Given  \(\alpha\) and \(\beta\) are the zeroes of 2x² - 4x + 5
\(\Rightarrow \quad \alpha +\beta =\frac { -\left( -4 \right) }{ 2 } =2\)
and \(\alpha\beta=\frac{5}{2}\)
Now, 
(i) \(\left( { \alpha }^{ 2 }+{ \beta }^{ 2 } \right) =\left( \alpha +\beta \right) ^{ 2 }-2\alpha \beta ={ 2 }^{ 2 }-2\times \frac { 5 }{ 2 } =4-5\)
= - 1
(ii) \(\frac { 1 }{ { \alpha } } +\frac { 1 }{ { \beta } } =\frac { \alpha +\beta }{ \alpha \beta } =\frac { 2 }{ \frac { 5 }{ 2 } } =\frac { 4 }{ 5 } \)
(iii) \(\left( \alpha -\beta \right) ^{ 2 }=\left( \alpha +\beta \right) ^{ 2 }-4\alpha \beta ={ 2 }^{ 2 }-\frac { 4\times 5 }{ 2 } \)
= 4 - 10 = - 6
(iv) \(\frac { 1 }{ { { \alpha }^{ 2 } } } +\frac { 1 }{ { { \beta }^{ 2 } } } =\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 } }{ \left( \alpha \beta \right) ^{ 2 } } =\frac { -1 }{ \left( \frac { 5 }{ 2 } \right) ^{ 2 } } =\frac { -4 }{ 5 } \)
(v) \(\left( { \alpha }^{ 3 }+{ \beta }^{ 3 } \right) =\left( \alpha +\beta \right) ^{ 3 }-3\alpha \beta \left( \alpha +\beta \right) \)
\(={ 2 }^{ 3 }-3\times \frac { 5 }{ 2 } \times 2=8-15=-7\)

1

2 units
\(=\sqrt { { (-cos\alpha -cos\alpha ) }^{ 2 }+{ (sin\alpha +sin\alpha ) }^{ 2 } } \)
\(=\sqrt { (-2cos\alpha )^{ 2 }+(2sin\alpha )^{ 2 } } \)
\(=\sqrt { 4cos^{ 2 }\alpha +4{ sin }^{ 2 }\alpha } =2\sqrt { cos^{ 2 }\alpha +{ sin }^{ 2 }\alpha } \)

We know that, radius is perpendicular to the tangent at the point of contact.
\(\therefore \ \ \ OA\ \bot \ PA\\ \Rightarrow \ \angle OAP={ 90 }^{ \circ }\)

In , \(\triangle OAP\)
 PO² = PA² + AO²     [by Pythagoras theorem]
\(\Rightarrow \) (17)² = (PA)² + (8)²
\(\Rightarrow \) (PA)² = 289 - 64 = 225
\(\Rightarrow \) PA = \(\sqrt { 225 } \) = 15 cm
Hence, the length of the tangent from point P is 15 cm.

\(3\over13\)

Here, A and B are the centres of two circle having radii 5 cm and 3 cm respectively touch each other internally.

ஃ AB = 5 cm - 3 cm
= 2 cm
Since PQ is perpendicular bisector of the segment AB.
ஃ  \(\angle \)ARP = \(\angle \)ARQ = 90° and PR = RQ
AR = RB = 1 cm
Now, in rt. \(\angle \) ed \(\triangle\)ARP, by using Pythagoras theorem, we have
PR² = AP² - AR²
= 5² - 1² = 25 - 1 = 24
ஃ PR = \(\sqrt{24}\)=2\(\sqrt6\) cm
PQ = 2PR
= 2 x 2\(\sqrt6\) = 4\(\sqrt6\) cm

9.72 cm²

False.
Since two points lie on y-axis and third point lies in first quadrant.

In art. \(\triangle\)ABC,  \(\angle\)B = 90°
\(\therefore\)  AB = BC = 10 m, let   \(\angle\)ACB = \(\theta\)
\(\therefore\)   tan \(\theta={{AB}\over{BC}}={{10}\over{10}}=1\)
tan \(\theta\) = 1
\(\Rightarrow\)  \(\theta\) = 45°

k=-8

Total number of cards = 52
Card drawn is red and a king = 2(as there are 2 red kings)
\(\therefore\) probability of drawing a card which is red and a king = \(\frac{2}{52}=\frac{1}{26}\)

Area of circle = \(\pi\) x 50² = 2500\(\pi\) cm²
New radius = 50 - \(50\over100\) x 50 = 25 cm
Area of new circle = 625\(\pi\) cm²
Decrease in area = 2500\(\pi\) - 625\(\pi\)
= 1875\(\pi\) cm²
%decrease in area = \(1875\pi\over2500\pi\) x 100 = 75%

