\( \text {Hint } \lim _{x \rightarrow 0} f(x)=f(0) \)
\(\Rightarrow \lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \cos x=k \)
\(\Rightarrow 1+1=k \)

\(\left(A^{2}\right)^{-1}=(A A)^{-1}=\left(A^{-1} A^{-1}\right) \quad\left[\because(A B)^{-1}=B^{-1} A^{-1}\right] \)
\(=\left(A^{-1}\right)^{2}\)

\( \tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5} \)
\(\text {Now, } \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2} \)
\(\tan ^{-1} x=\frac{\pi}{2}-\cot ^{-1} x \)
\(\therefore \frac{\pi}{2}-\cot ^{-1} x+\frac{\pi}{2}-\cot ^{-1} y=\frac{4 \pi}{5} \)
\(-\cot ^{-1} x-\cot ^{-1} y+\pi=\frac{4 \pi}{5} \)
\(\cot ^{-1} x+\cot ^{-1} y=\pi-\frac{4 \pi}{5} \)
\(=\frac{\pi}{5} \)

x = \(\frac { \pi }{ 6 } \)

f is not onto, as for some y ∈ R from co-domain, there is no x ∈ R from domain such that y = f{x), e.g. for - 2 ∈ R (co-domain) there is no x ∈ R (domain) such that f{x) = -2, i.e. |x|= -2. Hence, not onto.

Given, \(y=e^{m \sin ^{-1} x}\)
On differentiating both sides of Eq.(i) w.r.t. x, we get \(\frac{d y}{d x}=e^{m \sin ^{-1} x} \frac{d}{d x}\left(m \sin ^{-1} x\right)\)
[by using chain rule of derivative]
\(=e^{m \sin ^{-1} x} \cdot m \frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow  \sqrt{1-x^2} \frac{d y}{d x}=m y\)
Now, on squaring both sides, we get
\(\left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=m^2 y^2\)
On differentiating both sides of Eq. (ii) w.r.t. x, we get
\(\left(1-x^2\right) 2\left(\frac{d y}{d x}\right) \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2(-2 x)=2 m^2 y\left(\frac{d y}{d x}\right)\)
\(\Rightarrow \quad\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=m^2 y\)
\( {\left[\text { dividing both sides by } 2\left(\frac{d y}{d x}\right)\right]}\)
\( \therefore \quad\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y=0\)
Hence proved.

Given, \(y=\sin (\sin x)\)
On differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=\cos (\sin x) \cdot \cos x\)
Again, on differentiating both sides w.r.t. x, we get
\(\frac{d^2 y}{d x^2}=\cos (\sin x) \cdot(-\sin x)\)
\(\Rightarrow \frac{d^2 y}{d x^2}=\frac{1}{\cos x} \cdot\left(\frac{d y}{d x}\right)\left(-\cos x(-\sin x)-y \cos ^2 x\right.\)
[using Eqs. (i) and (ii)]
\(\Rightarrow \quad \frac{d^2 y}{d x^2}=-\tan x \frac{d y}{d x}-y \cos ^2 x\)
\(\Rightarrow \quad \frac{d^2 y}{d x^2}+\tan x \frac{d y}{d x}+y \cos ^2 x=0\)
Hence proved. 

Given, A ² = 1
Now, (A - I)³ + (A + I)³ -7A = (A³ - 3A 21 + 3AI²- ]³)+(A³ + 3A² I + 3AI² + ]³) -7A
= (A³ - 3A² + 3AI - I) + (A³ + 3A² + 3AI + I) - 7A
[ஃ A 21 = A 2 and 13 = 12 = 1]
= 2A³ + 6AI - 7A = 2A 2A + 6A - 7A [ஃ AI = A]
= 2IA-A
= 2A-A = A
[from Eq. (i)]
[ஃ IA = A]

\( \cos ^{-1}\left(\cos \frac{7 \pi}{6}\right) =\cos ^{-1}\left\{\cos \left(2 \pi-\frac{5 \pi}{6}\right)\right\} \)
\([ \because \frac{7 \pi}{6} \notin[0, \pi], \text { so we write } \frac{7 \pi}{6} \text { as } \left.\left(2 \pi-\frac{5 \pi}{6}\right)\right] \)
\(=\cos ^{-1}\left(\cos \frac{5 \pi}{6}\right)[\because \cos (2 \pi-\theta)=\cos \theta] \)
\(=\frac{5 \pi}{6} \quad\left[\because \frac{5 \pi}{6} \in[0, \pi]\right] \)

