The function  This is defined for all real numbers such that \(\cos x \neq 0, \text { i.e., } x \neq(2 n+1) \frac{\pi}{2}\)
We have just proved that both sine and cosine functions are continuous. Thus tan x being a quotient of two continuous functions is continuous wherever it is defined.
An interesting fact is the behaviour of continuous functions with respect to composition of functions. Recall that if f and g are two real functions, then
(f o g) (x) = f (g(x))
is defined whenever the range of g is a subset of domain of f.

Here  \(f(x)={ x }^{ 17 } \cos ^{ 2 }{ x } .\)
\(\therefore \ f\left( -x \right) ={ (-x) }^{ 17 } \cos ^{ 4 }{ (-x) } \)
\( ={ (-x) }^{ 17 }\ { (\cos { x } ) }^{ 4 }=-{ x }^{ 17 }\cos ^{ 4 }{ x } =-f\left( x \right) \)
\(\Rightarrow f\left( x \right) \) is an odd function.
\(\therefore \ I= \int _{ -1 }^{ 1 }{ { x }^{ 17 } } \cos ^{ 4 }{ x } dx=0\)

\(I= \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \cos ^{ 2 }{ x } }{ \cos ^{ 2 }{ x } +4\sin ^{ 2 }{ x } } } dx...(1)\)
\( =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin ^{ 2 }{ x } dx }{ \sec ^{ 2 }{ x } +\tan ^{ 2 }{ x } +\sec ^{ 2 }{ x } } } .\)
[Dividng numerator and denominator by \(\cos ^{ 4 }{ x } \)]
Put \(\cos ^{ 4 }{ x } \)  so that  \(\sec ^{ 2 }{ x } dx=dt.\)
when \(x=0,t=0.\)  when \(x=\frac { \pi }{ 2 } . t\rightarrow \infty \)
\(\therefore \ I=\lim _{ k\rightarrow \infty }{ \int _{ 0 }^{ k }{ \frac { dt }{ (1+{ t }^{ 2 })+4{ t }^{ 2 }(1+{ t }^{ 2 }) } } } \)
\(=\lim _{ k\rightarrow \infty }{ \int _{ 0 }^{ k }{ \frac { dt }{ (1+{ t }^{ 2 })(1+{ 4t }^{ 2 }) } } } \)
\(=\lim _{ k\rightarrow \infty }{ \int _{ 0 }^{ k }{ \left( \frac { -\frac { 1 }{ 3 } }{ 1+{ t }^{ 2 } } +\frac { \frac { 4 }{ 3 } }{ 1+{ t }^{ 2 } } \right) } } dt\)
[Partial Fractions]
\(=-\frac { 1 }{ 3 } \lim _{ k\rightarrow \infty }{ { \left[ \tan ^{ -1 }{ t } \right] }_{ 0 }^{ k } } +\frac { 4 }{ 3 } .\frac { 1 }{ 4 } \lim _{ k\rightarrow \infty }{ \int _{ 0 }^{ k }{ \frac { dt }{ { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ t }^{ 2 } } } } \)
\(=-\frac { 1 }{ 3 } \left[ \frac { \pi }{ 2 } -0 \right] +\frac { 1 }{ 3 } .\frac { 1 }{ \frac { 1 }{ 2 } } \lim _{ k\rightarrow \infty }{ { \left[ \tan ^{ -1 }{ \frac { t }{ \frac { 1 }{ 2 } } } \right] }_{ 0 }^{ k } } \)
\(=-\frac { \pi }{ 6 } +\frac { 2 }{ 3 } \lim _{ k\rightarrow \infty }{ \left( \tan ^{ -1 }{ 2k-0 } \right) } \)
\(=-\frac { \pi }{ 6 } +\frac { 2 }{ 3 } \left( \frac { \pi }{ 2 } -0 \right) =-\frac { \pi }{ 6 } +\frac { \pi }{ 3 } =\frac { \pi }{ 6 } .\)

The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are O (0, 0), A (2, 0), B (0, 3), and C\(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \)
The values of Z at these corner points are as follows.

Therefore, the maximum value of Z is \(\frac { 235 }{ 19 } \) at \(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \).

