Search This Blog

Thursday 4 April 2024

MATHEMATICS I - FOR CLASS IX (Science Group) - TARGET PAPER 2024 - By Sir Sajjad Akber Chandio

GO TO INDEX
Mathematics I
For Class IX (Science Group)
Target Paper 2024

By Sir Sajjad Akber Chandio

Q.1: If Z1 = − 4 + 6𝑖 and Z2 = 5/2 2i, then verify Z1 + Z2 = Z1 + Z2.
OR If Z1 = 2 − 5𝑖  and Z2 = 2 + 3i. Verify that(Z1 Z2) = Z1.Z2.

Q.2: Find the value of with the help of logarithmic table:
a) 85.7 x 2.47/8.89 OR 85.7 x 2.47/8.89
b)431.5 x (1.2)23 36.98  OR431.5 x (1.2)/ 3 36.98
c) (86.2)2 x (37.37) 591  OR (86.2)2 x (37.37) / 591
d) 99.878.369 X 0.785  OR 99.87 / 8.369 X 0.785

Q.3: Find the value of:
1) a2 + b2 = ? , a + b = -9, a - b = 5
2) xy = ? , x - y = 1 , x + y = 9
3) x2 + 1/x2 , if x + 1/x = 3
4) x + y + z = 8 , xy + yz + zx = ? , x2 + y2 + z2 = 30.
5) x3 + y3, if x + y = 6, xy = 10
6) b3 + 1/b3 = ?, b + 1/b = 3
7) l3 + m3 + n3 - 3lmn, when l + m + n =15, lm + mn + nl = 74
8) Find the value of a2 + b2 + c2, when a + b + c = 1/3 and ab+bc + ca = − 2/9

Q.4: (Remainder Theorem)
  • a) x3 + 8x2 + 19x + 12
  • b) x3 - 21x + 20
  • c) x3 + 7x2 - x - 2
  • d) 6x3 + 7x2 – x – 2
e) Find the square root of the expression 25x4 + 40x3 + 26x2 + 8x + 1 by division method.

Q.5: (Perfect Square)
  • a) x4 + 4x3 + 10x2 + ax + b, find a, b
  • b) 4x4 + 12x3 + 25x2 + 24x + q, Find q.
  • c) What should be added to 4x4 + 4x3 + 17x2 + 8x+ 9 to make a perfect square?

Q.6: Solve graphically the following simultaneous linear equation:
  1. 3x - 11 = y, x - 3y = 9
  2. y = 3x – 5, x + y = 11
  3. 2y - 3x = 12, x + 6 = y
  4. x + 3 = 2y, 2x+y=14
  5. 3π‘₯ = 5𝑦 – 2, 3π‘₯ + 5𝑦 = 8

Q.7: Solve it with the help of quadratic formula:
a) x2 - x - 56 = 0,
b) 10x2 + 19x - 15 = 0
c) π‘₯2 = −π‘₯ + 1

Q.8: Find the solution set of the following equation. Also verify the answer:
1: √2x + 11 = √3x + 7
2: 6/2x – 5 - 4/x - 3 = 0   OR 6/2x-5  -  4/x-3  =0
3: - 6 + |5x - 3| = 3
4: |2x + 3/5| + 2 = 7 OR |2x+3/5|  +  2 = 7
5: 10√x + 20 = 100

Q.9: Simplify:
i) 2-6i/3-i = 4 + i/3 + i
ii) x + 3 + √x - 3 /x + 3 - √x - 3  ORx + 3 + √x - 3/x + 3 - √x - 3

Q.10: The opposite side of parallelogram is congruent. Prove it.

Q.11: construct triangle mPQ = 6 cm m PR = 5cm m∠Q= 50° write down its construction.

