(x) = \lim_{x \to \frac{\pi}{2}^{+}} f(x) = f\left(\frac{\pi}{2}\right)
\displaystyle \Rightarrow \frac{1}{2} = \frac{q}{8} = p
\displaystyle \text{Therefore, } p = \frac{1}{2} \text{ and } q = 4

\displaystyle \text{15. Show that the equation of normal at any point } t \text{ on the curve}
\displaystyle x = 3\cos t - \cos^{3}t \text{ and } y = 3\sin t - \sin^{3}t \text{ is }   4(y\cos^{3}t - x\sin^{3}t) = 3\sin 4t.
\displaystyle \text{Answer:}
\displaystyle \text{15. Given:}
\displaystyle x = 3\cos t - \cos^{3}t
\displaystyle y = 3\sin t - \sin^{3}t
\displaystyle \text{Slope of the tangent,}
\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3\cos t - 3\sin^{2}t \cos t}{-3\sin t + 3\cos^{2}t \sin t}
\displaystyle = \frac{3\cos t(1 - \sin^{2}t)}{-3\sin t(1 - \cos^{2}t)} = \frac{3\cos t \cos^{2}t}{-3\sin t \sin^{2}t} = -\frac{\cos^{3}t}{\sin^{3}t}
\displaystyle \therefore \text{Slope of the normal } = -\frac{1}{\frac{dy}{dx}} = \frac{\sin^{3}t}{\cos^{3}t}
\displaystyle \text{Equation of the normal is given by}
\displaystyle \frac{y - (3\sin t - \sin^{3}t)}{x - (3\cos t - \cos^{3}t)} = \frac{\sin^{3}t}{\cos^{3}t}
\displaystyle \Rightarrow y\cos^{3}t - 3\sin t \cos^{3}t + \sin^{3}t \cos^{3}t = x\sin^{3}t - 3\cos t \sin^{3}t + \sin^{3}t \cos^{3}t
\displaystyle \Rightarrow y\cos^{3}t - x\sin^{3}t = 3(\sin t \cos^{3}t - \cos t \sin^{3}t)
\displaystyle = 3\sin t \cos t(\cos^{2}t - \sin^{2}t)
\displaystyle = \frac{3}{2}\sin 2t \cos 2t
\displaystyle = \frac{3}{4} \cdot 2\sin 2t \cos 2t
\displaystyle \Rightarrow 4(y\cos^{3}t - x\sin^{3}t) = 3\sin 4t
\displaystyle \text{Hence, proved.}

\displaystyle \text{16. Find } \int \frac{(3\sin\theta - 2)\cos\theta}{5 - \cos^{2}\theta - 4\sin\theta} \, d\theta
\displaystyle \text{OR}
\displaystyle \text{Evaluate } \int_{0}^{\pi} e^{2x}\sin\left(\frac{\pi}{4} + x\right) dx.
\displaystyle \text{Answer:}
\displaystyle \text{16. Let } I = \int \frac{(3\sin\theta - 2)\cos\theta}{5 - \cos^{2}\theta - 4\sin\theta} \, d\theta
\displaystyle \Rightarrow I = \int \frac{(3\sin\theta - 2)\cos\theta}{5 - (1 - \sin^{2}\theta) - 4\sin\theta} \, d\theta
\displaystyle \Rightarrow I = \int \frac{(3\sin\theta - 2)\cos\theta}{\sin^{2}\theta - 4\sin\theta + 4} \, d\theta
\displaystyle \text{Now, substitute } \sin\theta = t \Rightarrow \cos\theta \, d\theta = dt
\displaystyle \therefore I = \int \frac{(3t - 2)\,dt}{t^{2} - 4t + 4}
\displaystyle \text{Let } 3t - 2 = A\frac{d}{dt}(t^{2} - 4t + 4) + B
\displaystyle \Rightarrow 3t - 2 = A(2t - 4) + B
\displaystyle \Rightarrow 3t - 2 = (2A)t + B - 4A
\displaystyle \text{Comparing coefficients, we get } 2A = 3 \Rightarrow A = \frac{3}{2}
\displaystyle \text{and } B - 4A = -2 \Rightarrow B - 4\left(\frac{3}{2}\right) = -2 \Rightarrow B = 4
\displaystyle \text{Substituting the values of } A \text{ and } B, \text{ we get}
\displaystyle I = \int \frac{\frac{3}{2}(2t - 4) + 4}{t^{2} - 4t + 4} \, dt
\displaystyle = \frac{3}{2}\int \frac{2t - 4}{t^{2} - 4t + 4} \, dt + 4\int \frac{dt}{t^{2} - 4t + 4}
\displaystyle = \frac{3}{2}I_{1} + 4I_{2}
\displaystyle \text{Here, } I_{1} = \int \frac{(2t - 4)\,dt}{t^{2} - 4t + 4} \text{ and } I_{2} = \int \frac{dt}{t^{2} - 4t + 4}
\displaystyle \text{Now, let } t^{2} - 4t + 4 = p \Rightarrow (2t - 4)\,dt = dp
\displaystyle \Rightarrow I_{1} = \int \frac{dp}{p} = \log|p| + C_{1} = \log|t^{2} - 4t + 4| + C_{1}