121

Radius of the sphere cut = \(\frac{1}{2}\)radius of the hemisphere = \(\frac{7}{2}\)cm

Voulme of hemisphere: volume of sphere \(=\frac{2}{3} \pi(7)^{3}: \frac{4}{3} \times \pi \times\left(\frac{7}{2}\right)^{3}\)
\(=1: 2 \times \frac{1}{8}=1: \frac{1}{4}=4: 1\)

a + b , a - b

\({1\over x+1}+{2\over x+2}={5\over x+4}\)
\({x+2+2(x+1)\over (x+1)(x+2)}={5\over x+4}\)
\({x+2+2x+2\over x^2+3x+2}={5\over x+4}\)
(3x+4)(x+4)=5(x²+3x+2)
3x²+12x+4x+16=5x2+15x+10
3x²+16x+16=5x²+15x+10
2x²-x-6=0
2x²-4x+3x-6=0
2x(x-2)+3(x-2)=0
(2x+3)(x-2)=0
x=2 or \(x={-3\over 2}\)

We have,
\(\sum _{ i=1 }^{ n }{ \left( { x }_{ i }-50 \right) } =-10\) and \( \sum _{ i=1 }^{ n }{ \left( { x }_{ i }-46 \right) } =70\)
\(\sum _{ i=1 }^{ n }{ { x }_{ i }-50n } =-10\)
and    \(\sum _{ i=1 }^{ n }{ { x }_{ i }-46n } =70\)
Subtracting (ii) from (i), 
- 4n = - 80
\(\Rightarrow \)  n = 20
\(\sum _{ i=1 }^{ n }{ { x }_{ i }-50\times 20 } =-10\)
\(\Rightarrow \ \ \ \ \ \ \ \sum _{ i=1 }^{ n }{ { x }_{ i } } =990\)
\(\therefore\)  Mean = \(\frac { 1 }{ n } \left( \sum _{ i=1 }^{ n }{ { x }_{ i } } \right) =\frac { 990 }{ 20 } =49.5\)
Hence,  n = 20 and mean = 49.5

\(\frac { 5\cos ^{ 2 }{ { 60 }^{ ° }+4 } \cos ^{ 2 }{ { 30 }^{ ° }-\tan ^{ 2 }{ { 45 }^{ ° } } } }{ \sin ^{ 2 }{ { 30 }^{ ° }+\cos ^{ 2 }{ { 60 }^{ ° } } } } \)
\(=\frac { 5{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }+4{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }.{ \left( 1 \right) }^{ 2 } }{ { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } \)
\(=\frac { \frac { 5 }{ 4 } +3-1 }{ \frac { 1 }{ 4 } +\frac { 1 }{ 4 } } \)
\(=\frac { \frac { 5 }{ 4 } +2 }{ \frac { 1 }{ 2 } } =\frac { \frac { 13 }{ 4 } }{ \frac { 1 }{ 2 } } \)
\(=\frac{26}{4}=\frac{13}{2}\)

(a) \(\frac { x+y-8 }{ 2 } =\frac { x+2y-14 }{ 3 } \)
\(\Rightarrow \quad 3x+3y-24=2x+4y-28\)
\(\Rightarrow \quad x-y+4=0\)           ......(i)
and \(\frac { x+2y-14 }{ 3 } =\frac { 3x+y-12 }{ 11 } \)
\(\Rightarrow \quad 11x+22y-154=9x+3y-36\)
\(\Rightarrow \quad 2x+19y-118=0\)   .....(ii)
Multiply eqn. (i) by 2 and subtract eqn. (ii) from (iii)
2x - 2y + 8 = 0                .....(iii)
\(\Rightarrow\) 2x + 19y - 118 = 0
   \(\underline { -\quad -\quad + } \)
\(-21y+126=0\Rightarrow y=\frac { 126 }{ 21 } \)
\(\therefore\) y = 6
\(\therefore\) from (i), x + 6 - 8 = 0
\(\therefore\) x = 2
\(\therefore\) x = 2 and y = 6
(b) 7(y + 3) - 2(x + 2) =14,
\(\Rightarrow\) 7y + 21 - 2x - 4 = 14
\(\Rightarrow\) 2x - 7y - 3 = 0         .....(i)
and 4(y - 2)+ 3(x - 3) = 2
\(\Rightarrow\) 4y - 8 + 3x - 9 = 2
\(\Rightarrow\) 3x + 4y - 19 = 0     .....(ii)
Multiply equation (i) by 3 and (ii) by 2,
6x - 21y - 9 = 0
6x + 8y - 38 = 0
\(\underline { -\quad -\quad + } \)
- 29y + 29 = 0
\(\Rightarrow\) y = 1
From (i), 2x -7(1) - 3 = 0
\(\Rightarrow\) 2x = 10
\(\therefore\) x = 5
\(\therefore\) x = 5 and y = 1.

