\(\sqrt { \left( 6{ x }^{ 2 }+x-1 \right) \left( 3{ x }^{ 2 }+2x-1 \right) \left( 2{ x }^{ 2 }+3x+1 \right) } \)
= \(\sqrt { \left( 3x-1 \right) \left( 2x+1 \right) \left( 3x-1 \right) \left( x+1 \right) \left( 2x+1 \right) \left( x+1 \right) } \) 
= |(3x - 1)(2x + 1)(x + 1)|

\(A=\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]\)
A² = A.A
\(=\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]\)
\(=\left[\begin{array}{rr} 9-1 & 3+2 \\ -3-2 & -1+4 \end{array}\right]=\left[\begin{array}{rr} 8 & 5 \\ -5 & 3 \end{array}\right]\)
\(5 A=5\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]=\left[\begin{array}{rr} 15 & 5 \\ -5 & 10 \end{array}\right]\)
\(7 I_{2}=7\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]\)
\(\therefore A^{2}-5 A+7 I_{2}=\left[\begin{array}{rr} 8 & 5 \\ -5 & 3 \end{array}\right]-\left[\begin{array}{rr} 15 & 5 \\ -5 & 10 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]\)
\(=\left[\begin{array}{cc} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{array}\right]\)
\(=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0\)
Hence Proved.

Let A be the event of getting electrification contract
\(P(A)=\frac{3}{5}\)
Let B be the event of getting plumbing contract
\( P(\bar{B}) =\frac{5}{8} \)
\(1-P(B) =\frac{5}{8} \)
\(P(B) =1-\frac{5}{8}=\frac{3}{8} \)
\(Also\ P(A \cup B)=\frac{5}{7}\)
\(\mathrm{P}(A \cup B)=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B) \)
\(\frac{5}{7} =\frac{3}{5}+\frac{3}{8}-\mathrm{P}(A \cap B) \)
\(\mathrm{P}(A \cap B) =\frac{3}{5}+\frac{3}{8}-\frac{5}{7} \)
\(=\frac{168+105-200}{280}=\frac{73}{280} \)
Probability of getting both contracts \(=\frac{73}{280}\)

When a, b, c are divided by 13 leaves the remainder 9, 7, 10 respectively
a = 13q₁ + 9
b = 13q₂ + 7
c = 13q₃ + 10
Now a + 2b + 3c = (13q₁ + 9) + 2 (13q₂ + 7) + 3 (13q₃ + 10)
= 13q₁ + 9 + 26q₂ + 14 + 39q₃ + 30
= 13 (q₁ + 2q₂ + 3q₃) + 53
= 13 (q₁ + 2q₂ + 3q₃) + (4 x 13 + 1)
= 13 (q₁ + 2q₂ + 3q₃ + 4) + 1
a + 2b + 3c is divided by 13, the remainder is 1.

PQ = 20 m
PX - XQ = 4 m ...(1)

Squaring both sides,
PX² + XQ² - 2PX . QX = 16 (∵ PQ²-2P x QX = 16
400-16 = 2PX x QX
384 = 2PX· QX
\(\therefore { (PX+QX) }^{ 2 }=\underbrace { { PX }^{ 2 }+{ QX }^{ 2 } } +2PX.QX\)
= 400 + 2 x 192
= 784 = 282.
∴ PX + QX = 28
From(1) & (2)² PX = 32 ⇒ PX = 16 m
QX = 12 m
∴ The distance from the two gates to the pole PX and QX is 12m, 16m

Let \(\triangle\)ABC be B triangle points D, E, F are angular bisectors of \(\angle A, \angle B \text { and } \angle C\) respectively. By angular bisector theorem we have
\( \frac{B D}{D C}=\frac{A B}{A C} \Rightarrow \mathrm{AB}=\frac{B D \times A C}{D C} \)
\(\frac{A C}{B C}=\frac{A F}{F B} \Rightarrow \mathrm{AC}=\frac{A F \times B C}{F B} \)
\(\frac{A E}{E C}=\frac{A B}{B C} \Rightarrow \mathrm{AB}=\frac{A E \times B C}{E C}\)
From (1) and (3), we have
\(\frac{B D \times A C}{D C}=\frac{A E \times B C}{E C}\)
Now substituting (2) in (4) we have
\( \frac{B D \times\left(\frac{A F \times B C}{F B}\right)}{D C} =\frac{A E \times B C}{E C} \)
\(\frac{B D \times A F \times B C}{D C \times F B} =\frac{A E \times B C}{E C} \)
\(B D \times A F \times E C =\frac{A E \times B C \times D C \times F B}{B C} \)
\(B D \times A F \times C E =E A \times F B \times D C \)
\(\therefore \frac{B D \times A F \times C E}{E A \times F B \times D C}=1\)
Hence by Ceva's theorem we conclude that the angle bisectors of a triangle are concurrent.

