It is known that cos^-1 x : [-1, 1]\(\rightarrow\)[0, \(\pi\)] is given by
cos⁻¹x = y if and only if x = cos y for -1\(\le x\le1\ and\ 0\le y \le\pi\) 
Thus, we have
cos^-1\((cos(-\frac{7\pi}{6}))\) = \(\frac{5\pi}{6}\), since cos\((\frac{7x}{6})=cos(\pi+\frac{\pi}{6})=-\frac{\sqrt3}{2}=cos(\frac{5\pi}{6})\in[0, \pi].\)

(( p ∧ q) ∨ (¬r ∨¬s)) ∧ (¬ t ∧ v)) contains 6 variables p, q, r, s, t, and v. Hence the corresponding truth table will contain 2⁶ = 64 rows.

Given that I =\(\int ^\frac {\pi}{2}_{0} \)( sin²x + cos⁴x)dx =\(\int ^\frac {\pi}{2}_{0} \) sin²x dx+\(\int ^\frac {\pi}{2}_{0} \)cos⁴x dx\(\frac {1}{2} \times \frac {\pi}{2} + \frac {3}{4} \times \frac {1}{2} \times \frac {\pi}{2} = \frac {7\pi}{16} \)

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 equation of curve is
xy = ae^x + be^−x + x² ...(1)
where a &b are aribitrary constant. differentiate equation (1) twice successively, because we have two arbitray constant.
\(x \frac{d y}{d x}+y(1)=a \mathrm{e}^{x}-\mathrm{be}^{-x}+2 x\) ...(2)
\( x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}(1)+\frac{d y}{d x}=\mathrm{ae}^{x}+\mathrm{be}^{-x}+2 \)
\( x \frac{d^{2} y}{d x^{2}}+\frac{2 d y}{d x}=\mathrm{ae}^{x}+\mathrm{be}^{-x}+2 \) ...(3)
From (1), we get \(x y-x^{2}=\mathrm{ae}^{x}+\mathrm{be}^{-x}\) ...(4)
Substituting equation (a) in (3), we get
\(\therefore x \frac{d^{2} y}{d x^{2}}+\frac{2 d y}{d x}-x y+x^{2}-2=0\) which is the required differential equaiton. 

The normal vectors of the two given planes \(\vec { r } .(2\hat { i } +2\hat { j } +2\hat { k } )\)= 11 and 4x+2y+2z = 15 are \(\vec { { n }_{ 1 } } =2\hat { i } +2\hat { j } +2\hat { k } \) and \(\vec { { n }_{ 2 } } =4\hat { i } -2\hat { j } +2\hat { k } \) respectively.
If θ is the acute angle between the planes, then we have
\(\theta =cos^{ -1 }\left( \frac { |\vec { { n }_{ 1 } } .\vec { { n }_{ 2 } } | }{ |\vec { { n }_{ 1 } } .\vec { { n }_{ 2 } } | } \right) =cos^{ -1 }\left( \frac { |((2\hat { i } +2\hat { j } +2\hat { k } ).(4\hat { i } -2\hat { j } +2\hat { k } ))| }{ |(2\hat { i } +2\hat { j } +2\hat { k } )||4\hat { i } -2\hat { j } +2\hat { k } | } \right) =cos^{ -1 }\left( \frac { \sqrt { 2 } }{ 3 } \right) \).

ang \(z=\frac { -2 }{ 1+i\sqrt { 3 } } \)
= arg(-2)-arg \(\left( 1+i\sqrt { 3 } \right) \) \(\left( \because arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =arg{ z }_{ 1 }=g_{ 2 } \right) \)
= \(\left( \pi -{ tan }^{ -1 }\left( \frac { 0 }{ 2 } \right) \right) -tan^{ -1 }\left( \frac { \sqrt { 3 } }{ 1 } \right) \)
= \(\pi -\frac { \pi }{ 3 } =\frac { 2\pi }{ 3 } \)
This implies that one of the values of arg z is \(\frac { 2\pi }{ 3 } \)
Since \(\frac { 2\pi }{ 3 } \) lies between \(-\pi \), the principal argument Argz is \(\frac { 2\pi }{ 3 } \)

Here Δ = b²−4ac = 0 for equal roots. This implies 4(k + 2)² = 4(9)k. This implies k = 4 or 1.

Centre of the circle is (3,-2) which lies on x + y − 1 = 0. So the line x + y − 1 = 0 passes through the centre and therefore the line x + y −1 = 0 is a diameter of the circle for all possible values of c .

