Given equation are \({ x }^{ 2 }px+q=0\)  ............(1)
and \({ x }^{ 2 }+p'x+q'=0\)    ..........(2)
Let ∝ be the common root for (1) and (2)
∴ ∝² + p∝ + q = 0   .........(3)
and ∝²+ p'∝ + q' = 0    .............(4)
Solving (3) and (4) by cross multiplication method we get
p  q 1 p
p' q' 1 p'
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ pq'-p'q } =\frac { \alpha }{ q-q' } =\frac { 1 }{ p'-p } \)
consider \(\frac { { \alpha }^{ 2 } }{ pq'-p'q } =\frac { \alpha }{ q-q' } \)
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ \alpha } =\frac { pq'-p'q }{ q-q' } \)
\(\alpha =\frac { pq'-p'q }{ q-q } \)
Consider \(\frac { \alpha }{ q-q' } =\frac { 1 }{ p'-p } \Rightarrow \alpha =\frac { q-q' }{ p'-p } \)
Hence its roots are \(\frac { pq'-p'q }{ q-q' } or\quad \frac { q-q' }{ p'-p } \)

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.

The path y = ax² + bx + c passes through the points (10, 8), (20, 16), (40, 22). So, we get the system of equations 100a + 10b + c = 8, 400a + 20b + c = 16, 1600a + 40b + c = 22. To apply Cramer’s rule, we find
Δ = \(\left| \begin{matrix} 100 & 10 & 1 \\ 400 & 20 & 1 \\ 1600 & 40 & 1 \end{matrix} \right| =1000\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 16 & 4 & 1 \end{matrix} \right| \) = 1000 [-2 + 12 - 6] = -6000,
Δ₁ = \(\left| \begin{matrix} 8 & 10 & 1 \\ 16 & 20 & 1 \\ 22 & 40 & 1 \end{matrix} \right| =20\left| \begin{matrix} 4 & 1 & 1 \\ 8 & 2 & 1 \\ 11 & 4 & 1 \end{matrix} \right| \) = 20[-8 + 3 + 10] = 100,
Δ₂ = \(\left| \begin{matrix} 100 & 8 & 1 \\ 400 & 16 & 1 \\ 1600 & 22 & 1 \end{matrix} \right| =200\left| \begin{matrix} 1 & 4 & 1 \\ 4 & 8 & 1 \\ 16 & 11 & 1 \end{matrix} \right| \) = 200[-3 + 48 - 84] = -7800,
Δ₃ = \(\left| \begin{matrix} 100 & 10 & 8 \\ 400 & 20 & 16 \\ 1600 & 40 & 22 \end{matrix} \right| =2000\left| \begin{matrix} 1 & 1 & 4 \\ 4 & 2 & 8 \\ 16 & 4 & 11 \end{matrix} \right| \) = 2000[-10 + 84 - 64] = 20000.
By Cramer’s rule, we get a = \(\frac { { \Delta }_{ 1 } }{ \Delta } =-\frac { 1 }{ 60 } \), b = \(\frac { { \Delta }_{ 2 } }{ \Delta } =\frac { 7800 }{ 6000 } =\frac { 78 }{ 60 } =\frac { 13 }{ 10 } \), c = \(\frac { { \Delta }_{ 3 } }{ \Delta } =\frac { 20000 }{ 6000 } =-\frac { 20 }{ 6 } =-\frac { 10 }{ 3 } \).
So, the equation of the path is y = \(\frac { 1 }{ 60 } { x }^{ 2 }+\frac { 13 }{ 10 } x-\frac { 10 }{ 3 } \).
When x = 70, we get y = 6. So, the ball went by 6 metres high over the boundary line and it is impossible for a fielder standing even just before the boundary line to jump and catch the ball. Hence the ball went for a super six and the team won the match.

