\displaystyle \Large {\textbf{Series PQ1RS/1 \quad Set - 1 \quad Q.P. Code 65/1/1}}

\displaystyle \text{Candidates must write the Q.P. Code on the title page of the answer-book.}
\displaystyle \text{Please check that this question paper contains 23 printed pages.}
\displaystyle \text{Please check that this question paper contains 38 questions.}
\displaystyle \text{Q.P. Code given on the right hand side of the question paper should be written}
\displaystyle \text{on the title page of the answer-book by the candidate.}
\displaystyle \text{Please write down the serial number of the question in the answer-book before}
\displaystyle \text{attempting it.}
\displaystyle \text{15 minute time has been allotted to read this question paper. The question}
\displaystyle \text{paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the}
\displaystyle \text{students will read the question paper only and will not write any answer on}
\displaystyle \text{the answer-book during this period.}

\displaystyle \textbf{MATHEMATICS}
\displaystyle \text{Time allowed : 3 hours \quad 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 contains 38 questions. All questions are compulsory.}
\displaystyle \text{(ii) This question paper is divided into five Sections - A, B, C, D and E.}
\displaystyle \text{(iii) In Section A, Questions no. 1 to 18 are multiple choice questions (MCQs)}
\displaystyle \text{and questions number 19 and 20 are Assertion-Reason based questions of}
\displaystyle \text{1 mark each.}
\displaystyle \text{(iv) In Section B, Questions no. 21 to 25 are very short answer (VSA) type}
\displaystyle \text{questions, carrying 2 marks each.}
\displaystyle \text{(v) In Section C, Questions no. 26 to 31 are short answer (SA) type questions,}
\displaystyle \text{carrying 3 marks each.}
\displaystyle \text{(vi) In Section D, Questions no. 32 to 35 are long answer (LA) type questions}
\displaystyle \text{carrying 5 marks each.}
\displaystyle \text{(vii) In Section E, Questions no. 36 to 38 are case study based questions}
\displaystyle \text{carrying 4 marks each.}
\displaystyle \text{(viii) There is no overall choice. However, an internal choice has been provided}
\displaystyle \text{in 2 questions in Section B, 3 questions in Section C, 2 questions in Section D}
\displaystyle \text{and 2 questions in Section E.}
\displaystyle \text{(ix) Use of calculators is not allowed.}

\displaystyle \textbf{SECTION A}
\displaystyle \text{This section comprises multiple choice questions (MCQs) of 1 mark each.}

\displaystyle \textbf{1. } \text{A function } f : R_{+} \rightarrow R \text{ (where } R_{+} \text{ is the set of all non-negative real numbers)}
\displaystyle \text{defined by } f(x)=4x+3 \text{ is :}
\displaystyle \text{(A) one-one but not onto}
\displaystyle \text{(B) onto but not one-one}
\displaystyle \text{(C) both one-one and onto}
\displaystyle \text{(D) neither one-one nor onto}
\displaystyle \text{Answer:}
\displaystyle f(x)=4x+3,\quad x\in R_{+}
\displaystyle \text{To check one-one:}
\displaystyle f(x_{1})=f(x_{2})\Rightarrow 4x_{1}+3=4x_{2}+3
\displaystyle \Rightarrow\ 4x_{1}=4x_{2}\Rightarrow x_{1}=x_{2}
\displaystyle \therefore\ f\text{ is one-one}
\displaystyle \text{Range of }f(x)\text{ for }x\geq 0\text{ is }[3,\infty)
\displaystyle \text{Since codomain is }R,\text{ all real numbers are not attained}
\displaystyle \therefore\ f\text{ is not onto}
\displaystyle \therefore\ \text{Correct option is (A) one-one but not onto}
\\
\displaystyle \textbf{2. } \text{If a matrix has 36 elements, the number of possible orders it can have, }   \text{is :}
\displaystyle \text{(A) 13 \quad (B) 3 \quad (C) 5 \quad (D) 9}
\displaystyle \text{Answer:}
\displaystyle \text{Let the order of the matrix be }m\times n
\displaystyle \text{Number of elements }=mn=36
\displaystyle \text{Thus, possible pairs }(m,n)\text{ are the factor pairs of }36
\displaystyle (1,36),(2,18),(3,12),(4,9),(6,6),(9,4),(12,3),(18,2),(36,1)
\displaystyle \text{Hence total possible orders }=9
\displaystyle \therefore\ \text{Correct option is (D) }9
\\
\displaystyle \textbf{3. } \text{Which of the following statements is true for the function}
\displaystyle f(x)=\begin{cases} x^{2}+3, & x\neq 0 \\ 1, & x=0 \end{cases} \text{ ?}
\displaystyle \text{(A) } f(x) \text{ is continuous and differentiable } \forall x \in R
\displaystyle \text{(B) } f(x) \text{ is continuous } \forall x \in R
\displaystyle \text{(C) } f(x) \text{ is continuous and differentiable } \forall x \in R-\{0\}
\displaystyle \text{(D) } f(x) \text{ is discontinuous at infinitely many points}
\displaystyle \text{Answer:}
\displaystyle f(x)=\begin{cases}x^{2}+3,&x\neq 0\\1,&x=0\end{cases}
\displaystyle \text{For }x\neq 0,\ f(x)=x^{2}+3\text{ which is continuous and differentiable}
\displaystyle \text{At }x=0,\ \lim_{x\to 0}f(x)=\lim_{x\to 0}(x^{2}+3)=3
\displaystyle \text{But }f(0)=1
\displaystyle \therefore\ \lim_{x\to 0}f(x)\neq f(0)\Rightarrow f\text{ is discontinuous at }x=0
\displaystyle \text{Hence }f\text{ is continuous and differentiable for all }x\in R-\{0\}
\displaystyle \therefore\ \text{Correct option is (C)}
\\
\displaystyle \textbf{4. } \text{Let } f(x) \text{ be a continuous function on } [a,b] \text{ and differentiable}
\displaystyle \text{on } (a,b). \text{ Then, this function } f(x) \text{ is strictly increasing in } (a,b)
\displaystyle \text{if :}
\displaystyle \text{(A) } f'(x)<0, \forall x \in (a,b)
\displaystyle \text{(B) } f'(x)>0, \forall x \in (a,b)
\displaystyle \text{(C) } f'(x)=0, \forall x \in (a,b)
\displaystyle \text{(D) } f(x)>0, \forall x \in (a,b)
\displaystyle \text{Answer:}
\displaystyle \text{If }f(x)\text{ is differentiable on }(a,b)\text{ and }f'(x)>0\ \forall x\in(a,b),
\displaystyle \text{then }f(x)\text{ is strictly increasing on }(a,b)
\displaystyle \therefore\ \text{Correct option is (B)}
\\
\displaystyle \textbf{5. } \text{If } \begin{bmatrix} x+y & 2 \\ 5 & xy \end{bmatrix}=\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix},   \text{ then the value of } \left(\frac{24}{x}+\frac{24}{y}\right) \text{ is :}
\displaystyle \text{(A) 7 \quad (B) 6 \quad (C) 8 \quad (D) 18}
\displaystyle \text{Answer:}
\displaystyle \begin{pmatrix}x+y&2\\5&xy\end{pmatrix}= \begin{pmatrix}6&2\\5&8\end{pmatrix}
\displaystyle \Rightarrow\ x+y=6,\quad xy=8
\displaystyle \Rightarrow\ \frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{6}{8}=\frac{3}{4}
\displaystyle \therefore\ \frac{24}{x}+\frac{24}{y}=24\left(\frac{1}{x}+\frac{1}{y}\right)=24\cdot\frac{3}{4}=18
\displaystyle \therefore\ \text{Correct option is (D)}
\\
\displaystyle \textbf{6. } \int_{a}^{b} f(x)\,dx \text{ is equal to :}
\displaystyle \text{(A) } \int_{a}^{b} f(a-x)\,dx
\displaystyle \text{(B) } \int_{a}^{b} f(a+b-x)\,dx
\displaystyle \text{(C) } \int_{a}^{b} f(x-(a+b))\,dx
\displaystyle \text{(D) } \int_{a}^{b} f((a-x)+(b-x))\,dx
\displaystyle \text{Answer:}
\displaystyle \int_{a}^{b}f(x)\,dx
\displaystyle \text{Let }x=a+b-t\Rightarrow dx=-dt
\displaystyle \text{When }x=a,\ t=b;\quad x=b,\ t=a
\displaystyle \therefore\ \int_{a}^{b}f(x)\,dx=\int_{b}^{a}f(a+b-t)(-dt)
\displaystyle =\int_{a}^{b}f(a+b-t)\,dt
\displaystyle =\int_{a}^{b}f(a+b-x)\,dx
\displaystyle \therefore\ \text{Correct option is (B)}
\\
\displaystyle \textbf{7. } \text{Let } \theta \text{ be the angle between two unit vectors } \widehat{a} \text{ and}   \widehat{b} \text{ such that } \sin\theta=\frac{3}{5}.