\( \int_{-\infty}^{\infty} f(x) \mathrm{dx}=1 \)
\(\int_{-\infty}^{0} 0 \mathrm{dx}+\int_{0}^{1} \mathrm{kx}(1-x)^{10} \mathrm{dx}+\int_{1}^{\infty} 0 \mathrm{dx}=1 \)
\(\int_{0}^{1} k x(1-x)^{10} \mathrm{dx}=1 \)
\(\int_{0}^{1}(1-x)(1-1+x)^{10} \mathrm{dx}=1 \)
\(k \int_{0}^{1}(1-x)(\mathrm{x})^{10} \mathrm{dx}=1 \)
\([\therefore \left.k \int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_{0}^{10} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{dx}\right] \)
\( k \int_{0}^{1}(1-x)(x)^{10} d x =1 \)
\(k \int_{0}^{1}\left(x^{10}-x^{11}\right) d x =1 \)
\(k\left[\frac{x^{11}}{11}-\frac{x^{12}}{12}\right]_{0}^{1} =1 \)
\(k\left[\frac{1}{11}-\frac{1}{12}\right] =1 \)
\(k\left[\frac{1}{132}\right] =1\)
k = 132

The entries in the columns corresponding q ➝ p and ~p ➝ ~q are identical and hence they are equivalent.
q ➝ p ≡ ~p ➝ ~q
Hence proved

Given  W(x, y, z) = x² - xy + 3 sin z, x, y, z ∈ R
W(x₀, y₀, z₀) = W (2, -1, 0)
= 2²-2(-1)+3 sin 0
\(\frac { \partial W }{ \partial x } \) = 2x - y + 0 = 2x - y
\(\left( \frac { \partial W }{ \partial x } \right) _{(2, -1, 0)}\)= 2(2)-(-1) = 5
\(\frac { \partial W }{ \partial y } \) = 0-x + 0 = -x
\(\left( \frac { \partial W }{ \partial y } \right) _{(2, -1, 0)}\) = -2
\(\frac { \partial W }{ \partial z } \) = 0 - 0 + 3 cos z
 \(\left( \frac { \partial W }{ \partial z} \right) _{(2, -1, 0)}\)= 3 cos 0 = 3(1) = 3.
Linear approximation is given by
L(x, y, z) = w(x₀, y₀, z₀) + \(\left( \frac { \partial W }{ \partial x } \right) _{(x_0, y_0, z_0)}\) (x - x₀) + \(\left( \frac { \partial W }{ \partial y} \right) _{(x_0, y_0, z_0)}\) (y - y₀) + \(\left( \frac { \partial W }{ \partial z} \right) _{(x_0, y_0, z_0)}\)(z₀ - z₀)
∴ L(x,y,z) = 6 + 5(x - 2) - 2(y+1) + 3(z - 0)
= 6 + 5x - 10 - 2y - 2 + 3z
L (x, y, z) = 5x - 2y + 3z - 6

\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 3 }\theta { cos }^{ 5 }\theta d\theta } \)
\( \mathrm{m}=3, \mathrm{n}=5 \)
\( \int_{0}^{\pi / 2} (cos ^5 \theta - cos^7 \theta) sin \theta d\theta\)
 \(t = cos \theta\)
\(dt = -sin \theta d\theta\)
\(= \int ^0_1(t^2-t^7)(-dt)\\ = \int ^0_1(t^5-t^7)(dt)= [\frac{t^6}{6}-\frac{t^8}{8}]\)
\(\frac{1}{6}-\frac{1}{8}= \frac{8-6}{48}\)
\( =\frac{1}{24} \)
Aliter method:
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 3 }\theta { cos }^{ 5 }\theta d\theta } \)
Here m = 3, which is odd and n = 5, which is odd
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ m }\theta { cos }^{n } }x dx \)
\( \frac{n-1}{m+n} \cdot \frac{n-3}{m+n-2} \cdot \frac{n-5}{m+n-4} \cdots \frac{2}{m+3} \cdot \frac{1}{m+1} \\ \)
\( \int_{0}^{\pi / 2} \sin ^{3} \theta \cos ^{5} \theta d \theta=\frac{A}{8} \times \frac{2}{6} \times \frac{1}{A} \\ =\frac{1}{24} \)

Let f (t) be the temperature at time t. By the mean value theorem, we have
\(f'(c)=\frac{f(b)-f(a)}{b-a}\)
= \(\frac{100-(-10)}{22}\)
= \(\frac{110}{22}\)
= 5°C per second.
Hence the instantaneous rate of change of temperature at some time t should be 5°C per second.

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.