Since the centre is located at (0, 300), midway between the two foci, which are the coast guard stations, the equation is \(\frac { { \left( y-300 \right) }^{ 2 } }{ { a }^{ 2 } } -\frac { { \left( x-0 \right) }^{ 2 } }{ { b }^{ 2 } } =1\).... (1)
To determine the values of a and b, select two points known to be on the hyperbola and substitute each point in the above equation.
The point(0, 400) lies on the hyperbola, since it is 200 km further from Station A than from station B.
\(\frac { { \left( 400-300 \right) }^{ 2 } }{ { a }^{ 2 } } -\frac { O }{ { b }^{ 2 } } =1\frac { { 100 }^{ 2 } }{ a^{ 2 } } =1,{ a }^{ 2 }=10000.\) There is also a point (x, 600) on the hyperbola such that 600²+ x² = (x + 200)²    
360000 + x² = x²+400x+40000
x = 800
Substituting in (1), we have \(\frac { { \left( 600-300 \right) }^{ 2 } }{ 10000 } -\frac { { \left( 800-0 \right) }^{ 2 } }{ { b }^{ 2 } } =1\)
\(9-\frac { 640000 }{ { b }^{ 2 } } =1\)
b² = 80000
Thus the required equation of the hyperbola is \(\frac { { \left( y-300 \right) }^{ 2 } }{ 10000 } -\frac { { x }^{ 2 } }{ 80000 } =1\)
The ship lies somewhere on this hyperbola. The exact location can be determined using data from a third station.

Let \({ z }^{ 3 }+8i=0\)
\(\Rightarrow\) z³ = -8i
= \(8(-i)=8\left( cos\left( -\frac { \pi }{ 2 } +2k\pi \right) isin\left( -\frac { \pi }{ 2 } +2k\pi \right) \right) \),k\(\in Z\)
\(z=\sqrt [ 3 ]{ 8 } \left( cos\left( \frac { -\pi +4k\pi }{ 6 } \right) +isin\left( \frac { -\pi +4k\pi }{ 6 } \right) \right) \)
Taking k = 0, 1, 2 we get,
k = 0, \(z=2\left( cos\left( -\frac { \pi }{ 6 } \right) +isin\left( -\frac { \pi }{ 6 } \right) \right) =2\left( -\frac { 1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) =2\left( \frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) \)
k = 1, \(z=2\left( cos\left( \frac { \pi }{ 2 } \right) +isin\left( \frac { \pi }{ 2 } \right) \right) =2=\left( 0+i \right) =0+2i=2i\)
k = 2,\(z=2\left( xcos\left( \frac { 7\pi }{ 6 } \right) +isim\left( \frac { 7\pi }{ 6 } \right) \right) =2\left( cos\left( \pi +\frac { \pi }{ 6 } \right) \right) +isin\left( \pi +\frac { \pi }{ 6 } \right) \)
= \(2\left( -cos\left( \frac { \pi }{ 6 } \right) -isin\left( \frac { \pi }{ 6 } \right) \right) =2\left( -\frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) =-\sqrt { 3 } -i\)
The values of z are \(\sqrt { 3 } -i,2i\) and \(-\sqrt { 3 } -i\)

\(2{ tan }^{ -1 }x={ cos }^{ -1 }\frac { 1-{ a }^{ 2 } }{ 1+{ a }^{ 2 } } -{ cos }^{ -1 }\frac { 1-{ b }^{ 2 } }{ 1+{ b }^{ 2 } } ,a>0,b>0\)
Let \(a=tan\ \theta \ b=tan\phi \)
\(\therefore { cos }^{ -1 }\left( \frac { 1-{ a }^{ 2 } }{ 1+a^{ 2 } } \right) ={ cos }^{ -1 }\left( \frac { 1-{ tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \right) \)
= \({ cos }^{ -1 }\left( cos2\theta \right) =2\theta \)    ...(1)
\(\left[ \because cos2\theta =\frac { 1-{ tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \right] \)
Also \({ cos }^{ -1 }\left( \frac { 1-{ b }^{ 2 } }{ 1+{ b }^{ 2 } } \right) ={ cos }^{ -1 }\left( \frac { 1-{ tan }^{ 2 }\phi }{ 1+tan^{ 2 } } \right) \)
= \(cos^{ -1 }\left( cos2\Phi \right) \)
\(\therefore 2{ tan }^{ -1 }x=2\theta -2\phi =2(\theta -\phi )\)
[using (1) and (2)]
\(\Rightarrow tan^{ -1 }x=\theta -\phi \)
\(\Rightarrow x=tan\left( \theta -\phi \right) =\frac { tan\theta -tan\phi }{ 1+tan\theta tan\phi } \)
\(\left[ \because tan\left( A-B \right) =\frac { tanA-tanB }{ 1+tanAtanB } \right] \)
\(\Rightarrow x=\frac { a-b }{ 1+ab } \), a > 0