\displaystyle \text{and } I_{2} = \int \frac{dt}{t^{2} - 4t + 4} = \int \frac{dt}{(t - 2)^{2}}
\displaystyle = \int (t - 2)^{-2} \, dt = \frac{(t - 2)^{-1}}{-1} + C_{2} = -\frac{1}{t - 2} + C_{2}
\displaystyle \text{From the above, we get}
\displaystyle I = \frac{3}{2}\log|t^{2} - 4t + 4| + 4\left(-\frac{1}{t - 2}\right) + C
\displaystyle = \frac{3}{2}\log|\sin^{2}\theta - 4\sin\theta + 4| + \frac{4}{2 - \sin\theta} + C
\displaystyle = \frac{3}{2}\log(\sin\theta - 2)^{2} + \frac{4}{2 - \sin\theta} + C
\displaystyle = 3\log|2 - \sin\theta| + \frac{4}{2 - \sin\theta} + C
\displaystyle = 3\log(2 - \sin\theta) + \frac{4}{2 - \sin\theta} + C
\displaystyle \text{OR}
\displaystyle \text{Let } I = \int_{0}^{\pi} e^{2x}\sin\left(\frac{\pi}{4} + x\right) dx
\displaystyle \text{Integrating by parts, we get}
\displaystyle I = \frac{1}{2}\left[e^{2x}\sin\left(\frac{\pi}{4} + x\right)\right]_{0}^{\pi} - \frac{1}{2}\int_{0}^{\pi} e^{2x}\cos\left(\frac{\pi}{4} + x\right) dx
\displaystyle \text{Now, integrating the second term again by parts,}
\displaystyle \Rightarrow I = \frac{1}{2}\left[e^{2x}\sin\left(\frac{\pi}{4} + x\right)\right]_{0}^{\pi} - \frac{1}{4}\left[e^{2x}\cos\left(\frac{\pi}{4} + x\right)\right]_{0}^{\pi} - \frac{1}{4}I
\displaystyle \Rightarrow I + \frac{1}{4}I = \frac{1}{2}\left[e^{2x}\sin\left(\frac{\pi}{4} + x\right)\right]_{0}^{\pi} - \frac{1}{4}\left[e^{2x}\cos\left(\frac{\pi}{4} + x\right)\right]_{0}^{\pi}
\displaystyle \Rightarrow \frac{5}{4}I = \frac{1}{2}\left[e^{2\pi}\sin\left(\pi + \frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)\right] - \frac{1}{4}\left[e^{2\pi}\cos\left(\pi + \frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right)\right]
\displaystyle \Rightarrow \frac{5}{4}I = \frac{1}{2}\left[-\frac{e^{2\pi}}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right] - \frac{1}{4}\left[-\frac{e^{2\pi}}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right]
\displaystyle \Rightarrow \frac{5}{4}I = -\frac{e^{2\pi} + 1}{2\sqrt{2}} + \frac{e^{2\pi} + 1}{4\sqrt{2}} = -\frac{e^{2\pi} + 1}{4\sqrt{2}}
\displaystyle \Rightarrow I = -\frac{e^{2\pi} + 1}{5\sqrt{2}}

\displaystyle \text{17. Find } \int \frac{\sqrt{x}}{\sqrt{a^{3} - x^{3}}} dx.
\displaystyle \text{Answer:}
\displaystyle \text{17. } I = \int \frac{\sqrt{x}}{\sqrt{a^{3} - x^{3}}} \, dx
\displaystyle \text{Substitute } x^{\frac{3}{2}} = t \Rightarrow \frac{3}{2}x^{\frac{1}{2}} dx = dt
\displaystyle \Rightarrow x^{\frac{1}{2}} dx = \frac{2}{3} dt
\displaystyle \text{and } x^{\frac{3}{2}} = t \Rightarrow (x^{3})^{\frac{1}{2}} = t \Rightarrow x^{3} = t^{2}
\displaystyle \text{Putting the values in } I, \text{ we get}
\displaystyle I = \int \frac{\sqrt{x}}{\sqrt{a^{3} - x^{3}}} \, dx = \frac{2}{3}\int \frac{1}{\sqrt{a^{3} - t^{2}}} \, dt
\displaystyle \text{Using the formula } \int \frac{dx}{\sqrt{a^{2} - x^{2}}} = \sin^{-1}\left(\frac{x}{a}\right) + C
\displaystyle \therefore I = \frac{2}{3}\sin^{-1}\left(\frac{t}{a^{\frac{3}{2}}}\right) + C
\displaystyle \text{Again, putting the value of } t, \text{ we get}
\displaystyle I = \frac{2}{3}\sin^{-1}\left(\frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}}\right) + C
\displaystyle \text{Here, } C \text{ is a constant of integration.}

\displaystyle \text{18. Evaluate } \int_{-1}^{2} |x^{3} - x| dx.