\(\frac { 2\left( cos\frac { 9\pi }{ 4 } +isin\frac { 9\pi }{ 4 } \right) }{ 4\left( cos\left( \frac { -3\pi }{ 2 } + \right) sin\left( \frac { -3\pi }{ 2 } \right) \right) } \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { 9\pi }{ 4 } -\left( \frac { -3\pi }{ 2 } \right) \right) +isin\left( \frac { 9\pi }{ 4 } -\left( \frac { -3\pi }{ 2 } \right) \right) \right) \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { 9\pi }{ 4 } +\frac { 3\pi }{ 2 } \right) +isin\left( \frac { 9\pi }{ 4 } +\frac { 3\pi }{ 2 } \right) \right) \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { 15\pi }{ 4 } \right) +isin\left( \frac { 15\pi }{ 4 } \right) \right) =\frac { 1 }{ 2 } \left( cos\left( 4\pi -\frac { \pi }{ 4 } \right) +isin\left( 4\pi -\frac { \pi }{ 4 } \right) \right) \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { \pi }{ 4 } \right) -isin\left( \frac { \pi }{ 4 } \right) \right) =\frac { 1 }{ 2 } \left( \frac { 1 }{ \sqrt { 2 } } +i\frac { 1 }{ \sqrt { 2 } } \right) \)
\(\frac { 2\left( cos\frac { 9\pi }{ 4 } +isin\frac { 9\pi }{ 4 } \right) }{ 4\left( cos\left( \frac { -3\pi }{ 2 } + \right) sin\left( \frac { -3\pi }{ 2 } \right) \right) } \) = \(\frac { 1 }{ 2\sqrt { 2 } } -i\frac { 1 }{ 2\sqrt { 2 } } =\frac { \sqrt { 2 } }{ 4 } +i\frac { \sqrt { 2 } }{ 4 } \) Which is in rectangular form.

Given that sin^-1x > cos^-1x. Note that -1\(\le x\le\)
Adding both sides by sin^-1x, we get
sin^-1 x + sin^-1 x > x cos^-1 x + sin^-1x, Which resucess to 2 sin^-1 x > \(\frac{\pi}{2}\)
As sine function increases in the interval \(\left[ -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right] \), we have x > sin\(\frac{\pi}{4} or x> \frac{1}{\sqrt2}\)
Thus, the solution set is the interval \(\left[ \frac { 1 }{ \sqrt { 2 } } ,1 \right] \)

Given centre is (2, -1) and passing through the point (3, 6)
∴ r = distance between (2, -1) and (3, 6)
= \(\sqrt { { (2-3) }^{ 2 }+(-1{ -6) }^{ 2 } } \)
= \(\sqrt { { (-1) }^{ 2 }+({ -7) }^{ 2 } } \)
= \(\sqrt { 1+49 } =\sqrt { 50 } \)
∴ Equation of the circle is
(x - h)² + (y - k)² = r²
(x−2)²+(y+1)² = \({ (\sqrt { 50 } })^{ 2 }\)
(x−2)²+(y+1)² = 50

Given A = \(\left[ \begin{matrix} 1 & \tan { x } \\ -\tan { x } & 1 \end{matrix} \right] \)
|A| = 1 + tan² x
∴ A^-1 = \(\frac { 1 }{ |A| } \)adjA
= \(\frac { 1 }{ 1+tan^{ 2 } } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
[Interchange the elements in the leading diagonal and change the sign of elements in the off diagonal]
A^T= \(\\ \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
∴ A^TA^-1
=\(\left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
=\(\frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
=\(\frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1-tan^{ 2 }x & -tanx-tanx \\ tanx+tanx & -tan^{ 2 }x+1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} \frac { 1-{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } & \frac { -2tanx }{ 1+{ tan }^{ 2 }x } \\ \frac { 2tanx }{ 1+{ tan }^{ 2 }x } & \frac { 1-{ tan }^{ 2 }x }{ 1+tan^{ 2 }x } \end{matrix} \right] \)
A^TA^-1 =\(\left[ \begin{matrix} cos2x & -sin2x \\ sin2x & cos2x \end{matrix} \right] \)
Hence proved.