Q.12: Find out the distance between the point:
1: P(2,3) and Q(- 4,5)
2: L(0,√3), M(-1, 0), N(1,0) equilateral triangle.
3: Find the perimeter of the triangle formed by the points A(0,0), B(4,0) and C(2, 2√3)
4: Do the point A(-1, 2), B(7, 5) and C(2,-6 ) from a right angled triangle

Q.13: Factorization:
1: 108 x4 – 256xz3
2: a3b3 + 27b6
3: x12 – y12
4: a6 – b6
5: 1000 – x3y3/125 OR 1000 x3y3/125
6: 8x6 – 1/729 OR 8x6 1/729
7: (π‘₯2 + 2π‘₯𝑦 + 𝑦2) − 9𝑧4
8: π‘Ž8 + π‘Ž4 + 1
9: 8π‘₯3 + 36π‘₯2 + 54π‘₯ + 27
10: π‘Ž6 + 1
11: 3π‘₯2 − 38 π‘₯𝑦 − 13𝑦2
12: 4𝑑4 + 1

Q.14: Construct a Ξ”STU in which m∠T = 60° ,m∠U = 30° and mTU = 7cm.

THEOREM

  1. If two sides of a triangle are unequal in length, the longer side has an angle of greater measure opposite to it. Prove it.
  2. Parallelograms on the same base and lying between the same parallel lines(or of the same altitude) are equal in area. Prove it.
  3. Any point on the right bisector of a line segment is equidistant from its end points. Prove it.
  4. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and it is equal to one half of its length. Prove it.
  5. In any triangle, the square on the side opposite to an acute angle is equal to sum of the squares on the sides containing that acute angle diminished by twice the rectangle contained by one of those sides and the projection on it of the other. Prove it.
LONG:
  1. If in the correspondence of two right-angled triangle triangles, the hypotenuse and one side of one are congruent to the hypotenuse and the corresponding sides of the other, then the triangles are congruent. Prove it.
  2. In a corresponding of two triangle if three sides of one triangle are congruent to the corresponding three sides of the other then the two triangles are congruent. Prove it.




TARGET PAPER 2023

SECTION 'B' (40%)
SHORT ANSWER QUESTION (30 MARKS)


Q.10: Find the number of digits in 725. (By using logarithms)
Q.11: Find the value of a2 + b2 + c2, when a+ b + c = 1/3 and ab + bc + ca = - 2/9
Q.12: Find the value of a2 + b2 + c2, when a+ b + c = 9 and ab + bc + ca = 13
Q.13: Find the value of a2 + b2 and ab, when a+ b = 5 and a - b = 3
Q.14: Find the factors by using factor theorem. 6x3 + 7x2- x - 2
Q.15: Find the square root of the expression 25x4 + 40x3 + 26x2+ 8x + 1 by division method.
Q.16: By using synthetic division method, divide P(x) = x3 - 6x2 - 11x - 6 by (x+2) and also find the quotient and remainder.
Q.17: Factoriie the following. (x2 + 5 + 6)(x2 + 5x + 4) - 3 Q.18 Find the HCF of the following expressions by factorization method 4x2 - 9 and 2x2 - 5x + 3 Q.19: Solve the equation by using quadratic formula:
(i) x2 = - x + 1
(ii) x2 - 2x = 15
Q.20: For what value of 'm', 9x4 12x3 + 34x2 + mx +25 will be a perfect square?
Q.21: Find the solution set of the equation and also verify the answer. 10 √x + 20 = 100
Q.22: When 25 is added to a number. the result halved, the answer is 3 times the original number. What is the number?
Q.23 Find the solution set of the following equation and also n verify the answer.
4x + 5 = √3x - 7
Q.24: Find the LCM of the follow ing expressions by factorization method:
21x2 - 14x and 3x2 - 5x -2
Q.25: The centre of a circle is (3, 4) and one of its end points of a diameter as (4, 6). Find the other end point.
Q.26: Find the solution set of 2x-7 > 6+x. ∀x ∈ N. also represent on number line.
Q.27: Draw a Ξ”ABC by joining the points A(-1, 4), B(-3, -6) and C (3, -2) on a graph paper.
Q.28: If two sides of a triangle are unequal in length, the longer side has an angle of greater measure opposite to it. Prove it.
Q.29: Parallelograms on the same base and lying between the same parallel lines (or of the same altitude) are equal in area. Prove it.
Q.30: Any point on the right bisector of a line segment is equidistant from its end points. Prove it.
OR Any point on the bisector of an angle is equidistant from its arms. Prove it.
OR Construct a Ξ”STU in which m∠T = 60°, m∠U = 30° and mTU = 7cm.
Q.31: Construct the Ξ”DEF such that: to mDE = 5cm, m∠D = 45° and m∠E = 60° and also write the steps of constructions.
Q.32: Construct the Ξ”ABC such that, mAB = mAC = 5.1cm, m∠A = 65° and also write the steps of constructions.