\(M_{11}=\begin{vmatrix} 0&4\\5&-7\end{vmatrix}=-0-20=-20\)
\(A_{11}=(-1)^{1+1}M_{11}=(-1)^2(-20)=-20\)
\(M_{12}=\begin{vmatrix}6&4\\1&-7 \end{vmatrix}=-42-4=-46\)
\(A_{12}=(-1)^{1+2}M_{12}=(-1)^3(-46)=(-1)(-46)=46\)
\(M_{13}=\begin{vmatrix}6&0\\1&5 \end{vmatrix}=30-0=30\)
\(A_{13}=(-1)^{1+3}M_{13}=(-1)^4(30)=30\)
\(M_{21}=\begin{vmatrix} -3&5\\5&-7\end{vmatrix}=21-25=-4\)
\(A_{21}=(-1)^{ 2+1}M_{21}=(-1)^3(-4=(-1 )(-4)=4)\)
\(M_{22}=\begin{vmatrix} 2&5\\1&-7\end{vmatrix}=-14-15=-19\)
\(A_{22}=(-1)^{2+2}M_{22}=(1)^4(-19)=-19\)
\(M_{23}=\begin{vmatrix} 2&-3\\1&5\end{vmatrix}=10+3=13\)
\(A_{23}=(-1)^{2+3}M_{23}=(-1)^513=-13\)
\(M_{31}=\begin{vmatrix}-3&5\\0&4 \end{vmatrix}=-12-0=-12\)
\(A_{31}=(-1)^{3+1}M_{31}=(-1)^4(-12)=-12\)
\(M_{32}=\begin{vmatrix} 2&5\\6&4\end{vmatrix}=8-30=-22\)
\(A_{32}=(-1)^{3+2}M_{32}=(-1)^5(-22)=(-1)(-22)=22\)
\(M_{33}=\begin{vmatrix} 2&-3\\6&0\end{vmatrix}=0+18=18\)
\(A_{33}=(-1)^{3+3}M_{33}=(-1)^6(18)=18\)
(ii)\(a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{32}\)
\(=(2)(-12)+(-3)(22)+(5)(18)=-24-66+90=0\)

f is not one-one, as f(1) = f(2) = 1. But f is onto, as given any y \(\in \) N, y \(\ne\) 1, we can choose x as y + 1 such that f (y + 1) = y + 1 – 1 = y. Also for 1 \(\in \) N, we have f(1) = 1.

\(\text { From figure, } \frac{5}{10}=\frac{r}{h} \Rightarrow r=\frac{h}{2} ; \frac{d V}{d t}=\frac{3}{2} \mathrm{cc} / \mathrm{s}\)
\(\text { Volume of the cone, } V=\frac{1}{3} \pi r^{2} h\)
\(=\frac{1}{3} \pi\left(\frac{h}{2}\right)^{2} h=\frac{1}{12} \pi h^{3} \)
\(\frac{d V}{d t}=\frac{1}{12} \pi \cdot 3 h^{2} \cdot \frac{d h}{d t} \Rightarrow \frac{3}{2}=\frac{\pi}{4} h^{2} \cdot \frac{d h}{d t} \)
\(\left.\Rightarrow \frac{d h}{d t}=\frac{6}{\pi h^{2}} \Rightarrow \frac{d h}{d t}\right]_{h=4}=\frac{6}{16 \pi}=\frac{3}{8 \pi} \mathrm{cm} / \mathrm{s}\)

At x = 0, we have f(0) = a.
\(\mathrm{LHL} =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} \frac{1-\cos 4 h}{h^{2}}=\lim _{h \rightarrow 0} \frac{4 \cdot 2 \sin ^{2} 2 h}{4 \cdot h^{2}} \)
\(=\lim _{h \rightarrow 0} 8\left(\frac{\sin ^{2} h}{2 h}\right)^{2} \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \) 
\( \mathrm{RHL} =\lim _{h \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} \frac{\sqrt{h}}{\sqrt{16+\sqrt{h}-4}} \times \frac{\sqrt{16+\sqrt{h}+4}}{\sqrt{16+\sqrt{h}+4}} \)
\(=\lim _{h \rightarrow 0} \frac{\sqrt{h}[\sqrt{16+\sqrt{h}}+4]}{16+\sqrt{h}-16}=4+4=8 \) 
\(\therefore \mathrm{LHL}=\mathrm{RHL}=f(0) \text { [Ans. 8] }\)