Given polynomial equation is 
2x⁴+ 5x³−7x² + 8 = 0
Here a = 2, b = 5, c = -7, d = 0, e = 8
By Vieta's formula, 
\(\alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -5 }{ 2 } \)
\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { -7 }{ 2 } \)
\(\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =0\)
\(\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { 8 }{ 2 } =4\)
Given roots of quadratic equation are 
∝ + β + ૪ + \(\delta \) and ∝β૪\(\delta \)
∴ sum of the roots = (∝+β+૪+\(\delta \)) (∝β૪\(\delta \))
\(=\left( \frac { -5 }{ 2 } +4 \right) =\frac { -5+8 }{ 2 } =\frac { 3 }{ 2 } \)
\(=\left( \alpha +\beta +\gamma +\delta \right) (\alpha \beta \gamma \delta )\)
\(=\left( \frac { -5 }{ 2 } \right) (4)=\frac { -20 }{ 2 } =-10\)
∴ The required quadratic equation is x²-x 
(sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x\left( \frac { 3 }{ 2 } \right) -10=0\)
\(\Rightarrow { 2x }^{ 2 }-3x-20=0\)

The table for the operation x₁₁ is as follows.

Following the same kind of procedure as explained in the previous example, a brief outline of the process of verification of the properties of ×₁₁ on A is given below.
(i) Since each box has an unique element of A, ×₁₁ is a binary operation on A.
(ii) The entries are symmetrical about the main diagonal. Hence ×₁₁ has commutative property.
(iii) As usual, the associative property can be seen to be true.
(iv) The entries of both the row and column headed by the element 1 are identical. Hence 1 is the identity element.
(v) Since the identity 1 exists in each row and each column, the existence of inverse property is assured for ×₁₁. The inverse of 1 is 1, that of 3 is 4, that of 4 is 3, 5 is 9, and, that of 9 is 5.

Given v (x, y) = log \(\left( \frac { { x }^{ 2 }+{ y }^{ 2 } }{ x+y } \right) \)
Since log \(\left( \frac { { x }^{ 2 }+{ y }^{ 2 } }{ x+y } \right) \) is not homogeneous,
left f (x, y) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ x+y } \)
⇒ v = log f
⇒ ev = f
Now, f(tx, ty) = \(\frac { { t }^{ 2 }{ x }^{ 2 }+{ t }^{ 2 }{ y }^{ 2 } }{ tx+ty } =\frac { { t }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) }{ t(x+y) } \)
= t.f(x, y)
∴ f is a homogeneous function of degree 1.
∴ By Euler's theorem,
⇒ \(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = 1.f
From (1), \(x.\frac { \partial }{ \partial x } \left( { e }^{ v } \right) +y.\frac { \partial }{ \partial y } \left( { e }^{ v } \right) ={ e }^{ v }\) [using (1)]
⇒ \(x.{ e }^{ v }.\frac { \partial }{ \partial x } +y.{ e }^{ v }.\frac { \partial }{ \partial y } ={ e }^{ v }\)
⇒  \(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } \) = 1 [Divided by e^v]

Equation of the given circle is x² + y² = 16 ...(1)
and the parabola is y² = 6x. ...(2)
Substituting (2) in (1) we get,
x² + 6x - 16 = 0 \(\Rightarrow\) (x + 8) (x - 2) = 0
\(\Rightarrow\) (x-4) (x-2) = 0 \(\Rightarrow\) x = 2, 4
\(\Rightarrow\) x = -8, 2

\(\therefore\) Required area = 2
\(\\ \\ \\ \\ \\ \\ \\ \int _{ 0 }^{ 2 }{ Area\ below\ the\ parabola } +\int _{ 2 }^{ 4 }{ Area\ below\ the\ circle } \)
\(=2\left[ \int _{ 0 }^{ 2 }{ \sqrt { 6x } } dx+\int _{ 2 }^{ 4 }{ \sqrt { 16-{ x }^{ 2 } } dx } \right] \)
\(=2\left[ { \left( \frac { \sqrt { 6 } .{ x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right) }_{ 0 }^{ 2 }+{ \left( \frac { x }{ 2 } \sqrt { 16-{ x }^{ 2 } } +\frac { 16 }{ 2 } { sin }^{ -1 }\left( \frac { x }{ 4 } \right) \right) }_{ 0 }^{ 4 } \right] \)
\(=2\left[ \frac { 2 }{ 3 } \sqrt { 6. } 2\sqrt { 2 } +8{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right] \)
\(=2\left[ \frac { 4\sqrt { 12 } }{ 3 } +8\left( \frac { \pi }{ 2 } \right) -\sqrt { 12 } -8\left( \frac { \pi }{ 6 } \right) \right] \)
\(=2\left[ \frac { 4\sqrt { 12 } }{ 3 } +4\pi -2\sqrt { 3 } -\frac { 4\pi }{ 3 } \right] \)
\(=2\left[ \frac { 8\sqrt { 3 } -6\sqrt { 3 } }{ 3 } +\frac { 12\pi -4\pi }{ 3 } \right] \)
\(=\left[ \frac { 2\sqrt { 3 } }{ 3 } +\frac { 8\pi }{ 3 } \right] =2\times \frac { 2 }{ 3 } \left[ \sqrt { 3 } +4\pi \right] \)
\(=\frac { 4 }{ 3 } \left[ 4\pi +\sqrt { 3 } \right] \)sq.units

