Let us put I = \(\int ^{3}_{2} \frac{\sqrt {x}}{\sqrt {5-x}+\sqrt {x}}\) dx --- (1)
Applying the formula \(\int ^{a}_{b}\) f(x) dx =\(\int ^{a}_{b}\) f(a+b -x) dx, we get
I =  \(\int ^{3}_{2} \frac{\sqrt {(2+3-x)}}{\sqrt {5-(2+3-x)}+\sqrt { {(2+3-x)}}}\) dx = \(\int ^{3}_{2} \frac{\sqrt {(5-x)}}{\sqrt {x}+\sqrt { {(5-x}}}\) dx     -- (2)
Adding (1) and (2), we get 
2I =  \(\int ^{3}_{2} \frac{\sqrt {x} + \sqrt {5-x} }{\sqrt {x}+\sqrt { {5-x}}}\) dx =\(\int ^{3}_{2} \) dx = \([x]^{3}_{2}\) = 3 - 2 = 1
Hence, we get I = \(\frac{1}{2}\)

log₁₀e = 0.4343 to find log₁₀g 1003
f(1000) = log₁₀1000 = log₁₀10³ = 3log₁₀10³ = 3 log₁₀10
= 3(1) = 3
f'(x) = \(\frac1x\). log10^e
f'(1000) = \(\frac{1}{1000}\)(0.4343)
∴ L(x) = f(x₀) f'(x₀) (x - x₀)
= 3 + \(\frac{1}{1000}\) (0.4343) (3)
= 3 + \(\frac{1.3029}{1000}\)
= 3 + 0.0013029
log₁₀1003 = 3.0013029

We need to find an approximate value of \(\sqrt { 9.2 } \) using linear approximation. Now by (3), we have f(x₀+Δx) ≈ f(x₀)+f'(x₀)Δx. To do this, we have to identify an appropriate function f, a point x₀ and Δx. Our choice should be such that the right side of the above approximate equality, should be computable without the help of a calculator. So, we choose  
f(x) = \(\sqrt { x,{ x }_{ 0 } } \) = 9  and Δx = 0.2. Then f'(x₀) = \(\frac { 1 }{ 2\sqrt { 9 } } \) and hence.
 \(\sqrt { 9.2 } \) ≈ f(9) + f'(9)(0.2) = 3+\(\frac { 0.2 }{ 6 } \) = 3.03333
Now if we use a calculator, just to compare, we find \(\sqrt { 9.2 } \) = 3.03315. We see that our approximation is accurate to three decimal places and the error is 3.03315 - 3.03333 = 0.00018. [Also note that one could choose f (x) = \(\sqrt { 1+x,{ x }_{ 0 } } =8\) and Δx = 0.2. So the choice of f and x₀ - are not necessarily unique].
 So in the above example, the absolute error is 3.03315-3.03333 = -0.00018. Note that the absolute error says how much the error; but it does not say how good the approximation is. For instance, let us consider two simple cases
Case 1 : Suppose that the actual value of something is 5 and its approximated value is 4, then the absolute error is 5 − 4 = 1.
Case 2 : Suppose that the actual value of something is 100 and its approximated value is 95. In this case, the absolute error is 100 − 95 = 5. So the absolute error in the first case is smaller when compared to the second case.
Among these two approximations, which is a better approximation; and why? The absolute error does not give a clear picture about whether an approximation is a good one or not. On the other hand, if we calculate relative error or percentage of error (defined below), it will be easy to see how good an approximation is. If the actual value is zero, then we do know how close our approximate answer is to the actual value. So if the actual value is not zero.

\({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }(-1)\)
Let \({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) =x\) and -1 = sin y
\(\Rightarrow cosx=\frac { 1 }{ 2 } =cos\frac { \pi }{ 3 } \)
\(x=\frac { \pi }{ 3 } \)
\(siny=-1=-sin\left( \frac { -\pi }{ 2 } \right) \)
= \(-sin\left( \frac { -\pi }{ 2 } \right) \)
\(\left[ \because sin\left( -\theta \right) =-sin\ \theta\ and\ \frac { \pi }{ 2 } \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
\(\Rightarrow y=\frac { -\pi }{ 2 } \)
\(\therefore { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }\left( -1 \right) =x+y\)
= \(\frac { \pi }{ 3 } -\frac { \pi }{ 2 } =\frac { 2\pi -3\pi }{ 6 } =-\frac { \pi }{ 6 } \)
\(\therefore { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }(-1)=-\frac {\pi }{ 6 } \)

