\displaystyle \textbf{Time Allowed : 3 Hours} \hfill \textbf{Maximum Marks : 80}
\displaystyle \textbf{General Instructions:}
\displaystyle \text{Read the following instructions very carefully and strictly follow them:}
\displaystyle \text{(i) This question paper comprises four Sections A, B, C and D. This question}
\displaystyle \text{paper carries 36 questions. All questions are compulsory.}
\displaystyle \text{(ii) Section A: Questions no. 1 to 20 comprises of 20 questions of 1 mark each.}
\displaystyle \text{(iii) Section B: Questions no. 21 to 26 comprises of 6 questions of 2 marks each.}
\displaystyle \text{(iv) Section C: Questions no. 27 to 32 comprises of 6 questions of 4 marks each.}
\displaystyle \text{(v) Section D: Questions no. 33 to 36 comprises of 4 questions of 6 marks each.}
\displaystyle \text{(vi) There is no overall choice in the question paper. However, an internal choice}
\displaystyle \text{has been provided in 3 questions of one mark, 2 questions of two marks,}
\displaystyle \text{2 questions of four marks and 2 questions of six marks. Only one of the choices}
\displaystyle \text{in such questions have to be attempted.}
\displaystyle \text{(vii) In addition to this, separate instructions are given with each section and}
\displaystyle \text{question, wherever necessary.}
\displaystyle \text{(viii) Use of calculators is not permitted.}

\displaystyle \textbf{SECTION - A}
\displaystyle \text{Question numbers 1 to 20 carry 1 mark each.}
\displaystyle \text{Question numbers 1 to 10 are multiple choice type questions.}
\displaystyle \text{Select the correct option.}

\displaystyle \textbf{1. } \text{If } A \text{ is a square matrix of order 3 and } |A| = 5, \text{ then the} \text{value of } |2A| \text{ is}
\displaystyle \text{(a) } -10 \qquad \text{(b) } 10 \qquad \text{(c) } -40 \qquad \text{(d) } 40
\displaystyle \text{Answer:}
\displaystyle \text{(d) } |A| = 5
\displaystyle \text{We know that } |A| = |A|'
\displaystyle \therefore |A|' = 5
\displaystyle |2A|' = 2^{n} |A|'
\displaystyle = 2^{3} \times 5 \quad (n = 3)
\displaystyle = 8 \times 5 = 40

\displaystyle \textbf{2. } \text{If } A \text{ is a square matrix such that } A^{2} = A, \text{ then}  (I - A)^{3} + A \text{ is equal to}
\displaystyle \text{(a) } I \qquad \text{(b) } 0 \qquad \text{(c) } I - A \qquad \text{(d) } I + A
\displaystyle \text{Answer:}
\displaystyle \text{(a) } A^{2} = A \quad (\text{given})
\displaystyle (I - A)^{3} + A = I^{3} - A^{3} - 3IA(I - A) + A
\displaystyle \text{[using } (a - b)^{3} = a^{3} - b^{3} - 3ab(a - b)\text{]}
\displaystyle = I - A^{2}A - 3I^{2}A + 3IA^{2} + A \quad (\because I^{3} = I)
\displaystyle = I - AA A - 3IA + 3A^{2} + A \quad (\because I^{2} = I, IA = A)
\displaystyle = I - A^{2} - 3A + 3A + A
\displaystyle = I - A + A = I

\displaystyle \textbf{3. } \text{The principal value of } \tan^{-1}\left(\tan \frac{3\pi}{5}\right) \text{ is}
\displaystyle \text{(a) } \frac{2\pi}{5} \qquad \text{(b) } \frac{-2\pi}{5}
\displaystyle \text{(c) } \frac{3\pi}{5} \qquad \text{(d) } \frac{-3\pi}{5}
\displaystyle \text{Answer:}
\displaystyle \text{(b) Let } y = \tan^{-1}\left(\tan \frac{3\pi}{5}\right)
\displaystyle \Rightarrow \tan y = \tan \frac{3\pi}{5}
\displaystyle \Rightarrow \tan y = \tan \left(\pi - \frac{2\pi}{5}\right)
\displaystyle \Rightarrow \tan y = -\tan \frac{2\pi}{5} \quad [\because \tan(\pi - \theta) = -\tan \theta]
\displaystyle \Rightarrow \tan y = \tan \left(-\frac{2\pi}{5}\right) \quad [\because \tan(-\theta) = -\tan \theta]
\displaystyle \Rightarrow y = -\frac{2\pi}{5}

\displaystyle \textbf{4. } \text{If the projection of } \overrightarrow{a} = i - 2j + 3k \text{ on }  \overrightarrow{b} = 2i + \lambda k \text{ is zero, then the value of } \lambda \text{ is}
\displaystyle \text{(a) } 0 \qquad \text{(b) } 1 \qquad \text{(c) } \frac{-2}{3} \qquad \text{(d) } \frac{-3}{2}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Projection of } \overrightarrow{a} \text{ on } \overrightarrow{b} = 0 \quad (\text{given})
\displaystyle \text{Projection of } \overrightarrow{a} \text{ on } \overrightarrow{b} = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|}
\displaystyle \therefore \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{b}|} = 0
\displaystyle \Rightarrow \overrightarrow{a} \cdot \overrightarrow{b} = 0
\displaystyle \Rightarrow (i - 2j + 3k) \cdot (2i + \lambda k) = 0
\displaystyle \Rightarrow 2 + 0 + 3\lambda = 0
\displaystyle \Rightarrow 2 + 3\lambda = 0
\displaystyle \Rightarrow 3\lambda = -2
\displaystyle \Rightarrow \lambda = \frac{-2}{3}

\displaystyle \textbf{5. } \text{The vector equation of the line passing through the point } (-1, 5, 4) \text{ and}
\displaystyle  \text{perpendicular to the plane } z = 0 \text{ is}
\displaystyle \text{(a) } \overrightarrow{r} = -i + 5j + 4k + \lambda (i + j)
\displaystyle \text{(b) } \overrightarrow{r} = -i + 5j + (4 + \lambda)k
\displaystyle \text{(c) } \overrightarrow{r} = i - 5j - 4k + \lambda k
\displaystyle \text{(d) } \overrightarrow{r} = \lambda k
\displaystyle \text{Answer:}
\displaystyle \text{(b) } \text{Line is perpendicular to } z = 0 \text{ plane.}
\displaystyle \text{So it will be along } \overrightarrow{k} \text{ and passes through } (-1, 5, 4)
\displaystyle \therefore \overrightarrow{r} = -\overrightarrow{i} + 5\overrightarrow{j} + 4\overrightarrow{k} + \lambda \overrightarrow{k}
\displaystyle \Rightarrow \overrightarrow{r} = -\overrightarrow{i} + 5\overrightarrow{j} + (4 + \lambda)\overrightarrow{k}

\displaystyle \textbf{6. } \text{The number of arbitrary constants in the particular solution of a differential }
\displaystyle \text{equation of second order is (are)}
\displaystyle \text{(a) } 0 \qquad \text{(b) } 1 \qquad \text{(c) } 2 \qquad \text{(d) } 3
\displaystyle \text{Answer:}
\displaystyle \text{(a) } \text{In particular solution of any differential equation, arbitrary constants are removed}
\displaystyle \text{by substituting particular values. Hence } 0 \text{.}

\displaystyle \textbf{7. } \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^{2} x \, dx \text{ is equal to}
\displaystyle \text{(a) } -1 \qquad \text{(b) } 0 \qquad \text{(c) } 1 \qquad \text{(d) } 2
\displaystyle \text{Answer:}
\displaystyle \text{(d) } \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^{2} x \, dx = [\tan x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}
\displaystyle = \tan \frac{\pi}{4} - \tan \left(-\frac{\pi}{4}\right)
\displaystyle = 1 - (-1) \quad [\because \tan(-\theta) = -\tan \theta]
\displaystyle = 1 + 1 = 2

