As area is above the x-axis
∴ area = \(2\int _{ 0 }^{ 7 }{ \sqrt { 49-{ x }^{ 2 } } } \)
= \({ \left[ \frac { x }{ 2 } \sqrt { 49-{ x }^{ 2 } } +\frac { 49 }{ 2 } { sin }^{ -1 }\frac { x }{ 7 } \right] }_{ 0 }^{ 7 }\)
= \(2\left[ \left( \frac { 7 }{ 2 } \times 0+\frac { 49 }{ 2 } { sin }^{ -1 }1 \right) -(0) \right] \)
= \(\frac { 49 }{ 2 } \pi \) sq units

As \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 1+cos\left( \frac { \pi }{ 2 } -x \right) } } \)
= \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 1 }{ 2 } { sec }^{ 2 }\left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) dx } \)
\(=\frac { 1 }{ 2 } .{ \left[ \frac { tan\left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) }{ -\frac { 1 }{ 2 } } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(=-tan\left( \frac { \pi }{ 4 } -\frac { \pi }{ 4 } \right) +tan\left( \frac { \pi }{ 4 } -0 \right) =1\)

As \(\int { \frac { 1 }{ \sqrt { 4-{ 9x }^{ 2 } } } dx } \)
\(=\frac { 1 }{ 3 } \int { \frac { 1 }{ \sqrt { { \left( \frac { 2 }{ 3 } \right) }^{ 2 }-{ x }^{ 2 } } } dx } \)
\(=\frac { 1 }{ 3 } \int { \frac { 1 }{ \sqrt { { \left( \frac { 2 }{ 3 } \right) }^{ 2 }-{ x }^{ 2 } } } dx } \)
⇒ a = \(\frac32\)

As \(\frac { d }{ dx } \left( \frac { { 2 }^{ x } }{ { log }_{ e }2 } \right) \)
\(=\frac { 1 }{ { log }_{ e }2 } .{ 2 }^{ x }.{ log }_{ e }2 ={ 2 }^{ x }\)

As y = cos^-1 (x² - 4)
⇒ cos y = x² - 4
Since - 1 ≤ cos y ≤ 1
i.e - 1  ≤ x² -4  ≤1
⇒ 3 ≤ x² ≤ 5
⇒ \(\sqrt3\)≤ |x|≤ \(\sqrt5\)
⇒ x ∈  [-\(\sqrt5\) ,-\(\sqrt3\)] ∪ [-\(\sqrt5\),\(\sqrt3\)]

Let θ = sin^-1 \(-(\frac12)\)
⇒ sin^-1 = \(-\frac12\) = sin \((-\frac{\pi}{6})\) = θ = \(-\frac{\pi}{6}\)

A relation R is an identity relation in set A if for all a ∈ A, (a, a) ∈ R.

T₁ and T₃ are similar as their sides are proportional.

(i) b : Curves y = cos x and y = x + 1 meet at point C(O, 1).

(ii) C : curve y=cosx meet the x axis at \(A^{\prime}\left(\frac{-\pi}{2}, 0\right)\) and \(A\left(\frac{\pi}{2}, 0\right)\) .
(iii) (a) : \(\int_{-1}^{0}(x+1) d x=\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}=0-\left(\frac{1}{2}-1\right)=\frac{1}{2}\) 
(iv) (d) : \(\int_{0}^{\pi / 2} \cos x d x=[\sin x]_{0}^{\pi / 2}=\sin \frac{\pi}{2}-\sin 0=1\) 
(v) (b) : Required area \(\int_{-1}^{0}(x+1) d x+\int_{0}^{\pi / 2} \cos x d x\) 
\(=\frac{1}{2}+1=\frac{3}{2} \text { sq. units }\)

(i) (c) : We have, x² +y²= 16  ..(i)
and y = x ...(ii)
From (i) and (ii), \(2 x^{2}=16 \Rightarrow x^{2}=8 \Rightarrow x=2 \sqrt{2}\)  (\(\therefore\) x lies in first quadrant)
\(\therefore\) Point of intersection of (i) and (ii) in first quadrant is \((2 \sqrt{2}, 2 \sqrt{2})\) .
(ii) (b) : The shaded region which represent the areabounded by two given curves in first quadrant is shown below.
 
