A = {1, 2, 3 ...13, 14}
R = {(x, y) : 3x - y = 0}
∴ R = {(1, 3), (2, 6), (3, 9),(4, 12)}
R is not reflexive since (1, 2), (2, 2)...(14, 14) ∉ R
Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R.[3(3) - 1 ≠ 0]
Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.
[3(1) - 9 ≠ 0]
Hence R is neither reflexive, nor symmetric, nor transitive. 

\(\text {Consider, } \mathrm{RHS}=\sin ^{-1}\left(3 x-4 x^{3}\right) \)
\(\text {Let } x=\sin \theta\)
\(\text {then } \theta=\sin ^{-1} x\)
Now, from Eq. (i), we get
\(\mathrm{RHS}=\sin ^{-1}\left(3 \sin \theta-4 \sin ^{3} \theta\right) \)
\(=\sin ^{-1}(\sin 3 \theta) \quad\left[\because \sin 3 x=3 \sin x-4 \sin ^{3} x\right] \)
\(=3 \theta \because \sin ^{-1}(\sin \theta)=\theta, \forall \theta \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \)
\(\text { and here } \left.-\frac{1}{2} \leq x \leq \frac{1}{2} \Rightarrow-\frac{\pi}{6} \leq x \leq \frac{\pi}{6}\right] \)
\(=3 \sin ^{-1} x {\left[\because \theta=\sin ^{-1} x\right]} =\mathrm{LHS}\)

\(=\lim _{h \rightarrow 0} f(x)=\lim _{h \rightarrow 0} \cos x=\cos a[\therefore f(x)=\cos x]\)
f(a) = cos a
\(\therefore \lim _{x \rightarrow a} f(x)=f(a)\)
Thus, f(x) is continuous at x=a. But a is an arbitrary points so f(x) is continuous at all points.
So, f(x) is continuous at all points.
(ii) Here, f(x) = cosec x. Since, f(x) is not defined at $\times =n\pi ,nϵZ.$Thus, f(x) is continuous at all points except $n\pi ,n\mathrm{ϵ\; Z}$(iii) Here, f(x)sec x
Since, f(x) is not defined at
Thus, f(x) is continuous at all points except \(x=(2 n+1) \frac{\pi}{2}, n \in Z\)
(iv) Here, f(x) = cotx
SInce, f(x) is not defined at $n\pi ,nϵZ$
Thus, f(x) is continuous at all points excepts $x=n\pi ,nϵZ$

\((A+{ A }^{ T })^{ T }={ A }^{ T }+({ A }^{ T })^{ T }={ A }^{ T }+A=A+{ A }^{ T }. \) Hence, symmetric.

1368 Concern for children health and nutrient food for every child.

We have, \(A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\)
Clearly, the cofactors of elements of |A| are given by
\(A_{11}=\cos \alpha ; A_{12}=-\sin \alpha ; A_{13}=0 ; \)
\( A_{21}=\sin \alpha ; A_{22}=\cos \alpha ; A_{23}=0\)
\( A_{31}=0 ; A_{32}=0 \text { and } A_{33}=1\)
\(\therefore \operatorname{adj}(A)=\left[\begin{array}{lll} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}\right]^T\)
\(=\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right]^T=\left[\begin{array}{ccc} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right]\)

the required trapezium is as given in figure. Draw perpendiculars DP and CQ on AB. Let AP = x cm. Note that \(\Delta \)APD \(\cong \) \(\Delta \) BQC.
Therefore, QB = x cm. Also, by pythagoras theorem DP = QC = \(\sqrt { 100-{ x }^{ 2 } } \)
Let A be the area of the trapezium.