\displaystyle \textbf{8. } \text{The length of the perpendicular drawn from the point }  (4, -7, 3) \text{ on the } y\text{-axis is}
\displaystyle \text{(a) } 3 \text{ units} \qquad \text{(b) } 4 \text{ units}
\displaystyle \text{(c) } 5 \text{ units} \qquad \text{(d) } 7 \text{ units}
\displaystyle \text{Answer:}
\displaystyle \text{(c) } \text{Let the point on the } y\text{-axis be } A(0, y, 0) \text{ and } B(4, -7, 3).
\displaystyle \text{As the perpendicular is drawn from } B \text{ to the } y\text{-axis, the } y\text{-coordinate}
\displaystyle \text{of } A \text{ will be the same as that of } B \text{, i.e. } y = -7.
\displaystyle \therefore A(0, -7, 0)
\displaystyle \text{Distance } AB = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}}
\displaystyle \Rightarrow AB = \sqrt{(0 - 4)^{2} + (-7 + 7)^{2} + (0 - 3)^{2}}
\displaystyle \Rightarrow AB = \sqrt{16 + 0 + 9} = \sqrt{25} = 5
\displaystyle \Rightarrow AB = 5 \text{ units}

\displaystyle \textbf{9. } \text{If } A \text{ and } B \text{ are two independent events with } P(A) = \frac{1}{3} \text{and } P(B) = \frac{1}{4}\text{, then }
\displaystyle P(B'|A) \text{ is equal to}
\displaystyle \text{(a) } \frac{1}{4} \qquad \text{(b) } \frac{1}{3} \qquad \text{(c) } \frac{3}{4} \qquad \text{(d) } 1
\displaystyle \text{Answer:}
\displaystyle \text{(c) } P(B'|A) = \frac{P(B' \cap A)}{P(A)} = \frac{P(B')P(A)}{P(A)}
\displaystyle \text{[since } A \text{ and } B \text{ are independent events]}
\displaystyle = P(B') = 1 - P(B)
\displaystyle = 1 - \frac{1}{4} = \frac{3}{4}

\displaystyle \textbf{10. } \text{The corner points of the feasible region determined by the system of linear}
\displaystyle \text{inequalities are } (0, 0), (4, 0), (2, 4) \text{ and } (0, 5) \text{. If the maximum value of }
\displaystyle z = ax + by, \text{ where } a, b > 0  \text{occurs at both } (2, 4) \text{ and } (4, 0) \text{, then}
\displaystyle \text{(a) } a = 2b \qquad \text{(b) } 2a = b \qquad \text{(c) } a = b \qquad \text{(d) } 3a = b
\displaystyle \text{Answer:}
\displaystyle \text{(a) } Z = ax + by
\displaystyle \text{Maximum value at } (2, 4) \text{ is}
\displaystyle Z = a(2) + b(4) = 2a + 4b \qquad \ldots (1)
\displaystyle \text{Maximum value at } (4, 0) \text{ is}
\displaystyle Z = a(4) + b(0)
\displaystyle Z = 4a \qquad \ldots (2)
\displaystyle \text{Maximum value occurs at both points.}
\displaystyle \text{So, both values should be equal.}
\displaystyle 2a + 4b = 4a \qquad \text{[from (1) and (2)]}
\displaystyle \Rightarrow 4b = 4a - 2a
\displaystyle \Rightarrow 4b = 2a
\displaystyle \Rightarrow a = 2b

\displaystyle \textbf{Fill in the blanks in question numbers 11 to 15.}

\displaystyle \textbf{11. } \text{A relation } R \text{ in a set } A \text{ is called } \underline{\qquad\qquad} \text{, if}  (a_{1}, a_{2}) \in R \text{ implies } (a_{2}, a_{1}) \in R \text{, for all } a_{1}, a_{2} \in A \text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Symmetric}

\displaystyle \textbf{12. } \text{The greatest integer function defined by}  f(x) = [x], \, 0 < x < 2 \text{ is not} \\ \text{differentiable at } x = \underline{\qquad\qquad}.
\displaystyle \text{Answer:}
\displaystyle \text{At } \mathrm{L.H.D.} = \lim_{h \to 0} \frac{f(x) - f(x - h)}{h}
\displaystyle = \lim_{h \to 0} \frac{f(1) - f(1 - h)}{h} = \lim_{h \to 0} \frac{[1] - [(1 - h)]}{h}
\displaystyle = \lim_{h \to 0} \frac{1 - 0}{h} = \lim_{h \to 0} \frac{1}{h} = \frac{1}{0}, \text{ not defined.}
\displaystyle \mathrm{R.H.D.} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
\displaystyle = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0} \frac{[(1 + h)] - [1]}{h}
\displaystyle = \lim_{h \to 0} \frac{1 - 1}{h} = \lim_{h \to 0} \frac{0}{h} = \lim_{h \to 0} 0 = 0
\displaystyle \mathrm{L.H.D.} \neq \mathrm{R.H.D.}
\displaystyle \therefore f(x) \text{ is not differentiable at } x = 1

\displaystyle \textbf{13. } \text{If } A \text{ is a matrix of order } 3 \times 2 \text{, then the order of the matrix }  A' \text{ is } \underline{\qquad\qquad}.

\displaystyle \text{OR}

\displaystyle \text{A square matrix } A \text{ is said to be skew-symmetric, if}  \underline{\qquad\qquad}.
\displaystyle \text{Answer:}
\displaystyle  [2 \times 3]

\displaystyle \text{OR}

\displaystyle [A^{T} = -A]

\displaystyle \textbf{14. } \text{The equation of the normal to the curve } y^{2} = 8x \text{ at the origin }  \text{is } \underline{\qquad\qquad}.

\displaystyle \text{OR}

\displaystyle \text{The radius of a circle is increasing at the uniform rate of } 3 \text{ cm/sec. At the instant when} \\ \text{the radius of the circle is } 2 \text{ cm, its area increases at the rate of } \underline{\qquad\qquad} \text{ cm}^{2}\text{/s.}
\displaystyle \text{Answer:}
\displaystyle [y = 0] \text{ for } y^{2} = 8x
\displaystyle \text{Differentiating on both sides,}
\displaystyle \frac{d}{dx}(y^{2}) = \frac{d}{dx}(8x)
\displaystyle 2y \frac{dy}{dx} = 8
\displaystyle \frac{dy}{dx} = \frac{8}{2y} = \frac{4}{y}
\displaystyle \text{Slope of normal } = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{\frac{4}{y}} = -\frac{y}{4}
\displaystyle \text{At origin } (0, 0) \text{, slope of normal } = 0
\displaystyle \therefore \text{Equation of normal } \frac{y - 0}{x - 0} = 0
\displaystyle \Rightarrow y = 0

\displaystyle \text{OR}

\displaystyle [12\pi]
\displaystyle \text{Area of circle } = \pi r^{2}
\displaystyle A = \pi r^{2}
\displaystyle \frac{dA}{dt} = \pi (2r) \frac{dr}{dt} \text{ [since } \frac{d}{dx}(x^{n}) = nx^{n-1} \text{]}
\displaystyle = 2\pi r \frac{dr}{dt}
\displaystyle = 2\pi (2)(3)
\displaystyle = 12\pi \text{ cm}^{2}\text{/s}
\displaystyle \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \qquad \ldots (1)
\displaystyle \text{As radius is increasing at uniform rate of } 3 \text{ cm/sec,}
\displaystyle \frac{dr}{dt} = 3 \text{ cm/sec}
\displaystyle \text{From (1), } \frac{dA}{dt} = 2\pi r (3) = 6\pi r
\displaystyle \text{At } r = 2,
\displaystyle \frac{dA}{dt} = 6\pi (2) = 12\pi \text{ cm}^{2}\text{/sec}

\displaystyle \textbf{15. } \text{The position vectors of two points } A \text{ and } B \text{ are} \overrightarrow{OA} = 2\overrightarrow{i} - \overrightarrow{j} - \overrightarrow{k} \text{ and } \\  \overrightarrow{OB} = 2\overrightarrow{i} - \overrightarrow{j} + 2\overrightarrow{k} \text{, respectively.}
\displaystyle \text{The position vector of a point } P \text{ which divides the line segment}  \text{joining } A \text{ and } B \\ \text{ in the ratio } 2 : 1 \text{ is } \underline{\qquad\qquad}.
\displaystyle \text{Answer:}
\displaystyle [2\overrightarrow{i} - \overrightarrow{j} + \overrightarrow{k}] \quad \text{Point } A(2, -1, -1), \text{ Point } B(2, -1, 2)
\displaystyle \text{By section formula,}
\displaystyle P = \left(\frac{mx_{2} + nx_{1}}{m + n}, \frac{my_{2} + ny_{1}}{m + n}, \frac{mz_{2} + nz_{1}}{m + n}\right)
\displaystyle = \left(\frac{4 + 2}{2 + 1}, \frac{-2 - 1}{2 + 1}, \frac{4 - 1}{2 + 1}\right) = (2, -1, 1)
\displaystyle \text{Position vector, } \overrightarrow{OP} = 2\overrightarrow{i} - \overrightarrow{j} + \overrightarrow{k}