Let x and y be the length, breadth of the printed rectangular page

Given xy = 24
\(\Rightarrow y=\frac { 24 }{ x } \)    ..(1)
Length of the page with margin
= x+1+1 = x+2
breadth of the page with margin
= y + 1.5 + 1.5 = y + 3
Area of the page = (x + 2) (y + 3)
Let f(x) = (x + 2) (y + 3)
= \((x+2)\left( \frac { 24 }{ x } +3 \right) \) 
= \(24+3x+\frac { 48 }{ x } +30\) 
= \(3x+\frac { 48 }{ x } +30\) 
\(f'\left( x \right) =3-\frac { 48 }{ { x }^{ 2 } } \)
\(\Rightarrow { x }^{ 2 }=16\Rightarrow x=\pm 4\)
ஃ The critical number are 4,-4
\(f''\left( x \right) =-48\left( \frac { -2 }{ { x }^{ 3 } } \right) =\frac { 96 }{ { x }^{ 3 } } \) 
\(f''\left( 4 \right) =\frac { 96 }{ 64 } >0\) 
ஃ f(x) is minimum when x = 4
When \(x=4,y=\frac { 24 }{ 4 } =6\)  [From (1)]
ஃ Length of the page = x + 2 = 4 + 2 = 9 cm
Breadth of the page = y + 3 = 6 + 3 = 6 cm

\(\frac { dy }{ dx } +\frac { { 3x }^{ 2 }y }{ 1+{ x }^{ 3 } } =\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
This is a linear differential equation 
\(\therefore P=\frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } ;Q=\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
\(\therefore \int { pdx } =\int { \frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } dx } =log(1+{ x }^{ 3 })\)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log(1+{ x }^{ 3 }) }=(1+{ x }^{ 3 })\)
\(\therefore\)The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } (1+{ x }^{ 3 })dx+c } \)
\(cos2x=1-2{ sin }^{ 2 }x\)
\(sin2x=\frac { 1-cos2x }{ 2 } =\int { { sin }^{ 2 }xdx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { 1-cos2x }{ 2 } } dx+c\)
\(\Rightarrow y(1+{ x }^{ 3 })=\frac { x }{ 2 } -\frac { sin2x }{ 4 } +c\)

Given \(\begin{cases} \begin{matrix} k & 200\le x\le 600 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(i) Since f{x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x)d=1\Rightarrow \int _{ 200 }^{ 600 }{ kda } =1 } \) 
\(\Rightarrow k[x]_{ 200 }^{ 600 }=1\Rightarrow k(600-200)=1\) 
400 k = 1
\(\Rightarrow k=\frac { 1 }{ 400 } \) 
(ii) The distribution function  
= \(\int _{ -\infty }^{ x }{ f(u) } du\) 
Case 1: x < 200
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
Case 1: x < 200 ≤ x ≤  600
\(\int _{ -\infty }^{ x }{ f(u) } du\)
\(F(x)=\int _{ -\infty }^{ 200 }{ f(u)du } =+\int _{ 200 }^{ x }{ f(u)du } \)  
= \( =\frac { 1 }{ 400 }(x-200) =\frac { x }{ 400 } =\frac { 1 }{ 2 } \) 
Case 3: x > 600
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
\(f(x)= \begin{cases}0, & x<200 \\ \frac{x}{400}-\frac{1}{2}, & 200 \leq x \leq 600 \\ 0, & x>600\end{cases}\)
(iii) P(300 < x < 500)
= \(\int _{ 300 }^{ 500 }{ kdx=\frac { 1 }{ 400 } \left[ x \right] _{ 300 }^{ 500 } } \) 
= \(\frac { 1 }{ 400 } \left[ 500-300 \right] =\frac { 200 }{ 400 } =\frac { 1 }{ 2 } \)

\(put\sqrt { t } =z\Rightarrow \frac { 1 }{ 2\sqrt { t } } dt=dz\)
\(\Rightarrow \frac { 1 }{ 2z } dt=dz\Rightarrow dt=2zdz\)