We know that the slope of the tangent to the curve is \(\frac { dy }{ dx } .\)
By the question, \(\frac { dy }{ dx } =x+y\)
\(\Rightarrow\) \(\frac { dy }{ dx } -y=x\) .(1)
Here  \('P'=-1\) and \('Q'=x.\)
\(\therefore\) I.F. \(={ e }^{ \int { Pdx } }={ e }^{ \int { -1.dx } }={ e }^{ -x }\)
Multiplying (1) by \({ e }^{ -x },\) we get:
\({ e }^{ -x }.\frac { dy }{ dx } -y.{ e }^{ -x }=x.{ e }^{ -x }.\)
\(\Rightarrow\) \(\frac { d }{ dx } \left( y.{ e }^{ -x } \right) =x{ e }^{ -x }.\)
Integrating,
\(y.{ e }^{ -x }=\int { x.{ e }^{ -x } } dx+C\)
\(=x.\frac { { e }^{ -x } }{ -1 } -\int { \left( 1 \right) } \frac { { e }^{ -x } }{ -1 } dx+C\)
[Integrating by Parts]
\(=-x{ e }^{ -x }+\int { { e }^{ -x } } dx+C\)
\(=-x{ e }^{ -x }+\frac { { e }^{ -x } }{ -1 } +C\)
\(\Rightarrow\) \(y=-x-1+C\quad { e }^{ x }\)
Since the curve passes through of the origin(0,0),
\(\therefore\) \(0=-0-1+C\quad { e }^{ 0 }\Rightarrow C=1.\)
Putting in (2),\(y=-x-1+{ e }^{ x }\)
\(\Rightarrow x+y+1={ e }^{ x },\)
Which is the requested equation of the curve.

The sample space has 216 outcomes.
\(\text { Now } \quad \mathrm{A}=\left\{\begin{array}{lllll} (1,1,4) & (1,2,4) & \ldots & (1,6,4) & (2,1,4) & (2,2,4) & \ldots (2,6,4) \\ (3,1,4) & (3,2,4) & \ldots &(3,6,4) & (4,1,4) & (4,2,4) & \ldots(4,6,4) \\ (5,1,4) & (5,2,4) & \ldots & (5,6,4) & (6,1,4) & (6,2,4) & \ldots(6,6,4) \end{array}\right\}\)
B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)} and A ∩ B = {(6,5,4)}.
\(\text { Now }P(B)=\frac{6}{216} \text { and } P(A \cap B)=\frac{1}{216} \)
Then  \(P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac { \frac { 1 }{ 216 } }{ \frac { 6 }{ 216 } } =\frac { 1 }{ 6 }\)

\(\text { Put } \tan ^{-1} x^{4}=t \Rightarrow \frac{1}{1+x^{8}} \cdot 4 x^{3} d x=d t\) 
\(-\frac{1}{4} \cos \left(\tan ^{-1} x^{4}\right)+C\)

The area of a circle (A) with radius (r) is given by.
Therefore, the rate of change of area (A) with respect to time (t) is given by,
[By chain rule]
It is given that.
Thus, when r = 8 cm,

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80 π cm²/s.

Given,y= sin x
Here,\(\frac{d y}{d x}=\cos x \)
\(\therefore\) Slope of tangent at (0,0) = 1
Hence, the equation of tangent is y = Xl.

Given expression \(\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)
Putting \(\sin ^{-1}\left(\frac{3}{5}\right)=x \text { and } \cot ^{-1}\left(\frac{3}{2}\right)=y\)
Or \( \sin (x)=3 / 5 \ and \ \cot y=3 / 2 \)
\(Now, \sin (x)=3 / 5 \Rightarrow \cos x=\sqrt{1-\sin ^2 x}=4 / 5\ and \ \sec x=5 / 4 \)
(Using identities \(\cos \mathrm{x}=\sqrt{1-\sin ^2 x} \text { and } \sec \mathrm{x}=1 / \cos\))
\(\tan x=\sqrt{\sec ^2 x-1}=\sqrt{\frac{25}{16}-1}=3 / 4 \text { and } \tan y=1 / \cot (y)=2 / 3\)
we can written as
\( \tan \left(\sin ^{-1}\left(\frac{3}{5}\right)+\cot ^{-1} \frac{3}{2}\right)=\tan (x+y) \)
\(=\frac{\tan x+\tan y}{1-\tan x \tan y}=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}} \)
\(=17 / 6\)