SECTION 'C' (40%)
DESCRIPTIVE ANSWER QUESTION (30 MARKS)

Q.33: Factorize the following. (any four)
  1. (x2 + 2xy + y2) - 9z4
  2. a8+ a4 + 1
  3. 8x3 + 36x2 + 54x + 27
  4. a6 + 1 OR a6 - b6
  5. 3x2 - 38 xy - 13y2
  6. 4t4 + 1
  7. 4a2 + 12ab + 9b2
  8. (4a2 + 8ab + 4b2) - 9c2 
  9. x4+4y4 
  10. y2 + 7y - 98
Q.34: Find the solution set of the following simultaneous equations by graphical method. (Find four ordered pairs for each equation).
(i) 3x = 5y - 2 ; 3x + 5y = 8
(ii) x + y = 4; 2x - 1 - 5y
Q.35: If in the correspondence of two right angled triangle, the hypotenuse and one side of one are congruent to the hypotenuse and the corresponding side of the other, then the triangle are congruent (H.S ≅ H.S). Prove it.
OR
In a correspondence of two triangles, if three sides of e triangle are congruent to the corresponding three sides of the other then, prove that the two triangles are congruent. Q.36: The line segment joining the mid-points of two sides of a triangle, is parallel to the third side and it is equal to one half of its length. Prove it.
Q.37: Using distance formula, find the perimeter of the triangle formed by the points A(0,0), B(4,0) and C(2, 2√3)
OR
Show that the points P(-3, -4), Q(2, 6) and R(0, 2) are collinear (by using distance formula).
Q.38: Take a Ξ”, draw the internal bisectors of the angles and prove that they are concurrent and also write the steps of construction.
OR
In any triangle, the square on the side opposite to an acute angle is equal to sum of the squares on the sides containing that acute angle diminished by twice the rectangle contained by one of those sides and the projection on it of the other. Prove it.

MATHEMATICS Solved Target PAPER 2022



SOLUTION

Q1: If Z1 = 2 - 5i and Z2 = 2 + 3i, then verify Z1 . Z2 = Z .  Z2
Solution:
Z1 = 2 - 5i
Z2 = 2 + 3i

L.H.S
Z1 . Z
= (2 - 5i)(2 + 3i)
= 2(2 + 3i) - 5i(2 + 3i)
= 4 + 6i - 10i -15i2
= 4 - 4i - 15i2
= 4 - 15(-1) - 4i (As i = √ -1)
= 4 + 15 - 4i
= 19 - 4i ..... (A)
Taking conjugate
Z1 . Z2=Z1 = 19 - 4i
= 19 + 4i


Now Solve
R.H.S
Z1  .  Z2
= 2 - 5i . 2 + 3i
Taking conjugate
= (2 + 5i)(2 - 3i)
= 2(2 - 3i) + 5i(2 - 3i)
= 4 - 6i + 10i - 15i2
= 4 + 4i - 15i2
= 4 - 15(-1) + 4i (As i = √ -1)
= 4 + 15 + 4i
= 19 + 4i ..... (B)
L.H.S = R.H.S (Hence proved)


Q.1: Find the value of with the help of logarithmic table.
a) 85.7 x 2.47 / 8.89
b) √ 431.5 x (1.2)2 /  336.98
c) (86.2)2 (37.37) / 591


b) √ 431.5 x (1.2)2 /  336.98

c) (86.2)2 (37.37) / 591

Q.3: Find the value of:
  1.  a2 + b2 = ? , a + b = -9, a - b = 5
  2.  xy = ? , x - y = 1 , x + y = 9
  3.  x2 + 1 / x2; x + 1 / x = 3
  4. x + y + z = 8 , xy + yz + zx = ? , x2 + y2 + z2 = 30
  5.  x3 + y3, if x + y = 6, xy = 10 
  6. b3 + 1 / b3 = ?, b + 1 / b = 3 
  7. l3 + m3+ n3 - 3lmn, when l + m + n =15, lm + mn + nl = 74

1. a2 + b2 = ?, a + b = -9, a - b = 5
Solution:
Data:
a + b = -9
a - b = 5
a2 + b2 = ?