\displaystyle \textbf{Question numbers 16 to 20 are very short answer type questions.}

\displaystyle \textbf{16. } \text{If } A = \begin{bmatrix} 2 & 0 & 0 \\ -1 & 2 & 3 \\ 3 & 3 & 5 \end{bmatrix} \text{, then find } A(\mathrm{adj}\,A).
\displaystyle \text{Answer:}
\displaystyle A = \begin{bmatrix} 2 & 0 & 0 \\ -1 & 2 & 3 \\ 3 & 3 & 5 \end{bmatrix}
\displaystyle |A| = 2(10 - 9) = 2
\displaystyle A^{-1} = \frac{1}{|A|}\mathrm{adj}(A)
\displaystyle \Rightarrow \mathrm{adj}(A) = |A|A^{-1}
\displaystyle \text{Multiplying both sides by } A,
\displaystyle A(\mathrm{adj}\,A) = |A|AA^{-1}
\displaystyle \Rightarrow A(\mathrm{adj}\,A) = |A|I \qquad [\because AA^{-1} = I]
\displaystyle = 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}

\displaystyle \textbf{17. } \text{Find : } \int x^{4} \log x \, dx
\displaystyle \text{OR}
\displaystyle \text{Find : } \int \frac{2x}{\sqrt[3]{x^{2} + 1}} \, dx
\displaystyle \text{Answer:}
\displaystyle \int x^{4}\log x \, dx
\displaystyle \therefore \int u \, dv = uv - \int \frac{du}{dx}\left(\int v \, dx\right) dx
\displaystyle \Rightarrow \int \log x \cdot x^{4} \, dx = \log x \int x^{4} \, dx - \int \left[\frac{d}{dx}(\log x)\int x^{4} \, dx\right] dx
\displaystyle = \log x \left(\frac{x^{5}}{5}\right) - \int \frac{1}{x} \cdot \frac{x^{5}}{5} \, dx \qquad \left[\because \frac{d}{dx}\log x = \frac{1}{x}\right]
\displaystyle = \frac{x^{5}}{5}\log x - \frac{1}{5}\int x^{4} \, dx
\displaystyle = \frac{x^{5}}{5}\log x - \frac{1}{5}\cdot \frac{x^{5}}{5} + C
\displaystyle = \frac{x^{5}}{5}\log x - \frac{x^{5}}{25} + C
\displaystyle = \frac{x^{5}}{5}\left(\log x - \frac{1}{5}\right) + C
\displaystyle \text{OR}
\displaystyle \int \frac{2x}{(x^{2} + 1)^{3/2}} \, dx
\displaystyle \text{Let } x^{2} + 1 = t
\displaystyle \Rightarrow 2x \, dx = dt
\displaystyle \Rightarrow \int \frac{dt}{t^{3/2}} + C
\displaystyle = \int t^{-3/2} \, dt + C
\displaystyle = \frac{t^{-1/2}}{-\frac{1}{2}} + C \qquad \left[\because \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C\right]
\displaystyle = -2t^{-1/2} + C
\displaystyle = \frac{-2}{\sqrt{x^{2} + 1}} + C

\displaystyle \textbf{18. } \text{Evaluate: } \int_{1}^{3} |2x - 1| \, dx
\displaystyle \text{Answer:}
\displaystyle \int_{1}^{3} |2x - 1| \, dx
\displaystyle \text{For } x \in [1, 3], \, 2x - 1 > 0
\displaystyle \therefore \int_{1}^{3} |2x - 1| \, dx = \int_{1}^{3} (2x - 1) \, dx
\displaystyle = \left[\frac{2x^{2}}{2} - x\right]_{1}^{3}
\displaystyle = [x^{2} - x]_{1}^{3} = (9 - 3) - (1 - 1) = 6

\displaystyle \textbf{19. } \text{Two cards are drawn at random and one-by-one without replacement from a}
\displaystyle \text{well-shuffled pack of 52 playing cards. Find the probability that one card is red and the }
\displaystyle \text{other is black.}
\displaystyle \text{Answer:}
\displaystyle \text{Two cards are drawn one by one without replacement.}
\displaystyle \text{Number of red cards } = 26, \quad \text{Number of black cards } = 26
\displaystyle \text{Total number of cards } = 52
\displaystyle P(RB) = \text{Probability that first card is red and second is black}  = \frac{26}{52} \times \frac{26}{51}
\displaystyle P(BR) = \text{Probability that first card is black and second is red}  = \frac{26}{52} \times \frac{26}{51}
\displaystyle \therefore P(\text{one red and one black}) = P(RB) + P(BR)
\displaystyle = \frac{26}{52} \times \frac{26}{51} + \frac{26}{52} \times \frac{26}{51} = \frac{2 \times 26 \times 26}{52 \times 51} = \frac{26}{51}

\displaystyle \textbf{20. } \text{Find: } \int \frac{dx}{\sqrt{9 - 4x^{2}}}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{\sqrt{9 - 4x^{2}}}
\displaystyle I = \frac{1}{2} \int \frac{dx}{\sqrt{\left(\frac{3}{2}\right)^{2} - x^{2}}}
\displaystyle \text{Using } \int \frac{dx}{\sqrt{a^{2} - x^{2}}} = \sin^{-1}\left(\frac{x}{a}\right) + C
\displaystyle I = \frac{1}{2}\sin^{-1}\left(\frac{x}{\frac{3}{2}}\right) + C
\displaystyle = \frac{1}{2}\sin^{-1}\left(\frac{2x}{3}\right) + C

\displaystyle \textbf{SECTION - B}
\displaystyle \text{Question numbers 21 to 26 carry 2 marks each.}

\displaystyle \textbf{21. } \text{Prove that } \sin^{-1}(2x\sqrt{1 - x^{2}}) = 2\cos^{-1}x,  \frac{1}{\sqrt{2}} \leq x \leq 1.
\displaystyle \text{OR}
\displaystyle \text{Consider a bijective function } f = R_{+} \rightarrow (7, \infty) \text{ given by} \\ f(x) = 16x^{2} + 24x + 7,  \text{ where } R_{+} \text{ is the set of all positive real numbers. Find the}  \\ \text{inverse function of } f.
\displaystyle \text{Answer:}
\displaystyle \mathrm{L.H.S.} = \sin^{-1}(2x\sqrt{1 - x^{2}}), \quad \frac{1}{\sqrt{2}} \leq x \leq 1
\displaystyle \text{Let } x = \cos \theta \Rightarrow \theta = \cos^{-1}x, \quad 0 \leq \theta \leq \frac{\pi}{4}
\displaystyle \therefore \mathrm{L.H.S.} = \sin^{-1}(2\sin \theta \sqrt{1 - \sin^{2}\theta})
\displaystyle = \sin^{-1}(2\sin \theta \cos \theta) \quad [\because \sin^{2}\theta + \cos^{2}\theta = 1]
\displaystyle = \sin^{-1}(\sin 2\theta) \quad [\because \sin 2\theta = 2\sin \theta \cos \theta]
\displaystyle = 2\theta
\displaystyle = 2\cos^{-1}x = \mathrm{R.H.S.}
\displaystyle \text{OR}
\displaystyle f: R_{+} \rightarrow (7, \infty)
\displaystyle f(x) = 16x^{2} + 24x + 7
\displaystyle = (4x + 3)^{2} - 2 > 0, \forall x \in R_{+}
\displaystyle \therefore f(x) \text{ is strictly increasing in } R_{+}
\displaystyle \Rightarrow f(x) \text{ is injective}
\displaystyle \text{Now, } f(x) = (4x + 3)^{2} - 2
\displaystyle \Rightarrow (4x + 3)^{2} = f(x) + 2
\displaystyle \Rightarrow 4x + 3 = \sqrt{f(x) + 2}
\displaystyle \Rightarrow x = \frac{\sqrt{f(x) + 2} - 3}{4}
\displaystyle \text{For every } f(x) \in (7, \infty), \text{ there exists unique } x \in R_{+}
\displaystyle \therefore f(x) \text{ is surjective}
\displaystyle \therefore f(x) \text{ is bijective}
\displaystyle \therefore f^{-1}(x) = \frac{\sqrt{x + 2} - 3}{4}