\displaystyle \text{ Then, } \widehat{a}\cdot\widehat{b} \text{ is equal to :}
\displaystyle \text{(A) } \pm \frac{3}{5} \quad \text{(B) } \pm \frac{3}{4}
\displaystyle \text{(C) } \pm \frac{4}{5} \quad \text{(D) } \pm \frac{4}{3}
\displaystyle \text{Answer:}
\displaystyle \text{Given }\sin\theta=\frac{3}{5}
\displaystyle \Rightarrow\ \cos\theta=\pm\sqrt{1-\sin^{2}\theta}
\displaystyle =\pm\sqrt{1-\frac{9}{25}}
\displaystyle =\pm\sqrt{\frac{16}{25}}
\displaystyle =\pm\frac{4}{5}
\displaystyle \text{Since }\hat{a},\hat{b}\text{ are unit vectors, }\hat{a}\cdot\hat{b}=\cos\theta
\displaystyle \therefore\ \hat{a}\cdot\hat{b}=\pm\frac{4}{5}
\displaystyle \therefore\ \text{Correct option is (C)}
\\
\displaystyle \textbf{8. } \text{The integrating factor of the differential equation } \\ (1-x^{2})\frac{dy}{dx}+xy=ax,   -1<x<1, \text{ is :}
\displaystyle \text{(A) } \frac{1}{x^{2}-1} \quad \text{(B) } \frac{1}{\sqrt{x^{2}-1}}
\displaystyle \text{(C) } \frac{1}{1-x^{2}} \quad \text{(D) } \frac{1}{\sqrt{1-x^{2}}}
\displaystyle \text{Answer:}
\displaystyle (1-x^{2})\frac{dy}{dx}+xy=ax
\displaystyle \Rightarrow\ \frac{dy}{dx}+\frac{x}{1-x^{2}}y=\frac{ax}{1-x^{2}}
\displaystyle \text{Integrating factor }=\exp\left(\int \frac{x}{1-x^{2}}\,dx\right)
\displaystyle \int \frac{x}{1-x^{2}}\,dx=-\frac{1}{2}\log(1-x^{2})
\displaystyle \therefore\ \mathrm{I.F.}=e^{-\frac{1}{2}\log(1-x^{2})}
\displaystyle =\frac{1}{\sqrt{1-x^{2}}}
\displaystyle \therefore\ \text{Correct option is (D)}
\\
\displaystyle \textbf{9. } \text{If the direction cosines of a line are } \sqrt{3}k,\sqrt{3}k,\sqrt{3}k,   \text{ then the value of } k \text{ is :}
\displaystyle \text{(A) } \pm 1 \quad \text{(B) } \pm \sqrt{3}
\displaystyle \text{(C) } \pm 3 \quad \text{(D) } \pm \frac{1}{3}
\displaystyle \text{Answer:}
\displaystyle \text{Direction cosines }l,m,n=\sqrt{3}k,\ \sqrt{3}k,\ \sqrt{3}k
\displaystyle \text{For direction cosines, }l^{2}+m^{2}+n^{2}=1
\displaystyle \Rightarrow\ (\sqrt{3}k)^{2}+(\sqrt{3}k)^{2}+(\sqrt{3}k)^{2}=1
\displaystyle \Rightarrow\ 3k^{2}+3k^{2}+3k^{2}=1
\displaystyle \Rightarrow\ 9k^{2}=1
\displaystyle \Rightarrow\ k^{2}=\frac{1}{9}
\displaystyle \Rightarrow\ k=\pm\frac{1}{3}
\displaystyle \therefore\ \text{Correct option is (D)}
\\
\displaystyle \textbf{10. } \text{A linear programming problem deals with the optimization of a/an :}
\displaystyle \text{(A) logarithmic function \quad (B) linear function}
\displaystyle \text{(C) quadratic function \quad (D) exponential function}
\displaystyle \text{Answer:}
\displaystyle \text{A linear programming problem deals with the optimization of a linear function}
\displaystyle \therefore\ \text{Correct option is (B)}
\\
\displaystyle \textbf{11. } \text{If } P(A|B)=P(A'|B), \text{ then which of the following statements is true ?}
\displaystyle \text{(A) } P(A)=P(A') \quad \text{(B) } P(A)=2P(B)
\displaystyle \text{(C) } P(A\cap B)=\frac{1}{2}P(B) \quad \text{(D) } P(A\cap B)=2P(B)
\displaystyle \text{Answer:}
\displaystyle P(A\mid B)=P(A'\mid B)
\displaystyle \Rightarrow\ \frac{P(A\cap B)}{P(B)}=\frac{P(A'\cap B)}{P(B)}
\displaystyle \Rightarrow\ P(A\cap B)=P(A'\cap B)
\displaystyle \text{But }P(A\cap B)+P(A'\cap B)=P(B)
\displaystyle \Rightarrow\ 2P(A\cap B)=P(B)
\displaystyle \Rightarrow\ P(A\cap B)=\frac{1}{2}P(B)
\displaystyle \therefore\ \text{Correct option is (C)}
\\
\displaystyle \textbf{12. } \begin{vmatrix} x+1 & x-1 \\ x^{2}+x+1 & x^{2}-x+1 \end{vmatrix} \text{ is equal to :}
\displaystyle \text{(A) } 2x^{3} \quad \text{(B) } 2
\displaystyle \text{(C) } 0 \quad \text{(D) } 2x^{3}-2
\displaystyle \text{Answer:}
\displaystyle \begin{vmatrix}x+1&x-1\\x^{2}+x+1&x^{2}-x+1\end{vmatrix}
\displaystyle =(x+1)(x^{2}-x+1)-(x-1)(x^{2}+x+1)
\displaystyle =\left(x^{3}-x^{2}+x+x^{2}-x+1\right)-\left(x^{3}+x^{2}+x-x^{2}-x-1\right)
\displaystyle =(x^{3}+1)-(x^{3}-1)
\displaystyle =x^{3}+1-x^{3}+1
\displaystyle =2
\displaystyle \therefore\ \text{Correct option is (B)}
\\
\displaystyle \textbf{13. } \text{The derivative of } \sin(x^{2}) \text{ w.r.t. } x, \text{ at } x=\sqrt{\pi} \text{ is :}
\displaystyle \text{(A) } 1 \quad \text{(B) } -1
\displaystyle \text{(C) } -2\sqrt{\pi} \quad \text{(D) } 2\sqrt{\pi}
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dx}\left(\sin(x^{2})\right)=\cos(x^{2})\cdot 2x
\displaystyle \text{At }x=\sqrt{\pi},\ \frac{d}{dx}\left(\sin(x^{2})\right)=2\sqrt{\pi}\cos(\pi)
\displaystyle =2\sqrt{\pi}\cdot(-1)
\displaystyle =-2\sqrt{\pi}
\displaystyle \therefore\ \text{Correct option is (C)}
\\
\displaystyle \textbf{14. } \text{The order and degree of the differential equation } \left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{3}=\frac{d^{2}y}{dx^{2}}
\displaystyle \text{ respectively are :}
\displaystyle \text{(A) } 1,2 \quad \text{(B) } 2,3
\displaystyle \text{(C) } 2,1 \quad \text{(D) } 2,6
\displaystyle \text{Answer:}
\displaystyle \left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{3}=\frac{d^{2}y}{dx^{2}}
\displaystyle \text{Highest order derivative present is }\frac{d^{2}y}{dx^{2}}\Rightarrow \text{order }=2
\displaystyle \text{The equation is polynomial in derivatives and the highest order derivative appears to power }1
\displaystyle \therefore\ \text{degree }=1
\displaystyle \therefore\ \text{order }=2,\ \text{degree }=1
\displaystyle \therefore\ \text{Correct option is (C)}
\\
\displaystyle \textbf{15. } \text{The vector with terminal point A (2,-3,5) and initial point B (3,-4,7)}   \text{ is :}
\displaystyle \text{(A) } \widehat{i}-\widehat{j}+2\widehat{k} \quad \text{(B) } \widehat{i}+\widehat{j}+2\widehat{k}
\displaystyle \text{(C) } -\widehat{i}-\widehat{j}-2\widehat{k} \quad \text{(D) } -\widehat{i}+\widehat{j}-2\widehat{k}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{BA}=A-B=(2-3,\,-3-(-4),\,5-7)
\displaystyle =(-1,\,1,\,-2)
\displaystyle =-\widehat{i}+\widehat{j}-2\widehat{k}
\displaystyle \therefore\ \text{Correct option is (D)}
\\
\displaystyle \textbf{16. } \text{The distance of point P(a,b,c) from y-axis is :}
\displaystyle \text{(A) } b \quad \text{(B) } b^{2}
\displaystyle \text{(C) } \sqrt{a^{2}+c^{2}} \quad \text{(D) } a^{2}+c^{2}
\displaystyle \text{Answer:}
\displaystyle \text{Distance of }P(a,b,c)\text{ from the }y\text{-axis is the perpendicular distance}
\displaystyle \text{to the line }(0,y,0),\ \text{i.e., distance in the }xz\text{-plane}
\displaystyle =\sqrt{a^{2}+c^{2}}
\displaystyle \therefore\ \text{Correct option is (C)}
\\
\displaystyle \textbf{17. } \text{The number of corner points of the feasible region determined by}
\displaystyle \text{ constraints } x\geq 0, y\geq 0, x+y\geq 4 \text{ is :}
\displaystyle \text{(A) } 0 \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 3
\displaystyle \text{Answer:}
\displaystyle x\geq 0,\ y\geq 0,\ x+y\geq 4
\displaystyle \text{The feasible region lies in the first quadrant above the line }x+y=4
\displaystyle \text{Corner points occur at intersections of the boundary lines}
\displaystyle \text{Intersection of }x+y=4\text{ with }x\text{-axis: }(4,0)
\displaystyle \text{Intersection of }x+y=4\text{ with }y\text{-axis: }(0,4)
\displaystyle \text{Thus, the feasible region has two corner points}
\displaystyle \therefore\ \text{Correct option is (C)}
\\
\displaystyle \textbf{18. } \text{If A and B are two non-zero square matrices of same order such that}
\displaystyle (A+B)^{2}=A^{2}+B^{2}, \text{ then :}
\displaystyle \text{(A) } AB=O \quad \text{(B) } AB=-BA
\displaystyle \text{(C) } BA=O \quad \text{(D) } AB=BA
\displaystyle \text{Answer:}
\displaystyle (A+B)^{2}=A^{2}+AB+BA+B^{2}
\displaystyle \text{Given }(A+B)^{2}=A^{2}+B^{2}
\displaystyle \Rightarrow\ A^{2}+AB+BA+B^{2}=A^{2}+B^{2}
\displaystyle \Rightarrow\ AB+BA=0
\displaystyle \Rightarrow\ AB=-BA
\displaystyle \therefore\ \text{Correct option is (B)}
\\
\displaystyle \text{Questions number 19 and 20 are Assertion and Reason based questions. Two}
\displaystyle \text{statements are given, one labelled Assertion (A) and the other labelled Reason}
\displaystyle \text{(R). Select the correct answer from the codes (A), (B), (C) and (D) as given}
\displaystyle \text{below.}
\displaystyle \text{(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the}
\displaystyle \text{correct explanation of the Assertion (A).}
\displaystyle \text{(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not}
\displaystyle \text{the correct explanation of the Assertion (A).}
\displaystyle \text{(C) Assertion (A) is true, but Reason (R) is false.}
\displaystyle \text{(D) Assertion (A) is false, but Reason (R) is true.}
\\
\displaystyle \textbf{19. } \text{Assertion (A) : For matrix } A=\begin{bmatrix} 1 & \cos\theta & 1 \\ -\cos\theta & 1 & \cos\theta \\ -1 & -\cos\theta & 1 \end{bmatrix},
\displaystyle \text{ where } \theta \in [0,2\pi], \text{ } |A| \in [2,4].