\(\therefore I=\int _{ 0 }^{ sin\quad x }{ 2z{ sin }^{ -1 }zdz } +\int _{ 0 }^{ cosx }{ 2z{ cos }^{ -1 }zdz } ...(1)\)
= I₁+ I₂ 
\(=2\int _{ 0 }^{ x }{ \theta sin\theta cos\theta d\theta +2\int _{ \frac { \pi }{ 2 } }^{ x }{ \theta cos\theta (-sin\theta )d\theta } } \)
\(=\int _{ 0 }^{ x }{ \theta 2sin\theta cos\theta d\theta -\int _{ \frac { \pi }{ 2 } }^{ x }{ \theta 1sin\theta cos\theta d\theta } } \)
\(=\int _{ 0 }^{ x }{ \theta sin2\theta d\theta -\int _{ \frac { \pi }{ 2 } }^{ x }{ \theta sin2\theta d\theta } } \)
\([\because sin2\theta =sin\theta cos\theta ]\)
\(=\int _{ 0 }^{ x }{ \theta sin2\theta d\theta \int _{ x }^{ \frac { \pi }{ 2 } }{ \theta sin2\theta d\theta } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \theta sin2\theta d\theta } } \)
\(\left[ \because \int _{ a }^{ c }{ f(x)dx+\int _{ c }^{ b }{ f(x)dx=\int _{ a }^{ b }{ f(x)dx } } } \right] \)
\(={ \left[ -\frac { \theta cos2\theta }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }+\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { cos2\theta }{ 2 } d\theta } \)

\(={ \left[ -\frac { \theta cos2\theta }{ 2 } +\frac { sin2\theta }{ 4 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(=\frac { -\frac { \pi }{ 2 } cos\pi }{ 2 } +\frac { sin\pi }{ 4 } -(0+0)\)
\(=\frac { -\frac { \pi }{ 2 } (-1) }{ 2 } +\frac { 0 }{ 4 } =\frac { \pi }{ 4 } \)

Note that the function u is not homogeneous. So we cannot apply Euler’s Theorem for u.
However, note that f(x,y) = \(\frac { x+y }{ \sqrt { x } +\sqrt { y } }\) = sin u is homogeneous; because
f(tx,ty) = \(\frac { tx+ty }{ \sqrt { tx } +\sqrt { ty } } \) = t^1/2 f(x, y), \(\forall \) x, y, t\(\ge \)0
Thus f is homogeneous with degree \(\frac { 1 }{ 2 } \) and so by Euler’s Theorem we have
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } =\frac { 1 }{ 2 } f(x,y)\).
Now substituting f = sin u in the above equation, we obtain
\(x\frac { \partial (sinu) }{ \partial x } +y\frac { \partial (sinu) }{ \partial y } =\frac { 1 }{ 2 } sin \ u\)
\(x\quad cosu\frac { \partial u }{ \partial x } +y\quad cosu\frac { \partial u }{ \partial x } =\frac { 1 }{ 2 } sin \ u\)  ...(19)
Dividing both sides by cosu we obtain
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 1 }{ 2 } tan \ u\)
Note:
Solving this problem by direct calculation will be possible; but will involve lengthy calculations.

Prove that p⟶(q⟶r) = (p ∧ q)⟶r without using truth table. From example we know that p⟶ q = ~p V q
Consider LHS = p⟶(q⟶r)
= p ⟶ (~q V r) [Implication Law]
= ~p V (~q V r) [Implication Law]
= (~p V ~q) V r [associative property]
= ~(p ∧ q) V r [using Demorgan's law]
= (p ∧ q) ⟶ r [Implication Law]
Hence proved.

Take the midpoint of the base as the centre c(0, 0)
Since the base wide is 48 feet, the vertices A and A' are (24, 0) and (-24, 0) respectively. Clearly 2a = 48 and b = 20
The corresponding equation is
\(\frac{x^{2}}{24^{2}}+\frac{y^{2}}{20^{2}}=1\) .............(1)
Let x₁ be the distance between the pole whose height is 10 m and the  center.
Then (x₁, 10) satisfies the equation (1)
\(\therefore \frac{x_{1}^{2}}{24^{2}}+\frac{10^{2}}{20^{2}}=1 \Rightarrow x_{1}=12 \sqrt{3}\)
Width of the arch = 2x₁ = 24√3 feet