\displaystyle \textbf{22. } \text{If } x = at^{2}, y = 2at, \text{ then find } \frac{d^{2}y}{dx^{2}}.
\displaystyle \text{Answer:}
\displaystyle x = at^{2}, \quad y = 2at
\displaystyle \frac{dx}{dt} = 2at, \quad \frac{dy}{dt} = 2a
\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2at} = \frac{1}{t}
\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left(\frac{1}{t}\right) = \frac{d}{dt}\left(\frac{1}{t}\right)\cdot \frac{dt}{dx}
\displaystyle = \left(-\frac{1}{t^{2}}\right)\cdot \frac{1}{2at} = -\frac{1}{2at^{3}}
\displaystyle \text{Since } t = \sqrt{\frac{x}{a}},
\displaystyle \frac{d^{2}y}{dx^{2}} = -\frac{1}{2a\left(\frac{x}{a}\right)^{3/2}} = -\frac{a^{1/2}}{2x^{3/2}}

\displaystyle \textbf{23. } \text{Find the points on the curve } y = x^{3} - 3x^{2} - 4x \text{ at which the tangent lines are}
\displaystyle \text{parallel to the line } 4x + y - 3 = 0.
\displaystyle \text{Answer:}
\displaystyle y = x^{3} - 3x^{2} - 4x
\displaystyle \frac{dy}{dx} = 3x^{2} - 6x - 4
\displaystyle \text{Slope of tangent } = 3x^{2} - 6x - 4 \qquad (1)
\displaystyle \text{Given line: } 4x + y - 3 = 0 \Rightarrow y = -4x + 3
\displaystyle \text{Slope } = -4 \qquad (2)
\displaystyle \text{From (1) and (2): } 3x^{2} - 6x - 4 = -4
\displaystyle \Rightarrow 3x^{2} - 6x = 0 \Rightarrow 3x(x - 2) = 0
\displaystyle \Rightarrow x = 0, 2
\displaystyle \text{At } x = 0, y = 0 \Rightarrow (0, 0)
\displaystyle \text{At } x = 2, y = 8 - 12 - 8 = -12 \Rightarrow (2, -12)
\displaystyle \text{Required points: } (0, 0), (2, -12)

\displaystyle \textbf{24. } \text{Find a unit vector perpendicular to each of the vectors } \overrightarrow{a} \text{and } \overrightarrow{b} \text{ where } \\ \overrightarrow{a} = 5\overrightarrow{i} + 6\overrightarrow{j} - 2\overrightarrow{k} \text{ and }  \overrightarrow{b} = 7\overrightarrow{i} + 6\overrightarrow{j} + 2\overrightarrow{k}.
\displaystyle \text{OR}
\displaystyle \text{Find the volume of the parallelepiped whose adjacent edges are represented}
\displaystyle \text{by } 2\overrightarrow{a}, -\overrightarrow{b} \text{ and } 3\overrightarrow{c}, \text{ where } \overrightarrow{a} = \overrightarrow{i} - \overrightarrow{j} + 2\overrightarrow{k},
\displaystyle \overrightarrow{b} = 3\overrightarrow{i} + 4\overrightarrow{j} - 5\overrightarrow{k}, \text{ and } \overrightarrow{c} = 2\overrightarrow{i} - \overrightarrow{j} + 3\overrightarrow{k}.
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{c} = \overrightarrow{a} \times \overrightarrow{b}
\displaystyle = \begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ 5 & 6 & -2 \\ 7 & 6 & 2 \end{vmatrix}
\displaystyle = \overrightarrow{i}(12 + 12) - \overrightarrow{j}(10 + 14) + \overrightarrow{k}(30 - 42)
\displaystyle = 24\overrightarrow{i} - 24\overrightarrow{j} - 12\overrightarrow{k} = 12(2\overrightarrow{i} - 2\overrightarrow{j} - \overrightarrow{k})
\displaystyle |\overrightarrow{c}| = 12\sqrt{(2)^{2} + (-2)^{2} + (-1)^{2}} = 12\sqrt{9} = 36
\displaystyle \text{Unit vector } = \frac{\overrightarrow{c}}{|\overrightarrow{c}|} = \frac{1}{3}(2\overrightarrow{i} - 2\overrightarrow{j} - \overrightarrow{k})
\displaystyle = \frac{2}{3}\overrightarrow{i} - \frac{2}{3}\overrightarrow{j} - \frac{1}{3}\overrightarrow{k}
\displaystyle \text{OR}
\displaystyle \overrightarrow{l_{1}} = 2\overrightarrow{a} = 2(\overrightarrow{i} - \overrightarrow{j} + 2\overrightarrow{k}) = 2\overrightarrow{i} - 2\overrightarrow{j} + 4\overrightarrow{k}
\displaystyle \overrightarrow{l_{2}} = -\overrightarrow{b} = -(3\overrightarrow{i} + 4\overrightarrow{j} - 5\overrightarrow{k}) = -3\overrightarrow{i} - 4\overrightarrow{j} + 5\overrightarrow{k}
\displaystyle \overrightarrow{l_{3}} = 3\overrightarrow{c} = 3(2\overrightarrow{i} - \overrightarrow{j} + 3\overrightarrow{k}) = 6\overrightarrow{i} - 3\overrightarrow{j} + 9\overrightarrow{k}
\displaystyle \text{Volume } = |(\overrightarrow{l_{1}} \times \overrightarrow{l_{2}})\cdot \overrightarrow{l_{3}}|
\displaystyle \overrightarrow{l_{1}} \times \overrightarrow{l_{2}} = \begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ 2 & -2 & 4 \\ -3 & -4 & 5 \end{vmatrix}
\displaystyle = \overrightarrow{i}(-10 + 16) - \overrightarrow{j}(10 + 12) + \overrightarrow{k}(-8 - 6)
\displaystyle = 6\overrightarrow{i} - 22\overrightarrow{j} - 14\overrightarrow{k}
\displaystyle (\overrightarrow{l_{1}} \times \overrightarrow{l_{2}})\cdot \overrightarrow{l_{3}}
\displaystyle = (6, -22, -14)\cdot (6, -3, 9) = 36 + 66 - 126 = -24
\displaystyle \text{Volume } = 24 \text{ cubic units}

\displaystyle \textbf{25. } \text{Find the value of } k \text{ so that the lines } x = -y = kz \text{ and}
\displaystyle x - 2 = 2y + 1 = -z + 1 \text{ are perpendicular to each other.}
\displaystyle \text{Answer:}
\displaystyle x = -y = kz
\displaystyle \Rightarrow \frac{x}{1} = \frac{y}{-1} = \frac{z}{\frac{1}{k}}
\displaystyle \text{Direction vector of first line, } \overrightarrow{a} = \overrightarrow{i} - \overrightarrow{j} + \frac{1}{k}\overrightarrow{k} \qquad (1)
\displaystyle x - 2 = 2y + 1 = -z + 1
\displaystyle \Rightarrow \frac{x - 2}{1} = \frac{y + \frac{1}{2}}{\frac{1}{2}} = \frac{z - 1}{-1}
\displaystyle \text{Direction vector of second line, } \overrightarrow{b} = \overrightarrow{i} + \frac{1}{2}\overrightarrow{j} - \overrightarrow{k} \qquad (2)
\displaystyle \text{For the two lines to be perpendicular, } \overrightarrow{a}\cdot \overrightarrow{b} = 0
\displaystyle \Rightarrow \left(\overrightarrow{i} - \overrightarrow{j} + \frac{1}{k}\overrightarrow{k}\right)\cdot \left(\overrightarrow{i} + \frac{1}{2}\overrightarrow{j} - \overrightarrow{k}\right) = 0
\displaystyle \Rightarrow 1 - \frac{1}{2} - \frac{1}{k} = 0
\displaystyle \Rightarrow \frac{1}{2} = \frac{1}{k} \Rightarrow k = 2

\displaystyle \textbf{26. } \text{The probability of finding a green signal on a busy crossing} X \text{ is } 30\%.
\displaystyle  \text{What is the probability of finding a green signal on }  X \text{ on two consecutive days} \\ \text{out of three?}
\displaystyle \text{Answer:}
\displaystyle \text{Probability of green signal } = 30\% = \frac{30}{100} = 0.3
\displaystyle \therefore P(G) = 0.3
\displaystyle \text{Probability of other signals } = 1 - 0.3 = 0.7
\displaystyle \text{Let } P(X) \text{ be the probability of getting green signal on two consecutive days out of 3.}
\displaystyle P(X) = P(G)P(G)P(\text{other}) + P(\text{other})P(G)P(G)
\displaystyle = (0.3)(0.3)(0.7) + (0.7)(0.3)(0.3)
\displaystyle = 2(0.3)(0.3)(0.7)
\displaystyle = 2 \times 0.09 \times 0.7
\displaystyle = 0.126

