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Maths. Solutions

The document provides solutions to mathematical equations. In 3 sentences: Part 1 solves various equations for unknown variables, finding x=7 for part a, the remainder is 24 for part b, and (x,y,z)=(-1,3,-1) for part c. Part 2 involves solving equations derived from setting algebraic expressions equal to each other to find values for variables. This includes finding (A,B,C)=(3,2,-1) for part a and (A,B)=(-1,2) for part b. Part 3 calculates areas under curves, finding the area between the given curves and axes is 100 square units for part c.

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35 views8 pages

Maths. Solutions

The document provides solutions to mathematical equations. In 3 sentences: Part 1 solves various equations for unknown variables, finding x=7 for part a, the remainder is 24 for part b, and (x,y,z)=(-1,3,-1) for part c. Part 2 involves solving equations derived from setting algebraic expressions equal to each other to find values for variables. This includes finding (A,B,C)=(3,2,-1) for part a and (A,B)=(-1,2) for part b. Part 3 calculates areas under curves, finding the area between the given curves and axes is 100 square units for part c.

Uploaded by

sabaa.shafeeq
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Solutions of Mathematical Equations.

( 1 ) a . 2x =128 x
⇒2 =
2
7
⇒ x=7 (Ans)

(b) x−2√ 5 x 3−4 x 2−3 x +6 = 5 x 2-6x+9

± 5 x 3 ∓ 10 x 2

6 x 2 - 3x

±6 x 2 ∓ 12 x

9x +6

±9x∓18
24
So, remainder is 24 Ans.

(c). 2x +y+3z= -2 ______ ( i)

x - y - z = -3 _______ (ii)

3x-2y+3z= - 12 _______ (iii)

eq. i+ eq. ii

2x +y+3Z= -2
x - y - z = -3
3x +2z= -5 _______ (iv)

eq. i multiply ✖ 2+ (iii) eq.

4x+2y+6z = -4

3x -2y+ 3z =-12

7x+9z =-16 __________ v


Eq. iv ✖ 7-eq. (v) ✖ 3

21x+14z=-35

±21x±27z=∓48

-13z= 13
⇒ -z =1
⇒ z=-1
Sub. z into (4) eq.

3x+2z=-5
⇒ 3x + 2(-1) =-5
⇒3x-2 =-5
⇒ 3x=-3
⇒ x =-1
Sub. x, z value into eq. (ii)

x-y-z=-3
⇒ -1-y-(-1) = -3 -
⇒ -1-y+1=-3 -
⇒ -y= -3
⇒ y =3
(x, y, z) = (-1, 3, -1) Ans

3. Section B

2
5 x + 7 x+ 8 A Bx+ c
a. 2 = + 2
( x +1)(x +2 x +3) x+1 x +2 x+ 3
2
5 x 2+ 7x+ 8=A ( x + 2 x +3 ¿+(Bx +c ) (x+1)

Expand. RHS and compare co. efficient

5 x 2+ 7x+ 8=A x 2+ 2 Ax+ 3 A+ B x 2+ Bx +cx +c


2
5 x 2+ 7x+ 8= x (A + B)+ x (2 A+ B+C )+3 A+ c

A+B=5 ________ i

2A+B+C=7 _________ii

3A+C=8 _________iii
Eq. ii – eq. i

2A+B+C=7

±A±B =±5

A+ C=2 __________ iv

Eq. iii – eq iv

3A+C=8

±A±C=±2

2A =6
⇒ A =3

Substitute A into Eq. iv

A+C= 2
⇒ 3+C =2
⇒ C = -1

Substitute A, C into Eq. ii

2A+B+C=7
⇒ 2 ✖ 3 +B-1 =7
⇒ 5+B=7
⇒ B =2

(A, B, C) = (3, 2, -1) Ans.