Given
\(f(x)=\begin{cases} \begin{matrix} { Ke }^{ \frac { -x }{ 3 } } & \begin{matrix} for & x>0 \end{matrix} \end{matrix} \\ \begin{matrix} 0 & \begin{matrix} for & x\le 0 \end{matrix} \end{matrix} \end{cases}\)
(i) Since f(x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
\(\Rightarrow \int _{ 0 }^{ \infty }{ K.{ e }^{ \frac { -x }{ 3 } } } dx=1\Rightarrow k\frac { \left[ { e }^{ \frac { -x }{ 3 } } \right] ^{ \infty } }{ -\frac { 1 }{ 3 } } \) 
\(\Rightarrow -3k\left[ { e }^{ -\infty }-{ e }^{ 0 } \right] =1\) [∵ e^∞ = 0, e⁰ = 1]
\(\Rightarrow 3k=1\Rightarrow k=\frac { 1 }{ 3 } \) 
\(\therefore k=\cfrac { 1 }{ 3 } \) 
(ii) The distribution function F(x) = \(\int _{ -\infty }^{ x }{ f(u)du } \)
Case 1: x < 0, 
\(F(x)=\int _{ -\infty }^{ x }{ f(x)dx=0 } \) 
Case 2: x > 0, 
\(f(x)=\int _{ -\infty }^{ x }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ x }{ f(x) } dx } \) 
= \(0+k\int _{ 0 }^{ x }{ { e }^{ \frac { -x }{ 3 } } } dx\) 
= \(\frac { 1 }{ 3 } \left[ \cfrac { { e }^{ \frac { -x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] =-\left[ { e }^{ -\frac { x }{ 3 } }-{ e }^{ o } \right] \) 
= \(-[{ e }^{ -\frac { x }{ 3 } }-1]\) 
= \(1-{ e }^{ -\frac { x }{ 3 } }\) 
\(\therefore F(x)=\begin{cases} \begin{matrix} 0 & x\le 0 \end{matrix} \\ \begin{matrix} 1-{ e }^{ \frac { -x }{ 3 } } & x>0 \end{matrix} \end{cases}\) 
(iii) p(X < 3)
= \(\int _{ 0 }^{ 3 }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \int _{ 0 }^{ 3 }{ { e }^{ -\frac { x }{ 3 } }dx } \)
= \(\cfrac { 1 }{ 3 } \left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ \frac { -1 }{ 3 } } \right] \) 
= -[e^-1-e⁰] = -[e^-1-1]
(iv) \(p(5\le X)=p(X\ge 5)=\int _{ 5 }^{ \infty }{ f(x)dx } \) 
= \(\int _{ 5 }^{ \infty }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 5 }^{ \infty } }{ \frac { -1 }{ 3 } } \) 
= \(-\left[ { e }^{ -\infty }-e^{ \frac { -3 }{ 5 } } \right] =\left[ 0-{ e }^{ \frac { -5 }{ 3 } } \right] \) 
= \({ e }^{ \frac { -5 }{ 3 } }\) 
(v) \(p(X\le 4)=\int _{ -\infty }^{ 4 }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ 4 }{ f(x)dx } } \) 
= \(0+\int _{ 0 }^{ 4 }{ { ke }^{ -\frac { x }{ 3 } }dx } =k\left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] _{ 0 }^{ 4 }\) 
= \(\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 0 }^{ 4 } }{ -\frac { 1 }{ 3 } } =-\left[ { e }^{ \frac { -4 }{ 3 } }-{ e }^{ o } \right] \) 
= \(-\left[ { e }^{ \frac { -4 }{ 3 } }-1 \right] =1-{ e }^{ \frac { -4 }{ 3 } }\)