\displaystyle \text{Answer:}
\displaystyle \text{18. Let } I = \int_{-1}^{2} |x^{3} - x| \, dx
\displaystyle \text{Let } f(x) = x^{3} - x
\displaystyle \Rightarrow f(x) = x(x - 1)(x + 1)
\displaystyle \text{The signs of } f(x) \text{ for different values are as follows:}
\displaystyle f(x) > 0 \text{ for all } x \in (-1,0) \cup (1,2)
\displaystyle f(x) < 0 \text{ for all } x \in (0,1)
\displaystyle \text{Therefore, } |x^{3} - x| = \begin{cases} x^{3} - x, & x \in (-1,0) \cup (1,2) \\ -(x^{3} - x), & x \in (0,1) \end{cases}
\displaystyle \therefore I = \int_{-1}^{2} |x^{3} - x| \, dx
\displaystyle = \int_{-1}^{0} |x^{3} - x| \, dx + \int_{0}^{1} |x^{3} - x| \, dx + \int_{1}^{2} |x^{3} - x| \, dx
\displaystyle = \int_{-1}^{0} (x^{3} - x) \, dx - \int_{0}^{1} (x^{3} - x) \, dx + \int_{1}^{2} (x^{3} - x) \, dx
\displaystyle = \left[\frac{x^{4}}{4} - \frac{x^{2}}{2}\right]_{-1}^{0} - \left[\frac{x^{4}}{4} - \frac{x^{2}}{2}\right]_{0}^{1} + \left[\frac{x^{4}}{4} - \frac{x^{2}}{2}\right]_{1}^{2}
\displaystyle = -\left(\frac{1}{4} - \frac{1}{2}\right) - \left(\frac{1}{4} - \frac{1}{2}\right) + \left(\frac{16}{4} - \frac{4}{2}\right) - \left(\frac{1}{4} - \frac{1}{2}\right)
\displaystyle = \frac{3}{4} + (4 - 2) = \frac{11}{4}

\displaystyle \text{19. Find the particular solution of the differential equation}
\displaystyle (1 - y^{2})(1 + \log x)\, dx + 2xy\, dy = 0, \text{ given that } y = 0   \text{ when } x = 1.
\displaystyle \text{Answer:}
\displaystyle \text{19. The given differential equation is}
\displaystyle (1 - y^{2})(1 + \log x)\,dx + 2xy\,dy = 0
\displaystyle \text{Separate the variables}
\displaystyle \Rightarrow (1 - y^{2})(1 + \log x)\,dx = -2xy\,dy
\displaystyle \Rightarrow \left(\frac{1 + \log x}{2x}\right) dx = -\left(\frac{y}{1 - y^{2}}\right) dy
\displaystyle \text{Integrating both sides}
\displaystyle \int \frac{1 + \log x}{2x} \, dx = -\int \frac{y}{1 - y^{2}} \, dy
\displaystyle \text{Substitute } 1 + \log x = t \text{ and } 1 - y^{2} = p
\displaystyle \Rightarrow \frac{1}{x} dx = dt \text{ and } -2y\,dy = dp
\displaystyle \text{Therefore, it becomes}
\displaystyle \int \frac{t}{2} \, dt = \int \frac{1}{2p} \, dp
\displaystyle \Rightarrow \frac{t^{2}}{4} = \frac{\log p}{2} + C
\displaystyle \text{Substituting the values of } t \text{ and } p, \text{ we get}
\displaystyle \frac{(1 + \log x)^{2}}{4} = \frac{\log(1 - y^{2})}{2} + C
\displaystyle \text{At } x = 1 \text{ and } y = 0, \text{ the above equation becomes}
\displaystyle C = \frac{1}{4}
\displaystyle \text{Substituting the value of } C, \text{ we get}
\displaystyle \frac{(1 + \log x)^{2}}{4} = \frac{\log(1 - y^{2})}{2} + \frac{1}{4}
\displaystyle \Rightarrow (1 + \log x)^{2} = 2\log(1 - y^{2}) + 1
\displaystyle \Rightarrow 1 + (\log x)^{2} + 2\log x = 2\log(1 - y^{2}) + 1
\displaystyle \Rightarrow (\log x)^{2} + 2\log x = 2\log(1 - y^{2})
\displaystyle \text{This is the required particular solution.}

\displaystyle \textbf{SECTION - C}
\displaystyle \textbf{Question numbers 20 to 26 carry 6 Mark each}

\displaystyle \text{20. Find the coordinate of the point P where the line through } A(3, -4, -5) \text{ and }
\displaystyle B(2, -3, 1) \text{ crosses the plane passing } \text{through three points } L(2, 2, 1),  M(3, 0, 1) \text{ and }
\displaystyle N(4, -1, 0).   \text{Also, find the ratio in which P divides the line segment AB.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through three given points can be given by}
\displaystyle \left|\begin{matrix} x - 2 & y - 2 & z - 1 \\ 3 - 2 & 0 - 2 & 1 - 1 \\ 4 - 2 & -1 - 2 & 0 - 1 \end{matrix}\right| = 0
\displaystyle \Rightarrow \left|\begin{matrix} x - 2 & y - 2 & z - 1 \\ 1 & -2 & 0 \\ 2 & -3 & -1 \end{matrix}\right| = 0
\displaystyle \text{Solving the above determinant, we get}
\displaystyle (x - 2)(2 - 0) - (y - 2)(-1 - 0) + (z - 1)(-3 + 4) = 0
\displaystyle \Rightarrow (2x - 4) + (y - 2) + (z - 1) = 0
\displaystyle \Rightarrow 2x + y + z - 7 = 0
\displaystyle \text{Therefore, the equation of the plane is } 2x + y + z - 7 = 0
\displaystyle \text{Now, the equation of the line passing through two given points is}
\displaystyle \frac{x - 3}{2 - 3} = \frac{y + 4}{-3 + 4} = \frac{z + 5}{1 + 5} = \lambda
\displaystyle \Rightarrow \frac{x - 3}{-1} = \frac{y + 4}{1} = \frac{z + 5}{6} = \lambda
\displaystyle \Rightarrow x = -\lambda + 3,\; y = \lambda - 4,\; z = 6\lambda - 5
\displaystyle \text{At the point of intersection these points satisfy the equation of the plane } 2x + y + z - 7 = 0
\displaystyle \text{Putting the values of } x, y \text{ and } z \text{ in the equation of the plane, we get}
\displaystyle 2(-\lambda + 3) + (\lambda - 4) + (6\lambda - 5) - 7 = 0
\displaystyle \Rightarrow -2\lambda + 6 + \lambda - 4 + 6\lambda - 5 - 7 = 0
\displaystyle \Rightarrow 5\lambda - 10 = 0