We know that the line \(\vec { r } =(\hat { i } +2\hat { j } +4\hat { k } )+t(2\hat { i } +2\hat { j } +\hat { k } )\) is parallel to the vector \(2\hat { i } +2\hat { j } +\hat { k } \).
Direction ratios of the straight line joining the two given points (5, 1, 4) and (9, 2, 12) are 4,1,8 and hence this line is parallel to the vector \(\hat { 4i } +\hat { j } +8\hat { k } \)
Therefore, the angle between the given two straight lines is
\(\theta ={ cos }^{ -1 }\left( \frac { \left| \vec { b } .\vec { d } \right| }{ \left| \vec { b } \right| \left| \vec { d } \right| } \right) \), where \(​​\vec { b } \) = \(2\hat { i } +2\hat { j } +\hat { k } \) and \(\vec { d } \) = \(\hat { 4i } +\hat { j } +8\hat { k } \)
Therefore, \(\theta ={ cos }^{ -1 }\left( \frac { \left| (2\hat { i } +2\hat { j } +\hat { k } ).(4\hat { i } +\hat { j } +8\hat { k } ) \right| }{ \left| 2\hat { i } +2\hat { j } +\hat { k } \right| \left| 4\hat { i } +\hat { j } +8\hat { k } \right| } \right) ={ cos }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)

LHS = \([\vec { a } -\vec { b } ,\vec { b } -\vec { c } ,\vec { c } -\vec { a } ]\) = 0
[∵ cross product is distributive]
\((\vec { a } -\vec { b } ).[(\vec { b } -\vec { c } )\times (\vec { c } -\vec { a } )]\)
= \((\vec { a } -\vec { b } ).[(\vec { b } \times \vec { c } -\vec { b } \times \vec { a } -\vec { c } \times \vec { c } +\vec { c } \times \vec { a } )\)
= \((\vec { a } -\vec { b } ).[\vec { b } \times \vec { c } -\vec { b } \times \vec { a } -0+\vec { c } \times \vec { a } ]\)
\([\because \vec { c } \times \vec { c } =0]\)
= \([\vec { a } \vec { b } \vec { c } ]-[\vec { a } \vec { b } \vec { a } ]+[\vec { a } \vec { c } \vec { a } ]-[\vec { b } \vec { b } \vec { c } ]+[\vec { b } \vec { b } \vec { a } ]-[\vec { b } \vec { c } \vec { a } ]\)
= \([\vec { a } \vec { b } \vec { c } ]-0+0-0+0-[\vec { b } \vec { c } \vec { a } ]\)
= \([\because [\vec { a } \vec { b } \vec { a } ]=[\vec { b } \vec { b } \vec { c } ]=0]\)
= \([\vec { a } \vec { b } \vec { c } ]-[\vec { a } \vec { b } \vec { c } ]\)
= 0 = RHS.

Let O be the centre of the semi-circle and AA¹ be the diameter.
Let P be any point on the circumference of the semi circle.
Taking O as the origin, let the position vectors of A and P be a and \(\vec { r } \) respectively.
Let us prove that \(\angle A P B=90^{\circ}\)
W.K.T OA = OB = OP ( because of radius)
\( \overrightarrow{P A} =\overrightarrow{P O}+\overrightarrow{O A} \)
\(\overrightarrow{P B} =\overrightarrow{P O}+\overrightarrow{O B} \)
\( =\overrightarrow{P O}-\overrightarrow{O A} \)
\(\overrightarrow{P A} \cdot \overrightarrow{P B} =(\overrightarrow{P O}+\overrightarrow{O A})(\overrightarrow{P O}-\overrightarrow{O A}) \)
\( =\overrightarrow{P O}^{2}-\overrightarrow{O A}^{2}=0 \)
\( \overrightarrow{P A} \perp \overrightarrow{P B} \)
\( \Rightarrow \ \angle A P B=90^{\circ} \). Hence proved.