Let x be the number of C₁ cars produced, y be the number of C₂ cars produced and z be the number of C₃ cars produced.
Now, according to the given conditions, we have
2x + 3y +4 z = 29, x + y + 2z =13 and 3x + 2y + z = 16
[Ans. x = 2, y = 3 and z = 4]

Firstly, assume the given inverse trigonometric function equal to y( or x) Let \(y=\cos ^{-1}\left(-\frac{1}{2}\right)\)
\(\Rightarrow \cos y=\frac{-1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right) \)
\(\Rightarrow \cos y=\cos \left(\frac{2 \pi}{3}\right) \Rightarrow y=\frac{2 \pi}{3}\)
Since, the principal value branch of \(\cos ^{-1} is [0, \pi]\) and
\(\frac{2 \pi}{3} \in[0, \pi]\) 
So, the principal value of \(\cos ^{-1}\left(-\frac{1}{2}\right)\) is \(\frac{2 \pi}{3}\).
Now, find the principal value of \(tan ^{-1}(-\sqrt{3})\ Let\ x=\tan ^{-1}(-\sqrt{3})\)
\(\Rightarrow \tan x=-\sqrt{3}=-\tan \frac{\pi}{3} \quad\left[\because \tan \frac{\pi}{3}=\sqrt{3}\right]\)
\(\Rightarrow \tan x=\tan \left(\frac{-\pi}{3}\right) \quad[\because \tan (-\theta)=-\tan \theta]\)
\(\Rightarrow x=-\frac{\pi}{3}\)
Since, the principal value branch of \(\tan ^{-1}\) is \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) and \(-\frac{\pi}{3} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) 
Now, find the principal value of \(\operatorname{cosec}^{-1}\) 
Let \(z=\operatorname{cosec}^{-1}(2) \Rightarrow \operatorname{cosec} z=2=\operatorname{cosec} \frac{\pi}{6} \Rightarrow z=\frac{\pi}{6}\)
Since, the principal value branch of \(\operatorname{cosec}^{-1}\) is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}\) and \(\frac{\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)
So, the principal value of \(\operatorname{cosec}^{-1}(2)\) is \(\frac{\pi}{6}\).
\(\therefore \cos ^{-1}\left(-\frac{1}{2}\right)+\tan ^{-1}(-\sqrt{3})-\operatorname{cosec}^{-1} 2\)
\(=\frac{2 \pi}{3}-\frac{\pi}{3}-\frac{\pi}{6}=\frac{(4-2-1) \pi}{6}=\frac{\pi}{6}\)

f(x) = sin3x - cos3x, 0 < x < π 
\(\Rightarrow f'(x)=3\cos3x+3\sin3x\)
\(=3(\cos3x+\sin3x)\)
Put \(f'(x)=0\)
\(\Rightarrow \cos3x+\sin3x=0\)
\(\Rightarrow \sin3x=-\cos3x\)
\(\Rightarrow -\tan3x=1\)
\(\Rightarrow \tan3x=-1\)
As 0\(\pi\), 0 < 3x > 3\(\pi\)
ஃ tan 3x is negative for the following values:
\(3x=\frac{3\pi}{4}\)
\(\Rightarrow x=\frac{\pi}{4}\)
\(3x=\pi+\frac{3\pi}{4}=\frac{7\pi}{4}\)
\(\Rightarrow x=\frac{7\pi}{12}\)
\(3x=\frac{7\pi}{4}+\pi=\frac{11\pi}{4}\)
\(\Rightarrow x=\frac{11\pi}{12}\)
Hence we have intervals:
Hence, \(f(x)-\sin3x-\cos3x\) is strictly increasing in the intervals \((0,\frac{\pi}{4})\cup(\frac{7\pi}{12},\frac{11\pi}{12}) \) and strictly decreasing in intervals \((\frac{\pi}{4},\frac{7\pi}{12})\cup(\frac{11\pi}{12},\pi)\)

\(\text {Consider } \tan ^{-1}\left[2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right] \)
\(=\tan ^{-1}\left[2 \sin \left(2 \cdot \frac{\pi}{6}\right)\right]=\tan ^{-1}\left[2 \sin \frac{\pi}{3}\right] \)
\(=\tan ^{-1}\left(2 \cdot \frac{\sqrt{3}}{2}\right)=\tan ^{-1}(\sqrt{3})=\frac{\pi}{3} \)

We know that, if and g are continuous functions, then
(a) f + g is continuous
(b) f - g is continuous.
(c) fg is continuous
(d) \(\frac{f}{g}\) is continuous at these points, where \(g(x) \neq 0\) 
Here,\(\frac{g(x)}{f(x)}=\frac{\frac{x^{2}}{2}+1}{2 x}=\frac{x^{2}+2}{4 x}\) 
which is discontinuous at x = o

By definition of collinearity.