By Euclid's division algorithm, for two positive integers a and b, we have
 a =bq + r, 
Let    b=6,
 r = 0, 1, 2, 3,4,5
So         a = 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4,6q + 5
Clearly, a = 6q, 6q + 2,6q + 4 are even, as they are divisible by 2.
But 6q + 1, 6q + 3, 6q + 5 are odd, as they are not divisible by 2. 1
∴  Any positive odd integer is of the form 6q + 1, 6q + 3 or 6q + 5

Given A \(\triangle ABC\) in which \(\angle ABC=\) 90° and AB = BC.\(\triangle ABD\) and \(\triangle ACE\) are equilateral triangles.

To Prove \(ar\left( \triangle ABD \right) =\frac { 1 }{ 2 } ar\left( \triangle CAE \right) \)
Proof Let AB = BC = x units
Now, \(CA=\sqrt { { AB }^{ 2 }+{ AC }^{ 2 } } =\sqrt { { x }^{ 2 }+{ x }^{ 2 } } =x\sqrt { 2 } \) units.
In \(\triangle ABD\) and \(\triangle CAE\), each angle is 60° as they are equilateral triangle.
\(\therefore \triangle ABD\sim \triangle CAE\)
Since, the ratio of the area of two similar triangles is equal to the ratio of the squares of their corresponding sides.
\(\therefore \frac { ar\left( \triangle ABD \right) }{ ar\left( \triangle CAE \right) } =\frac { { AB }^{ 2 } }{ { CE }^{ 2 } } =\frac { { x }^{ 2 } }{ { \left( x\sqrt { 2 } \right) }^{ 2 } } =\frac { { x }^{ 2 } }{ 2{ x }^{ 2 } } =\frac { 1 }{ 2 } \)
Hence, \(ar\left( \triangle ABD \right) =\frac { 1 }{ 2 } ar\left( \triangle CAE \right) \)

100

161.5 cm²

(i) \(\frac { 2 }{ 3 } or-1\)
(ii) \(2or\frac { -10 }{ 9 } \)

\(\frac { 11 }{ 2 } \)

Rs.163.28, Rs.171.13

AB²=(-2-7)²+(5-10)² = (-9)²+(-5)² =81+25 =106
BC²=(3-(-2))²+(-4-5)²=(5)²+(-9)²=25+81=106
AC²=(3-7)²+(-4-10)²=(4)^2₊(14)²=16+196=212
Since AB²+BC²=AC²
∴ ABC is a right triangle.
AB = \(\sqrt { 106 } \) and BC = \(\sqrt { 106 } \)
∵ AB = BC
∴ ABC is an isosceles right triangle. 

In rt. ΔABC, tan60^o = \(\frac { AB }{ BC } =\frac { h }{ x } \)
⇒ \(\sqrt { 3 } =\frac { h }{ x } \Rightarrow h=\sqrt { 3 } x\)
In rt. ∆ABD, tanፀ = \(\frac { AB }{ BD } \)
⇒  tanፀ = \(\frac { h }{ 3x } \)
⇒  tanፀ = \(\frac { \sqrt { 3 } x }{ 3x } =\frac { 1 }{ \sqrt { 3 } } \)⇒ ፀ = 30^o​​​

Here,    AB = AC (Given)      .......(i)
AF = AE (Tangent from A)    ......(ii)
AB - AF = AC - AE
⇒ BF = CE
Now,   BF = BD  (Tangent from B)
Also,  CE = CD  (Tangent from C)
⇒ BD = CD