\displaystyle \text{Reason (R) : } \cos\theta \in [-1,1], \forall \theta \in [0,2\pi].
\displaystyle \text{Answer:}
\displaystyle A=\begin{pmatrix}1&\cos\theta&1\\-\cos\theta&1&\cos\theta\\-1&-\cos\theta&1\end{pmatrix}
\displaystyle |A|=\begin{vmatrix}1&\cos\theta&1\\-\cos\theta&1&\cos\theta\\-1&-\cos\theta&1\end{vmatrix}
\displaystyle =1\begin{vmatrix}1&\cos\theta\\-\cos\theta&1\end{vmatrix} -\cos\theta\begin{vmatrix}-\cos\theta&\cos\theta\\-1&1\end{vmatrix} +1\begin{vmatrix}-\cos\theta&1\\-1&-\cos\theta\end{vmatrix}
\displaystyle =(1+\cos^{2}\theta)-\cos\theta(0)+(\cos^{2}\theta+1)
\displaystyle =2(1+\cos^{2}\theta)
\displaystyle \text{Since }\cos\theta\in[-1,1],\ \cos^{2}\theta\in[0,1]
\displaystyle \Rightarrow\ |A|=2(1+\cos^{2}\theta)\in[2,4]
\displaystyle \text{Thus Assertion (A) is true}
\displaystyle \text{Reason (R): }\cos\theta\in[-1,1]\text{ is also true}
\displaystyle \text{and it explains the range of }|A|
\displaystyle \therefore\ \text{Correct option is (A)}
\\
\displaystyle \textbf{20. } \text{Assertion (A) : A line in space cannot be drawn perpendicular to x, y}
\displaystyle \text{and z axes simultaneously.}
\displaystyle \text{Reason (R) : For any line making angles } \alpha,\beta,\gamma \text{ with the}   \text{positive directions} \\ \text{of x, y and z axes respectively, } \cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1.
\displaystyle \text{Answer:}
\displaystyle \text{Assertion (A): A line cannot be perpendicular to }x,y,z\text{ axes simultaneously}
\displaystyle \text{If a line is perpendicular to all three axes, its direction cosines }l,m,n=0
\displaystyle \text{But }l^{2}+m^{2}+n^{2}=1\Rightarrow 0\neq 1
\displaystyle \text{Hence Assertion (A) is true}
\displaystyle \text{Reason (R): }\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1\text{ is true}
\displaystyle \text{and it explains why all three cannot be zero simultaneously}
\displaystyle \therefore\ \text{Correct option is (A)}

\displaystyle \textbf{SECTION B}
\displaystyle \text{This section comprises very short answer (VSA) type questions of 2 marks each.}

\displaystyle \textbf{21. } \text{(a) Check whether the function } f(x)=x^{2}|x| \text{ is differentiable at } x=0   \text{or not.}
\displaystyle \textbf{OR}
\displaystyle \text{(b) If } y=\sqrt{\tan\sqrt{x}}, \text{ prove that } \sqrt{x}\frac{dy}{dx}=\frac{1+y^{4}}{4y}.
\displaystyle \text{Answer:}
\displaystyle \textbf{(a) } \text{Checking differentiability of }f(x)=x^{2}|x|\text{ at }x=0:
\displaystyle f(0)=0
\displaystyle f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}
\displaystyle =\lim_{h\to 0}\frac{h^{2}|h|}{h}
\displaystyle =\lim_{h\to 0}h|h|
\displaystyle =0
\displaystyle \therefore\ f'(0)\text{ exists}
\displaystyle \therefore\ f(x)=x^{2}|x|\text{ is differentiable at }x=0
\displaystyle \textbf{OR}
\displaystyle \textbf{(b) } \text{Given }y=\sqrt{\tan\sqrt{x}}
\displaystyle \Rightarrow\ y^{2}=\tan\sqrt{x}
\displaystyle \text{Differentiating both sides w.r.t. }x,
\displaystyle 2y\frac{dy}{dx}=\sec^{2}\sqrt{x}\cdot\frac{1}{2\sqrt{x}}
\displaystyle \Rightarrow\ 4y\sqrt{x}\frac{dy}{dx}=\sec^{2}\sqrt{x}
\displaystyle \text{Now, }\tan\sqrt{x}=y^{2}
\displaystyle \Rightarrow\ \sec^{2}\sqrt{x}=1+\tan^{2}\sqrt{x}
\displaystyle =1+y^{4}
\displaystyle \therefore\ 4y\sqrt{x}\frac{dy}{dx}=1+y^{4}
\displaystyle \therefore\ \sqrt{x}\frac{dy}{dx}=\frac{1+y^{4}}{4y}
\\
\displaystyle \textbf{22. } \text{Show that the function } f(x)=4x^{3}-18x^{2}+27x-7 \text{ has neither maxima}
\displaystyle \text{nor minima.}
\displaystyle \text{Answer:}
\displaystyle f(x)=4x^{3}-18x^{2}+27x-7
\displaystyle f'(x)=12x^{2}-36x+27
\displaystyle =3(4x^{2}-12x+9)
\displaystyle =3(2x-3)^{2}
\displaystyle \therefore\ f'(x)\geq 0\ \text{for all }x
\displaystyle \text{and }f'(x)=0\text{ only at }x=\frac{3}{2}
\displaystyle \text{Since }f'(x)\text{ does not change sign at }x=\frac{3}{2},\text{ the function is increasing on both sides of }x=\frac{3}{2}
\displaystyle \text{Hence }x=\frac{3}{2}\text{ is neither a point of maxima nor a point of minima}
\displaystyle \text{Also, }f''(x)=24x-36
\displaystyle \Rightarrow\ f''\left(\frac{3}{2}\right)=0
\displaystyle \text{Thus }x=\frac{3}{2}\text{ is a stationary point of inflexion}
\displaystyle \therefore\ \text{the function }f(x)=4x^{3}-18x^{2}+27x-7\text{ has neither maxima nor minima}
\\
\displaystyle \textbf{23. } \text{(a) Find :}
\displaystyle \int x\sqrt{1+2x}\,dx
\displaystyle \textbf{OR}
\displaystyle \text{(b) Evaluate :}
\displaystyle \int_{0}^{\frac{\pi^{2}}{4}} \frac{\sin\sqrt{x}}{\sqrt{x}}\,dx
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \int x\sqrt{1+2x}\,dx
\displaystyle \text{Let }t=1+2x\Rightarrow dt=2dx,\ x=\frac{t-1}{2}
\displaystyle \Rightarrow\ \int x\sqrt{1+2x}\,dx=\int \frac{t-1}{2}\cdot\sqrt{t}\cdot\frac{dt}{2}
\displaystyle =\frac{1}{4}\int (t-1)t^{\frac{1}{2}}\,dt
\displaystyle =\frac{1}{4}\int (t^{\frac{3}{2}}-t^{\frac{1}{2}})\,dt
\displaystyle =\frac{1}{4}\left(\frac{2}{5}t^{\frac{5}{2}}-\frac{2}{3}t^{\frac{3}{2}}\right)+c
\displaystyle =\frac{1}{10}(1+2x)^{\frac{5}{2}}-\frac{1}{6}(1+2x)^{\frac{3}{2}}+c
\displaystyle \textbf{OR}
\displaystyle \textbf{(b)}
\displaystyle \int_{0}^{\frac{\pi^{2}}{4}}\frac{\sin\sqrt{x}}{\sqrt{x}}\,dx
\displaystyle \text{Let }t=\sqrt{x}\Rightarrow x=t^{2},\ dx=2t\,dt
\displaystyle \Rightarrow\ \int_{0}^{\frac{\pi^{2}}{4}}\frac{\sin\sqrt{x}}{\sqrt{x}}\,dx=\int_{0}^{\frac{\pi}{2}}\frac{\sin t}{t}\cdot 2t\,dt
\displaystyle =2\int_{0}^{\frac{\pi}{2}}\sin t\,dt
\displaystyle =2\left[-\cos t\right]_{0}^{\frac{\pi}{2}}
\displaystyle =2(1-0)
\displaystyle =2
\\
\displaystyle \textbf{24. } \text{If } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ are two non-zero vectors such that}   (\overrightarrow{a}+\overrightarrow{b})\perp \overrightarrow{a}  \\ \text{ and } (2\overrightarrow{a}+\overrightarrow{b})\perp \overrightarrow{b},   \text{ then prove that } |\overrightarrow{b}|=\sqrt{2}\,|\overrightarrow{a}|.