Then, A \(\equiv \) A(x)
\(=\frac { 1 }{ 2 } (sum\ of\ parallel\ sides)\times (height)\)
\(=\frac { 1 }{ 2 } (2x+10+10)\sqrt { 100-{ x }^{ 2 } } \)
= (x + 10)\(\sqrt { 100-{ x }^{ 2 } } \)
\(or\ A'(x)=(x+10)\frac { (-2x) }{ 2\sqrt { 100-{ x }^{ 2 } } } +(\sqrt { 100-{ x }^{ 2 }) } \)
\(=\frac { -2{ x }^{ 2 }-10x+100 }{ \sqrt { 100-{ x }^{ 2 } } } \)
Now, A'(x) = 0 gives 2x² + 10x - 100 = 0,
i.e., x = 5 and x = -10
So, x = 5.
Now, A''(x) = \(\frac { \sqrt { 100-{ x }^{ 2 } } (-4x-10)-(-2x^{ 2 }-10x+100)\frac { (-2x) }{ 2\sqrt { 100-{ x }^{ 2 } } } }{ 100-{ x }^{ 2 } } \)
\(=\frac { { 2x }^{ 3 }-300x-1000 }{ { (100-{ x }^{ 2 }) }^{ \frac { 1 }{ 2 } } } \)
(on simplification)
or \(A''(5)=\frac { { 2(5) }^{ 3 }-300(5)-1,000 }{ (100-(5{ ) }^{ 2 })^{ \frac { 3 }{ 2 } } } \)
\(=\frac { -2,250 }{ 75\sqrt { 75 } } =\frac { -30 }{ \sqrt { 75 } } <0\)
Thus, area of trapezium is maximum at x = 5 and the maximum area is given by
A(5) = (5 + 10)\(\sqrt { 100-{ (5) }^{ 2 } } \)
= 15\(\sqrt { 75 } =75\sqrt { 3 } { cm }^{ 2 }\)

Given
\( \tan ^{-1}\left(\frac{x-2}{x-4}\right)+\tan ^{-1}\left(\frac{x+2}{x+4}\right)=\frac{\pi}{4} \)
\(\Rightarrow \tan ^{-1}\left[\frac{\frac{x-2}{x-4}+\frac{x+2}{x+4}}{1-\left(\frac{x-2}{x-4}\right)\left(\frac{x+2}{x+4}\right)}\right]=\frac{\pi}{4} \) 
\(\Rightarrow \tan ^{-1}\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ {\left[\frac{(x-2)(x+4)+(x+2)(x-4)}{\frac{(x-4)(x+4)}{(x-4)(x+4)-(x-2)(x+2)}}{(x-4)(x+4)}\right]} \end{array}\right]=\frac{\pi}{4} \)
\(\Rightarrow \frac{x^{2}-2 x+4 x-8+x^{2}-4 x+2 x-8}{\left(x^{2}-16\right)-\left(x^{2}-4\right)}=\tan \frac{\pi}{4} \) 
\( \Rightarrow \frac{x^{2}-2 x+4 x-8+x^{2}-4 x+2 x-8}{\left(x^{2}-16\right)-\left(x^{2}-4\right)}=\tan \frac{\pi}{4} \)
\(\Rightarrow \frac{2 x^{2}-16}{-12}=1 \)
\(\Rightarrow 2 x^{2}-16=-12 \)
\(\Rightarrow 2 x^{2}=-12+16 \)
\(\Rightarrow 2 x^{2}=4 \Rightarrow x^{2}=2 \ \therefore \ x=\pm \sqrt{2}\)
\(\text {Hence, } \sqrt{2} \text { and }-\sqrt{2} \text { are the required value } \)

\(\Rightarrow \) \(\left( { 1-x }^{ 2 } \right) ^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -x\frac { { d }y }{ d{ x } } =0\)\(\left[ On\quad dividing\quad by\quad 2\left( \frac { dy }{ dx } \right) \right] \)

Side of a square base is x and height is y.
\(V=x^{2} y \text { and cost }=2 x^{2} \times 5+4 x y \times 2.50 \text { [Ans. Rs } \left.1920\right]\)

Given points \((2,-3),(\lambda,-1) \text { and }(0,4)\) are collinear So, area of triangle formed by these three points will be zero.
\(\therefore \left|\begin{array}{rrr} 2 & -3 & 1 \\ \lambda & -1 & 1 \\ 0 & 4 & 1 \end{array}\right|=0\) 
Applying \(R_{2} \rightarrow R_{2}-R_{1}\) and \(R_{3} \rightarrow R_{3}-R_{1}\) we get
\(\left|\begin{array}{ccc} 2 & -3 & 1 \\ \lambda-2 & 2 & 0 \\ -2 & 7 & 0 \end{array}\right|=0\) 
Now, expanding along C₃, we get \(\left|\begin{array}{cc} \lambda-2 & 2 \\ -2 & 7 \end{array}\right|=0\) 
\( \Rightarrow 7(\lambda-2)-2(-2)=0 \)
\(\Rightarrow 7 \lambda-14+4=0 \)
\(\Rightarrow 7 \lambda-10=0 \Rightarrow \lambda \mid=\frac{10}{7} \) 
Hence, required value of \(\lambda \text { is } \frac{10}{7}\).