The line y = x lies above the curve y = x³ in the first quadrant y = x³ lies above the line y = x in the third quadrant.
To get the point of intersection solve the curves y = x³, y = x
\( \mathbf{x}^{3} =\mathrm{x} \)
\(\mathrm{x}^{3}-\mathrm{x} =0 \)
\(\mathbf{x}\left(\mathrm{x}^{2}-\mathrm{x}\right) =0 \quad \Rightarrow \quad \mathrm{x}=0, \mathrm{x}=\pm 1\)

The required Area
\( =\int_{-1}^{0}[g(x)-f(x)] d x+\int_{0}^{1}[f(x)-g(x)] d x \)
\(=\int_{-1}^{0}\left(x^{3}-x\right) d x+\int_{0}^{1}\left(x-x^{3}\right) d x \)
\(=\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{-1}^{0}+\left[\frac{x^{2}}{2}-\frac{x^{4}}{4}\right]_{0}^{1} \)
\(=0-\frac{1}{4}-\frac{1}{2}+\frac{1}{2}-0-\frac{1}{4} \)
\(=-\frac{1}{4}+\frac{1}{2}+\frac{1}{2}-\frac{1}{4}=\frac{1}{2} \)

Since \(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) is a root, x-\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) is a factor. To remove the outermost square root, we take x +\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) as another factor and find their product.
\(\left( x+\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \right) \left( x-\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \right) ={ x }^{ 2 }-\frac { \sqrt { 2 } }{ \sqrt { 3 } } \)
Still we didn’t achieve our goal. So we include another factor x²+\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) and get the product.
\(\left( { x }^{ 2 }-\frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) \left( { x }^{ 2 }+\frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) ={ x }^{ 4 }-\frac { 2 }{ 3 } \)
So, 3x⁴- 2 = 0 is a required polynomial equation with the integer coefficients.
Now we identify the nature of roots of the given equation without solving the equation. The idea comes from the negativity, equality to 0, positivity of Δ = b²- 4ac.

Let ABCD be a parallelogram such that its diagonals AC and BD are equal.
Taking A as the origin, let the p.v. of B and D be \(\vec { b } \)and \(\vec { d } \) respectively.
Then \(\vec { AB } =\vec { b } \) and \(\vec { AD } =\vec { d } \)
Using triangle law of addition of vectors is ΔABC, we get
\(\vec { AB } +\vec { BC } =\vec { AC } \)
⇒ \(\vec { AB } +\vec { AD } =\vec { AC } \)
⇒ \(\vec { b } +\vec { d } =\vec { AC } \)
Using triangle law of addition of vectors in ΔABD, we get \(\vec { AB } +\vec { BD } =\vec { AD } \)
⇒ \(\vec { b } +\vec { BD } =\vec { d } \)
⇒ \(\vec { BD } =\vec { d } -\vec { b } \)
In parallelogram ABCD we have AC = BD
⇒ \(|\vec { AC } |=|\vec { BD } |\)
⇒ \(|\vec { AC } |^{ 2 }=|\vec { BD } |^{ 2 }\)
⇒ \(|\vec { b } +\vec { d } |^{ 2 }=|\vec { d } -\vec { b } |^{ 2 }\)

⇒ \(4(\vec { b } .\vec { d } )=0\Rightarrow \vec { b } .\vec { d } =0\)=0
⇒ \(\vec { b } \bot \vec { d } \)
⇒ \(\vec { AB } \bot \vec { AD } \)
Hence, ABCD is a rectangle.

The latus rectum LL' of the hyperbola passes through S(ae, 0)