\displaystyle \text{Answer:}
\displaystyle (\overrightarrow{a}+\overrightarrow{b})\perp \overrightarrow{a}\text{ and }(2\overrightarrow{a}+\overrightarrow{b})\perp \overrightarrow{b}
\displaystyle \Rightarrow\ (\overrightarrow{a}+\overrightarrow{b})\cdot \overrightarrow{a}=0
\displaystyle \Rightarrow\ \overrightarrow{a}\cdot \overrightarrow{a}+\overrightarrow{b}\cdot \overrightarrow{a}=0
\displaystyle \Rightarrow\ |\overrightarrow{a}|^{2}+\overrightarrow{a}\cdot \overrightarrow{b}=0
\displaystyle \Rightarrow\ \overrightarrow{a}\cdot \overrightarrow{b}=-|\overrightarrow{a}|^{2}
\displaystyle \text{Also, }(2\overrightarrow{a}+\overrightarrow{b})\cdot \overrightarrow{b}=0
\displaystyle \Rightarrow\ 2\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{b}=0
\displaystyle \Rightarrow\ 2\overrightarrow{a}\cdot \overrightarrow{b}+|\overrightarrow{b}|^{2}=0
\displaystyle \text{Substituting }\overrightarrow{a}\cdot \overrightarrow{b}=-|\overrightarrow{a}|^{2},
\displaystyle 2(-|\overrightarrow{a}|^{2})+|\overrightarrow{b}|^{2}=0
\displaystyle \Rightarrow\ |\overrightarrow{b}|^{2}=2|\overrightarrow{a}|^{2}
\displaystyle \Rightarrow\ |\overrightarrow{b}|=\sqrt{2}\,|\overrightarrow{a}|
\displaystyle \therefore\ \text{Proved that }|\overrightarrow{b}|=\sqrt{2}\,|\overrightarrow{a}|
\\
\displaystyle \textbf{25. } \text{In the given figure, ABCD is a parallelogram. If } \overrightarrow{AB}=2\widehat{i}-4\widehat{j}+5\widehat{k} \text{ and}
\displaystyle \overrightarrow{DB}=3\widehat{i}-6\widehat{j}+2\widehat{k}, \text{ then find } \overrightarrow{AD} \text{ and hence find the area of }   \text{parallelogram ABCD.} \displaystyle \text{Answer:}
\displaystyle \overrightarrow{DB}=\overrightarrow{DA}+\overrightarrow{AB}
\displaystyle \text{But }\overrightarrow{DA}=-\overrightarrow{AD}
\displaystyle \therefore\ \overrightarrow{DB}=\overrightarrow{AB}-\overrightarrow{AD}
\displaystyle \Rightarrow\ \overrightarrow{AD}=\overrightarrow{AB}-\overrightarrow{DB}
\displaystyle \overrightarrow{AD}=(2\widehat{i}-4\widehat{j}+5\widehat{k})-(3\widehat{i}-6\widehat{j}+2\widehat{k})
\displaystyle \therefore\ \overrightarrow{AD}=-\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \text{Area of parallelogram }ABCD=\left|\overrightarrow{AB}\times\overrightarrow{AD}\right|
\displaystyle \overrightarrow{AB}\times\overrightarrow{AD}= \begin{vmatrix} \widehat{i}&\widehat{j}&\widehat{k}\\ 2&-4&5\\ -1&2&3 \end{vmatrix}
\displaystyle =\widehat{i}\left((-4)(3)-5(2)\right)-\widehat{j}\left(2(3)-5(-1)\right)+\widehat{k}\left(2(2)-(-4)(-1)\right)
\displaystyle =\widehat{i}(-12-10)-\widehat{j}(6+5)+\widehat{k}(4-4)
\displaystyle =-22\widehat{i}-11\widehat{j}
\displaystyle \therefore\ \left|\overrightarrow{AB}\times\overrightarrow{AD}\right|=\sqrt{(-22)^{2}+(-11)^{2}}
\displaystyle =\sqrt{484+121}
\displaystyle =\sqrt{605}=11\sqrt{5}
\displaystyle \therefore\ \overrightarrow{AD}=-\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \therefore\ \text{Area of parallelogram }ABCD=11\sqrt{5}\text{ square units}

\displaystyle \textbf{SECTION C}
\displaystyle \text{This section comprises short answer (SA) type questions of 3 marks each.}

\displaystyle \textbf{26. } \text{(a) A relation } R \text{ on set } A=\{1,2,3,4,5\} \text{ is defined as}
\displaystyle R=\{(x,y):|x^{2}-y^{2}|<8\}. \text{ Check whether the relation } R \text{ is reflexive,}
\displaystyle \text{symmetric and transitive.}
\displaystyle \textbf{OR}
\displaystyle \text{(b) A function } f \text{ is defined from } R\rightarrow R \text{ as } f(x)=ax+b, \text{ such that } \\ f(1)=1   \text{and } f(2)=3. \text{ Find function } f(x). \text{ Hence, check whether function}\\  f(x) \text{ is}   \text{one-one and onto or not.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(a) } \text{Checking whether }R\text{ is reflexive, symmetric and transitive:}
\displaystyle A=\{1,2,3,4,5\},\quad R=\{(x,y):|x^{2}-y^{2}|<8\}
\displaystyle \textbf{Reflexive:}
\displaystyle \text{For every }x\in A,\ |x^{2}-x^{2}|=0<8
\displaystyle \therefore\ (x,x)\in R\text{ for all }x\in A
\displaystyle \therefore\ R\text{ is reflexive}
\displaystyle \textbf{Symmetric:}
\displaystyle \text{If }(x,y)\in R,\text{ then }|x^{2}-y^{2}|<8
\displaystyle \text{But }|y^{2}-x^{2}|=|x^{2}-y^{2}|<8
\displaystyle \therefore\ (y,x)\in R
\displaystyle \therefore\ R\text{ is symmetric}
\displaystyle \textbf{Transitive:}
\displaystyle (1,2)\in R\text{ since }|1^{2}-2^{2}|=|1-4|=3<8
\displaystyle (2,3)\in R\text{ since }|2^{2}-3^{2}|=|4-9|=5<8
\displaystyle \text{But }(1,3)\notin R\text{ since }|1^{2}-3^{2}|=|1-9|=8\not<8
\displaystyle \therefore\ R\text{ is not transitive}
\displaystyle \therefore\ R\text{ is reflexive and symmetric, but not transitive}
\displaystyle \textbf{OR}
\displaystyle \textbf{(b) } \text{Finding }f(x)\text{:}
\displaystyle f(x)=ax+b
\displaystyle f(1)=1\Rightarrow a+b=1
\displaystyle f(2)=3\Rightarrow 2a+b=3
\displaystyle \text{Subtracting, }a=2
\displaystyle \text{Hence }b=1-a=1-2=-1
\displaystyle \therefore\ f(x)=2x-1
\displaystyle \text{Checking one-one:}
\displaystyle \text{Let }f(x_{1})=f(x_{2})
\displaystyle \Rightarrow\ 2x_{1}-1=2x_{2}-1
\displaystyle \Rightarrow\ 2x_{1}=2x_{2}
\displaystyle \Rightarrow\ x_{1}=x_{2}
\displaystyle \therefore\ f(x)\text{ is one-one}
\displaystyle \text{Checking onto:}
\displaystyle \text{Let }y\in R
\displaystyle \text{We need }x\in R\text{ such that }y=2x-1
\displaystyle \Rightarrow\ 2x=y+1
\displaystyle \Rightarrow\ x=\frac{y+1}{2}
\displaystyle \text{Since }\frac{y+1}{2}\in R\text{ for every }y\in R,\text{ such an }x\text{ exists}
\displaystyle \therefore\ f(x)\text{ is onto}
\displaystyle \therefore\ f(x)=2x-1\text{ is both one-one and onto}
\displaystyle \textbf{(a) } \text{Checking whether }R\text{ is reflexive, symmetric and transitive:}
\displaystyle A=\{1,2,3,4,5\},\quad R=\{(x,y):|x^{2}-y^{2}|<8\}
\displaystyle \textbf{Reflexive:}
\displaystyle \text{For every }x\in A,\ |x^{2}-x^{2}|=0<8