Factors of x² + 7x + 12 :
x² + 7x + 12 = 0
\(\Rightarrow\)  x² + 4x + 3x + 12 = 0
\(\Rightarrow\)  x(x + 4) + 3 (x + 4) = 0
\(\Rightarrow\)  (x + 4) (x + 3) = 0
\(\Rightarrow\)  x = - 4, - 3                .... (i)
Let p^'(x) = x⁴ + 7x³ + 7x² + px + q
If p(x) is exactly divisible by x² + 7x + 12, then x = - 4 and x = - 3 are zeroes of p(x)  [from eq (i)]
p(x) = x⁴ + 7x³ + 7x² + px + q
p(- 4) = (-4)⁴ + 7(-4)³ + 7(-4)² + p(-4) + q
but p(-4) = 0
\(\therefore\)  0 = 256 - 448 + 112 - 4p + q
\(\Rightarrow\)  0 = - 4p + q - 80
\(\Rightarrow\)  4p - q = 80                       ... (ii)
and  p(-3) = (-3)⁴ + 7 (-3)³ + 7(-3)² + p(-3) + q
but p(-3) = 0
\(\therefore\)   0 = 81 - 189 + 63 - 3p + q
\(\Rightarrow\)  0 = -3 p+ q - 45
\(\Rightarrow\)  3p - q = - 45
On solving eq.(ii) and eq. (iii) by elimination method, we get
4p - q = - 80
3p - q = - 45
p = - 35
On putting the value of p in eq. (i),
4(- 35) - q = - 80
\(\Rightarrow\)  - 140 - q = - 80
\(\Rightarrow\)  - q = 140 - 80
\(\Rightarrow\)  - q = 60
\(\Rightarrow\)   q = - 60
Hence, p = - 35, q = - 60

Given, LCM (26, 91) = 182
\(\therefore \quad HCF(26,91)=\frac { 26\times 91 }{ LCM(26,91) } =\frac { 26\times 91 }{ 182 } =13\)

1.594 m

Let SQ be the height and PQ be the shadow of a man.
According to the question, SQ = PQ

Again, let the angle of elevation of the sun be \(\theta \) 
In right angled \(\\ \Delta PQS\)
\(tan\theta =\frac { perpendicular }{ Base } =\frac { QS }{ PQ }\)
\( \Rightarrow tan\theta =\frac { QS }{ QS } \left[ \because PQ = QS \right]\)
\( \\ \Rightarrow tan\theta =1=tan 45° \left[ \because tan 45°=1 \right] \)
\( \therefore \theta =45°\)
Hence, the angle of elevation of the sun is 45\(°\)

Here,  \({ a }_{ 1 }=-\frac { 1 }{ 2 } ,{ a }_{ 2 }=-\frac { 1 }{ 2 } ,{ a }_{ 3 }=-\frac { 1 }{ 2 } ,{ a }_{ 4 }=-\frac { 1 }{ 2 } \)
Now,   \({ a }_{ 2 }-{ a }_{ 1 }=-\frac { 1 }{ 2 } -\left( -\frac { 1 }{ 2 } \right) =-\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =0\)
 \({ a }_{ 3 }-{ a }_{ 2 }=-\frac { 1 }{ 2 } -\left( \frac { 1 }{ 2 } \right) =-\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =0\)
 \({ a }_{ 4 }-{ a }_{ 3 }=-\frac { 1 }{ 2 } -\left( \frac { 1 }{ 2 } \right) =-\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =0\)
Clearly, the difference of successive terms is same. Therefore, the given list of numbers forms an AP and its common difference,(d) is 0.
Now, next three terms of this AP are
 \({ a }_{ 5 }={ a }_{ 4 }+d=-\frac { 1 }{ 2 } +0=-\frac { 1 }{ 2 } \)
 \({ a }_{ 6 }={ a }_{ 5 }+d=-\frac { 1 }{ 2 } +0=-\frac { 1 }{ 2 } \)
 \({ a }_{ 7 }={ a }_{ 6 }+d=-\frac { 1 }{ 2 } +0=-\frac { 1 }{ 2 } \)

Given equation is 
\(x^{ 3 }-4x^{ 2 }-x+1=(x-2)^{ 3 }\)
\(\Rightarrow x^{ 3 }-4x^{ 2 }-x+1=x^{ 3 }-3(x^{ 2 })2+3x(2)^{ 2 }-(2)^{ 3 }\)
\(\left[ \because (a-b) \right] ^{ 3 }=a^{ 3 }-3a^{ 2 }b+3ab^{ 2 }-b^{ 3 }]\)
\(\Rightarrow x^{ 3 }-4x^{ 2 }-x+1=x^{ 3 }-6x^{ 2 }-12x+8=0\)
\(\Rightarrow 2x^{ 2 }-13x+9=0\)
Which is of the form \(ax^{ 2 }+bx+x=0\) Hence, it is a quadratic equation. where \(a\neq 0\) and b,c are any real numbers.

(i) 28 cm
(ii) 88 cm

Let QS = x; SR = 12 - x
∴  PT = 5 - x; PM = PT
ஃ PM = 5 - x
Also   SR = MR ⇒ MR = 12 - x
Also  PQ² + QR² = PR²
⇒ PR = 13 ⇒ PM + MR = 13
⇒ 5 - x + 12 - x 13 ⇒ 2x  - 4 ⇒ x = 2
Also OSQT is a square
ஃ  OS = QS ⇒ OS = 2 cm
ஃ  Radius of incircle = 2 cm.