Let us assume that f(−5) = f(5).
Then ′f′ satisfies all the conditions of Rolle′s Theorem in [−5, 5].
Then there must exist atleast once cϵ(−5, 5) such that f′(c) = 0 Thus our supposition is wrong.
Hence, f(−5) ≠ f(5)

One-one function from {1, 2, 3} to itself is simply a permutation on three symbols 1, 2, 3. Therefore, total number of one-one maps from {1, 2, 3} to itself is same as total number of permutations on three symbols 1, 2, 3 which is 3! = 6.

\(Chem.\quad \quad \quad Phy.\quad \quad \quad Eco.\\ A=\left[ 10\times 12\quad \quad 8\times 12\quad \quad 10\times 12 \right] \)
\(Selling\ Price\ matrix,\ B=\left[ \begin{matrix} 80 \\ 60 \\ 40 \end{matrix} \right] .\)
\(\therefore \ Reqd.\ amount=AB\)
\(=\left[ 120\quad 96\quad 120 \right] \left[ \begin{matrix} 80 \\ 60 \\ 40 \end{matrix} \right] \)
\(=\left[ 120\times 80+96\times 60+120\times 40 \right] \)
\(=\left[ 9600+5760+4800 \right] =\left[ 20160 \right] .\)
Hence, the bookshop will receive Rs. 20, 160 by selling all the books.

Let \({ cos }^{ -1 }\left( -\frac { 1 }{ \sqrt { 2 } } \right) =y\), where  \(y\in [0,\pi ]\)
\(\Rightarrow cosy=\frac { 1 }{ \sqrt { 2 } } \)
\(\Rightarrow cosy=-cos\frac { \pi }{ 4 } =cos\left( \pi -\frac { \pi }{ 4 } \right) \)
\(\Rightarrow cosy=cos\frac { 3\pi }{ 4 } \Rightarrow y=\frac { 3\pi }{ 4 } \)
Hence, the required principal value = \(\frac { 3\pi }{ 4 } \)

x - [x] = 0 when x is an integer, so that f(x) is discontinuous for all x ∈ I i.e. f(x) is discontinuous at infinite number of points.

If square matrix in which all diagonals elements are 1 and rest are 0, is called unit matrix.

\( 2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}\)
\(\therefore \ 2 \tan ^{-1} \frac{2}{3}=\sin ^{-1} \frac{2\left(\frac{2}{3}\right)}{1+\left(\frac{2}{3}\right)^{2}}=\sin ^{-1} \frac{12}{13}\)
\(\cos \left(2 \tan ^{-1} x\right)=\cos \left(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right) \)

 Let \(f(f(y))=y\), for all \(y \neq-1\) \(\Rightarrow f\left(\frac{a y}{y+1}\right)=y\), for all \(y \neq-1\)
\(\Rightarrow \frac{a\left(\frac{a y}{y+1}\right)}{\frac{a y}{y+1}+1}=y,for\ all\ y \neq-1\)
\(\Rightarrow \frac{a^{2} y}{a y+y+1}=y, for\ all \ y \neq-1\)
\(\Rightarrow a^{2} y=(a+1) y^{2}+y, for\ all \ y \neq-1\)
\(\Rightarrow (a+1) y^{2}+\left(1-a^{2}\right) y=0, for \ all \ y \neq-1\)
\(\Rightarrow a+1=0 and 1-a^{2}=0 \Rightarrow a=-1\)

As \(\frac{dy}{dt}\) = 2x.\(\frac{dx}{dt}\)
⇒ 6.\(\frac{dx}{dt}\) = 2x.\(\frac{dx}{dt}\) ⇒ x = 3
From curve, y = 9. Point is (3, 9)