\displaystyle \therefore\ (x,x)\in R\text{ for all }x\in A
\displaystyle \therefore\ R\text{ is reflexive}
\displaystyle \textbf{Symmetric:}
\displaystyle \text{If }(x,y)\in R,\text{ then }|x^{2}-y^{2}|<8
\displaystyle \text{But }|y^{2}-x^{2}|=|x^{2}-y^{2}|<8
\displaystyle \therefore\ (y,x)\in R
\displaystyle \therefore\ R\text{ is symmetric}
\displaystyle \textbf{Transitive:}
\displaystyle (1,2)\in R\text{ since }|1^{2}-2^{2}|=|1-4|=3<8
\displaystyle (2,3)\in R\text{ since }|2^{2}-3^{2}|=|4-9|=5<8
\displaystyle \text{But }(1,3)\notin R\text{ since }|1^{2}-3^{2}|=|1-9|=8\not<8
\displaystyle \therefore\ R\text{ is not transitive}
\displaystyle \therefore\ R\text{ is reflexive and symmetric, but not transitive}
\displaystyle \textbf{OR}
\displaystyle \textbf{(b) } \text{Finding }f(x)\text{:}
\displaystyle f(x)=ax+b
\displaystyle f(1)=1\Rightarrow a+b=1
\displaystyle f(2)=3\Rightarrow 2a+b=3
\displaystyle \text{Subtracting, }a=2
\displaystyle \text{Hence }b=1-a=1-2=-1
\displaystyle \therefore\ f(x)=2x-1
\displaystyle \text{Checking one-one:}
\displaystyle \text{Let }f(x_{1})=f(x_{2})
\displaystyle \Rightarrow\ 2x_{1}-1=2x_{2}-1
\displaystyle \Rightarrow\ 2x_{1}=2x_{2}
\displaystyle \Rightarrow\ x_{1}=x_{2}
\displaystyle \therefore\ f(x)\text{ is one-one}
\displaystyle \text{Checking onto:}
\displaystyle \text{Let }y\in R
\displaystyle \text{We need }x\in R\text{ such that }y=2x-1
\displaystyle \Rightarrow\ 2x=y+1
\displaystyle \Rightarrow\ x=\frac{y+1}{2}
\displaystyle \text{Since }\frac{y+1}{2}\in R\text{ for every }y\in R,\text{ such an }x\text{ exists}
\displaystyle \therefore\ f(x)\text{ is onto}
\displaystyle \therefore\ f(x)=2x-1\text{ is both one-one and onto}
\\
\displaystyle \textbf{27. } \text{(a) If } \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y), \text{ prove that } \frac{dy}{dx}=   \sqrt{\frac{1-y^{2}}{1-x^{2}}}.
\displaystyle \textbf{OR}
\displaystyle \text{(b) If } y=(\tan x)^{x}, \text{ then find } \frac{dy}{dx}.
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)
\displaystyle \text{Differentiating both sides w.r.t. }x,
\displaystyle \frac{-x}{\sqrt{1-x^{2}}}+\frac{-y}{\sqrt{1-y^{2}}}\frac{dy}{dx}=a\left(1-\frac{dy}{dx}\right)
\displaystyle \Rightarrow\ a+\frac{x}{\sqrt{1-x^{2}}}=\left(a-\frac{y}{\sqrt{1-y^{2}}}\right)\frac{dy}{dx}
\displaystyle \therefore\ \frac{dy}{dx}=\frac{a+\frac{x}{\sqrt{1-x^{2}}}}{a-\frac{y}{\sqrt{1-y^{2}}}}
\displaystyle \text{From the given equation, }a=\frac{\sqrt{1-x^{2}}+\sqrt{1-y^{2}}}{x-y}
\displaystyle \text{Substituting this in the numerator,}
\displaystyle a+\frac{x}{\sqrt{1-x^{2}}}=\frac{\sqrt{1-x^{2}}+\sqrt{1-y^{2}}}{x-y}+\frac{x}{\sqrt{1-x^{2}}}
\displaystyle =\frac{(1-x^{2})+\sqrt{(1-x^{2})(1-y^{2})}+x(x-y)}{(x-y)\sqrt{1-x^{2}}}
\displaystyle =\frac{1-xy+\sqrt{(1-x^{2})(1-y^{2})}}{(x-y)\sqrt{1-x^{2}}}
\displaystyle \text{Similarly,}
\displaystyle a-\frac{y}{\sqrt{1-y^{2}}}=\frac{\sqrt{1-x^{2}}+\sqrt{1-y^{2}}}{x-y}-\frac{y}{\sqrt{1-y^{2}}}
\displaystyle =\frac{\sqrt{(1-x^{2})(1-y^{2})}+(1-y^{2})-y(x-y)}{(x-y)\sqrt{1-y^{2}}}
\displaystyle =\frac{1-xy+\sqrt{(1-x^{2})(1-y^{2})}}{(x-y)\sqrt{1-y^{2}}}
\displaystyle \therefore\ \frac{dy}{dx}=\frac{\frac{1-xy+\sqrt{(1-x^{2})(1-y^{2})}}{(x-y)\sqrt{1-x^{2}}}}{\frac{1-xy+\sqrt{(1-x^{2})(1-y^{2})}}{(x-y)\sqrt{1-y^{2}}}}
\displaystyle =\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}
\displaystyle =\sqrt{\frac{1-y^{2}}{1-x^{2}}}
\displaystyle \therefore\ \frac{dy}{dx}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}
\displaystyle \textbf{OR}
\displaystyle \textbf{(b)}
\displaystyle y=(\tan x)^{x}
\displaystyle \log y=x\log(\tan x)
\displaystyle \text{Differentiating both sides w.r.t. }x,
\displaystyle \frac{1}{y}\frac{dy}{dx}=\log(\tan x)+x\frac{1}{\tan x}\sec^{2}x
\displaystyle \therefore\ \frac{dy}{dx}=y\left(\log(\tan x)+x\frac{\sec^{2}x}{\tan x}\right)
\displaystyle \therefore\ \frac{dy}{dx}=(\tan x)^{x}\left(\log(\tan x)+x\frac{\sec^{2}x}{\tan x}\right)
\\
\displaystyle \textbf{28. } \text{(a) Find :}
\displaystyle \int \frac{x^{2}}{(x^{2}+4)(x^{2}+9)}\,dx
\displaystyle \textbf{OR}
\displaystyle \text{(b) Evaluate :}
\displaystyle \int_{1}^{3}\left(|x-1|+|x-2|+|x-3|\right)\,dx
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \int \frac{x^{2}}{(x^{2}+4)(x^{2}+9)}\,dx
\displaystyle \text{Let }\frac{x^{2}}{(x^{2}+4)(x^{2}+9)}=\frac{A}{x^{2}+4}+\frac{B}{x^{2}+9}
\displaystyle \Rightarrow\ x^{2}=A(x^{2}+9)+B(x^{2}+4)
\displaystyle \Rightarrow\ x^{2}=(A+B)x^{2}+(9A+4B)
\displaystyle \Rightarrow\ A+B=1,\quad 9A+4B=0
\displaystyle \Rightarrow\ A=-\frac{4}{5},\quad B=\frac{9}{5}
\displaystyle \therefore\ \int \frac{x^{2}}{(x^{2}+4)(x^{2}+9)}\,dx=\int \left(-\frac{4}{5(x^{2}+4)}+\frac{9}{5(x^{2}+9)}\right)dx
\displaystyle =-\frac{4}{5}\int \frac{dx}{x^{2}+4}+\frac{9}{5}\int \frac{dx}{x^{2}+9}
\displaystyle =-\frac{4}{5}\cdot\frac{1}{2}\tan^{-1}\frac{x}{2}+\frac{9}{5}\cdot\frac{1}{3}\tan^{-1}\frac{x}{3}+c
\displaystyle =-\frac{2}{5}\tan^{-1}\frac{x}{2}+\frac{3}{5}\tan^{-1}\frac{x}{3}+c
\displaystyle \textbf{OR}
\displaystyle \textbf{(b)}
\displaystyle \int_{1}^{3}\left(|x-1|+|x-2|+|x-3|\right)dx
\displaystyle \text{For }1\leq x\leq 2,\ |x-1|=x-1,\ |x-2|=2-x,\ |x-3|=3-x
\displaystyle \Rightarrow\ |x-1|+|x-2|+|x-3|=(x-1)+(2-x)+(3-x)=4-x
\displaystyle \text{For }2\leq x\leq 3,\ |x-1|=x-1,\ |x-2|=x-2,\ |x-3|=3-x
\displaystyle \Rightarrow\ |x-1|+|x-2|+|x-3|=(x-1)+(x-2)+(3-x)=x
\displaystyle \therefore\ \int_{1}^{3}\left(|x-1|+|x-2|+|x-3|\right)dx=\int_{1}^{2}(4-x)\,dx+\int_{2}^{3}x\,dx
\displaystyle =\left[4x-\frac{x^{2}}{2}\right]_{1}^{2}+\left[\frac{x^{2}}{2}\right]_{2}^{3}
\displaystyle =(8-2)-\left(4-\frac{1}{2}\right)+\frac{9}{2}-2
\displaystyle =6-\frac{7}{2}+\frac{5}{2}
\displaystyle =5
\\
\displaystyle \textbf{29. } \text{Find the particular solution of the differential equation given by}
\displaystyle x^{2}\frac{dy}{dx}-xy=x^{2}\cos^{2}\left(\frac{y}{2x}\right), \text{ given that when } x=1, y=\frac{\pi}{2}.