Clearly,
\( f(a) =\left|\begin{array}{ccc} 0 & 0 & a-b \\ 2 a & 0 & a-c \\ a+b & a+c & 0 \end{array}\right| \)
\(=[(a-b)\{2 a \cdot(a+c)\}] \neq 0 \) 
\( \therefore \ f(b) =\left|\begin{array}{ccc} 0 & b-a & 0 \\ b+a & 0 & b-c \\ 2 b & b+c & 0 \end{array}\right|\)
\( =-(b-a)[2 b(b-c)] \)
\( =-2 b(b-a)(b-c) \neq 0 \) 
 

As \(\frac { dx }{ dt } =2\) cm/s, x is side of equiolatral triangle.
A = \(\frac { \sqrt { 3 } }{ 4 } { x }^{ 2 }\)
\(\Rightarrow \frac { dA }{ dx } =\frac { \sqrt { 3 } }{ 2 } { x }\frac { dx }{ dt } \)
= \(\frac { \sqrt { 3 } }{ 2 } x2=\sqrt { 3 } x\)
∴\(|\frac{dA}{dx}|\)x=10 = 10\(\sqrt3\) cm²/s

As differentiable functions is continuous also

as sec^-1x + sec^-1 y = \(\frac{\pi}{2}\)
⇒ \(\frac{\pi}{2}\) - cosec^-1 x + \(\frac{\pi}{2}\) - cosec^-1 y = \(\frac{\pi}{2}\)
⇒ cosec^-1 x + cosec^-1 y = \(\frac{\pi}{2}\)

(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) Given, R = {(V₁,V₂) :  V₁,V₂ ∈ l}and both use their voting right in general election-2019. Since, X,Y∈ I⋅X exercised his voting right while Y did not cast her vote in general election-2019.
∴ Clearly, (X,Y)∉R
(ii) (a) Relation is symmetric.
∴(X,W)∈R
⇒(W,X)∈R
(iii) (a) Since, (F,F)∈R,F ∈ I and F use their voting right.
⇒R is reflexive.
⇒(F1,F2)∈R
⇒(F2,F1)∈R
R is symmetric.
and (F₁,F₂) ∈ R 
and  (F₂,F₃) ∈ R
⇒ (F₁,F₃) ∈ R 
(By transitive property)
(iv) (c) Given, relation RR is reflexive, symmetric and transitive.
∴R is equivalence relation.
(v) (a) Clearly, Mr. Shyam exercised his voting right in general election-2019, then Mr. Shyam is related to all those eligible voters who cast their votes.

(i) (c) : Let Rs. A, Rs. B and Rs.C be the cost incurred by the organisation for villages X, Y and Z respectively. Then A, B, C will be given by the following matrix equation.
\(\left[\begin{array}{ccc} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{array}\right]\left[\begin{array}{l} 50 \\ 20 \\ 40 \end{array}\right]=\left[\begin{array}{c} A \\ B \\ C \end{array}\right]\) 
\(\Rightarrow\left[\begin{array}{l} A \\ B \\ C \end{array}\right]=\left[\begin{array}{c} 400 \times 50+300 \times 20+100 \times 40 \\ 300 \times 50+250 \times 20+75 \times 40 \\ 500 \times 50+400 \times 20+150 \times 40 \end{array}\right]\) 
(ii) (c)
(iii) (b)
(iv) (c) : Total number of toilets that can be expected in each village is given by the following matrix.
\(\begin{array}{l} X \\ Y \\ Z \end{array}\left[\begin{array}{ccc} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{array}\right]\)\(\left[\begin{array}{c} 2 / 100 \\ 4 / 100 \\ 20 / 100 \end{array}\right]\) 
\(\begin{array}{l} X \\ Y \\ Z \end{array}\)\(\left[\begin{array}{c} 8+12+20 \\ 6+10+15 \\ 10+16+30 \end{array}\right]\)=\(\begin{array}{l} X \\ Y \\ Z \end{array}\)\(\left[\begin{array}{c} 40 \\ 31 \\ 56 \end{array}\right]\) 
(v) (d)

(c) Assertion is correct, Reason is incorrect

(a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.