Let there be N radioactive nuclei in a sample at any time t and let N_o be the initial number of radioactive nuclei.
Then \(\frac{dN}{dt}\infty N\)
\(\Rightarrow \frac { dN }{ dt } =-\lambda N\)
Where \(\lambda>0\) is a constant
\(\Rightarrow \frac { dN }{ N } =-\lambda dt\)
\(\int { \frac { dN }{ N } } =-\lambda dt\)
\(\int { \frac { dN }{ N } =-\lambda \int { dt } } \)
\(\Rightarrow log\ N=-\lambda t+C\ ...(1)\)
\(T\quad t=0,\ we\ have\ N={ N }_{ 0 }\)
\(\therefore log{ N }_{ 0 }=0+C\)
\(\Rightarrow C=log{ N }_{ 0 }\)
\(\therefore\)(1) becomes, log N = -\(\lambda t+log{ N }_{ 0 }\)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } =-\lambda t\quad ...(2)\)
It is given that 10% of the original number of nuclei have undergone disintegration in a period of 100 years.
Whent= 100 \(N={ N }_{ 0 }-\frac { 10 }{ 100 } \times { N }_{ 0 }=\frac { { 9N }_{ 0 } }{ 10 } \)
Substituting in (2) we get
\(log\quad \frac { 9 }{ 10 } =-100\lambda \)
\(\Rightarrow \lambda =-\frac { 1 }{ 100 } log\frac { 9 }{ 10 } \)
Substituting in (2) we get,
\(log\frac { N }{ { N }_{ 0 } } =\left( \frac { 1 }{ 100 } log\frac { 9 }{ 10 } \right) t\)
when t = 1000,
\(log\frac { N }{ { N }_{ 0 } } =\frac { 1 }{ 100 } log\left( \frac { 9 }{ 10 } \right) \times 1000\)
\(=10log\left( \frac { 9 }{ 10 } \right) \)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } \times 100={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\times 100=\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \)
Hence, \(\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \%\) of radioactive nuclei will remain after 1000 years,

Let the point on the circle be

Let the length of the rectangle be 'x' cm
Breadth of the rectangle be 'y' cm
\( x^{2}+y^{2} =(20)^{2} \)
\(y^{2} =400-x^{2}\)
[radius of the circle is 10 cm]
\(y=\sqrt{400-x^{2}}\)
Now, Area of the rectangle A = xy
\( A =x \sqrt{400-x^{2}} \)
\(\frac{d A}{d x} =x \frac{(-2 x)}{2 \sqrt{400-x^{2}}}+\sqrt{400-x^{2}} \) .....(1)
\(=\frac{-x^{2}+400-x^{2}}{\sqrt{400-x^{2}}}=\frac{-2 x^{2}+400}{\sqrt{400-x^{2}}}\)
For maximum or minimum,
\( \frac{d A}{d x} =0 \Rightarrow \frac{-2 x^{2}+400}{\sqrt{400-x^{2}}}=0 \)
\(x^{2} =200 \)
\(x =\pm 10 \sqrt{2} \)
\(x =-10 \sqrt{2} \) (is not possible)
\(\therefore x =10 \sqrt{2} \)
Now, \( \frac{d^{2} A}{d x^{2}}= \frac{\sqrt{400-x^{2}}(-4 x)-\left(-2 x^{2}+400\right)\left(-\frac{2 x}{2 \sqrt{400-x^{2}}}\right)}{400-x^{2}} \)
\(=\frac{2 x^{3}-1200 x}{\left(400-x^{2}\right) \sqrt{400-x^{2}}}\)
At \(x=10 \sqrt{2}, \quad \frac{d^{2} A}{d x^{2}}<0\)
Area of the rectangle is maximum when x \(=10 \sqrt{2}\)
\(y=\sqrt{400-200}=\sqrt{200}=10 \sqrt{2}\)
\(\therefore x=y=10 \sqrt{2}\)
Length of the rectangls = 10\(\sqrt{2}\) cm
Breadth of the rectangls = 10\(\sqrt{2}\) cm

Let AB be the position of the ladder at any time t such that OA = x and OB = y
Then OA² + OB² = AB²
⇒ x² + y² = 17²
Given \(\frac { dx }{ dt } \) = 5 and x = 8
When x = 8, 8² + y² = 17²
 ⇒ y² = 289 - 64 = 225
 ⇒ y = 15
Differentiating (1) with respect to 't' we get,
\(2x\frac { dx }{ dt } +2y\frac { dy }{ dt } =0\)
⇒ 8(5) + 15 \(\frac { dy }{ dt } \) = 0 [∵ x = 8, \(\frac { dx }{ dt } \) = 15, y = 15]
⇒ 40 + 15\(\frac { dy }{ dt } \) = 0
⇒ \(\frac { dy }{ dt } =\frac { -40 }{ 15 } =\frac { -8 }{ 3 } \) m/sec
∴ The rate of top of the ladder moving down the wall is \(\frac{-8}{3}\) m/sec
(ii) The ladder, the wall and the floor forms a right angled triangle.
 Area = \(\frac12\)xy
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } =\frac { 1 }{ 2 } \left[ x\frac { dy }{ dx } +y\frac { dx }{ dt } \right] \)
\(=\frac { 1 }{ 2 } \left[ 8\left( -\frac { 8 }{ 3 } \right) +15(5) \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64 }{ 3 } +75 \right] =\frac { 1 }{ 2 } \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64+225 }{ 3 } \right] =\frac { 1 }{ 2 } \left( \frac { 161 }{ 3 } \right) \)
= \(\frac { dA }{ dt } \) = 26.83 sq.m/sec