\displaystyle \text{Answer:}
\displaystyle x^{2}\frac{dy}{dx}-xy=x^{2}\cos^{2}\left(\frac{y}{2x}\right)
\displaystyle \Rightarrow\ \frac{dy}{dx}-\frac{y}{x}=\cos^{2}\left(\frac{y}{2x}\right)
\displaystyle \text{Let }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \Rightarrow\ v+x\frac{dv}{dx}-v=\cos^{2}\left(\frac{v}{2}\right)
\displaystyle \Rightarrow\ x\frac{dv}{dx}=\cos^{2}\left(\frac{v}{2}\right)
\displaystyle \Rightarrow\ \sec^{2}\left(\frac{v}{2}\right)\,dv=\frac{dx}{x}
\displaystyle \int \sec^{2}\left(\frac{v}{2}\right)\,dv=\int \frac{dx}{x}
\displaystyle 2\tan\left(\frac{v}{2}\right)=\log x+C
\displaystyle \text{Since }v=\frac{y}{x},\ 2\tan\left(\frac{y}{2x}\right)=\log x+C
\displaystyle \text{Given }x=1,\ y=\frac{\pi}{2}
\displaystyle \Rightarrow\ 2\tan\left(\frac{\pi}{4}\right)=\log 1+C
\displaystyle \Rightarrow\ 2=C
\displaystyle \therefore\ 2\tan\left(\frac{y}{2x}\right)=\log x+2
\displaystyle \text{Hence, the particular solution is }2\tan\left(\frac{y}{2x}\right)=\log x+2
\\
\displaystyle \textbf{30. } \text{Solve the following linear programming problem graphically :}
\displaystyle \text{Maximise } z=500x+300y,
\displaystyle \text{Subject to constraints}
\displaystyle x+2y\leq 12
\displaystyle 2x+y\leq 12
\displaystyle 4x+5y\geq 20
\displaystyle x\geq 0, y\geq 0
\displaystyle \text{Answer:}
\displaystyle \text{Given }z=500x+300y
\displaystyle \text{subject to }x+2y\leq 12,\ 2x+y\leq 12,\ 4x+5y\geq 20,\ x\geq 0,\ y\geq 0
\displaystyle \text{The boundary lines are }x+2y=12,\ 2x+y=12,\ 4x+5y=20
\displaystyle \text{Now find the corner points of the feasible region}
\displaystyle \text{Intersection of }x+2y=12\text{ and }2x+y=12:
\displaystyle x+2y=12,\quad 2x+y=12
\displaystyle \text{Multiplying the first by }2,\ 2x+4y=24
\displaystyle \text{Subtracting }(2x+y=12),\ 3y=12
\displaystyle \Rightarrow\ y=4
\displaystyle \Rightarrow\ 2x+4=12\Rightarrow x=4
\displaystyle \text{So one corner point is }(4,4)
\displaystyle \text{On }x\text{-axis, }y=0
\displaystyle x+2y\leq 12\Rightarrow x\leq 12
\displaystyle 2x+y\leq 12\Rightarrow 2x\leq 12\Rightarrow x\leq 6
\displaystyle 4x+5y\geq 20\Rightarrow 4x\geq 20\Rightarrow x\geq 5
\displaystyle \text{Hence points on }x\text{-axis are from }(5,0)\text{ to }(6,0)
\displaystyle \text{So two corner points are }(5,0)\text{ and }(6,0)
\displaystyle \text{On }y\text{-axis, }x=0
\displaystyle x+2y\leq 12\Rightarrow 2y\leq 12\Rightarrow y\leq 6
\displaystyle 2x+y\leq 12\Rightarrow y\leq 12
\displaystyle 4x+5y\geq 20\Rightarrow 5y\geq 20\Rightarrow y\geq 4
\displaystyle \text{Hence points on }y\text{-axis are from }(0,4)\text{ to }(0,6)
\displaystyle \text{So two corner points are }(0,4)\text{ and }(0,6)
\displaystyle \text{Therefore the corner points of the feasible region are }(0,4),(0,6),(4,4),(6,0),(5,0)\displaystyle \text{Now compute }z=500x+300y\text{ at these points}
\displaystyle z(0,4)=500(0)+300(4)=1200
\displaystyle z(0,6)=500(0)+300(6)=1800
\displaystyle z(4,4)=500(4)+300(4)=2000+1200=3200
\displaystyle z(6,0)=500(6)+300(0)=3000
\displaystyle z(5,0)=500(5)+300(0)=2500
\displaystyle \text{Hence the maximum value of }z\text{ is }3200
\displaystyle \therefore\ \text{Maximum }z=3200\text{ at }(x,y)=(4,4)
\\
\displaystyle \textbf{31. } \text{E and F are two independent events such that } P(\overline{E})=0.6 \text{ and}
\displaystyle P(E\cup F)=0.6. \text{ Find } P(F) \text{ and } P(\overline{E}\cup \overline{F}).
\displaystyle \text{Answer:}
\displaystyle P(\overline{E})=0.6\Rightarrow P(E)=0.4
\displaystyle P(E\cup F)=0.6
\displaystyle \text{Since }E\text{ and }F\text{ are independent, }P(E\cap F)=P(E)P(F)
\displaystyle \text{Also, }P(E\cup F)=P(E)+P(F)-P(E\cap F)
\displaystyle \Rightarrow\ 0.6=0.4+P(F)-0.4P(F)
\displaystyle \Rightarrow\ 0.6=0.4+0.6P(F)
\displaystyle \Rightarrow\ 0.2=0.6P(F)
\displaystyle \Rightarrow\ P(F)=\frac{1}{3}
\displaystyle \text{Now, }P(\overline{E}\cup\overline{F})=1-P(E\cap F)
\displaystyle =1-P(E)P(F)
\displaystyle =1-\frac{2}{5}\cdot\frac{1}{3}
\displaystyle =1-\frac{2}{15}
\displaystyle =\frac{13}{15}
\displaystyle \therefore\ P(F)=\frac{1}{3},\quad P(\overline{E}\cup\overline{F})=\frac{13}{15}

\displaystyle \textbf{SECTION D}
\displaystyle \text{This section comprises long answer type questions (LA) of 5 marks each.}

\displaystyle \textbf{32. } \text{(a) If } A=\begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix}, \text{ find } A^{-1} \text{ and use it to solve the following}
\displaystyle \text{system of equations: }   x-2y=10,\ 2x-y-z=8,\ -2y+z=7
\displaystyle \textbf{OR}
\displaystyle \text{(b) If } A=\begin{bmatrix} -1 & a & 2 \\ 1 & 2 & x \\ 3 & 1 & 1 \end{bmatrix} \text{ and } A^{-1}=\begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ b & y & 3 \end{bmatrix},
\displaystyle \text{find the value of } (a+x)-(b+y).