∴ L is (ae, y₁)
Substituting L in \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } \) = 1 we get,
Substituting L in \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { a }^{ 2 }{ e }^{ 2 } }{ { a }^{ 2 } } -\frac { { { y }_{ 1 } }^{ 2 } }{ { b }^{ 2 } } =1\Rightarrow { e }^{ 2 }-\frac { { { y }_{ 1 } }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\Rightarrow { e }^{ 2 }-1=\frac { { { y }_{ 1 } }^{ 2 } }{ { b }^{ 2 } } \) \([{ b }^{ 2 }={ a }^{ 2 }({ e }^{ 2 }-1)\)
\(\Rightarrow { { y }_{ 1 } }^{ 2 }={ b }^{ 2 }({ e }^{ 2 }-1)\) \(\Rightarrow \frac { { b }^{ 2 } }{ { a }^{ 2 } } ={ e }^{ 2 }-1]\)
\(\Rightarrow { { y }_{ 1 } }^{ 2 }={ b }^{ 2 }\left( \frac { { b }^{ 2 } }{ { a }^{ 2 } } \right) \) \(\Rightarrow { { y }_{ 1 } }^{ 2 }=\frac { { b }^{ 4 } }{ { a }^{ 2 } } \)
\(\Rightarrow { y }_{ 1 }=\pm \frac { { b }^{ 2 } }{ a } \)
∴ End points oflatus rectum Land L' are
\(\left( ae,\frac { { b }^{ 2 } }{ a } \right) \) and \(\left( ae,-\frac { { b }^{ 2 } }{ a } \right) \)
Hence, the length of latus rectum LL' = \(\frac { { b }^{ 2 } }{ a } +\frac { { b }^{ 2 } }{ a } =\frac { 2{ b }^{ 2 } }{ a } \) units.
Hence proved.

\(\frac { 3 }{ x } \) + 2y = 12, \(\frac { 2 }{ x } \) + 3y = 13
Let \(\frac { 1 }{ x } \)
∴ z+2y = 12, 2z+3z = 13
∴ Δ = \(\left| \begin{matrix} 3 & 2 \\ 2 & 3 \end{matrix} \right| \)= 9 - 4 = 5
Δ₁ = \(\left| \begin{matrix} 12 & 2 \\ 13 & 3 \end{matrix} \right| \)= 36 - 26 = 10
Δ₂ = \(\left| \begin{matrix} 3 & 12 \\ 2 & 13 \end{matrix} \right| \)= 39 - 26 = 10
∴ z = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 10 }{ 5 } =2\Rightarrow \frac { 1 }{ x } =2\Rightarrow x=\frac { 1 }{ 2 } \)
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 15 }{ 5 } \) = 3
∴ Solution set {\(\frac{1}{2}\), 3}

Given g(x) = sin^-1 x + cos^-1x
From the definition of sin^-1x.
\(-1\le x\le 1\) ...(1)
Also from the definition of cos^-1x
\(-1\le x\le 1\)   .........(2)
\(\therefore \) From (1) & (2),
Domain ofg(x) = [-1, 1] U [-1, 1]
= [-1, 1]
Hence the domain of g(x) is [-1, 1].

Given that \((1+{ x }^{ 2 })\frac { dy }{ dx } =1+{ y }^{ 2 }\) ..(1)
The given equation is written in the variables separable form 
\(\frac { dy }{ { 1+y }^{ 2 } } =\frac { dx }{ { 1+x }^{ x } } \) ...(2)
Integrating both sides of (2), we get tan⁻¹ tan⁻¹x +C.
But tan^-1 y - tan^-1 x = tan^-1 \(\left( \frac { y-x }{ 1+xy } \right) .\) ...(4)
Using (4) in (3) leads to tan^-1 \(\left( \frac { y-x }{ 1+xy } \right)\) = C, which implies \(\frac { y-x }{ 1+xy } \) = tan C = a (say).
Thus, y − x = a(1+ xy) gives the required solution

Let f (x) = sin x which is a differentiable function in any open interval. Consider an interval \([\alpha, \beta]\). Applying the mean value theorem there exists  \(c \in (\alpha, \beta)\) such that, 
\(\frac{sin \beta - sin \alpha}{\beta-\alpha}=f'(c)=cos(c)\)
Therefore, \(\frac{sin \beta - sin \alpha}{\beta-\alpha}=|cos(c)|\le1\)
Hence, \(|sin\alpha-sin\beta|\le |\alpha-\beta|\)
Remark
If we take \(\beta=0\) in the above problem, we ge \(|sin \alpha|\le |\alpha|\) 