Since E and F are independent, we have
P(E ∩ F) = P(E) . P(F) ....(1)
From the venn diagram in Fig 13.3, it is clear that E ∩ F and E ∩ F′ are mutually exclusive events and also E =(E ∩ F) ∪ (E ∩ F′). Therefore P(E) = P(E ∩ F) + P(E ∩ F′)
or P(E ∩ F′) = P(E) − P(E ∩ F)
= P(E) − P(E) . P(F) (by (1))
= P(E) (1−P(F))
= P(E). P(F′)
Hence, E and F′ are independent

\( \Delta_{1} =\left|\begin{array}{ccc} A & B & C \\ x & y & z \\ z y & z x & x y \end{array}\right| =\left|\begin{array}{ccc} A & x & z y \\ B & y & z x \\ C & z & x y \end{array}\right| \\ =\frac{1}{x y z}\left|\begin{array}{lll} A x & x^{2} & x y z \\ B y & y^{2} & x y z \\ C z & z^{2} & x y z \end{array}\right|=\frac{x y z}{x y z}\left|\begin{array}{ccc} A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1 \end{array}\right|=\Delta \)

R is reflexive, as 2 divides (a – a) for all a ∈ Z. Further, if (a, b) ∈ R, then 2 divides a – b. Therefore, 2 divides b – a. Hence, (b, a) ∈ R, which shows that R is symmetric. Similarly, if (a, b) ∈ R and (b, c) ∈ R, then a – b and b – c are divisible by 2. Now, a – c = (a – b) + (b – c) is even (Why?). So, (a – c) is divisible by 2. This shows that R is transitive. Thus, R is an equivalence relation in Z.
note that all even integers are related to zero, as (0, ± 2), (0, ± 4) etc., lie in R and no odd integer is related to 0, as (0, ± 1), (0, ± 3) etc., do not lie in R. Similarly, all odd integers are related to one and no even integer is related to one. Therefore, the set E of all even integers and the set O of all odd integers are subsets of Z satisfying following conditions:
(i) All elements of E are related to each other and all elements of O are related to each other.
(ii) No element of E is related to any element of O and vice-versa.
(iii) E and O are disjoint and Z = E \(\cup\) O.
The subset E is called the equivalence class containing zero and is denoted by [0]. Similarly, O is the equivalence class containing 1 and is denoted by [1]. Note that [0] \(\ne\)[1], [0] = [2r] and [1] = [2r + 1], r \(\in\) Z. Infact, what we have seen above is true for an arbitrary equivalence relation R in a set X. Given an arbitrary equivalence relation R in an arbitrary set X, R divides X into mutually disjoint subsets A_i called partitions or subdivisions of X satisfying:
(i) all elements of A_i are related to each other, for all i.
(ii) no element of A_i is related to any element of A_j, i \(\ne\) j.
(iii) \(\cup\) A_j = X and A_i \(\cap\) A_j = f, i \(\ne\) j.
The subsets Ai are called equivalence classes. The interesting part of the situation is that we can go reverse also. For example, consider a subdivision of the set Z given by three mutually disjoint subsets A₁, A₂ and A₃ whose union is Z with
A₁ = {x \(\in\) Z : x is a multiple of 3} = {..., – 6, – 3, 0, 3, 6, ...}
A₂ = {x \(\in\) Z : x – 1 is a multiple of 3} = {..., – 5, – 2, 1, 4, 7, ...}
A₃ = {x \(\in\) Z : x – 2 is a multiple of 3} = {..., – 4, – 1, 2, 5, 8, ...}
Define a relation R in Z given by R = {(a, b) : 3 divides a – b}. Following the arguments similar to those , we can show that R is an equivalence relation. Also, A₁ coincides with the set of all integers in Z which are related to zero, A₂ coincides with the set of all integers which are related to 1 and A₃ coincides with the set of all integers in Z which are related to 2. Thus, A₁ = [0], A₂ = [1] and A₃ = [2].
In fact, A₁ = [3r], A₂ = [3r + 1] and A₃ = [3r + 2], for all r \(\in\) Z.