Given vertex A(5, -2) and focus S(2, -2) and the focal distance
AS = a = 3  
Parabola is open left and symmetric about the line parallel to x -axis.
Then, the equation of the required parabola is
(y + 2)² = −4(3)(x − 5)
y² + 4y + 4 = −12x + 60
y² + 4y +12x − 56 = 0  

The condition for the line y = mx + c to be a tangent to the circle x² + y² = a² is c² = a²(1 + m²) from  
Then \(c=\pm \sqrt { 9\left( 1+16 \right) } \)
\(c=\pm 3\sqrt { 17 } \)

Given \(\frac { \pi }{ 2 } <{ cos }^{ -1 }(3x-1)<\pi \)
\((cos2\frac { \pi }{ 2 } <3x-1\))
\(\Rightarrow 0<3x-1<-1\)
\(0+1<3x<-1+1\)
\(1<3x<0\)
This inequality is true, only when 0 < x < \(\frac { 1 }{ 3 }\)

Given sin x = 0
\(\Rightarrow\) sin x = sin 0
\(\Rightarrow\) \(​​x=n\pi ,n\varepsilon z\)
Since \(-10\pi \le x\le 10\pi \) n can take the values only from -10 to +10.
\(\therefore\) \(x=n\pi ,\) When \(n=0,\pm ,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 7,\pm 8,\pm 9,\pm 10\)

Both the curves are symmetrical about y -axis
The curve y = |x| is \(y=\begin{cases} x\quad if\quad x\ge 0 \\ -x\quad if\quad x\le 0 \end{cases}\)
It intersects the parabola x² = y at (1, 1) and (−1, 1). The area of the region bounded by the curves is sketched. It lies in the first quadrant as well as in the second quadrant. By symmetry, the required area is twice the area in the first quadrant.

In the first quadrant, the upper curve is y = x, 0 \(\le x\le\)1 and the lower curve is y = x² \(\le x\le\),0 1. Hence, the required area is given by
\(A=2\int _{ 0 }^{ 1 }{ \left[ { y }_{ U }-{ y }_{ L } \right] } dx=2\int _{ 0 }^{ 1 }{ \left[ x-{ x }^{ 2 } \right] dx } \)
\(=2{ \left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1 }\)
\(=2\left( \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right) =\frac { 1 }{ 3 } \)

The given curve is \(y=\begin{cases} cosx,0\le x\le \frac { \pi }{ 2 } \\ -cosx,\frac { \pi }{ 2 } \le x\le \pi \end{cases}\)
It lies above the x − axis. The required area is sketched. So, the required area is given by

\(A=\int _{ 0 }^{ \pi }{ ydx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ cosxdx } +\int _{ \frac { \pi }{ 2 } }^{ \pi }{ (-cosx)dx } ={ [sin\quad x] }_{ 0 }^{ \frac { \pi }{ 2 } }-{ [sin\quad x] }_{ \frac { \pi }{ 2 } }^{ \pi } } \)
= [1-0]-[0-1] = 2

Recall that volume of a right circular cylinder is given by V = \(\pi \)r²h where r is the radius and h is the height. So we have V (r) = \(\pi \)r²h = 20\(\pi \)r²
V (10.1) −V (10)≈ \(\frac { dV }{ dr } { { | }_{ r=10 } }\) (10.1 10) = 20\(\pi \)²(10(0.1))
Thus the estimate for the change in the volume is 40 \(\pi \) cm³
Exact calculation of the volume change gives
V (10.1) −V (10) = 2040.2\(\pi \) -2000\(\pi \) = 40.2\(\pi \) cm³.
So relative error = \(\frac { 40.2\pi -40\pi }{ 40.2\pi } \) = \(\frac { 1 }{ 201 } \) = 0.00497 and hence
the percentage error = relative error x 100 = \(\frac { 1 }{ 201 } \)x100 = 0.497% 