\displaystyle \text{Answer:}
\displaystyle \textbf{(a) } \text{Finding }A^{-1}\text{:}
\displaystyle A=\begin{pmatrix}1&-2&0\\2&-1&-1\\0&-2&1\end{pmatrix}
\displaystyle |A|=1\begin{vmatrix}-1&-1\\-2&1\end{vmatrix}-(-2)\begin{vmatrix}2&-1\\0&1\end{vmatrix}+0
\displaystyle =1\big((-1)(1)-(-1)(-2)\big)+2\big(2\cdot 1-0\big)
\displaystyle =(-1-2)+4=1
\displaystyle \text{Cofactor matrix of }A\text{ is } \begin{pmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{pmatrix}
\displaystyle \therefore\ \mathrm{adj}(A)=\begin{pmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{pmatrix}
\displaystyle \therefore\ A^{-1}=\frac{1}{|A|}\mathrm{adj}(A)=\begin{pmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{pmatrix}
\displaystyle \text{Using }A^{-1}\text{ to solve the system:}
\displaystyle x-2y=10,\quad 2x-y-z=8,\quad -2y+z=7
\displaystyle \text{Matrix form is }AX=B,\ \text{where }X=\begin{pmatrix}x\\y\\z\end{pmatrix},\ B=\begin{pmatrix}10\\8\\7\end{pmatrix}
\displaystyle \therefore\ X=A^{-1}B
\displaystyle =\begin{pmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{pmatrix}\begin{pmatrix}10\\8\\7\end{pmatrix}
\displaystyle =\begin{pmatrix}-30+16+14\\-20+8+7\\-40+16+21\end{pmatrix}
\displaystyle =\begin{pmatrix}0\\-5\\-3\end{pmatrix}
\displaystyle \therefore\ x=0,\ y=-5,\ z=-3
\displaystyle \textbf{OR}
\displaystyle \textbf{(b) } \text{Given }A=\begin{pmatrix}-1&a&2\\1&2&x\\3&1&1\end{pmatrix},\quad A^{-1}=\begin{pmatrix}1&-1&1\\-8&7&-5\\b&y&3\end{pmatrix}
\displaystyle \text{Since }AA^{-1}=I,\ \text{compare suitable entries}
\displaystyle \text{From }(1,3)\text{ entry: }(-1)(1)+a(-5)+2(3)=0
\displaystyle \Rightarrow\ -1-5a+6=0
\displaystyle \Rightarrow\ 5-5a=0
\displaystyle \Rightarrow\ a=1
\displaystyle \text{From }(2,3)\text{ entry: }1(1)+2(-5)+3x=0
\displaystyle \Rightarrow\ 1-10+3x=0
\displaystyle \Rightarrow\ 3x=9
\displaystyle \Rightarrow\ x=3
\displaystyle \text{From }(3,1)\text{ entry: }3(1)+1(-8)+b=0
\displaystyle \Rightarrow\ 3-8+b=0
\displaystyle \Rightarrow\ b=5
\displaystyle \text{From }(3,2)\text{ entry: }3(-1)+1(7)+y=0
\displaystyle \Rightarrow\ -3+7+y=0
\displaystyle \Rightarrow\ y=-4
\displaystyle \therefore\ (a+x)-(b+y)=(1+3)-(5+(-4))
\displaystyle =4-1=3
\displaystyle \therefore\ (a+x)-(b+y)=3
\\
\displaystyle \textbf{33. } \text{(a) Evaluate :}
\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16\sin 2x}\,dx
\displaystyle \textbf{OR}
\displaystyle \text{(b) Evaluate :}
\displaystyle \int_{0}^{\frac{\pi}{2}} \sin 2x\,\tan^{-1}(\sin x)\,dx
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle I=\int_{0}^{\frac{\pi}{4}}\frac{\sin x+\cos x}{9+16\sin 2x}\,dx
\displaystyle \text{Let }t=\sin x-\cos x\Rightarrow dt=(\cos x+\sin x)\,dx
\displaystyle \text{Also, }(\sin x-\cos x)^{2}=1-2\sin x\cos x=1-\sin 2x
\displaystyle \Rightarrow\ \sin 2x=1-t^{2}
\displaystyle \therefore\ I=\int \frac{dt}{9+16(1-t^{2})}
\displaystyle =\int \frac{dt}{25-16t^{2}}
\displaystyle =\frac{1}{40}\log\left|\frac{5+4t}{5-4t}\right|
\displaystyle \text{Limits: }x=0\Rightarrow t=-1,\ x=\frac{\pi}{4}\Rightarrow t=0
\displaystyle \therefore\ I=\frac{1}{40}\left[\log\left(\frac{5+4t}{5-4t}\right)\right]_{-1}^{0}
\displaystyle =\frac{1}{40}\left[\log 1-\log\left(\frac{1}{9}\right)\right]
\displaystyle =\frac{1}{40}\log 9=\frac{1}{20}\log 3
\displaystyle \textbf{OR}
\displaystyle \textbf{(b)}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\sin 2x\tan^{-1}(\sin x)\,dx
\displaystyle =\int_{0}^{\frac{\pi}{2}}2\sin x\cos x\tan^{-1}(\sin x)\,dx
\displaystyle \text{Let }u=\tan^{-1}(\sin x),\ dv=2\sin x\cos x\,dx
\displaystyle \Rightarrow\ du=\frac{\cos x}{1+\sin^{2}x}\,dx,\ v=\sin^{2}x
\displaystyle \therefore\ I=\left[\sin^{2}x\tan^{-1}(\sin x)\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}x\cos x}{1+\sin^{2}x}\,dx
\displaystyle =\frac{\pi}{4}-\int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}x\cos x}{1+\sin^{2}x}\,dx
\displaystyle \text{Let }t=\sin x\Rightarrow dt=\cos x\,dx
\displaystyle \Rightarrow\ I=\frac{\pi}{4}-\int_{0}^{1}\frac{t^{2}}{1+t^{2}}\,dt
\displaystyle =\frac{\pi}{4}-\int_{0}^{1}\left(1-\frac{1}{1+t^{2}}\right)dt
\displaystyle =\frac{\pi}{4}-\left[1-\frac{\pi}{4}\right]
\displaystyle =\frac{\pi}{2}-1
\\
\displaystyle \textbf{34. } \text{Using integration, find the area of the ellipse } \frac{x^{2}}{16}+\frac{y^{2}}{4}=1, \text{ included}
\displaystyle \text{between the lines } x=-2 \text{ and } x=2.
\displaystyle \text{Answer:}
\displaystyle \frac{x^{2}}{16}+\frac{y^{2}}{4}=1
\displaystyle \Rightarrow\ \frac{y^{2}}{4}=1-\frac{x^{2}}{16}
\displaystyle \Rightarrow\ y^{2}=4\left(1-\frac{x^{2}}{16}\right)
\displaystyle \Rightarrow\ y=\pm 2\sqrt{1-\frac{x^{2}}{16}}
\displaystyle \text{Area included between }x=-2\text{ and }x=2
\displaystyle A=\int_{-2}^{2}\left[2\sqrt{1-\frac{x^{2}}{16}}-\left(-2\sqrt{1-\frac{x^{2}}{16}}\right)\right]dx
\displaystyle =\int_{-2}^{2}4\sqrt{1-\frac{x^{2}}{16}}\,dx
\displaystyle =\int_{-2}^{2}\sqrt{16-x^{2}}\,dx
\displaystyle =\left[\frac{x}{2}\sqrt{16-x^{2}}+8\sin^{-1}\left(\frac{x}{4}\right)\right]_{-2}^{2}
\displaystyle =\left[\frac{2}{2}\sqrt{12}+8\sin^{-1}\left(\frac{1}{2}\right)\right]-\left[\frac{-2}{2}\sqrt{12}+8\sin^{-1}\left(\frac{-1}{2}\right)\right]
\displaystyle =\left[2\sqrt{3}+8\cdot\frac{\pi}{6}\right]-\left[-2\sqrt{3}-8\cdot\frac{\pi}{6}\right]
\displaystyle =4\sqrt{3}+\frac{8\pi}{3}
\displaystyle \therefore\ \text{Required area is }\frac{8\pi}{3}+4\sqrt{3}\text{ square units}
\\
\displaystyle \textbf{35. } \text{The image of point } P(x,y,z) \text{ with respect to line } \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} \text{ is}
\displaystyle P'(1,0,7). \text{ Find the coordinates of the point } P.
\displaystyle \text{Answer:}
\displaystyle \text{Line: }\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=t
\displaystyle \Rightarrow\ x=t,\ y=1+2t,\ z=2+3t
\displaystyle \text{Let }A(t,1+2t,2+3t)\text{ be a point on the line (midpoint of }P\text{ and }P')
\displaystyle \text{Given }P'(1,0,7)
\displaystyle \text{Let }P(x,y,z)
\displaystyle \text{Since the line is the perpendicular bisector, }A\text{ is midpoint of }P\text{ and }P'
\displaystyle \Rightarrow\ \left(\frac{x+1}{2},\frac{y+0}{2},\frac{z+7}{2}\right)=(t,1+2t,2+3t)
\displaystyle \Rightarrow\ x+1=2t,\ y=2(1+2t),\ z+7=2(2+3t)
\displaystyle \Rightarrow\ x=2t-1,\ y=2+4t,\ z=4+6t-7=6t-3
\displaystyle \text{Also, }PP'\text{ is perpendicular to the line, so direction ratios satisfy:}
\displaystyle (x-1,\ y-0,\ z-7)\cdot(1,2,3)=0
\displaystyle \Rightarrow\ (2t-1-1,\,2+4t,\,6t-3-7)\cdot(1,2,3)=0
\displaystyle \Rightarrow\ (2t-2,\,2+4t,\,6t-10)\cdot(1,2,3)=0
\displaystyle \Rightarrow\ (2t-2)+2(2+4t)+3(6t-10)=0
\displaystyle \Rightarrow\ 2t-2+4+8t+18t-30=0
\displaystyle \Rightarrow\ 28t-28=0
\displaystyle \Rightarrow\ t=1
\displaystyle \therefore\ x=2(1)-1=1,\ y=2+4(1)=6,\ z=6(1)-3=3
\displaystyle \therefore\ \text{Coordinates of }P\text{ are }(1,6,3)

\displaystyle \textbf{SECTION E}
\displaystyle \text{This section comprises 3 case study based questions of 4 marks each.}

\displaystyle \textbf{Case Study - 1}
\displaystyle \textbf{36. } \text{The traffic police has installed Over Speed Violation Detection (OSVD)}
\displaystyle \text{system at various locations in a city. These cameras can capture a speeding}
\displaystyle \text{vehicle from a distance of 300 m and even function in the dark.} \displaystyle \text{A camera is installed on a pole at the height of 5 m. It detects a car travelling}
\displaystyle \text{away from the pole at the speed of 20 m/s. At any point, } x \text{ m away from}
\displaystyle \text{the base of the pole, the angle of elevation of the speed camera from the car C}
\displaystyle \text{is } \theta.