\displaystyle \Rightarrow \lambda = 2
\displaystyle \text{Thus, the point of intersection is } P(1, -2, 7)
\displaystyle \text{Now, let P divide the line AB in the ratio } m : n
\displaystyle \text{By the section formula, we have}
\displaystyle P = \left(\frac{2m + 3n}{m + n}, \frac{-3m - 4n}{m + n}, \frac{m - 5n}{m + n}\right)
\displaystyle (1, -2, 7) = \left(\frac{2m + 3n}{m + n}, \frac{-3m - 4n}{m + n}, \frac{m - 5n}{m + n}\right)
\displaystyle \text{Equating x-coordinate from both the sides,}
\displaystyle 1 = \frac{2m + 3n}{m + n}
\displaystyle \Rightarrow m + n = 2m + 3n \Rightarrow m + 2n = 0
\displaystyle \Rightarrow \frac{m}{n} = -\frac{2}{1}
\displaystyle \text{Hence, P divides the line segment AB externally in the ratio } 2 : 1

\displaystyle \text{21. An urn contains 3 white and 6 red balls. Four balls are drawn one by one }
\displaystyle \text{with replacement from the urn. Find the probability distribution of the number }
\displaystyle \text{of red balls drawn. Also find mean and variance of the distribution. Also find }
\displaystyle \text{mean and variance of the distribution.}
\displaystyle \text{Answer:}
\displaystyle \text{Let X denote the total number of red balls when four balls are drawn one by one with replacement.}
\displaystyle P(\text{getting a red ball in one draw}) = \frac{6}{9} = \frac{2}{3} = p
\displaystyle P(\text{getting a white ball in one draw}) = 1 - \frac{2}{3} = \frac{1}{3} = q
\displaystyle \text{We know that } P(X = x) = {}^{n}C_{x} p^{x} q^{n-x}
\displaystyle \text{where } p = \text{probability of success, } q = \text{probability of failure}
\displaystyle P(X = 0) = \left(\frac{1}{3}\right)^{4} = \frac{1}{81}
\displaystyle P(X = 1) = \left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^{3} \cdot {}^{4}C_{1} = \frac{8}{81}
\displaystyle P(X = 2) = \left(\frac{2}{3}\right)^{2}\left(\frac{1}{3}\right)^{2} \cdot {}^{4}C_{2} = \frac{24}{81}
\displaystyle P(X = 3) = \left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right) \cdot {}^{4}C_{3} = \frac{32}{81}
\displaystyle P(X = 4) = \left(\frac{2}{3}\right)^{4} = \frac{16}{81}
\displaystyle \text{Using the formula for mean, we have}
\displaystyle \overline{X} = \sum P_{i}X_{i}
\displaystyle \text{Mean}(X) = 0\left(\frac{1}{81}\right) + 1\left(\frac{8}{81}\right) + 2\left(\frac{24}{81}\right) + 3\left(\frac{32}{81}\right) + 4\left(\frac{16}{81}\right)
\displaystyle = \frac{1}{81}(8 + 48 + 96 + 64) = \frac{216}{81} = \frac{8}{3}
\displaystyle \text{Using the formula for variance, we have}
\displaystyle \mathrm{Var}(X) = \sum P_{i}X_{i}^{2} - \left(\sum P_{i}X_{i}\right)^{2}
\displaystyle \mathrm{Var}(X) = \left\{0\left(\frac{1}{81}\right) + 1\left(\frac{8}{81}\right) + 4\left(\frac{24}{81}\right) + 9\left(\frac{32}{81}\right) + 16\left(\frac{16}{81}\right)\right\} - \left(\frac{8}{3}\right)^{2}
\displaystyle = \frac{648}{81} - \frac{64}{9} = \frac{8}{9}
\displaystyle \text{Hence, the mean of the distribution is } \frac{8}{3} \text{ and the variance of the distribution is } \frac{8}{9}

\displaystyle \text{22. A manufacturer produces two products A and B. Both the products are }
\displaystyle \text{processed on two different machines. The available capacity of first machine is }
\displaystyle \text{12 hours and that of second machine is 9 hours per day. Each unit of product A }
\displaystyle \text{requires 3 hours on both machines and each unit of product B requires 2 hours on}
\displaystyle \text {first machine and 1 hour on second requires 2 hours on first machine and 1 hour}
\displaystyle \text{ on second machine. Each unit }   \text{of product A is sold at Rs }7 \text{ profit and B at a profit of }
\displaystyle \text{Rs }4. \text{ Find the production level per day for}   \text{maximum profit graphically.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of units of products A and B be } x \text{ and } y \text{ respectively}
\displaystyle \text{Machine I: A = 3h,\; B = 2h}
\displaystyle \text{Machine II: A = 3h,\; B = 1h}
\displaystyle \text{Total profit (objective function): } Z = 7x + 4y
\displaystyle \text{We have to maximise } Z = 7x + 4y \text{ subject to constraints}
\displaystyle 3x + 2y \leq 12
\displaystyle 3x + y \leq 9
\displaystyle x \geq 0,\; y \geq 0
\displaystyle \text{Corner points of the feasible region are } (0,6),\; (2,3),\; (3,0)\displaystyle \text{Value of objective function at corner points:}
\displaystyle Z(0,6) = 24
\displaystyle Z(2,3) = 26
\displaystyle Z(3,0) = 21
\displaystyle \text{Maximum value of } Z = 26 \text{ at } (2,3)
\displaystyle \text{Therefore, the manufacturer should produce 2 units of A and 3 units of B for maximum profit of Rs } 26

\displaystyle \text{23. Let } f : N \rightarrow N \text{ be a function defined as } f(x) = 9x^{2} + 6x - 5.