Given equation is y = - 0.1 x + 1 where 'x' is time (in seconds) and 'y' is percentage of megabytes remaining.
Graph of y = -0.1x + 1 
\(y=-\frac{x}{10}+1\)
10y = -x + 10
Points to be plotted

Therefore, \(\sqrt { 64{ x }^{ 2 }-16{ x }^{ 3 }+17{ x }^{ 2 }-2x+1 } \) = |8x² - x + 1|

Clearly the ladder AC make a right triangle with the wall AB force at a distance BC. \(\angle\)B - 90^o

By Pythagoras theorem
AC² = AB² + BC²
5² = 4² + BC²
BC² = 25 - 16
BC² = 9
BC = 3m
If C moves 1.6 m towards the wall BC becomes
3m - 1.6m = 1.4m
Now in \(\triangle\)ABC
AC² = AB² + BC²
5² = AB2 +(1.4)²
25 - 1.96 = AB²
AB² = 23.04
AB = 4.8m
The new height of the wall = 4.8 m
Difference =4.8 - 4 = -0.8m
The ladder would be placed 0.8 m upward the wall.

From the figure,
slope = \(\frac { change\quad in\quad y\quad coordinate }{ change\quad is\quad x\quad coordinate } \)
=\(\frac { 68-32 }{ 20-0 } =\frac { 36 }{ 20 } =\frac { 9 }{ 5 } \)= 1.8
The line crosses the Y axis at (0, 32)
So the slope is \(\frac { 9 }{ 5 } \) and y intercept is 32.

Here the value of n is not given. But the last term is given. From this, we can find the value of n.
Given a = 0.40 and l = 1, we find d = 0.43 - 0.40 = 0.03
Therefore, n = \(\left( \frac { l-a }{ d } \right) +1\)
= \(\left( \frac { 1-0.40 }{ 0.03 } \right) +1=21\)
Sum of first n terms of an A.P S_n = \(\frac { n }{ 2 } \left[ a+l \right] \)
Here, n = 21. Therefore, S₂₁ = \(\frac { 21 }{ 2 } \left[ 0.40+1 \right] =14.7\)
So, the sum of 21 term of the given series is 14,7.

f: [-5,9] ⟶ R
(i) f(-3) + f(2)
= [6(-3) + 1 ] + [ 5(2)² - 1]
= ( -18 + 1) + ( 20 - 1)
 = -17 + 19 = 2.
(ii) f(7) - f(1)
= [ 3(7) - 4 ] - [6(1) + 1 ]
= (21 - 4) - (6 + 1)
=17 - 7 = 10
(iii) 2 f(4) + f(8)
= 2 [ 5(4)² - 1] + [3(8) - 4]
= 2[80 - 1] + [ 24 - 4]
= 158 + 20 = 178
(iv) \(\frac { 2f(-2)-f(6) }{ f(4)+f(-2) } \)
f(-2) = 6x + 1 = 6(-2) + 1 = -11
f(6) = 3x - 4 = 3(6) - 4 = 14
f(4) = 5x² - 1 = 5(4²) - 1 = 79
f(-2) = 6x + 1 = 6(-2) + 1 =-11
\(\frac { 2f(-2)-f(6) }{ f(4)+f(-2) } =\frac { 2(-11)-14 }{ 79+(-11) } =\frac { -22-14 }{ 68 } \)
= \(\frac { -36 }{ 68 } =\frac { -9 }{ 17 } \)