Here the number of unknowns is 3. So, if the system is consistent and has a non-trivial solution, then the rank of the coefficient matrix is equal to the rank of the augmented matrix and is less than 3.
So the determinant of the coefficient matrix should be 0.
Hence we get
\(\left| \begin{matrix} 3\lambda -8 & 3 & 3 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 or \(\left| \begin{matrix} 3\lambda -2 & 3\lambda -2 & 3\lambda -2 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 (by applying R₁ ➝ R₁ + R₂ + R₃)
or (3λ - 2) \(\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 (by taking out (3λ − 2) from R₁)
or (3λ - 2) \(\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 3\lambda -11 & 3 \\ 3 & 3 & 3\lambda -11 \end{matrix} \right| \) = 0 (by applying R₂ ➝ R₂ - 3R₁, R₃ ➝ R₃ - 3R₁)
or (3λ - 2)(3λ - 11)² 0. So λ = \(\frac { 2 }{ 3 } \) and λ = \(\frac { 11 }{ 3 } \).
We now give an application of system of linear homogeneous equations to chemistry. You are already aware of balancing chemical reaction equations by inspecting the number of atoms present on both sides.

The plane containing the line
\(\vec { r } =(\hat { i } -\hat { j } +3\hat { k } )+t(2\hat { i } -\hat { j } +4\hat { k } )\)
∴ The required plane is passing through the point \(\vec { a } =\hat { i } -\hat { j } +3\hat { k } \) and parallel to a vector \(\vec { b } =2\hat { i } -\hat { j } +4\hat { k } \) Also, the plane is perpendicular to the plane 
\(\vec { c } =\hat { i } +2\hat { j } +\hat { k } \) 
∴ The parametric form of vector equation of the plane passing through one point (\(\vec { a } \)) and parallel to two vectors \(\vec { b } \) and \(\vec { c } \) 
\(\vec { r } =\vec { a } +s\vec { b } +t\vec { c } \) where s, t ∈ R
⇒ \(\vec { r } .(\hat { i } -\hat { j } +3\hat { k } )+s(2\hat { i } -\hat { j } +4\hat { k } )+t(\hat { i } +2\hat { j } +\hat { k } )\) s, t  ∈ R 
Cartesian equation is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 & y+1 & z-3 \\ 2 & -1 & 4 \\ 1 & 2 & 1 \end{matrix} \right| =0\)
⇒ (x - 1)(-1 - 8) - (y + 1)(2 - 4) + (z - 3)(4 + 1) = 0
⇒ (x - 1)(-9) - (y + 1)(-2) + (z - 3)5 = 0
⇒ -9x + 9 +2y + 2 + 5z - 15 = 0
⇒ -9x + 2y + 5z - 4 = 0
⇒ 9x - 2y - 5z + 4 = 0

Let the cross section of the satellite dish be an right open parabola.

Its equation is y² = 4ax
Since focus is placed 1.2 m from the vertex OA = 1.2 m and BC = 2.5 m since the width of the dish is 5 m.
From the diagram, a = 1.2 m
∴ y² = 4(1.2)x
(a) ⇒ y² = 4.8x   ...(1)
(b) Since (x₁, 2.5) lines on (1)(2.5)² = 4.8(x₁)
x₁ = \(\frac { 2.5\times 2.5 }{ 4.8 } \)
x₁ = 1.3 m
∴ Depth of the satellite dish at the vertex is 1.3 m.