\displaystyle \text{Show that } f : N \rightarrow S, \text{ where S is the range of f, is invertible.}
\displaystyle \text{Find the inverse of f and hence find } f^{-1}(43) \text{ and } f^{-1}(163).
\displaystyle \text{Answer:}
\displaystyle \text{Let } y = 9x^{2} + 6x - 5
\displaystyle \text{Complete the square:}
\displaystyle y = (9x^{2} + 6x + 1) - 1 - 5
\displaystyle \Rightarrow y = (3x + 1)^{2} - 6
\displaystyle \text{To show } f \text{ is one-one, let } f(x_{1}) = f(x_{2})
\displaystyle (3x_{1} + 1)^{2} - 6 = (3x_{2} + 1)^{2} - 6
\displaystyle \Rightarrow (3x_{1} + 1)^{2} = (3x_{2} + 1)^{2}
\displaystyle \Rightarrow 3x_{1} + 1 = 3x_{2} + 1
\displaystyle \Rightarrow x_{1} = x_{2}
\displaystyle \therefore f \text{ is one-one}
\displaystyle \text{To show } f \text{ is onto, let } y \in \text{Range}(f)
\displaystyle y = (3x + 1)^{2} - 6
\displaystyle \Rightarrow (3x + 1)^{2} = y + 6
\displaystyle \Rightarrow 3x + 1 = \sqrt{y + 6}
\displaystyle \Rightarrow x = \frac{\sqrt{y + 6} - 1}{3}
\displaystyle \text{Thus, for every } y \text{ there exists } x
\displaystyle \therefore f \text{ is onto}
\displaystyle \text{Hence, } f \text{ is invertible}
\displaystyle f^{-1}(y) = \frac{\sqrt{y + 6} - 1}{3}
\displaystyle f^{-1}(43) = \frac{\sqrt{43 + 6} - 1}{3} = 2
\displaystyle f^{-1}(163) = \frac{\sqrt{163 + 6} - 1}{3} = 4

\displaystyle \text{24. Prove that } \left| \begin{matrix} yz - x^{2} & zx - y^{2} & xy - z^{2} \\ zx - y^{2} & xy - z^{2} & yz - x^{2} \\ xy - z^{2} & yz - x^{2} & zx - y^{2} \end{matrix} \right| \text{ is divisible by } (x + y + z)
\displaystyle  \text{ and hence find the quotient.}
\displaystyle \text{OR}
\displaystyle \text{Using elementary transformations, find the inverse of the}   \text{matrix } A = \begin{bmatrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix}
\displaystyle \text{and use it to solve the following system of linear equations:}
\displaystyle 8x + 4y + 3z = 19, \quad 2x + y + z = 5, \quad x + 2y + 2z = 7.
\displaystyle \text{Answer:}
\displaystyle \text{Let } \Delta = \left|\begin{matrix} yz - x^{2} & zx - y^{2} & xy - z^{2} \\ zx - y^{2} & xy - z^{2} & yz - x^{2} \\ xy - z^{2} & yz - x^{2} & zx - y^{2} \end{matrix}\right|
\displaystyle \text{Applying } C_{1} \rightarrow C_{1} + C_{2} + C_{3}, \text{ we get}
\displaystyle \Delta = \left|\begin{matrix} xy + yz + zx - x^{2} - y^{2} - z^{2} & zx - y^{2} & xy - z^{2} \\ xy + yz + zx - x^{2} - y^{2} - z^{2} & xy - z^{2} & yz - x^{2} \\ xy + yz + zx - x^{2} - y^{2} - z^{2} & yz - x^{2} & zx - y^{2} \end{matrix}\right|
\displaystyle \Rightarrow \Delta = (xy + yz + zx - x^{2} - y^{2} - z^{2}) \left|\begin{matrix} 1 & zx - y^{2} & xy - z^{2} \\ 1 & xy - z^{2} & yz - x^{2} \\ 1 & yz - x^{2} & zx - y^{2} \end{matrix}\right|
\displaystyle \text{Applying } R_{2} \rightarrow R_{2} - R_{1} \text{ and } R_{3} \rightarrow R_{3} - R_{1}, \text{ we get}
\displaystyle \Delta = (xy + yz + zx - x^{2} - y^{2} - z^{2}) \left|\begin{matrix} 1 & zx - y^{2} & xy - z^{2} \\ 0 & (x + y + z)(y - z) & (x + y + z)(z - x) \\ 0 & (x + y + z)(y - x) & (x + y + z)(z - y) \end{matrix}\right|
\displaystyle \Rightarrow \Delta = (x + y + z)^{2}(xy + yz + zx - x^{2} - y^{2} - z^{2}) \left|\begin{matrix} 1 & zx - y^{2} & xy - z^{2} \\ 0 & y - z & z - x \\ 0 & y - x & z - y \end{matrix}\right|
\displaystyle \text{Expanding along the first column,}
\displaystyle \Delta = (x + y + z)^{2}(xy + yz + zx - x^{2} - y^{2} - z^{2})\left[(y - z)(z - y) - (z - x)(y - x)\right]
\displaystyle \Rightarrow \Delta = (x + y + z)^{2}(xy + yz + zx - x^{2} - y^{2} - z^{2})^{2}
\displaystyle \text{We can see that } (x + y + z) \text{ is a factor of } \Delta
\displaystyle \text{So, } \Delta \text{ is divisible by } (x + y + z)