\displaystyle \text{On the basis of the above information, answer the following questions :}
\displaystyle \text{(i) Express } \theta \text{ in terms of height of the camera installed on the pole}   \text{and } x.
\displaystyle \text{(ii) Find } \frac{d\theta}{dx}.
\displaystyle \text{(iii) (a) Find the rate of change of angle of elevation with respect to time at}
\displaystyle \text{an instant when the car is 50 m away from the pole.}
\displaystyle \textbf{OR}
\displaystyle \text{(iii) (b) If the rate of change of angle of elevation with respect to time of}
\displaystyle \text{another car at a distance of 50 m from the base of the pole is } \frac{3}{101} \text{ rad/s,}
\displaystyle \text{then find the speed of the car.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle \tan \theta=\frac{5}{x}
\displaystyle \therefore\ \theta=\tan^{-1}\left(\frac{5}{x}\right)
\displaystyle \textbf{(ii)}
\displaystyle \theta=\tan^{-1}\left(\frac{5}{x}\right)
\displaystyle \frac{d\theta}{dx}=\frac{1}{1+\left(\frac{5}{x}\right)^{2}}\cdot\left(-\frac{5}{x^{2}}\right)
\displaystyle =\frac{1}{1+\frac{25}{x^{2}}}\cdot\left(-\frac{5}{x^{2}}\right)
\displaystyle =\frac{x^{2}}{x^{2}+25}\cdot\left(-\frac{5}{x^{2}}\right)
\displaystyle \therefore\ \frac{d\theta}{dx}=-\frac{5}{x^{2}+25}
\displaystyle \textbf{(iii)(a)}
\displaystyle \frac{d\theta}{dt}=\frac{d\theta}{dx}\cdot\frac{dx}{dt}
\displaystyle =-\frac{5}{x^{2}+25}\cdot 20
\displaystyle =-\frac{100}{x^{2}+25}
\displaystyle \text{At }x=50,
\displaystyle \frac{d\theta}{dt}=-\frac{100}{50^{2}+25}
\displaystyle =-\frac{100}{2525}
\displaystyle =-\frac{4}{101}
\displaystyle \therefore\ \frac{d\theta}{dt}=-\frac{4}{101}\ \text{rad/s}
\displaystyle \textbf{OR}
\displaystyle \textbf{(iii)(b)}
\displaystyle \frac{d\theta}{dt}=\frac{d\theta}{dx}\cdot\frac{dx}{dt}
\displaystyle -\frac{3}{101}=-\frac{5}{50^{2}+25}\cdot\frac{dx}{dt}
\displaystyle -\frac{3}{101}=-\frac{5}{2525}\cdot\frac{dx}{dt}
\displaystyle -\frac{3}{101}=-\frac{1}{505}\cdot\frac{dx}{dt}
\displaystyle \therefore\ \frac{dx}{dt}=15
\displaystyle \therefore\ \text{Speed of the car }=15\ \text{m/s}
\\
\displaystyle \textbf{Case Study - 2}
\displaystyle \textbf{37. } \text{According to recent research, air turbulence has increased in various}
\displaystyle \text{regions around the world due to climate change. Turbulence makes flights}
\displaystyle \text{bumpy and often delays the flights.}
\displaystyle \text{Assume that, an airplane observes severe turbulence, moderate turbulence}
\displaystyle \text{or light turbulence with equal probabilities. Further, the chance of an}
\displaystyle \text{airplane reaching late to the destination are } 55\%, 37\% \text{ and } 17\% \text{ due to}
\displaystyle \text{severe, moderate and light turbulence respectively.} \displaystyle \text{On the basis of the above information, answer the following questions :}
\displaystyle \text{(i) Find the probability that an airplane reached its destination late.}
\displaystyle \text{(ii) If the airplane reached its destination late, find the probability that it}
\displaystyle \text{was due to moderate turbulence.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }S,\ M,\ L\text{ denote severe, moderate and light turbulence respectively}
\displaystyle P(S)=P(M)=P(L)=\frac{1}{3}
\displaystyle P(\text{late}\mid S)=55\%=0.55,\quad P(\text{late}\mid M)=37\%=0.37,\quad P(\text{late}\mid L)=17\%=0.17
\displaystyle \textbf{(i) } \text{Probability that the airplane reached late:}
\displaystyle P(\text{late})=P(S)P(\text{late}\mid S)+P(M)P(\text{late}\mid M)+P(L)P(\text{late}\mid L)
\displaystyle =\frac{1}{3}(0.55)+\frac{1}{3}(0.37)+\frac{1}{3}(0.17)
\displaystyle =\frac{1}{3}(1.09)
\displaystyle =\frac{109}{300}
\displaystyle \therefore\ P(\text{late})=\frac{109}{300}
\displaystyle \textbf{(ii) } \text{Probability that it was due to moderate turbulence, given that the airplane reached late:}
\displaystyle P(M\mid \text{late})=\frac{P(M)P(\text{late}\mid M)}{P(\text{late})}
\displaystyle =\frac{\frac{1}{3}\cdot 0.37}{\frac{109}{300}}
\displaystyle =\frac{\frac{37}{300}}{\frac{109}{300}}
\displaystyle =\frac{37}{109}
\displaystyle \therefore\ P(M\mid \text{late})=\frac{37}{109}
\\
\displaystyle \textbf{Case Study - 3}
\displaystyle \textbf{38. } \text{If a function } f:X\rightarrow Y \text{ defined as } f(x)=y \text{ is one-one and onto, then we can}
\displaystyle \text{define a unique function } g:Y\rightarrow X \text{ such that } g(y)=x, \text{ where } x\in X \text{ and}
\displaystyle y=f(x), y\in Y. \text{ Function } g \text{ is called the inverse of function } f.
\displaystyle \text{The domain of sine function is } R \text{ and function sine : } R\rightarrow R \text{ is neither}
\displaystyle \text{one-one nor onto. The following graph shows the sine function.} \displaystyle \text{Let sine function be defined from set } A \text{ to } [-1,1] \text{ such that inverse of sine}
\displaystyle \text{function exists, i.e., } \sin^{-1}x \text{ is defined from } [-1,1] \text{ to } A.
\displaystyle \text{On the basis of the above information, answer the following questions :}
\displaystyle \text{(i) If } A \text{ is the interval other than principal value branch, give an}
\displaystyle \text{example of one such interval.}
\displaystyle \text{(ii) If } \sin^{-1}(x) \text{ is defined from } [-1,1] \text{ to its principal value branch, find}
\displaystyle \text{the value of } \sin^{-1}\left(-\frac{1}{2}\right)-\sin^{-1}(1).
\displaystyle \text{(iii) (a) Draw the graph of } \sin^{-1}x \text{ from } [-1,1] \text{ to its principal value }   \text{branch.}
\displaystyle \textbf{OR}
\displaystyle \text{(iii) (b) Find the domain and range of } f(x)=2\sin^{-1}(1-x).
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle \text{An interval other than the principal value branch on which }\sin x\text{ is one-one and onto }[-1,1]\text{ is }\left[\frac{\pi}{2},\frac{3\pi}{2}\right]
\displaystyle \textbf{(ii)}
\displaystyle \sin^{-1}\left(-\frac{1}{2}\right)-\sin^{-1}(1)=-\frac{\pi}{6}-\frac{\pi}{2}
\displaystyle =-\frac{\pi+3\pi}{6}
\displaystyle =-\frac{4\pi}{6}
\displaystyle =-\frac{2\pi}{3}
\displaystyle \textbf{(iii)(a)}
\displaystyle \text{For the principal value branch, }\sin^{-1}x\text{ is defined on }[-1,1]\text{ with range }\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
\displaystyle \text{Its graph is the reflection of }y=\sin x,\ x\in \left[-\frac{\pi}{2},\frac{\pi}{2}\right],\text{ in the line }y=x
\displaystyle \text{It passes through }\left(-1,-\frac{\pi}{2}\right),\ (0,0),\ \left(1,\frac{\pi}{2}\right)
\displaystyle \textbf{OR}
\displaystyle \textbf{(iii)(b)}
\displaystyle f(x)=2\sin^{-1}(1-x)
\displaystyle \text{For }f(x)\text{ to be defined, }1-x\in[-1,1]
\displaystyle \Rightarrow\ -1\leq 1-x\leq 1
\displaystyle \Rightarrow\ -2\leq -x\leq 0
\displaystyle \Rightarrow\ 0\leq x\leq 2
\displaystyle \therefore\ \text{Domain of }f(x)=[0,2]
\displaystyle \text{Since }1-x\in[-1,1],\ \sin^{-1}(1-x)\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
\displaystyle \Rightarrow\ 2\sin^{-1}(1-x)\in[-\pi,\pi]
\displaystyle \therefore\ \text{Range of }f(x)=[-\pi,\pi]

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