Given that equation of line is
x -1 = 2y -4 = 3z -12
\(\Rightarrow \frac{x-1}{1}=\frac{y-2}{1 / 2}=\frac{z-4}{1 / 3}\) 
The plane \(\vec{r} \cdot(-5 \hat{i}+2 \hat{j}+12 \hat{k})=47\) contains the given line if 
(i) it passes through (1, 2, 4).
(ii) it is parallel to the line
We have, \((\hat{i}+2 \hat{j}+4 \hat{k})(-5 \hat{i}+2 \hat{j}+12 \hat{k})\) 
= - 5 + 4 + 48 =47
So, the plane passes through the point (1, 2, 4)
Also \(1(-5)+\frac{1}{2}(2)+\frac{1}{3}(12)\) 
= -5 + 1 + 4 = 0
Therefore, the plane is parallel to the line
= \(\vec{r} \cdot(-5 \hat{i}+2 \hat{j}+12 \hat{k})=47\)

As shown in the Figure, the line y = 3x + 2 meets x-axis at x\(=\frac{-2}{3}\) and its graph lies below x-axis for \(x \in\left(-1, \frac{-2}{3}\right)\) and above x-axis for \(x \in\left(\frac{-2}{3}, 1\right)\)
The required area = Area of the region ACBA + Area of the region ADEA
\(=\left|\int_{-1}^{\frac{-2}{3}}(3 x+2) d x\right|+\int_{\frac{-2}{3}}^1(3 x+2) d x\)
\(=\left|\left[\frac{3 x^2}{2}+2 x\right]_{-1}^{\frac{-2}{3}}\right|+\left[\frac{3 x^2}{2}+2 x\right]_{\frac{-2}{3}}^1=\frac{1}{6}+\frac{25}{6}=\frac{13}{3}\)

Given, determinant \(\left|\begin{array}{rrr} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right|\) 
We have \(M_{11}=\left|\begin{array}{rr} 0 & 4 \\ 5 & -7 \end{array}\right|=0-20=-20\) 
\(A_{11}=(-1)^{1+1} M_{11}=(-20)=-20\) 
\(\left[\because \text { cofactor } C_{i j}=(-1)^{i+j} M_{i j}\right]\) 
\(M_{12}=\left|\begin{array}{rr} 6 & 4 \\ 1 & -7 \end{array}\right|=-42-4=-46 \)
\(C_{12}=(-1)^{1+2} M_{12}=-(-46)=46 \)
\( M_{13}=\left|\begin{array}{ll} 6 & 0 \\ 1 & 5 \end{array}\right|=30-0=30\) 
\(C_{13}=(-1)^{1+3} M_{13}=(30)=30 \)
\(M_{21}=\left|\begin{array}{rr} -3 & 5 \\ 5 & -7 \end{array}\right|=21-25=-4 \)
\(C_{21}=(-1)^{2+1} M_{21}=-(-4)=4 \)
\(M_{22}=\left|\begin{array}{rr} 2 & 5 \\ 1 & -7 \end{array}\right|=-14-5=-19 \) 
\(C_{22}=(-1)^{2+2} M_{22}=(-19)=-19 \)
\(M_{23}=\left|\begin{array}{rr} 2 & -3 \\ 1 & 5 \end{array}\right|=10+3=13 \)
\(C_{23}=(-1)^{2+3} M_{23}=-(13)=-13 \)
\(M_{31}=\left|\begin{array}{rr} -3 & 5 \\ 0 & 4 \end{array}\right|=-12-0=-12 \)
\(C_{31}=(-1)^{3+1} M_{31}=(-12)=-12 \) 
\(M_{32}=\left|\begin{array}{rr} 2 & 5 \\ 6 & 4 \end{array}\right|=8-30=-22 \)
\(C_{32}=(-1)^{3+2} M_{32}=-(-22)=22 ;\)
\(M_{33}=\left|\begin{array}{rr} 2 & -3 \\ 6 & 0 \end{array}\right|=0+18=18\)
\(C_{33}=(-1)^{3+3} M_{33}=(18)=18 \) 
We have, \(a_{11}=2, a_{12}=-3, a_{13}=5, C_{31}=-12\) 
 \( C_{32} =22, C_{33}=18 \)
\(\therefore a_{11} C_{31} +a_{12} C_{32}+a_{13} C_{33} \)
\(=(2)(-12)+(-3)(22)+(5)(18) \)
\( =-24-66+90=-90+90=0 \) 