\displaystyle \text{The quotient when } \Delta \text{ is divided by } (x + y + z) \text{ is}
\displaystyle (x + y + z)(xy + yz + zx - x^{2} - y^{2} - z^{2})^{2}
\displaystyle \text{OR}
\displaystyle \text{Using elementary row transformations to find the inverse of } A
\displaystyle A = \begin{bmatrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix}
\displaystyle A = IA
\displaystyle \left[\begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix}\right] = \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]A
\displaystyle \text{Applying } R_{1} \leftrightarrow R_{3}, \text{ we get}
\displaystyle \left[\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 8 & 4 & 3 \end{matrix}\right] = \left[\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}\right]A
\displaystyle \text{Applying } R_{2} \rightarrow R_{2} - 2R_{1}, \text{ we get}
\displaystyle \left[\begin{matrix} 1 & 2 & 2 \\ 0 & -3 & -3 \\ 8 & 4 & 3 \end{matrix}\right] = \left[\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & -2 \\ 1 & 0 & 0 \end{matrix}\right]A
\displaystyle \text{Applying } R_{3} \rightarrow R_{3} - 8R_{1}, \text{ we get}
\displaystyle \left[\begin{matrix} 1 & 2 & 2 \\ 0 & -3 & -3 \\ 0 & -12 & -13 \end{matrix}\right] = \left[\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & -2 \\ 1 & 0 & -8 \end{matrix}\right]A
\displaystyle \text{Applying } R_{2} \rightarrow -\frac{1}{3}R_{2}, \text{ we get}
\displaystyle \left[\begin{matrix} 1 & 2 & 2 \\ 0 & 1 & 1 \\ 0 & -12 & -13 \end{matrix}\right] = \left[\begin{matrix} 0 & 0 & 1 \\ 0 & -\frac{1}{3} & \frac{2}{3} \\ 1 & 0 & -8 \end{matrix}\right]A
\displaystyle \text{Applying } R_{1} \rightarrow R_{1} - 2R_{2}, \text{ we get}
\displaystyle \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -12 & -13 \end{matrix}\right] = \left[\begin{matrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 0 & -\frac{1}{3} & \frac{2}{3} \\ 1 & 0 & -8 \end{matrix}\right]A
\displaystyle \text{Applying } R_{3} \rightarrow R_{3} + 12R_{2}, \text{ we get}
\displaystyle \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & -1 \end{matrix}\right] = \left[\begin{matrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 0 & -\frac{1}{3} & \frac{2}{3} \\ 1 & -4 & 0 \end{matrix}\right]A
\displaystyle \text{Applying } R_{3} \rightarrow -R_{3} \text{ and } R_{2} \rightarrow R_{2} - R_{3}, \text{ we get}
\displaystyle \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right] = \left[\begin{matrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 1 & -\frac{13}{3} & \frac{2}{3} \\ -1 & 4 & 0 \end{matrix}\right]A
\displaystyle \text{Thus, we have}
\displaystyle A^{-1} = \begin{bmatrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 1 & -\frac{13}{3} & \frac{2}{3} \\ -1 & 4 & 0 \end{bmatrix}
\displaystyle \text{The given system of equations is}
\displaystyle 8x + 4y + 3z = 19
\displaystyle 2x + y + z = 5
\displaystyle x + 2y + 2z = 7
\displaystyle \text{The given system can be written as } AX = B
\displaystyle \text{where } A = \begin{bmatrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix}, \; X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \; B = \begin{bmatrix} 19 \\ 5 \\ 7 \end{bmatrix}
\displaystyle \therefore X = A^{-1}B
\displaystyle \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 1 & -\frac{13}{3} & \frac{2}{3} \\ -1 & 4 & 0 \end{bmatrix}\begin{bmatrix} 19 \\ 5 \\ 7 \end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 + \frac{10}{3} - \frac{7}{3} \\ 19 - \frac{65}{3} + \frac{14}{3} \\ -19 + 20 + 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}
\displaystyle \therefore x = 1,\; y = 2 \text{ and } z = 1

\displaystyle \text{25. Show that the altitude of the right circular cone of maximum volume that can }