\displaystyle \textbf{SECTION - C}
\displaystyle \text{Question numbers 27 to 32 carry 4 marks each.}

\displaystyle \textbf{27. } \text{Let } N \text{ be the set of natural numbers and } R \text{ be the relation} \text{on } N \times N \text{ defined by } \\ (a, b) \, R \, (c, d) \text{ if ad = bc \text{ for all } a, b, c,}  d \in N. \text{ Show that } R \text{ is an equivalence relation.}
\displaystyle \text{Answer:}
\displaystyle (a, b)\, R \,(c, d) \Leftrightarrow ad = bc \qquad (\text{given relation})
\displaystyle \text{(i) Reflexivity:}
\displaystyle \text{Since } ab = ba,
\displaystyle (a, b)\, R \,(a, b) \text{ is true}
\displaystyle \therefore R \text{ is reflexive.}
\displaystyle \text{(ii) Symmetry:}
\displaystyle \text{Let } (a, b)\, R \,(c, d)
\displaystyle \Rightarrow ad = bc
\displaystyle \text{But } bc = cb \text{ and } ad = da
\displaystyle \Rightarrow cb = da \Rightarrow cd \text{? No, rather } cb = da
\displaystyle \Rightarrow (c, d)\, R \,(a, b)
\displaystyle \therefore (a, b)\, R \,(c, d) \Leftrightarrow (c, d)\, R \,(a, b)
\displaystyle \therefore R \text{ is symmetric.}
\displaystyle \text{(iii) Transitivity:}
\displaystyle \text{Let } (a, b)\, R \,(c, d) \text{ and } (c, d)\, R \,(e, f)
\displaystyle \Rightarrow ad = bc \text{ and } cf = de
\displaystyle \Rightarrow (ad)(cf) = (bc)(de)
\displaystyle \Rightarrow af(cd) = be(cd)
\displaystyle \Rightarrow af = be
\displaystyle \Rightarrow (a, b)\, R \,(e, f)
\displaystyle \therefore R \text{ is transitive.}
\displaystyle \text{Since } R \text{ is reflexive, symmetric and transitive,}
\displaystyle \therefore R \text{ is an equivalence relation.}

\displaystyle \textbf{28. } \text{If } y = e^{x^{2}\cos x} + (\cos x)^{x}, \text{ then find } \frac{dy}{dx}.
\displaystyle \text{Answer:}
\displaystyle y = e^{x^{2}\cos x} + (\cos x)^{x}
\displaystyle y = e^{x^{2}\cos x} + e^{\log (\cos x)^{x}}
\displaystyle y = e^{x^{2}\cos x} + e^{x\log (\cos x)}
\displaystyle \text{By chain rule, } \frac{d}{dx}(g(x)) = \frac{dg}{du}\cdot \frac{du}{dx}
\displaystyle \frac{dy}{dx} = \frac{d}{dx}\left(e^{x^{2}\cos x}\right) + \frac{d}{dx}\left(e^{x\log (\cos x)}\right)
\displaystyle = e^{x^{2}\cos x}(2x\cos x - x^{2}\sin x) + e^{x\log (\cos x)}\left[\log (\cos x) - \frac{x\sin x}{\cos x}\right]
\displaystyle \left[\because \frac{d}{dx}(e^{x}) = e^{x}, \frac{d}{dx}(\log x) = \frac{1}{x}, \frac{d}{dx}(\cos x) = -\sin x\right]
\displaystyle = e^{x^{2}\cos x}(2x\cos x - x^{2}\sin x) + (\cos x)^{x}\left[\log (\cos x) - x\tan x\right]

\displaystyle \textbf{29. } \text{Find: } \int \sec^{3} x \, dx
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \sec^{3} x \, dx
\displaystyle I = \int \sec x \sec^{2} x \, dx
\displaystyle \text{Using } \int u \, dv = u\int v \, dx - \int \left(\frac{du}{dx}\int v \, dx\right) dx
\displaystyle I = \sec x \int \sec^{2} x \, dx - \int \left(\frac{d}{dx}(\sec x)\int \sec^{2} x \, dx\right) dx
\displaystyle = \sec x \tan x - \int \sec x \tan x \tan x \, dx
\displaystyle \left[\because \int \sec^{2} x \, dx = \tan x + C,\ \frac{d}{dx}(\sec x) = \sec x \tan x\right]
\displaystyle = \sec x \tan x - \int \sec x \tan^{2} x \, dx
\displaystyle = \sec x \tan x - \int \sec x (\sec^{2} x - 1) \, dx
\displaystyle \left[\because \sec^{2} \theta = 1 + \tan^{2} \theta\right]
\displaystyle = \sec x \tan x - \int \sec^{3} x \, dx + \int \sec x \, dx
\displaystyle = \sec x \tan x - I + \log |\sec x + \tan x| + C
\displaystyle \left[\because \int \sec x \, dx = \log |\sec x + \tan x| + C\right]
\displaystyle 2I = \sec x \tan x + \log |\sec x + \tan x| + C
\displaystyle \Rightarrow I = \frac{1}{2}\sec x \tan x + \frac{1}{2}\log |\sec x + \tan x| + C

\displaystyle \textbf{30. } \text{Find the general solution of the differential equation}  y e^{y} \, dx = (y^{3} + 2x e^{y}) \, dy.
\displaystyle \text{OR}
\displaystyle \text{Find the particular solution of the differential equation}  x\frac{dy}{dx} = y - x\tan\left(\frac{y}{x}\right), \\ \text{given that } y = \frac{\pi}{4} \text{ at } x = 1.
\displaystyle \text{Answer:}
\displaystyle  y e^{y} \, dx = (y^{3} + 2x e^{y}) \, dy
\displaystyle \Rightarrow \frac{dx}{dy} = \frac{y^{3} + 2x e^{y}}{y e^{y}}
\displaystyle \Rightarrow \frac{dx}{dy} = y^{2}e^{-y} + \frac{2x}{y}
\displaystyle \Rightarrow \frac{dx}{dy} - \frac{2}{y}x = y^{2}e^{-y}
\displaystyle \text{This is a linear differential equation of the form } \frac{dx}{dy} + Px = Q
\displaystyle \text{where } P = -\frac{2}{y}, \quad Q = y^{2}e^{-y}
\displaystyle \therefore \mathrm{I.F.} = e^{\int P \, dy} = e^{\int -\frac{2}{y} \, dy} = e^{-2\log y}
\displaystyle \mathrm{I.F.} = \frac{1}{y^{2}} \qquad \left[\because e^{\log m^{n}} = m^{n}\right]
\displaystyle x \cdot \mathrm{I.F.} = \int \mathrm{I.F.}\cdot Q \, dy + C
\displaystyle \Rightarrow \frac{x}{y^{2}} = \int \frac{1}{y^{2}}(y^{2}e^{-y}) \, dy + C
\displaystyle \Rightarrow \frac{x}{y^{2}} = \int e^{-y} \, dy + C
\displaystyle \Rightarrow \frac{x}{y^{2}} = -e^{-y} + C \qquad \left[\because \int e^{-y} \, dy = -e^{-y}\right]
\displaystyle \Rightarrow \frac{x}{y^{2}} + e^{-y} = C
\displaystyle \text{OR}
\displaystyle x\frac{dy}{dx} = y - x\tan\left(\frac{y}{x}\right)
\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y}{x} - \tan\left(\frac{y}{x}\right) \qquad \ldots (1)
\displaystyle \text{Let } z = \frac{y}{x}
\displaystyle \text{Given } y = \frac{\pi}{4} \text{ at } x = 1
\displaystyle \therefore z = \frac{\pi}{4}
\displaystyle z = \frac{y}{x}, \text{ differentiating w.r.t. } x
\displaystyle \frac{dz}{dx} = \frac{1}{x}\frac{dy}{dx} + y\frac{d}{dx}\left(\frac{1}{x}\right)
\displaystyle \Rightarrow \frac{dz}{dx} = \frac{1}{x}\frac{dy}{dx} - \frac{y}{x^{2}}
\displaystyle \Rightarrow \frac{dz}{dx} = \frac{1}{x}\left(\frac{dy}{dx} - \frac{y}{x}\right)
\displaystyle \Rightarrow \frac{dz}{dx} = \frac{1}{x}\left(-\tan\left(\frac{y}{x}\right)\right) \qquad \text{[from (1)]}
\displaystyle \Rightarrow \frac{dz}{dx} = -\frac{\tan z}{x}
\displaystyle \Rightarrow -\frac{dz}{\tan z} = \frac{dx}{x} \Rightarrow -\cot z \, dz = \frac{dx}{x}
\displaystyle \text{Integrating on both sides,}
\displaystyle \int \cot z \, dz = -\int \frac{dx}{x}
\displaystyle \log |\sin z| = -\log |x| + \log |c| \qquad \left[\because \int \cot z \, dz = \log |\sin z|\right]
\displaystyle \Rightarrow \log |\sin z| = \log \left|\frac{c}{x}\right|
\displaystyle \Rightarrow \sin z = \frac{c}{x}
\displaystyle \text{Putting } x = 1,\ y = \frac{\pi}{4}
\displaystyle \sin \frac{\pi}{4} = c
\displaystyle c = \frac{1}{\sqrt{2}}
\displaystyle \sin z = \frac{1}{\sqrt{2}x}
\displaystyle \Rightarrow z = \sin^{-1}\left(\frac{1}{\sqrt{2}x}\right)
\displaystyle \Rightarrow \frac{y}{x} = \sin^{-1}\left(\frac{1}{\sqrt{2}x}\right)