7 x−1
(b). f(x) =
( 1+ 3 x ) ( 3−x )
7 x−1 A B
= +
( 1+ 3 x ) ( 3−x ) 1+ 3 x 3−x
⇒ 7x-1 = A (3-x) + B (1 +3x)

Expand RHS and compare co. efficient

7x-1 = A (3-x) + B (1 +3x)

⇒ 7x-1 = 3A -Ax + B + 3Bx


⇒ 7x-1 = x(-A+3B) + 3A + B
-A + 3B =7 _______ i

3A + B =-1 ________ ii

Eq. i multiply ✖ 3 + eq. ii

-3A + 9B =21

3A + B = -1

10B= 20

⇒B =2

Subt. B into eq. i

-A + 3B =7

⇒-A + 6 =7
⇒-A =7 -6
⇒ A = -1

(A, B) = ( -1, 10) Ans.

2(a). Let f(x) = x 3+c x 2+7x+d

x+2=0
⇒ x = -2

f(-2) =0
⇒ (−2)3 +c (−2)2+7(-2) +d =0 -
⇒8+2c-14+d=0
⇒2c+d-22=0
⇒2c+d=22 ________ i

x-1 = 0

⇒x=1

F(1) = 3
⇒ (1)3+ c(1)2 +7 (1)+ d =3
⇒ 1 +c +7 +d =3

⇒ c+ d =3-8
⇒ c+ d =-5 __________ ii

Eq. i – eq. ii
2c+ d=22

±c± d =∓5

C =27

Subs. c into eq. ii

c+ d =-5

⇒ 27+d =-5
⇒ d =-32

(c, d) = ( 27, -32) Ans.

(ii) f(x) = x 3+c x 2+7x+d =0

By puting value of c and d

f(x) = x 3+27 x 2+7x -32 =0


y x+ 4 y x
(b) 5lo g a -2 lo g a = 2 lo g a + lo g a

⇒5lo g ay -2 lo g ay = lo g xa +2 lo g x+
a
4

y5 y2 x ( x +4 ) 2
⇒ lo g a -2 lo g a = lo g a + lo g a
y5 x
⇒ lo g y2 = lo g (x+ 4)2
a a

3 x
⇒ lo g ay = lo g (x+ 4)2
a
x
⇒ y 3=
( x+ 4 ) 2

⇒ y =3
√3 x
Ans.
√( x + 4 ) 2

i. 2.7 =10 x

5 (a). y=6- x 2

For y intercept x =0

y=6-¿

y=6
Thus (0,6)

For x intercept y =0

0=6- x 2
⇒ x 2= 6
⇒ x=√ 6

X= 2.44

Thus (2.44,0)

y=6- x 2

here a<0

In this case it has a maximum point

All the points are (0,0), (0,6), (2.44,0)

Hence the graph is:

(b)

y=6- x 2

y=2

area for x intercept y= 0


6- x 2 =2
⇒- x 2 =2-6

⇒- x 2 =-4
⇒ x = ±2

X = (2, -2)

for y intercept x= 0

y = 6- x 2

y=6-0

y =6

y= (0,6)

when x=2 when x = -2

y = 6- x 2 y = 6- x 2
2
y=6-4 y = 6-(−2)

y=2 y=2

(2,2) (-2,2)

∫ (6−x 2 ¿ )¿ dx
−2

[ ]
2
x3
= 6 x−
3 −2

[ ] [ ]
3
2 8
= 6.2− - ¿ = 12− -
3 3

[−12−
−8
3 ] = [ ]
36−8
3

- [
−36 +8
3 ]
28 −28
= -
3 3
28 28
= +
3 3
28+28
=
3
56
= square units.
3
(c)
3

∫ 3 x 3 +6 x+ 8
1

= ∫ ¿ ¿) dx
1

[ ]
3
3 x4 6 x2
= + +8 x
4 2 1

[ ]
3
3 x4
= 2
+3 x +8 x
4 1

=¿ - ¿

=[ 243
4 ][
3
+27+ 24 - +3+8
4 ]
=[ 243+108 +96
4
-][
3+12+32
4 ]
447 47
= -
4 4
447−47
=
4
400
=
4
=100 Ans.
3
x
(1) ∫ dx
√ 1−x 4

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