18x²+ 12y² - 144x + 48y + 120 = 0
Given equation is
18x² + 12y² - 144x + 48y + 120 = 0
18x² - 144x + 12y² + 48y = -120
⇒ 18(x² - 8x) + 12(y² + 4y) = -120
⇒ 18(x²-8x+ 16-16)+ 12(y² +4y+4-4) =-120
18(x - 4)² - 288 + 12 (y + 2)²- 48 = -120
⇒ 18(x - 4)²+ 12(y + 2)² = -120 + 288 + 48
⇒ 18(x - 4)²+ 12(y + 2)² = 216
Dividing by 216 we get,
\(\frac { { 18(x-4) }^{ 2 } }{ 216 } +\frac { 12({ y+2) }^{ 2 } }{ 216 } =1\)
\(\Rightarrow \frac { { (x-4) }^{ 2 } }{ 12 } +\frac { ({ y+2) }^{ 2 } }{ 18 } =1\)
This is an equation of the ellipse with major axis parallel to y-axis,
∴ a² = 18, b² = 12
∴ c² = a² - b² = 18 -12 = 6 ⇒ c = \(\sqrt { 6 } \)
e =\( \sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\frac { 12 }{ 18 } } =\sqrt { \frac { 18-12 }{ 18 } } \)
\(=\sqrt{\frac{\not 6^1}{\not{18}}_{3}}=\sqrt{\frac{1}{3}}\)= \(\frac{1}{\sqrt 3}\)
(a) Center is (4, -2)
⇒ h = 4, k = -2
(b) Vertices are (h, k-a), (h, k + a)
⇒ (4, -2 - 3\(\sqrt { 2 } \)), (4, -2 + 3\(\sqrt { 2 } \))
[∴ a² = 18 ⇒ a = \(\sqrt { 18 } \) = 3\(\sqrt { 2 } \)]
(c) Foci are (h, k - c), (h, k + c)
⇒ (4, -2 - \(\sqrt { 6 } \)), (4, -2 + \(\sqrt { 6 } \))
(d) Equation of directrices are y + 2 = \(\pm \frac { a }{ e } \)
⇒ y+ 2 = \(\pm \frac { a }{ e } \)
\(\Rightarrow y-2=\pm \frac { 3\sqrt { 2 } }{ \frac { 1 }{ \sqrt { 3 } } } =\pm 3\sqrt { 2 } \times \sqrt { 3 } =\pm 3\sqrt { 6 } \)
\(\Rightarrow y+2=\pm 3\sqrt { 6 } ,y+2=-3\sqrt { 6 } \)
\(\Rightarrow y=-2+3\sqrt { 6 } \) and \( y=-2-3\sqrt { 6 } \)

 x²-2x+ 8y+ 17 = 0
x² - 2x = -8y - 17
Adding 1 both sides, we get
x² - 2x + 1 = -8y - 17 + 1
⇒ (x - 1)² = -8y - 16 = -8(y + 2)
⇒ (x - 1)² = -8(y + 2)
This is a open downward parabola, latus
rectum 4a = 8 ⇒ a = 2.
(a) Vertex is (1, -2)
⇒ h = 1, k = -2
(b) focus is (0 + h, - a + k)
⇒(0 + 1, -2-2)
⇒ (1, -4)
(c) Equation of directrix is y = k + a
⇒ y = -2 + 2 ⇒ y = 0
(d) Length of latus rectum is 4a = 8 units.

The required plane is parallel to the given line and so it is parallel to the vector \(\vec { c } =\hat { i } +\hat { j } -\hat { k } \) and the plane passes through the points \(\vec { a } =-\hat { i } +2\hat { j } ,\vec { b } =2\hat { i } +2\hat { j } -\hat { k } \)
(i) vector equation of the plane in parametric form is \(\vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t\vec { c } \), where s, t ∈ R
which implies that \(\vec { r } =(-\hat { i } +2\hat { j } )+s(3\hat { i } -\hat { k } )+t(\hat { i } +\hat { j } -\hat { k } )\), where s, t ∈ R
(ii) vector equation of the plane in non-parametric form is \((\vec { r } -\vec { a } ).(\vec { b } -\vec { a } )\times \vec { c } )\) = 0
Now, \((\vec { b } -\vec { a } )\times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 0 & -1 \\ 1 & -1 & -1 \end{matrix} \right| =\hat { i } +2\hat { j } +3\hat { k } \)
we have \((\vec { r } -(-\hat { i } +2\hat { j } ).(\hat { i } +2\hat { j } +3\hat { k } )\) = 0 ⇒ \(\vec { r } .(\hat { i } +2\hat { j } +3\hat { k } )\) = 3
If  \(\vec { r } .(\hat { i } +2\hat { j } +3\hat { k } )\) is the position vector of an arbitrary point on the plane, then from the above equation, we get the Cartesian equation of the plane as x + 2y + 3z = 3