\displaystyle \textbf{31. } \text{A furniture trader deals in only two items - chairs and tables.} \text{He has Rs } 50{,}000
\displaystyle \text{ to invest and a space to store at most } 35  \text{items. A chair costs him Rs } 1000 \text{ and a table}
\displaystyle \text{costs him Rs } 2000.\text{The trader earns a profit of Rs } 150 \text{ on a chair and Rs } 250 \text{ on a table,}
\displaystyle \text{respectively. Formulate the above problem as an LPP to maximise the profit and}
\displaystyle \text{solve it graphically.}
\displaystyle \text{Answer:}
\displaystyle \text{Let number of chairs be } x \text{ and number of tables be } y\displaystyle \text{Maximize } Z = 150x + 250y
\displaystyle \text{Subject to: } 1000x + 2000y \leq 50000
\displaystyle x + 2y \leq 50,\quad x + y \leq 35,\quad x \geq 0,\ y \geq 0
\displaystyle \text{Corner points: } O(0,0),\ A(0,25),\ C(35,0)
\displaystyle \text{Intersection of } x + y = 35 \text{ and } x + 2y = 50
\displaystyle x + y = 35
\displaystyle x + 2y = 50
\displaystyle \Rightarrow y = 15,\ x = 20 \Rightarrow B(20,15)
\displaystyle \text{Evaluate } Z:
\displaystyle O(0,0):\ Z = 0
\displaystyle A(0,25):\ Z = 150(0) + 250(25) = 6250
\displaystyle C(35,0):\ Z = 150(35) = 5250
\displaystyle B(20,15):\ Z = 150(20) + 250(15) = 3000 + 3750 = 6750
\displaystyle \text{Maximum profit } = 6750 \text{ at } (20,15)
\displaystyle \text{31 chairs and tables interpretation: chairs } = 20,\ tables = 15

\displaystyle \textbf{32. } \text{There are two bags, I and II. Bag I contains } 3 \text{ red and } 5 \text{black balls and Bag II}
\displaystyle \text{contains } 4 \text{ red and } 3 \text{ black balls. One} \text{ball is transferred randomly from Bag I to Bag II}
\displaystyle \text{and then a ball is drawn randomly from Bag II. If the ball so drawn is found to be}
\displaystyle \text{black in colour, then find the probability that}  \text{the transferred ball is also black.}
\displaystyle \text{OR}
\displaystyle \text{An urn contains } 5 \text{ red, } 2 \text{ white and } 3 \text{ black balls. Three balls are drawn, one-by-one,}
\displaystyle \text{ at random without replacement. Find the probability distribution of the number of white}
\displaystyle \text{balls. Also, find the mean and the variance of the number of white balls drawn.}
\displaystyle \text{Answer:}
\displaystyle \text{Bag I: Red = 3,\ Black = 5}
\displaystyle \text{Bag II: Red = 4,\ Black = 3}
\displaystyle \text{Let } E_{1}: \text{transferred ball is black},\ E_{2}: \text{transferred ball is red}
\displaystyle F: \text{ball drawn from bag II is black}
\displaystyle P(E_{1}) = \frac{5}{8},\quad P(E_{2}) = \frac{3}{8}
\displaystyle P(F|E_{1}) = \frac{4}{8},\quad P(F|E_{2}) = \frac{3}{8}
\displaystyle \text{By Bayes' theorem:}
\displaystyle P(E_{1}|F) = \frac{P(F|E_{1})P(E_{1})}{P(F|E_{1})P(E_{1}) + P(F|E_{2})P(E_{2})}  = \frac{\frac{5}{8}\cdot \frac{4}{8}}{\frac{5}{8}\cdot \frac{4}{8} + \frac{3}{8}\cdot \frac{3}{8}}  = \frac{20}{29}
\displaystyle \text{OR}
\displaystyle \text{Red = 5,\ White = 2,\ Black = 3}
\displaystyle \text{Let } X = \text{number of white balls when 3 balls are drawn without replacement}
\displaystyle X = 0,1,2
\displaystyle P(X=0) = \frac{{8 \choose 3}}{{10 \choose 3}} = \frac{7}{15}
\displaystyle P(X=1) = \frac{{8 \choose 2}{2 \choose 1}}{{10 \choose 3}} = \frac{7}{15}
\displaystyle P(X=2) = \frac{{8 \choose 1}{2 \choose 2}}{{10 \choose 3}} = \frac{1}{15}
\displaystyle \text{Probability distribution:}
\displaystyle \begin{array}{c|ccc} X & 0 & 1 & 2 \\ \hline P(X) & \frac{7}{15} & \frac{7}{15} & \frac{1}{15} \end{array}
\displaystyle \text{Mean } = \sum XP(X) = \frac{9}{15} = 0.6
\displaystyle \text{Variance } = \sum X^{2}P(X) - [\sum XP(X)]^{2}
\displaystyle = \frac{11}{15} - \left(\frac{9}{15}\right)^{2} = \frac{28}{75} \approx 0.37

\displaystyle \textbf{SECTION - D}
\displaystyle \text{Question numbers 33 to 36 carry 6 marks each.}