Put x = sin \(\theta\). Then, dx = cos \(\theta\) d \(\theta\)
when x = 0, sin \(\theta\) = 0 and so \(\theta\) = 0. When x = 1, sin \(\theta\) = 1 and so \(\theta =\frac{\pi}{2}\)
Hence, we get
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 5 }\theta { (1-{ sin }^{ 2 }\theta ) }^{ 5 }cos\theta d\theta } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 5 }\theta { cos }^{ 11 }\theta d\theta =\frac { 10 }{ 16 } \times \frac { 8 }{ 14 } \times \frac { 6 }{ 12 } \times \frac { 4 }{ 10 } \times \frac { 2 }{ 8 } \times \frac { 1 }{ 6 } =\frac { 1 }{ 336 } } \)
By applying the reduction formula III iteratively, we get the following results (stated without proof):
\(\int _{ 0 }^{ 1 }{ { x }^{ m }{ (1-x) }^{ n }dx } =\frac { m!\times n! }{ (m+n+1)! } \), where m and n are positive integers

Let us denote white and red balls as w₁, w₂, r₁, r₂ and r₃
The sample space consists of 5C₃ = 10 different samples of size 3.
That is S = \(\left\{w_{1} w_{2} r_{1}, w_{1} w_{2} r_{2}, w_{1} w_{2} r_{3}, w_{1} r_{1} r_{2}, w_{1} r_{2} r_{3}, w_{1} r_{1} r_{3}, w_{2} r_{1} r_{2}, w_{2} r_{2} r_{3}, w_{2} r_{1} r_{3}, r_{1} r_{2} r_{3}\right\} .\)
The random variable X takes on the values 1, 2, and 3.

Clearly 1 is the identity element of (G₁)
Inverse of 1 is 1 [∴ (1)(1) = 1]
Inverse of i is -i [∴ (i) (-i) = -i² = 1]
Inverse of -1 is -1 [∴ (-1)(-1) = 1]
Inverse of is i [∴ (-i)(i) = -i² = 1]

Given adj (A) =\(\left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
We know that A^-1 = ±\(\frac { 1 }{ \sqrt { |adjA| } } \) (adj A) ...............(1)
|adj A| = 0 + 2\(\left| \begin{matrix} 6 & -6 \\ -3 & 6 \end{matrix} \right| \) + 0
[Expanded along R₁]
= 2(36-18) = 2(18) = 36
∴ A^-1 = \(\pm \frac { 1 }{ \sqrt { 36 } } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
= \(\pm \frac { 1 }{ 6 } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \).

Given ends of diameter are (3, 4)(2, -7)
∴ Equation of the circle is
(x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0
⇒ (x - 3)(x - 2) + (y - 4)(y + 7) = 0
⇒ x² - 2x - 3x + 6 + y² + 7y - 4y - 28 = 0
⇒ x² + y² − 5x + 3y − 22 = 0

let z = |4+3i|
= \(\\ \sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 16+9 } =\sqrt { 25 } \)
\(\sqrt { a+ib } =\pm \sqrt { \frac { |z|+a }{ 2 } } +i\frac { b }{ |b| } \sqrt { \frac { |z|-a }{ 2 } } \)
[Here |z| = 5, a = 4, b = 3]
\(\sqrt { 4+3i } =\pm \sqrt { \frac { 5+4 }{ 2 } } +i\frac { 3 }{ |3| } \sqrt { \frac { 5-4 }{ 2 } } \)
= \(\pm \sqrt { \frac { 9 }{ 2 } } +i\frac { 3 }{ 3 } \sqrt { \frac { 1 }{ 2 } } \)
= \(\pm \frac { 3 }{ \sqrt { 2 } } + \frac { i }{ \sqrt { 2 } } \)
Aliter :
Square root of 4 + 3i
Formula method
\(\sqrt{a+i b}=\pm\left[\sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}\right]\)
Now, \(|4+3 i|=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5\)
\(\therefore \sqrt{4+3 i}=\pm\left[\sqrt{\frac{5+4}{2}}+i \sqrt{\frac{5-4}{2}}\right]\)
\(=\pm\left[\frac{3}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right]\)