(i) Let the width of the printed part be x cm and height of the printed part be y cm.
Given, area of printed part = 24 cm²
\(\Rightarrow \quad x y=24 \Rightarrow y=\frac{24}{x}\)
From given condition,
Width of the page = (x+ 3) cm
Height of the page = (y + 2) cm
\(\therefore\) Area of paper, \(\begin{aligned} A & =(x+3)(y+2)=(x+3)\left(\frac{24}{x}+2\right) \end{aligned}\)
\(\begin{aligned} =24+2 x+\frac{72}{x}+6=30+3 x+\frac{48}{x} \end{aligned}\)
(ii) We have, \(A=24+2 x+\frac{72}{x}+6\)
\(\therefore \quad \frac{d A}{d x}=2-\frac{72}{x^2}\)
For maximum or minimum,
\(\frac{d A}{d x}=0 \Rightarrow 2 x^2-72=0\)
\(\begin{array}{ll} \Rightarrow & x^2=36 \end{array}\)
\(\begin{array}{ll} \Rightarrow & x= \pm 6 \Rightarrow x=6 \end{array}\)     [\(\because\) x cannot be negative]
Now, \(\frac{d^3 A}{d x^2}=\frac{44}{x^3}\)
At \(x=6 \frac{d^3 A}{d x^2}>0\)
\(\therefore\) Area is minimum when x = 6
\(\Rightarrow \quad y=\frac{24}{6}=4\)
\(\therefore\) dimensions of the page are
Width of the page = x + 3 = 6 + 3 = 9 cm
Height of the page = y + 2 = (4 + 2) = 6 cm

Here, \(\left|\vec{F}_{1}\right|=\sqrt{(4)^{2}+0^{2}}=4 \mathrm{KN}\)
\( \left|\vec{F}_{2}\right|=\sqrt{(-2)^{2}+4^{2}}=\sqrt{20} \mathrm{KN} \)
\(\left|\vec{F}_{3}\right|=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{18} \mathrm{KN}\)
(i) (a): Since, \(\sqrt 20\) is larger. So, team B will win the game.
(ii) (b): Let F be the combined force
\(\therefore \vec{F}=\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}=4 \hat{i}+0 \hat{j}-3 \hat{i}-3 \hat{j}-2 \hat{i}+4 \hat{j} \)
\(=-\hat{i}+\hat{j} \)
\(\therefore |\vec{F}|=\sqrt{(-1)^{2}+1^{2}}=\sqrt{2}=1.4 \mathrm{KN}\)
(iii) (c) : We have,  \(\vec{F}=-\hat{i}+\hat{j}\)
\(\therefore \theta=\tan ^{-1}\left(\frac{F_{y}}{F_{x}}\right)=\tan ^{-1}\left(\frac{1}{-1}\right)=\frac{3 \pi}{4} \text { radian }\)
= 0.75 x 3.14 radian = 2.3555 radian ≈ 2.4 radian 
(iv) (a): Magnitude of force of Team B = \(\sqrt 20\) KN
= 2\(\sqrt 5\) KN
(v) (b): 4 KN force is applied by team A.

Let B, R, Y and G denote the events that ball drawn is blue, red, yellow and green respectively.
\(\therefore P(B)=\frac{12}{35}, P(R)=\frac{8}{35}, P(Y)=\frac{10}{35} \text { and } P(G)=\frac{5}{35}\)
\((i) \ (c): P(G \cap B)=P(B) \cdot P(G \mid B)=\frac{12}{35} \cdot \frac{5}{34}=\frac{6}{119}\)
\((ii) \ (\mathbf{b}): P(R \cap Y)=P(Y) \cdot P(R \mid Y)=\frac{10}{35} \cdot \frac{8}{34}=\frac{8}{119}\)
(iii) (a): Let E = event of drawing a first red ball and
F = event of drawing a second red ball
Here, \(P(E)=\frac{8}{35} \text { and } P(E)=\frac{7}{34}\)
\(\therefore \P(F \cap E)=P(E) \cdot P(F \mid E)=\frac{8}{35} \cdot \frac{7}{34}=\frac{4}{85}\)
\(\text {(iv) }(c): P\left(Y^{\prime} \cap G\right)=P(G) \cdot\left(Y^{\prime} \mid G\right)=\frac{5}{35} \cdot \frac{24}{34}=\frac{12}{119}\)
(v) (d): Let E = event of drawing a first non-blue ball and F = event of drawing a second non-blue ball
Here, \(P(E)=\frac{23}{35} \text { and } P(F)=\frac{22}{34}\)
\(\therefore \ P(F \cap E)=P(E) \cdot P(F \mid E)=\frac{23}{35} \cdot \frac{22}{34}=\frac{253}{595}\)