Construction
Step 1 : Draw a line segment QR = 5 cm.
Step 2 : At Q draw QE such that \(\angle\)RQE = 30^o.
Step 3 : At Q draw QF such that \(\angle EQF\) = 90^o
Step 4 : Draw the perpendicular bisector XY to QR which intersects QF at O and QR at G.
Step 5 : With O as centre and OQ as radius draw a circle.
Step 6: From G mark an arc in the line XY at M, such that GM = 42. cm.
Step 7 : Draw AB through M which is parallel to QR.
Step 8 : AB meets the circle at P and S.
Step 9 : Join QP and RP. Then\(\triangle\)PQR is the required triangle

Let the present ages of Vani, her father and her grand father be x, y, z respectively
Given
\(\frac { x+y+z }{ 3 } =53\Rightarrow x+y+z=159\quad \quad ...(1)\)
\(\frac{z}{2}+\frac{y}{3}+\frac{x}{4}=65\)
6z + 4y + 3x = 780          ...(2)
(z - 4) = 4(x - 4)
4x - z = 12       ..........(3)
Consider (1) and (3


Substituting x = 24 in (4)
5(24) + y = 171
y = 171 - 120 = 51
Substituting x = 24, y = 51 in (1)
24 + 51 + z = 159
z = 159 - 75 = 84
Present age of Vani = 24 years
Present age of her father = 51 years
Present age of her grandfather = 84 years.

From the diagram,
The solid is a cuboid' volume of cuboid = length x breadth x height
where l = 24 - 2x, b - 24 - 2x,. h = x
Volume V (x) = (24 - 2x) (24 - 2x) x
V(x) = x(24 - 2x)², x > 0
= 4x³ - 96x² + 576x, x > 0
So, the domain is 0 < x < 12

(i) From the given graph
(a) f(0) = 9
(b) f(7) = 6
(c) = f(2)
(d) = f(10) = 0
(ii) From the graph, it is known that
when x = 9.5, f(x) = 1
(iii) (a) Domain = {x|0 ≤ x ≤ 10, x \(\in \) R}
(b) Range = {x|0 ≤ x ≤ 9, x \(\in \) R}
(iv) The image of '6' under f is '5'.

\(60-t=\frac { { t }^{ 2 } }{ 4 } -3\)
⇒ t²-12 = 240-4t
⇒ t²+4t-252 = 0
⇒ t²+18t-14t-252 = 0
⇒ t(t +18)-14(t +18) = 0
⇒ (t +18)(t-14) = 0
∴ t = 14 or t = -18 is not possible

Let the speed of boat in still water be 'v'
\(\because speed=\frac { distance }{ time } \Rightarrow time=\frac { distance }{ speed } \)
\(\therefore \frac { 36 }{ v-4 } -\frac { 36 }{ v+4 } =\frac { 96 }{ 60 } =\frac { 8 }{ 5 } (\because 1.6hrs=\frac { 96 }{ 60 } )\)
\(\Rightarrow 36(v+4)-36(v-4)=\frac { 8 }{ 5 } (v-4)(v+4)\)
\(\Rightarrow 36v+144-36v+144=\frac { 8 }{ 5 } ({ v }^{ 2 }+4v+4v-16)\)
\(\Rightarrow 288=\frac { 8 }{ 5 } { v }^{ 2 }-\frac { 128 }{ 5 } \Rightarrow 8{ v }^{ 2 }-128=1440\)
\(\Rightarrow 8{ v }^{ 2 }=1568\Rightarrow { v }^{ 2 }=196{ v }^{ 2 }=\pm 14\)
∴ Speed of the boat = 144m/hr.
(∵ speed cannot be -ve)

252525 = 3¹ x 5² x 7¹ x 13¹ x 37¹
363636 = 2² x 3³ x 7¹ x 13¹ x 37¹
H.C.F. = 3¹ x 7¹ x 13¹ x 37¹
= 3 x 3367
= 10101

Given, Speed = 1.2 m/s,
time = 4 seconds
Distance = speed x time
= 1.2 x 4
= 4.8 m
Let x be the length of the shadow after 4 seconds
\(\Delta ABE\sim \Delta CDE,\frac { BE }{ DE } =\frac { AB }{ CD } \) gives \(\frac { 4.8+x }{ x } =\frac { 3.6 }{ 0.9 } =\frac { 3.6 }{ 0.9 } =4\) (since 90 cm = 0.9 m)
4.8 + x = 4x, gives 3x = 4.8 so, x = 1.6m
The length of his DE = 1.6m