Given lines are \(\vec { r } =(4\hat { i } -\hat { j } )+t(\hat { i } +2\hat { j } -2\hat { k } )\) \([\vec { r } =\vec { a } +t\vec { b } ]\)
∴ \(\vec { b } =\hat { i } +2\hat { j } -2\hat { k } \)
and \(\vec { r } =(\hat { i } -2\hat { j } +4\hat { k } )+s(-\hat { i } -2\hat { j } +2\hat { k } )\)
∴ \(\vec { d } =-\hat { i } -2\hat { j } +2\hat { k } \)
Let θ be the angle between the given lines
Then cos θ = \(\frac { \vec { b } .\vec { d } }{ |\vec { b } ||\vec { d } | } \)
= \(\frac { (\hat { i } +2\hat { j } -2\hat { k } ).(-\hat { i } -2\hat { j } +2\hat { k } ) }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+(-2)^{ 2 } } .\sqrt { (-1)^{ 2 }+{ (-2) }^{ 2 }+{ (2) }^{ 2 } } } \)
= \(\frac { -1-4-4 }{ \sqrt { 9 } .\sqrt { 9 } } =\frac { -9 }{ 9 } \) = -1
∴ cos θ = -1
⇒ cos-1(-1)
⇒ θ = 0

 sin^-1 [sin5]
We know that \({ sin }^{ -1 }\left( sin5 \right) =0\quad if\quad \theta \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ e } \right] \)
Consiideerning the approximation \(\frac { \pi }{ 2 } =\frac { 11 }{ 7 } \)
\(5=5\times \frac { 11 }{ 7 } \times \frac { 7 }{ 11 } =5\times \frac { \pi }{ 2 } \times \frac { 7 }{ 11 } =\frac { 35\pi }{ 22 } \notin \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
But \(5-2\pi =5-2\times \frac { 22 }{ 7 } =5-\frac { 44 }{ 7 } =\frac { 35z44 }{ 7 } \)
= \(\frac { -11 }{ 7 } \varepsilon \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
\(5-2\pi \notin \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
\({ sin }^{ -1 }\left( sin5 \right) ={ sin }^{ -1 }\left( sin(5-2\pi ) \right) \)
= \(5-2\pi \quad \) 

Let r be the radius of the ripple and A be the area of the ripple.
GIven \(\frac { dr }{ dt } \) = 2 cm/sec and r = 5 cm ... (1)
We know A = πr²
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } \) = π(2r).\(\frac { dr }{ dt } \)
= π(2) (5) (2) [using (1)]
\(\frac { dA }{ dt } \) = 20 πsq.cm/sec.

\(\int _{ 0 }^{ 1 }{ [2x]dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ [2x] } dx+\int _{ \frac { 1 }{ 2 } }^{ 1 }{ [2x]dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ 0dx+ } \int _{ \frac { 1 }{ 2 } }^{ 1 }{ 1 dx} = 0+[x]^1_{\frac{1}{2}} = 1 -\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)

Volume of sphere = \(\frac43\) πr³
Given r = 5 mm
⇒ dr = (5.3 - 5) = 0.3 mm
\(\text { Approximate volume }=\frac{4}{\not 3} \pi \cdot \not 3 r^{2} d r\)
= 4π (5²) (0.3)
= 100 π (0.3)
= 30π mm³

A = {a+\(\sqrt5\) b:a,b ∈ z}
Let C = a+\(\sqrt5\) b
B = c+\(\sqrt5\)d∈A
where a, b, c, d ∈ Z
[∵ ac + 5bd∈Z and ad+bc ∈Z]
∴ B = (a+\(\sqrt5\)b).(c+\(\sqrt5\)d)
= ac+\(\sqrt5\)ad+cb\(\sqrt5\) + 5bd
= (ac+5bd)+\(\sqrt5\)(ad+bc)∈A
∴ C.B ∈A∀ a, b, c, d∈Z
∴ Usual multiplicaition is a binary operation on.

\( \frac{d y}{d x}+y=e^{-x} \)
\(\frac{d y}{d x}+P y=Q \)
comparing, we get P = 1, Q = e^-x
\( \text {I.F } =\mathrm{e}^{\int \mathrm{Pdx}} \)
\(=\mathrm{e}^{\int \mathrm{dx} x} \)
\(\text {I.F } =\mathrm{e}^{x} \)
The general solution, 
\( y(\text { I.F }) =\int Q(I . F) d x+C \)
\(y^{x} =\int e^{-x} \times e^{x} d x+c \)
\(y^{x} =\int d x+c \)
\(y^{x} =x+c \)
\(y =x e^{-x}+c e^{-x}(\text { or }) \)
\(y =e^{-x}(x+c) \)