\displaystyle \text{be inscribed in a sphere of radius r is } \frac{4r}{3}. \text{ Also find maximum volume in terms of}
\displaystyle \text{volume of the sphere.}
\displaystyle \text{OR}
\displaystyle \text{Find the intervals in which } f(x) = \sin 3x - \cos 3x, \; 0 < x < \pi, \text{ is strictly increasing or }
\displaystyle \text{strictly decreasing.}
\displaystyle \text{Answer:}
\displaystyle \text{A sphere of fixed radius } r \text{ is given.}
\displaystyle \text{Let } R \text{ and } h \text{ be the radius and the height of the cone, respectively.}
\displaystyle \text{The volume } (V) \text{ of the cone is given by}\displaystyle V = \frac{1}{3}\pi R^{2}h \qquad \ldots (i)
\displaystyle \text{Now, from the right triangle } BCD, \text{ we have}
\displaystyle BC = \sqrt{r^{2} - R^{2}} \Rightarrow h - r = \sqrt{r^{2} - R^{2}}
\displaystyle \therefore h = r + \sqrt{r^{2} - R^{2}}
\displaystyle \text{Put the value of } h \text{ in } (i)
\displaystyle \therefore V = \frac{1}{3}\pi R^{2}\left(r + \sqrt{r^{2} - R^{2}}\right) = \frac{1}{3}\pi rR^{2} + \frac{1}{3}\pi R^{2}\sqrt{r^{2} - R^{2}}
\displaystyle \text{Differentiating both sides w.r.t. } R
\displaystyle \therefore \frac{dV}{dR} = \frac{2}{3}\pi rR + \frac{2}{3}\pi R\sqrt{r^{2} - R^{2}} + \frac{\pi R^{2}}{3}\left(\frac{-2R}{2\sqrt{r^{2} - R^{2}}}\right)
\displaystyle = \frac{2}{3}\pi rR + \frac{2}{3}\pi R\sqrt{r^{2} - R^{2}} - \frac{\pi R^{3}}{3\sqrt{r^{2} - R^{2}}}
\displaystyle = \frac{2}{3}\pi rR + \frac{2\pi R(r^{2} - R^{2}) - \pi R^{3}}{3\sqrt{r^{2} - R^{2}}}
\displaystyle = \frac{2}{3}\pi rR + \frac{2\pi Rr^{2} - 3\pi R^{3}}{3\sqrt{r^{2} - R^{2}}}
\displaystyle \text{Now, } \frac{dV}{dR} = 0
\displaystyle \Rightarrow \frac{2\pi rR}{3} = \frac{3\pi R^{3} - 2\pi Rr^{2}}{3\sqrt{r^{2} - R^{2}}}
\displaystyle \Rightarrow 2r\sqrt{r^{2} - R^{2}} = 3R^{2} - 2r^{2}
\displaystyle \Rightarrow 4r^{2}(r^{2} - R^{2}) = (3R^{2} - 2r^{2})^{2}
\displaystyle \Rightarrow 4r^{4} - 4r^{2}R^{2} = 9R^{4} + 4r^{4} - 12R^{2}r^{2}
\displaystyle \Rightarrow 9R^{4} - 8r^{2}R^{2} = 0
\displaystyle \Rightarrow 9R^{2} = 8r^{2} \Rightarrow R^{2} = \frac{8r^{2}}{9}
\displaystyle \text{Now, when } R^{2} = \frac{8r^{2}}{9}, \text{ it can be shown that } \frac{d^{2}V}{dR^{2}} < 0
\displaystyle \therefore \text{The volume is maximum when } R^{2} = \frac{8r^{2}}{9}
\displaystyle \text{When } R^{2} = \frac{8r^{2}}{9}, \text{ height of the cone}
\displaystyle h = r + \sqrt{r^{2} - \frac{8r^{2}}{9}} = r + \sqrt{\frac{r^{2}}{9}} = r + \frac{r}{3} = \frac{4r}{3}
\displaystyle \text{Hence, the altitude of a right circular cone of maximum volume}
\displaystyle \text{that can be inscribed in a sphere of radius } r \text{ is } \frac{4r}{3}
\displaystyle \text{Let volume of the sphere be } V_{s} = \frac{4}{3}\pi r^{3}
\displaystyle \therefore \text{Volume of cone, } V = \frac{1}{3}\pi R^{2}h
\displaystyle \Rightarrow R = \frac{2\sqrt{2}}{3}r
\displaystyle V = \frac{1}{3}\pi \left(\frac{2\sqrt{2}}{3}r\right)^{2} \times \frac{4r}{3}
\displaystyle V = \frac{1}{3}\pi \cdot \frac{8r^{2}}{9} \cdot \frac{4r}{3}
\displaystyle V = \frac{32\pi r^{3}}{81} = \frac{32}{81}\pi \left(\frac{3V_{s}}{4\pi}\right)
\displaystyle \therefore \text{Volume of cone in terms of sphere } = \frac{8V_{s}}{27}
\displaystyle \text{OR}
\displaystyle \text{Consider the function } f(x) = \sin 3x - \cos 3x,\; 0 < x < \pi
\displaystyle f'(x) = 3\cos 3x + 3\sin 3x
\displaystyle = 3(\sin 3x + \cos 3x)
\displaystyle = 3\sqrt{2}\left(\sin 3x \cdot \frac{1}{\sqrt{2}} + \cos 3x \cdot \frac{1}{\sqrt{2}}\right)
\displaystyle = 3\sqrt{2}\left(\sin 3x \cos \frac{\pi}{4} + \cos 3x \sin \frac{\pi}{4}\right)
\displaystyle = 3\sqrt{2}\sin\left(3x + \frac{\pi}{4}\right)
\displaystyle \text{For the increasing interval } f'(x) > 0