We know that \(sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) =cos^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \)
Thus, \(cos\left( sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right) =\frac { 1 }{ \sqrt { 1+x^{ 2 } } } \) ...(1)
Let \(\cot ^{-1}\left(\frac{3}{4}\right)=\theta\). Then \(\cot \theta=\frac{3}{4}\) and so \(\theta\) is cute. 
From the diagram, we get,
Hence \(\sin \left\{\cot ^{-1}\left(\frac{3}{4}\right)\right\}=\sin \theta=\frac{4}{5}\) ................ (2)
Using (1) and (2) in the given equation, we \(\frac { 1 }{ \sqrt { 1+x^{ 2 } } } =\frac { 4 }{ 5 } \) \(\sqrt{1+x^2}=\frac{5}{4}\)
Thus, x = \(\pm\frac{3}{4}\)

Given z = x + iy and arg\(\left( \frac { z-i }{ z+2 } \right) =\frac { \pi }{ 4 } \)
⇒ arg(z-i) - arg(z+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x + iy-i) - arg(x+iy+2) = \(\frac { \pi }{ 4 } \)
⇒ arg(x+i(y-1)-arg((x+2)+iy) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { y-1 }{ x } \right) -tan^{ -1 }\left( \frac { y }{ x+2 } \right) \) = \(\frac { \pi }{ 4 } \)
⇒ \(tan^{ -1 }\left( \frac { \frac { y-1 }{ x } -\frac { y }{ x+2 } }{ 1+\frac { y-1 }{ x } .\frac { y }{ x+2 } } \right) \)
= \(\frac { \pi }{ 4 } \)\(\left[ \because tan^{ -1 }x-tan^{ -1 }y=tan^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
\(\Rightarrow \frac{\left(\frac{(x+2)(y-1)- x y}{\not {x (\not x+\not2)}}\right)}{\left(\frac{x(x+2)+y(y-1)}{\not x(\not x+\not 2)}\right)}=\tan \frac{\pi}{4}=1\)
⇒ \(\frac { (x+2)(y-1)-xy }{ x(x+2)+y(y-1) } \) = 1
⇒ -x + 2y-2 = x²+ 2x + y²-y
⇒ x² + 2x + y²-y + x-2y + 2 = 0
⇒ x² + y²+3x-3y + 2 = 0
Hence proved.

This equation is Type I even degree reciprocal equation. Hence it can be rewritten as
x²\(\left[ \left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -10 \left( x+\frac { 1 }{ x } \right) +26 \right] \)= 0 Since x \(\neq\) 0, we get \(\left(x^{2}+\frac{1}{x^{2}}\right)-10\left(x+\frac{1}{x}\right)+26=0\)
Let y = \(\left( x+\frac { 1 }{ x } \right) \)
[(y²-2)-10y+26] = 0 ⇒ (y²-10y+24) = 0 ⇒ (y-6)(y-4) = 0 ⇒ y = 6 or y = 4
Case (i)
y = 6 ⇒ x +\(\frac{1}{x}\) = 6 ⇒ x = 3+2\(\sqrt{2}\), x = 3 - 2\(\sqrt{2}\)
Case (ii)
y = 4 ⇒ x = 2+\(\sqrt{3}\), x = 2-\(\sqrt{3}\).
Hence, the roots are \(3 \pm 2 \sqrt{2}, 2 \pm \sqrt{3}\)

Consider a triangle ABC in which the two altitudes AD and BE intersect at O. Let CO be produced to meet AB at F. We take O as the origin and let \(\vec { OA } =\vec { a } \), \(\vec { OB } =\vec { b} \) and  \(\vec { OC } =\vec { c } \)

Since \(\vec { AD } \) is perpendicular to \(\vec { BC } \), we have \(\vec { OA } \) is perpendicular to \(\vec { BC } \), and
hence we get \(\vec { OA } \) . \(\vec { BC } \) = 0. That is, \(\vec { a } .(\vec { c } -\vec { b } )=0\), which means
\(\vec { a } .\hat{c}-\hat{a}.\hat{b}=0\)....(1)
Similarly, since \(\vec { BE } \) is perpendicular to \(\vec { CA } \), we have \(\vec { OB } \) is perpendicular to \(\vec { CA } \), and hence we get \(\vec { OB } .\vec { CA } \) = 0.
That is, \(\vec {b } .(\vec {a } -\vec { c } )=0\)
\(\vec { a } .\hat{c}-\hat{b}.\hat{c}=0\).......(2)
Adding equations (1) and (2), gives  \(\vec { a } .\hat{c}-\hat{b}.\hat{c}=0\). That is, \(\hat{c}(\hat{a}-\hat{b})=0\)
That is \(\vec { OC } \) . \(\vec { BA } \) = 0.
Therefore, \(\vec { BA } \) is perpendicular to \(\vec { OC} \). 
Which implies that \(\vec { CF} \) is perpendicular to \(\vec { AB } \). 
Hence, the perpendicular drawn from C to the side AB passes through O. Therefore, the altitudes are concurrent.