Given lines are \(\frac { x-3 }{ 3 } =\frac { y-3 }{ -1 } \)....(1)
and z-1 = 0 
\(\Rightarrow\) z = 1
and \(\frac { x-6 }{ 2 } =\frac { z-1 }{ 3 } \) ...(2)  
and y-2 = 0\(\Rightarrow\) y = 2
Substituting y = 2 and z = 1 in (1) we get
\(\frac { x-3 }{ 3 } =\frac { 2-3 }{ -1 } =\frac { -1 }{ -1 } =1\Rightarrow x-3=3\Rightarrow x=6\)
The point of intersection is (6, 2, 1)
Let us check whether (6, 2, 1) satisfies (1) and (2)
\((1)\rightarrow \frac { 6-6 }{ 2 } =\frac { 1-1 }{ 3 } \Rightarrow 0=0\)
\((2)\rightarrow \frac { 6-3 }{ 3 } =\frac { 1-3 }{ -1 } \Rightarrow -1=-1\)
Hence, the given two lines intersect and the point of intersection is (6, 2, 1).

By taking the vertex; at the origin, the parabola is open downward.

Its equation is x² = -4ay
It passes through (6, -4)
∴ 36 = -4a(-4) ⇒ 4a = - \(\frac { 36 }{ 4 } \) = 9
∴ (1) becomes, x² = -9y
To find the slope at (-6, -4)
Differentiating (1) with respect to 'x' we get,
2x = -9\(\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { -2x }{ 9 } \)
At (-6, -4), \(\frac { dy }{ dx } =-2\frac { (-6) }{ 9 } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)
∴ \(tan\theta =\frac { 4 }{ 3 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)
∴ The angle of projection is tan^-1 \(\left( \frac { 4 }{ 3 } \right) \)

Let the equation of the parabola be
(x - h)² = -4a(y - k).
Here the vertex is (0.5, 4)
Equation of the parabola is (x - 0.5)²
= -4a(y-4) ...(1)
O(0, 0) is a point on the parabola
(0 - 0.5)² = -4a (0 - 4)
⇒ \({ \left( \frac { -1 }{ 2 } \right) }^{ 2 }=-4a(-4)\)
⇒ \(\frac { 1 }{ 4 } =16a\Rightarrow a=\frac { 1 }{ 64 } \)
∴ (1) becomes as (x - 0.5)² = \(-4\times \frac { 1 }{ 64 } (y-4)\)
Also D(0.75, y₁) is a point on the parabola
∴ (0.75 - 0.5)² = \(\frac { -1 }{ 16 } ({ y }_{ 1 }-4)\)
⇒ \({ \left( \frac { 1 }{ 4 } \right) }^{ 2 }=\frac { -1 }{ 6 } ({ y }_{ 1 }-4)\)
\(\Rightarrow \frac{1}{\not 16}=\frac{1}{\not16}\left(y_{1}-4\right)\)
⇒ 1 = -y₁ + 4
⇒  y₁ = -1 + 4 = 3m
Height of the water at a horizontal distance of 0.75m is 3m

Let the equation of the ellipse be
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Length of semi minor axis b = 5
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Let BB' be the road width and AA' be the end points of the opening of the tunnel.
Let CB = 8, BD = 1,
D is (8, 4) lies on the ellipse
\( \frac{8^{2}}{a^{2}}+\frac{4^{2}}{5^{2}}=1 \)
\( \Rightarrow a^{2}=\frac{25}{9} \times 64 \)
\( \Rightarrow a=\frac{40}{3} \)
The width AA' = 2a
\(=\frac{80}{3}=26.66 \mathrm{~m}\)
The required width is 26.66 m.

Equation of normal at 't₁' to the parabola y² = 4 axis is
y + xt₁ = at₁³ + 2at₁ ...(1)
(1) meets the-parabola y₂ = 4ax at 't₂'.
At 't₂', the point on the parabola is x = at²₂, y = 2at₂ (2) lies on (1)
∴ Substituting (2) in (1) we get,
2at₂ + (at₂²)2t₁ = at₁³ + 2at₁
⇒ 2at₂ + at₁t₂² = at₁³ + 2at₁
⇒ 2a(t₂ - t₁) = -at₁[t₂² - t₁²]

⇒ 2 = -t₁(t₂+ t₁)
⇒ \(\frac { -2 }{ { t }_{ 1 } } ={ t }_{ 2 }+{ t }_{ 1 }\Rightarrow { t }_{ 2 }=\frac { -2 }{ { t }_{ 1 } } { -t }_{ 1 }\)
⇒ t₂ = -(t₁ + \(\frac { -2 }{ { t }_{ 1 } } \))
Hence proved.