Let r be the radius and h be the height of conical part.
Here, \(\tan 45^{\circ}=\frac{r}{h} \Rightarrow 1=\frac{r}{h} \Rightarrow h=r\)
The volume of water in the tank
\(V=\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi r^3 \mathrm{~cm}^3 \quad[\because h=r]\)
(ii) \( \frac{d V}{d t}=\frac{3}{3} \pi r^2 \frac{d r}{d t} \)
\(\Rightarrow \quad-2=\pi r^2 \frac{d r}{d t} \quad\left[\because \frac{d V}{d t}=-2\right]\)
\( \Rightarrow \quad \frac{d r}{d t}=\frac{-2}{\pi r^2}=\frac{-2}{\pi(2 \sqrt{2})^2}=\frac{-1}{4 \pi}\)
\({[r=2 \sqrt{2} \text { (given) }]}\)
Hence, the rate of decrease of radius is \(\frac{1}{4 \pi} \mathrm{cm} / \mathrm{s}\).
(iii) (b) \(\therefore S=\pi r l \Rightarrow S=\pi h l\) \([\because r=h]\)
\(\Rightarrow \quad S=\pi h \sqrt{h^2+r^2}\)
(a) \(\therefore S=\pi r l \)
\(=\pi r \sqrt{h^2+r^2}=\pi r \sqrt{2 r^2} \)
\({\left[\because l=\sqrt{h^2+r^2} ; \ \because h=r\right]}\)
\(=\sqrt{2} \pi r^2\)
\(\Rightarrow \frac{d S}{d t}=\sqrt{2} \pi(2 r) \frac{d r}{d t}\)
\(\Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi r \frac{d r}{d t}\)
\(\Rightarrow\left(\frac{d S}{d t}\right)_{r=2 \sqrt{2}}=2 \sqrt{2} \pi(2 \sqrt{2})\left(\frac{-1}{4 \pi}\right)\)
\(=-2 \mathrm{~cm}^2 / \mathrm{s} \)
Or \(\text { (b) } \therefore S=\pi r l \Rightarrow S=\pi h l\) \({[\because r=h]}\)
\( \Rightarrow S=\pi h \sqrt{h^2+r^2} \)
\(\Rightarrow \quad S=\sqrt{2} \pi h^2 \)
\(\Rightarrow \quad \frac{d S}{d t}=\sqrt{2} \pi(2 h) \frac{d h}{d t}\)
\(\Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi h \frac{d h}{d t}\)
\( \Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi(2 \sqrt{2}) \frac{d h}{d t}\)
\( \text { [since, } l^2=r^2+h^2 \Rightarrow l^2=2 h^2\)
\(\left.\Rightarrow 16=2 h^2 \Rightarrow h=2 \sqrt{2}\right]\)
\( \Rightarrow \quad-2=8 \pi \frac{d h}{d t}\) \({[\because \text { from Eq. (i) }]}\)
\( \Rightarrow \quad \frac{d h}{d t}=-\frac{1}{4 \pi} \mathrm{cm} / \mathrm{s}\)
\(\text { (b) } \therefore S=\pi r l \Rightarrow S=\pi h l\)
\([\because r=h]\)
\(\Rightarrow S=\pi h \sqrt{h^2+r^2}\)
\(\Rightarrow \quad S=\sqrt{2} \pi h^2\)
\(\Rightarrow \quad \frac{d S}{d t}=\sqrt{2} \pi(2 h) \frac{d h}{d t}\)
\(\Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi h \frac{d h}{d t}\)
\(\Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi(2 \sqrt{2}) \frac{d h}{d t}\)
\(\text { [since, } l^2=r^2+h^2 \Rightarrow l^2=2 h^2\)
\(\left.\Rightarrow 16=2 h^2 \Rightarrow h=2 \sqrt{2}\right]\)
\(\Rightarrow \quad-2=8 \pi \frac{d h}{d t}\)
\([\because \text { from Eq. (i) }]\)
\(\Rightarrow \quad \frac{d h}{d t}=-\frac{1}{4 \pi} \mathrm{cm} / \mathrm{s}\)

(i) (b) \(\ln \triangle A B D\)
\(\tan \alpha  =\frac{B D}{A D}\)
\( =\frac{30}{30 \sqrt{3}}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow  \alpha  =30^{\circ}\)
\(\therefore  \sin \alpha =\sin 30^{\circ}=\frac{1}{2}\)
\(\Rightarrow  \alpha  =\sin ^{-1}\left(\frac{1}{2}\right)\)
(ii) (c) \(\cos \alpha=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \alpha =\cos ^{-1}\left(\frac{\sqrt{3}}{2})\right.\)
(iii) (d) \(\ln \triangle B DC\)
\(\tan \beta  =\frac{(BD)}{(C D)}\)
\(=\frac{30}{10 \sqrt{3}}=\sqrt{3} \)
\(\Rightarrow \beta =\tan ^{-1}(\sqrt{3})\)
(iv) \( \text { (c) Since, } \alpha=\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \text {, } \)
\(\beta=\tan ^{-1}(\sqrt{3})=\frac{\pi}{3} \)
\(\therefore \quad \angle A B C=\pi-\left(\frac{\pi}{6}+\frac{\pi}{3}\right)\)
\( =\pi-\frac{\pi}{2}=\frac{\pi}{2}\)
(v) (c) We know that domain and range of \(\cos ^{-1} x\) are [-1,1] and \([0, \pi]\) respectively.

 (i) (d)
(ii) (a)
(iii) (b) ,
(iv) (a)
(v) (b)

(d) Assertion is incorrect, reason is correct.

(d) Assertion is incorrect, Reason is correct.