(iii)( d) : \(\int_{0}^{2 \sqrt{2}} x d x=\left[\frac{x^{2}}{2}\right]_{0}^{2 \sqrt{2}}=\frac{(2 \sqrt{2})^{2}}{2}=\frac{8}{2}=4\) 
(iv) (a) : \(\int_{2 \sqrt{2}}^{4} \sqrt{16-x^{2}} d x=\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \cdot \sin ^{-1}\left(\frac{x}{4}\right)\right]_{2 \sqrt{2}}^{4}\) 
 \(=8 \sin ^{-1}(1)-4-8 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\) 
\(=8\left(\frac{\pi}{2}\right)-4-8\left(\frac{\pi}{4}\right)=4 \pi-4-2 \pi=2 \pi-4=2(\pi-2)\) 
(v) (d) : Required area = Area (OLA) + Area (BAL)
\(=\int_{0}^{2 \sqrt{2}} x d x+\int_{2 \sqrt{2}}^{4} \sqrt{16-x^{2}} d x\) 
\(=4+2(\pi-2)=2 \pi \text { sq. units. }\)

(i) (a) : For f(x) to be defined \(x-2 \neq 0\) i.e.,\(x \neq 2\) 
\(\therefore\) Domain of f = R - {2}
(ii) (b) : Let y =J(x), then \(y=\frac{x-1}{x-2}\) 
\(\Rightarrow x y-2 y=x-1 \Rightarrow x y-x=2 y-1 \Rightarrow x=\frac{2 y-1}{y-1}\) 
Since, \(x \in R-\{2\}\),therefore \(y \neq 1\)
Hence, range of f = R-{1} 
(iii) (d): We have,g(x) = 2f(x) - 1
\(=2\left(\frac{x-1}{x-2}\right)-1=\frac{2 x-2-x+2}{x-2}=\frac{x}{x-2}\)
(iv) (a) : We have, \(g(x)=\frac{x}{x-2}\) ,
Let  \(g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow \frac{x_{1}}{x_{1}-2}=\frac{x_{2}}{x_{2}-2}\) 
 \(\Rightarrow x_{1} x_{2}-2 x_{1}=x_{1} x_{2}-2 x_{2} \Rightarrow 2 x_{1}=2 x_{2} \Rightarrow x_{1}=x_{2}\) 
Thus, \(g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow x_{1}=x_{2}\) 
Hence, g(x) is one-one.
(v) (c) 

(i) (a) : Clearly (1, 1), (2, 2), (3, 3), \(\in\) R. So, R is reflexive on A.
Since, \((1,2) \in R \text { but }(2,1) \notin R\) So, R is not symmetric on A.
Since, \((2,3), \in R\) and \((3,1) \in R\) but \((2,1) \notin R\) .So, R is not transitive on A.
(ii) (b) : Since, (1,1), (2, 2) and (3, 3) are not in R. So, R is not reflexive on A.
Now, \((1,2) \in R \Rightarrow(2,1) \in R\) 
and \((1,3) \in R \Rightarrow(3,1) \in R\) 
So, R is symmetric
Clearly,\((1,2) \in R \text { and }(2,1) \in R \text { but }(1,1) \notin R\) 
So, R is not transitive on A.
(iii) (c) : We have, \(R=\{(x, y): y=x+5 \text { and } x<4\}\) ,where \(x, y \in N\) .
\(\therefore R=\{(1,6),(2,7),(3,8)\}\) 
Clearly, (1, 1), (2, 2) etc. are not in R. So, R is not reflexive.
Since, \((1,6) \in R\) but \((6,1) \notin R\) So, R is not symmetric. 
Since, \((1,6) \in R\) R and there is no order pair in R which has 6 as the first element. Same is the case for (2, 7) and (3, 8). So, R is transitive.
(iv) (d) : We have,R = {(x, y) : 3x - y = 0}, where \(x, y \in A=\{1,2, \ldots \ldots, 14\}\) .
\(\therefore\) R = {(I, 3), (2, 6), (3, 9), (4, 12)}
Clearly,\((1,1) \notin R\) So, R is not reflexive on A.
Since, \((1,3) \in R\) but \((3,1) \notin R\) .So, R is not symmetric on A.
Since, \((1,3) \in R\) and \((3,9) \in R\)  but  \((1,9) \notin R\) So, R is not transitive on A.
(v) (d) : Clearly, (1, 1), (2, 2), (3, 3) ∈  R. So, R is reflexive on A.
We find that the ordered pairs obtained by interchanging the components of ordered pairs in R are also in R. So, R is symmetric on A.
For \(1,2,3 \in A\)  such that (1, 2) and (2, 3) are in Rimplies that (1, 3) is also, in R. So, R is transitive on A. Thus, R is an equivalence relation.