Rearranging terms in the equation of hyperbola to bring it to standard form,
we have, 11(x²-4x)-25(y²-2y)-256 = 0
11(x− 2)²−25(y−1)² = 256−44+25
11(x−2 )²− 25 (y−1)² = 275
\(\frac { { \left( x-2 \right) }^{ 2 } }{ 25 } -\frac { { \left( y-1 \right) }^{ 2 } }{ 11 } =1\)
Centre (2, 1) a² = 25, b² = 11
c² = a² +b²
= 25 +11 = 36
Therefore, c = ±6
and e = \(\frac { c }{ a } =\frac { 6 }{ 5 } \)and the coordinates of foci are(8, 1) and(-4, 1) from figure.

 f(x) = sin^-1 \((\frac{|x|-2}{3})+cos^-1(\frac{1-|x|}{4})\)
From the definition of sin^-1
\(-1\le \frac { \left| x \right| -2 }{ 3 } \le 1\)
\(\Rightarrow -3\le \left| x \right| -2\le 3\)
\(\Rightarrow -3+2\le \left| x \right| \le \left| x \right| \le 3+2\)
\(\Rightarrow -1\le \left| x \right| \le 5\)
It reduces to
\(0\le \left| x \right| \le 5\)
\(\Rightarrow 0\le \left| x \right| and\left| x \right| \le 5\)
\(\Rightarrow \left| x \right| \ge 0and-5\le x\le 5\)
From the definition of cos^-1x.
\(-1\le \frac { 1-\left| x \right| }{ 4 } \le 1\)
\(\Rightarrow -4\le 1-\left| x \right| \le 4\)
\(\Rightarrow -4-1\le \left| x \right| \le 4-1\)
\(\Rightarrow -5\le -\left| x \right| \le 3\)
\(\Rightarrow -3\le \left| x \right| >5\)
It reduces to
\(0\le \left| x \right| \le 5\)
\(-5\le |x|\le 5\)
From (1) & (2),
Domain is [-5, 5]

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors and which make angles α and β, respectively, with positive x-axis, where A and B are as in the diagram.
Draw AL and BM perpendicular to the x-axis. Then \(\left| \vec { OL } \right| =\left| \vec { OA } \right| \) cos α = cos α, \(\left| \vec { LA } \right| =\left| \vec { OA } \right| \) sin α =  sin α
So, \(\vec { OL } =\left| \vec { OL } \right| \)\(\hat { i } \) =  cos,α \(\hat { i } \), \(\overrightarrow { LA } \) = sin α (-\(\hat { j } \))
Therefore, \(\hat { a } =\overrightarrow { OA} = \overrightarrow { OL } +\overrightarrow { LA } \) = cos α \(\hat { i } \) - sin α \(\hat { j } \) ..(1)
Similarly \(\hat { b } \) = cos β \(\hat { i } \)+ sin β \(\hat { j } \) ....(2)
The angle between \(\hat { a } \) and \(\hat{b}\) is α + β and so,
\(\hat { a } .\hat { b } =\left| \hat { a } \right| \left| \hat { b } \right| \) cos (α + β) = cos (α + β) ... (3)

On the other hand, from (1) and (2)
\(\hat { a } .\hat { b } =(cos\alpha \hat { i } -sina\hat { j } )(cos\beta \hat { i } -sin\beta \hat { j } )\) = cos α cos β - sin α sin β....(4)
From (3) and (4), we get cos(α + β) = cos α cos β - sin α sin β

Given equation of line is 3x - 2y + 6 = 0
2y = 3x + 6 \(\Rightarrow\) y = \(\frac{3x+6}{2}\)

\(\therefore Area=\int _{ -3 }^{ -2 }{ -ydx+ydx } +\int _{ -2 }^{ 1 }{ ydx } \)
[\(\because\) the Area is below the x - axis]