\displaystyle 3\sqrt{2}\sin\left(3x + \frac{\pi}{4}\right) > 0
\displaystyle \Rightarrow \sin\left(3x + \frac{\pi}{4}\right) > 0
\displaystyle \Rightarrow 0 < 3x + \frac{\pi}{4} < \pi
\displaystyle \Rightarrow -\frac{\pi}{4} < 3x < \frac{3\pi}{4}
\displaystyle \Rightarrow 0 < x < \frac{\pi}{4} \text{ since } 0 < x < \pi
\displaystyle \text{Also, } \sin\left(3x + \frac{\pi}{4}\right) > 0
\displaystyle \text{when } 2\pi < 3x + \frac{\pi}{4} < 3\pi
\displaystyle \Rightarrow \frac{7\pi}{4} < 3x < \frac{11\pi}{4}
\displaystyle \Rightarrow \frac{7\pi}{12} < x < \frac{11\pi}{12}
\displaystyle \text{Therefore, the function is strictly increasing in } \left(0,\frac{\pi}{4}\right) \cup \left(\frac{7\pi}{12},\frac{11\pi}{12}\right)
\displaystyle \text{Similarly, for the decreasing interval } f'(x) < 0
\displaystyle 3\sqrt{2}\sin\left(3x + \frac{\pi}{4}\right) < 0
\displaystyle \Rightarrow \sin\left(3x + \frac{\pi}{4}\right) < 0
\displaystyle \Rightarrow \pi < 3x + \frac{\pi}{4} < 2\pi
\displaystyle \Rightarrow \frac{3\pi}{4} < 3x < \frac{7\pi}{4}
\displaystyle \Rightarrow \frac{\pi}{4} < x < \frac{7\pi}{12}
\displaystyle \text{Also, } \sin\left(3x + \frac{\pi}{4}\right) < 0
\displaystyle \text{when } 3\pi < 3x + \frac{\pi}{4} < 4\pi
\displaystyle \Rightarrow \frac{11\pi}{4} < 3x < \frac{15\pi}{4}
\displaystyle \Rightarrow \frac{11\pi}{12} < x < \frac{15\pi}{12}
\displaystyle \Rightarrow \frac{11\pi}{12} < x < \pi
\displaystyle \text{The function is strictly decreasing in } \left(\frac{\pi}{4},\frac{7\pi}{12}\right) \cup \left(\frac{11\pi}{12},\pi\right)

\displaystyle \text{26. Using integration find the area of the region}
\displaystyle \{(x, y) : x^{2} + y^{2} \leq 2ax,\; y^{2} \geq ax,\; x, y \geq 0\}.
\displaystyle \text{Answer:}
\displaystyle \text{Given: } x^{2} + y^{2} \leq 2ax,\; y^{2} \geq ax,\; x,\; y \geq 0
\displaystyle \Rightarrow x^{2} + y^{2} - 2ax \leq 0,\; y^{2} \geq ax,\; x,\; y \geq 0
\displaystyle \Rightarrow x^{2} + y^{2} - 2ax + a^{2} - a^{2} \leq 0,\; y^{2} \geq ax,\; x,\; y \geq 0
\displaystyle \Rightarrow (x - a)^{2} + y^{2} \leq a^{2},\; y^{2} \geq ax,\; x,\; y \geq 0
\displaystyle \text{To find the points of intersection of the circle } (x - a)^{2} + y^{2} = a^{2}\displaystyle \text{and the parabola } y^{2} = ax, \text{ we substitute } y^{2} = ax \text{ in } (x - a)^{2} + y^{2} = a^{2}
\displaystyle \Rightarrow (x - a)^{2} + ax = a^{2}
\displaystyle \Rightarrow x^{2} - 2ax + a^{2} + ax = a^{2}
\displaystyle \Rightarrow x^{2} - ax = 0
\displaystyle \Rightarrow x(x - a) = 0
\displaystyle \Rightarrow x = 0,\; a
\displaystyle \text{Therefore, the points of intersection are } (0,0) \text{ and } (a,a)
\displaystyle \text{Now, find the area of the shaded region.}
\displaystyle \text{Area of the shaded region from } x = 0 \text{ to } x = a
\displaystyle = \int_{0}^{a} \sqrt{a^{2} - (x - a)^{2}} \, dx - \int_{0}^{a} \sqrt{ax} \, dx
\displaystyle \text{Let } x - a = t \text{ for the first part of the integral}
\displaystyle \Rightarrow dx = dt
\displaystyle \therefore A_{1} = \int_{-a}^{0} \sqrt{a^{2} - t^{2}} \, dt - \sqrt{a}\int_{0}^{a} x^{\frac{1}{2}} \, dx
\displaystyle = \left[\frac{t}{2}\sqrt{a^{2} - t^{2}} + \frac{a^{2}}{2}\sin^{-1}\left(\frac{t}{a}\right)\right]_{-a}^{0} - \sqrt{a}\left[\frac{2}{3}x^{\frac{3}{2}}\right]_{0}^{a}
\displaystyle = \left[0 + 0\right] - \left[0 - \frac{\pi a^{2}}{4}\right] - \frac{2a^{2}}{3}
\displaystyle = \frac{\pi a^{2}}{4} - \frac{2a^{2}}{3}
\displaystyle = \left(\frac{\pi}{4} - \frac{2}{3}\right)a^{2}
\displaystyle \therefore \text{Area of the shaded region } = \left(\frac{\pi}{4} - \frac{2}{3}\right)a^{2} \text{ square units.}

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