\displaystyle \textbf{33. } \text{If } A = \begin{bmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{bmatrix} \text{, then find } A^{-1} \text{ and use it to solve the following system of}
\displaystyle \text{the equations:}
\displaystyle x + 2y - 3z = 6
\displaystyle 3x + 2y - 2z = 3
\displaystyle 2x - y + z = 2
\displaystyle \text{OR}
\displaystyle \text{Using properties of determinants, prove that}
\displaystyle \left|\begin{matrix} (b + c)^{2} & a^{2} & bc \\ (c + a)^{2} & b^{2} & ca \\ (a + b)^{2} & c^{2} & ab \end{matrix}\right|  = (a - b)(b - c)(c - a)(a + b + c)(a^{2} + b^{2} + c^{2}).
\displaystyle \text{Answer:}
\displaystyle A = \begin{bmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{bmatrix}
\displaystyle |A| = 1(2 - 2) - 2(3 + 4) - 3(-3 - 4)
\displaystyle = -14 + 21 = 7
\displaystyle \therefore |A| \neq 0,\ A^{-1} \text{ exists}
\displaystyle A_{11} = (-1)^{2}\begin{vmatrix} 2 & -2 \\ -1 & 1 \end{vmatrix} = 2 - 2 = 0
\displaystyle A_{12} = (-1)^{3}\begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix} = (-1)[3 + 4] = -7
\displaystyle A_{13} = (-1)^{4}\begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} = -3 - 4 = -7
\displaystyle A_{21} = (-1)^{3}\begin{vmatrix} 2 & -3 \\ -1 & 1 \end{vmatrix} = (-1)[2 - 3] = 1
\displaystyle A_{22} = (-1)^{4}\begin{vmatrix} 1 & -3 \\ 2 & 1 \end{vmatrix} = 1 + 6 = 7
\displaystyle A_{23} = (-1)^{5}\begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (-1)[-1 - 4] = 5
\displaystyle A_{31} = (-1)^{4}\begin{vmatrix} 2 & -3 \\ 2 & -2 \end{vmatrix} = -4 + 6 = 2
\displaystyle A_{32} = (-1)^{5}\begin{vmatrix} 1 & -3 \\ 3 & -2 \end{vmatrix} = (-1)[-2 + 9] = -7
\displaystyle A_{33} = (-1)^{6}\begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix} = 2 - 6 = -4
\displaystyle \mathrm{adj}\,A = \begin{bmatrix} 0 & -7 & -7 \\ 1 & 7 & 5 \\ 2 & -7 & -4 \end{bmatrix}^{T} = \begin{bmatrix} 0 & 1 & 2 \\ -7 & 7 & -7 \\ -7 & 5 & -4 \end{bmatrix}
\displaystyle A^{-1} = \frac{1}{|A|}\mathrm{adj}\,A  = \frac{1}{7}\begin{bmatrix} 0 & 1 & 2 \\ -7 & 7 & -7 \\ -7 & 5 & -4 \end{bmatrix}  = \begin{bmatrix} 0 & \frac{1}{7} & \frac{2}{7} \\ -1 & 1 & -1 \\ -1 & \frac{5}{7} & -\frac{4}{7} \end{bmatrix}
\displaystyle \text{Given equation:}
\displaystyle x + 2y - 3z = 6
\displaystyle 3x + 2y - 2z = 3
\displaystyle 2x - y + z = 2
\displaystyle A = \begin{bmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{bmatrix},\ B = \begin{bmatrix} 6 \\ 3 \\ 2 \end{bmatrix},\ X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}
\displaystyle AX = B
\displaystyle \text{Multiply } A^{-1} \text{ on both sides}
\displaystyle (A^{-1})AX = A^{-1}B
\displaystyle IX = A^{-1}B \qquad [\because A^{-1}A = I]
\displaystyle X = A^{-1}B
\displaystyle X = \begin{bmatrix} 0 & \frac{1}{7} & \frac{2}{7} \\ -1 & 1 & -1 \\ -1 & \frac{5}{7} & -\frac{4}{7} \end{bmatrix}\begin{bmatrix} 6 \\ 3 \\ 2 \end{bmatrix}
\displaystyle X = \begin{bmatrix} \frac{3}{7} + \frac{4}{7} \\ -6 + 3 - 2 \\ -6 + \frac{15}{7} - \frac{8}{7} \end{bmatrix}
\displaystyle X = \begin{bmatrix} 1 \\ -5 \\ -5 \end{bmatrix}
\displaystyle \therefore \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -5 \\ 7 \end{bmatrix}
\displaystyle \therefore x = 1,\ y = -5,\ z = 7
\displaystyle \text{OR}
\displaystyle \mathrm{L.H.S.}
\displaystyle \Delta = \begin{vmatrix} (b+c)^{2} & a^{2} & bc \\ (c+a)^{2} & b^{2} & ca \\ (a+b)^{2} & c^{2} & ab \end{vmatrix}
\displaystyle R_{1} \rightarrow R_{1} - R_{3},\ R_{2} \rightarrow R_{2} - R_{3}
\displaystyle \Delta = \begin{vmatrix} (b+c+a+b)(b+c-a-b) & (a+c)(a-c) & b(c-a) \\ (c+a+a+b)(c+a-a-b) & (b+c)(b-c) & a(c-b) \\ (a+b)^{2} & c^{2} & ab \end{vmatrix}
\displaystyle = \begin{vmatrix} (c-a)(2b+a+c) & (a+c)(a-c) & b(c-a) \\ (c-b)(2a+c+b) & (b+c)(b-c) & a(c-b) \\ (a+b)^{2} & c^{2} & ab \end{vmatrix}
\displaystyle \Delta = (c-a)(b-c)\begin{vmatrix} a+c+2b & -a-c & b \\ -2a-b-c & b+c & -a \\ (a+b)^{2} & c^{2} & ab \end{vmatrix}
\displaystyle R_{1} \rightarrow R_{1} + R_{2}
\displaystyle \Delta = (c-a)(b-c)\begin{vmatrix} -a+b & b-a & b-a \\ -2a-b-c & b+c & -a \\ (a+b)^{2} & c^{2} & ab \end{vmatrix}
\displaystyle \Delta = (a-b)(c-a)(b-c)\begin{vmatrix} -1 & -1 & -1 \\ -2a-b-c & b+c & -a \\ (a+b)^{2} & c^{2} & ab \end{vmatrix}
\displaystyle C_{1} \rightarrow C_{1} - C_{3},\ C_{2} \rightarrow C_{2} - C_{3}
\displaystyle \Delta = (a-b)(b-c)(c-a)\begin{vmatrix} 0 & 0 & -1 \\ -a-b-c & b+c+a & -a \\ (a+b)^{2} - ab & c^{2} - ab & ab \end{vmatrix}
\displaystyle \Delta = (a-b)(b-c)(c-a)\{-a[(a+b+c)(c^{2}-ab) + (a+b+c)((a+b)^{2}-ab)]\}
\displaystyle = (a-b)(b-c)(c-a)\{(a+b+c)[c^{2}-ab + a^{2}+2ab+b^{2}-ab]\}
\displaystyle \Delta = (a-b)(b-c)(c-a)(a+b+c)(a^{2}+b^{2}+c^{2}) = \mathrm{R.H.S.}
\displaystyle \text{Hence Proved.}

\displaystyle \textbf{34. } \text{Using integration, find the area of the region bounded by}  \text{the triangle whose} \\ \text{vertices are } (2, -2), (4, 5) \text{ and } (6, 2).
\displaystyle \text{Answer:}
\displaystyle A(2,-2),\ B(4,5),\ C(6,2)
\displaystyle \text{Equation of a line: } y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}(x - x_{1})
\displaystyle \text{Equation of line } AB:
\displaystyle y - 5 = \frac{-2 - 5}{2 - 4}(x - 4)
\displaystyle y - 5 = \frac{-7}{-2}(x - 4) = \frac{7}{2}(x - 4)
\displaystyle y = \frac{7}{2}x - 14 + 5
\displaystyle \Rightarrow y = \frac{7}{2}x - 9
\displaystyle \text{Equation of line } BC:
\displaystyle y - 5 = \frac{2 - 5}{6 - 4}(x - 4)
\displaystyle y - 5 = \frac{-3}{2}(x - 4)
\displaystyle y = -\frac{3}{2}x + 6 + 5
\displaystyle \Rightarrow y = -\frac{3}{2}x + 11
\displaystyle \text{Equation of line } AC:
\displaystyle y - 2 = \frac{-2 - 2}{2 - 6}(x - 6)
\displaystyle y - 2 = \frac{-4}{-4}(x - 6) = x - 6
\displaystyle \Rightarrow y = x - 4
\displaystyle \text{Area of } \triangle ABC
\displaystyle = \int_{2}^{4} \left(\frac{7}{2}x - 9\right) dx + \int_{4}^{6} \left(-\frac{3}{2}x + 11\right) dx - \int_{2}^{6} (x - 4) dx
\displaystyle = \left[\frac{7x^{2}}{4} - 9x\right]_{2}^{4} + \left[-\frac{3x^{2}}{4} + 11x\right]_{4}^{6} - \left[\frac{x^{2}}{2} - 4x\right]_{2}^{6}
\displaystyle = [(28 - 36) - (7 - 18)] + [(-27 + 66) - (-12 + 44)] - [(18 - 24) - (2 - 8)]
\displaystyle = [(-8) - (-11)] + [39 - 32] - [(-6) - (-6)]
\displaystyle = 3 + 7 - 0 = 10
\displaystyle \therefore \text{Area of } \triangle ABC = 10 \text{ sq. units}