\(=\frac { -1 }{ 2 } \int _{ -3 }^{ -2 }{ (3x+6)dx+\frac { 1 }{ 2 } \int _{ -2 }^{ 1 }{ (3x+6)dx } } \)
\(={ \left[ \frac { { 3x }^{ 2 } }{ 2 } +6x \right] }_{ -3 }^{ -2 }+\frac { 1 }{ 2 } { \left[ \frac { { 3x }^{ 2 } }{ 2 } +6x \right] }_{ -2 }^{ 1 }\)
\(=-\frac { 1 }{ 2 } \left[ \left( \frac { 12 }{ 2 } -12 \right) -\left( \frac { 27 }{ 2 } -18 \right) \right] +\frac { 1 }{ 2 } \left[ \left( \frac { 3 }{ 2 } +6 \right) -\left( \frac { 12 }{ 2 } -12 \right) \right] \)
\(=-\frac { 1 }{ 2 } \left[ (-6)-\left( \frac { 27-36 }{ 2 } \right) \right] +\frac { 1 }{ 2 } \left[ \left( \frac { 3+12 }{ 2 } \right) -(-6) \right] \)
\(=-\frac { 1 }{ 2 } \left[ -6+\frac { 9 }{ 2 } \right] +\frac { 1 }{ 2 } \left[ \frac { 15 }{ 2 } +6 \right] \)
\(\\ =-\frac { 1 }{ 2 } \left[ \frac { -3 }{ 2 } \right] +\frac { 1 }{ 2 } \left[ \frac { 27 }{ 2 } \right] =\frac { 3 }{ 4 } +\frac { 27 }{ 4 } =\frac { 30 }{ 4 } =\frac { 15 }{ 2 } \)
\(\therefore\)  A = 7.5 sq.units

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)

\(\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-4 } } \)
\(\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-4 } } =\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-{ 2 }^{ 2 } } } \)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| +c\right] \)
\(=\frac { 1 }{ 4 } \left[ log\left( \frac { 4-2 }{ 4+2 } \right) -log\left( \frac { 3-2 }{ 3+2 } \right) \right] \)
\(=\frac { 1 }{ 4 } log\left[ \left( \frac { 2 }{ 6 } \right) - log \ \frac { 1 }{ 5 } \right] \\ =\frac { 1 }{ 4 } log\left( \frac { 1 }{ 3 } \times 5 \right) \)
\(=\frac { 1 }{ 4 } log\left( \frac { 5 }{ 3 } \right) \)

Given r = 2 mm
dr = (2.1 - 2) = 0.1 mm
Area = πr²
Approximate area dA = 2πr dr
= 2π (2) (0.1)
= 4 π (0.1) = 0.4 π mm²

Given = (3 + sin(2x)) ^2/3 
Taking differentilas,
dy =  \(\frac23\)(3 + sin(2x)) ^2/3-1 (cos 2x) (2)dx
dy = \(\frac { 4 }{ 3 } .\frac { cos2x }{ { (3+sin2x) }^{ \frac { 1 }{ 3 } } } dx\)

Equation of the circle is
2x² + 2y² - 6x + 4y + 2 = 0
Dividing by 2, we get
x² + y² - 3x + 2y + 1 = 0
Here 2g = -3 ⇒ g = \(\frac { -3 }{ 2 } \)
2f = 2 ⇒ f = 1
and c = 1
∴ Centre is (-g, -f) = \(\left( \frac { 3 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \) = \(\sqrt { { \left( \frac { 3 }{ 2 } \right) }^{ 2 }+{ 1 }^{ 2 }-1 } \)
= \(\sqrt { \frac { 9 }{ 4 } } =\frac { 3 }{ 2 } \) units.

Since the given three vectors are coplanar, we have \(\left| \begin{matrix} 2 & -1 & 3 \\ 3 & 2 & 1 \\ 1 & m & 4 \end{matrix} \right| \) = 0 ⇒ m = -3

 Let \(f(x)={ tan }^{ -1 }\sqrt { 9-{ x }^{ 2 } } \)
\(\sqrt { 9-{ x }^{ 2 } } \varepsilon R\) but \(\sqrt { 9-{ x }^{ 2 } } \varepsilon R\)
\(\therefore \ 9-{ x }^{ 2 }\ge 0\)
\(\Rightarrow { x }^{ 2 }-9\ge 0\)
\(\Rightarrow (x+3)(x-3)\le 0\)
\(\Rightarrow \) Domain is [-3, 3]