To find the H.C.F. of (340 , 412)
412 > 340 = H.C.F. (412, 340)
Using Euclid's division algorithm we have
412 = 340 x 1 + 72
The remainder 72 ≠ 0
Again applying Euclid's division algorithm
340 = 72 x 4 + 52
The remainder 52 ≠  0.
Again applying Euclid's division algorithm
72 = 52 x 1 + 20
The remainder 20 ≠  0.
Again applying Euclid's division algorithm,
52 = 20 x 2 + 12
The remainder 12 ≠  0.
Again applying Euclid's division algorithm.
20 = 12 x 1 + 8
The remainder 8 ≠ 0.
Again applying Euclid's division algorithm
12 = 8 x 1 + 4
The remainder 4 ≠  0.
Again applying Euclid's division algorithm
8 = 4 x 2+0
The remainder is 0.
Therefore H.C.F. of 340 and 412 is 4.

Given A = {x \(\in \) W| x < 2} A = {0,1}
B = {x \(\in \) N| 1 < x ≤ 4} B = {2,3,4}
C = {3,5}
A x (B U C) = (A x B) U (A x C)
\(B\cup C\) = {2,3,4,5}
A x (B U C) = {0,1} x {2,3,4,5}
= {{0,2},(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}          ...(1)
A x B = {0,1} x {2,3,4}
= {(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)}
A x C = {0,1} x {3,5}
= {{0,3},(0,5),(1,3),(1,5)}
\((A\times B)\cup (A\cup C)\) = {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)} ...(2)
From (1) x (2),it is clear that
\(A\times (B\cup C)=(A\times B)\cup (A\times C)\)
Hence verified

Given A = {5,6} , B = {4,5,6} , C = {5,6,7}
L.H.S: A x A = {5,6} x {5,6}
= {(5,6),(5,6),(6,5),(6,6)}          
R.H.S: B x B = {4,5,6} x {4,5,6}
= {(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)}
C x C = {5,6,7} x {5,6,7}
= {(5,5),(5,6),(5,7),(6,5),(6,6),(6,7),(7,5),(7,5),(7,6),(7,7)}
(B x B) ∩ (C x C) = {(5,5),(5,6),(6,5),(6,6)}
LHS = RHS
A x A = (B x B) ∩ (C x C)
Hence proved

To find HCF of three given numbers, first we have to find HCF of the first two numbers.
To find HCF of 396 and 504
Using Euclid’s division algorithm we get 504 = 396  x  1 + 108
The remainder is 108 \(\neq \) 0
Again applying Euclid’s division algorithm 396  = 108  x 3 + 72 
The remainder is 72 \(\neq \) 0
Again applying Euclid’s division algorithm 108  = 72 x 1 + 36 
The remainder is 36 \(\neq \) 0
Again applying Euclid division algorithm 72 = 36 x 2 + 0
Here the remainder is zero. Therefore HCF of 396 , 504 = 36, To find the HCF of 636 and 36
Using Euclid’s division algorithm we get 636 = 36 x 17 + 24
The remainder is 24 \(\neq \) 0
Again applying Euclid's division algorithm 36 = 24 x 1 + 12
The remainder is 12 \(\neq \) 0
Again applying Euclid's division algorithm 24 = 12 x 2 + 0
Here the remainder is zero. Therefore HCF of 636,36 = 12
Therefore Highest Common Factor of 396, 504 and 636 is 12. 

We know that the tangents drawn from are external point to a circle are equal.
Therefore AD AF = x 
BD = BE = y
and CE = CF = z 
Now, AB = 12 cm, BC = 8 cm, and CA = 10 cm.
x + y = 12,y + z = 8 and z + x = 10
(x + y) + (y + z) + (z + x) = 12 + 8 + 10
2(x + y + z) = 30
x+  y + z = 15
Now, x + y = 12 and x + y + z = 15
12 + z  = 15
Z = 3
y + z = 8 and x + y + z = 15
\(x+8=15\Rightarrow x=7\)
and z + x  = 10 and x + y + z = 15
\(10+y=15\Rightarrow y=5\)
Hence, AD = x = 7cm,
BE = y = 5 cm and
CF = z = 3 cm