\displaystyle \textbf{35. } \text{Show that the height of the right circular cylinder of greatest volume which can}
\displaystyle \text{be inscribed in a right circular cone of} \text{height } h \text{ and radius } r \text{ is one-third of the}
\displaystyle \text{height of the cone, and the greatest volume of the cylinder is } \frac{4}{9} \text{ times the volume} \\ \text{of the cone.}
\displaystyle \text{Answer:} \displaystyle \text{Given height of cone } = h \text{ and radius of cone } = r
\displaystyle \text{Let height of cylinder } = H \text{ and radius of cylinder } = R
\displaystyle \text{Volume of cylinder } V = \pi R^{2}H \qquad \ldots (1)
\displaystyle \text{In right } \triangle ADE,
\displaystyle \tan \theta = \frac{R}{h - H}
\displaystyle \text{In right } \triangle ABC,
\displaystyle \tan \theta = \frac{r}{h}
\displaystyle \Rightarrow \frac{r}{h} = \frac{R}{h - H}
\displaystyle \Rightarrow rh - rH = hR
\displaystyle \Rightarrow R = r - \frac{rH}{h}
\displaystyle \Rightarrow R = r\left(1 - \frac{H}{h}\right) \qquad \ldots (2)
\displaystyle \text{From (1),}
\displaystyle V = \pi r^{2}\left(1 - \frac{H}{h}\right)^{2}H
\displaystyle \text{Differentiate both sides w.r.t. } H
\displaystyle \frac{dV}{dH} = \pi r^{2}\left[\left(1 - \frac{H}{h}\right)^{2} + H \cdot 2\left(1 - \frac{H}{h}\right)\left(-\frac{1}{h}\right)\right]
\displaystyle = \pi r^{2}\left(1 - \frac{H}{h}\right)\left[1 - \frac{H}{h} - \frac{2H}{h}\right]
\displaystyle = \pi r^{2}\left(1 - \frac{H}{h}\right)\left(1 - \frac{3H}{h}\right) = 0
\displaystyle \because 1 - \frac{H}{h} \neq 0
\displaystyle \therefore 1 - \frac{3H}{h} = 0 \Rightarrow H = \frac{h}{3} \qquad \ldots (3)
\displaystyle \frac{d^{2}V}{dH^{2}} = \pi r^{2}\left[-\frac{1}{h}\left(1 - \frac{3H}{h}\right) + \left(1 - \frac{H}{h}\right)\left(-\frac{3}{h}\right)\right]
\displaystyle \text{Put } H = \frac{h}{3}
\displaystyle \frac{d^{2}V}{dH^{2}} = \pi r^{2}\left[0 + \left(1 - \frac{1}{3}\right)\left(-\frac{3}{h}\right)\right]
\displaystyle = \pi r^{2}\left(-\frac{2}{h}\right) < 0
\displaystyle \therefore \text{Volume is maximum when } H = \frac{h}{3}
\displaystyle \text{Volume of cylinder } = \pi r^{2}\left(1 - \frac{H}{h}\right)^{2}H
\displaystyle \text{Putting } H = \frac{h}{3},\ V = \pi r^{2}\left(1 - \frac{1}{3}\right)^{2}\frac{h}{3}
\displaystyle = \pi r^{2}\cdot \frac{4}{9}\cdot \frac{h}{3}
\displaystyle = \frac{4}{27}\pi r^{2}h
\displaystyle = \frac{4}{9}\times \text{volume of cone}
\displaystyle \text{Hence proved.}

\displaystyle \textbf{36. } \text{Find the equation of the plane that contains the point}  A(2, 1, -1) \text{ and is} \\ \text{perpendicular to the line of intersection} \text{of the planes } 2x + y - z = 3 \text{ and } \\ x + 2y + z = 2. \text{ Also find} \text{the angle between the plane thus obtained and the } y\text{-axis.}
\displaystyle \text{OR}
\displaystyle \text{Find the distance of the point } P(-2, -4, 7) \text{ from the point of}  \text{intersection } Q \text{ of the line }
\displaystyle \overrightarrow{r} = (3\overrightarrow{i} - 2\overrightarrow{j} + 6\overrightarrow{k}) +
\displaystyle \lambda (2\overrightarrow{i} - \overrightarrow{j} + 2\overrightarrow{k}) \text{ and the plane}
\displaystyle \overrightarrow{r} \cdot (\overrightarrow{i} - \overrightarrow{j} + \overrightarrow{k}) = 6.
\displaystyle \text{Also write the vector equation of the line } PQ.
\displaystyle \text{Answer:}
\displaystyle \text{Plane passing through } A(2,1,-1) \text{ and perpendicular to the line of intersection of }
\displaystyle \text{the two planes } 2x+y-z=3 \text{ and } x+2y+z=2 \text{Since the plane is perpendicular}
\displaystyle \text{to the line of intersection, the normal of the required plane is parallel to the line}
\displaystyle \text{of intersection and perpendicular to the normals of both given planes.}
\displaystyle \overrightarrow{n_{1}} \rightarrow \text{normal vector of first given plane}
\displaystyle \overrightarrow{n_{2}} \rightarrow \text{normal vector of second given plane}
\displaystyle \overrightarrow{N} \rightarrow \text{normal vector of required plane}
\displaystyle \overrightarrow{N} \parallel \overrightarrow{n_{1}} \times \overrightarrow{n_{2}} \qquad [\because \overrightarrow{N} \perp \overrightarrow{n_{1}},\ \overrightarrow{N} \perp \overrightarrow{n_{2}}]
\displaystyle \overrightarrow{N} = \begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix} = 3\overrightarrow{i} - 3\overrightarrow{j} + 3\overrightarrow{k}
\displaystyle \therefore \text{Direction ratios of the normal are } 3,-3,3
\displaystyle \text{Equation of plane passing through } (x_{1},y_{1},z_{1}) \text{ is}
\displaystyle A(x-x_{1})+B(y-y_{1})+C(z-z_{1})=0
\displaystyle \text{Here } A,B,C \text{ are the direction ratios of the normal to the plane}
\displaystyle 3(x-2)-3(y-1)+3(z+1)=0
\displaystyle \Rightarrow (x-2)-(y-1)+(z+1)=0
\displaystyle \Rightarrow x-2-y+1+z+1=0
\displaystyle \Rightarrow x-y+z=0
\displaystyle \cos \theta = \frac{|\overrightarrow{N}\cdot \overrightarrow{b}|}{|\overrightarrow{N}|\,|\overrightarrow{b}|}
\displaystyle \text{Direction vector of } y\text{-axis is } \overrightarrow{b}=0\overrightarrow{i}+\overrightarrow{j}+0\overrightarrow{k}
\displaystyle \cos \theta = \frac{|(3\overrightarrow{i}-3\overrightarrow{j}+3\overrightarrow{k})\cdot (0\overrightarrow{i}+\overrightarrow{j}+0\overrightarrow{k})|}{\sqrt{3^{2}+(-3)^{2}+3^{2}}\sqrt{1^{2}}}
\displaystyle \Rightarrow \cos \theta = \frac{|-3|}{3\sqrt{3}} = \frac{1}{\sqrt{3}}
\displaystyle \Rightarrow \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)
\displaystyle \text{OR}
\displaystyle \overrightarrow{r} = (3\overrightarrow{i}-2\overrightarrow{j}+6\overrightarrow{k})+\lambda (2\overrightarrow{i}-\overrightarrow{j}+2\overrightarrow{k}) \text{ is the line equation}
\displaystyle \overrightarrow{r}\cdot (\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k})=6 \text{ is the plane equation}
\displaystyle \text{From the line equation,}
\displaystyle x=3+2\lambda,\quad y=-2-\lambda,\quad z=6+2\lambda
\displaystyle \text{Equation of plane can be written as } x-y+z=6
\displaystyle \Rightarrow (3+2\lambda)-(-2-\lambda)+(6+2\lambda)=6
\displaystyle \Rightarrow 3+2\lambda+2+\lambda+6+2\lambda=6
\displaystyle \Rightarrow 5\lambda+11=6
\displaystyle \Rightarrow \lambda=\frac{-5}{5}=-1
\displaystyle \text{Point of intersection } Q:
\displaystyle x=3+2(-1)=1,\quad y=-2-(-1)=-1,\quad z=6+2(-1)=4
\displaystyle \therefore Q(1,-1,4)
\displaystyle \text{Given } P(-2,-4,7)
\displaystyle \text{Distance } PQ=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}
\displaystyle =\sqrt{(-2-1)^{2}+(-4+1)^{2}+(7-4)^{2}}
\displaystyle =\sqrt{3^{2}+3^{2}+3^{2}}=3\sqrt{3}
\displaystyle \text{Vector equation of } PQ
\displaystyle \overrightarrow{r}=(\overrightarrow{i}-\overrightarrow{j}+4\overrightarrow{k})+\lambda [(1+2)\overrightarrow{i}+(-1+4)\overrightarrow{j}+(4-7)\overrightarrow{k}]
\displaystyle \overrightarrow{r}=(\overrightarrow{i}-\overrightarrow{j}+4\overrightarrow{k})+\lambda (3\overrightarrow{i}+3\overrightarrow{j}-3\overrightarrow{k})
\displaystyle \overrightarrow{r}=(\overrightarrow{i}-\overrightarrow{j}+4\overrightarrow{k})+\lambda (\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k})

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