Using Formula
(a + b)2 + (a - b)2 = 2(a2 + b2)

⇒ (-9)2 + (5)2 = 2(a2 + b2)
⇒ 81 + 25 = 2(a2 + b2)
⇒ 106 = 2(a2 + b2)
106 / 2 = (a2 + b2)
⇒ 53 = (a2 + b2)
⇒ a2 + b2) = 53 Ans

2. xy = ? , x - y = 1 , x + y = 9
Solution:
Data:
xy = ?
x - y = 1
x + y = 9

Using Formula
4ab = (a + b)2 - (a - b)2

⇒ 4xy = (x + y)2 - (x - y)2
⇒ 4xy = (9)2 - (1)2
⇒ 4xy = 81 - 1
⇒ 4xy = 80
⇒ xy = 80 / 4
⇒ xy = 20 Ans.

3. x2 + 1 / x2, if x + 1 / x = 3

4. x + y + z = 8 , xy + yz + zx = ? , x2 + y2 + z2 = 30.
Solution:
Data:
xy + yz + zx = ?
x + y + z = 8
x2 + y2 + z2 = 30

Squaring on both side
⇒ (x + y + z)2 = (8)2

Using Formula
(a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ac)

⇒ x2 + y2 + z2 + 2 (xy + yz + zx) = 64
x2 + y2 + z2 + 2 (xy + yz + zx) = 64
⇒ 30 + 2 (xy + yz + zx) = 64
⇒ 2 (xy + yz + zx) = 64 -30
⇒ 2 (xy + yz + zx) = 34
⇒ xy + yz + zx = 34 / 2
⇒ xy + yz + zx = 17 Ans

5. x3 + y3, if x + y = 6, xy = 10
Solution:
Data:
x3 + y3 = ?
x + y = 6
xy = 10

By cubing on both side
⇒ (x + y)3 = (6)3

Using Formula

(a + b)3 = a3 + 3ab(a + b) + b3

⇒ x3 + 3xy(x + y) + y3 = 216
⇒ x3 + 3(10)(6) + y3 = 216
⇒ x3 + 180 + y3 = 216
⇒ x3 + y3 = 216 - 180
⇒ x3 + y3 = 36 Ans

6. b3 + 1 / b3 = ?, b + 1 / b = 3


7. l3 + m3+ n3 - 3lmn, when l + m + n =15, lm + mn + nl = 74
Solution:
Data:
l + m + n =15
lm + mn + nl = 74
l3 + m3+ n3 - 3lmn = ?

Now by applying formula
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac

Squaring on both side
⇒ (l + m + n)2 = (15)2
⇒ l2 + m2 + n2 + 2lm + 2mn + 2nl = 225
⇒ l2 + m2 + n2 + 2(lm + mn + nl) = 225
⇒ l2 + m2 + n2 + 2(74) = 225
⇒ l2 + m2 + n2 + 148 = 225
⇒ l2 + m2 + n2 = 225 - 148
⇒ l2 + m2 + n2 = 77

Using Formula:
a3 + b3+ c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ac)
⇒ l3 + m3+ n3 - 3lmn = (l + m + n)(l2 + m2 + n2 - lm - mn - nl)
⇒ l3 + m3+ n3 - 3lmn = (15)(77 - (lm + mn + nl)
⇒ l3 + m3+ n3 - 3lmn = (15)(77- 74)
⇒ l3 + m3+ n3 - 3lmn = (15)(3)
⇒ l3 + m3+ n3 - 3lmn = 45 Ans.


Q.4: (a) x3 + 8x2 + 19x + 12 (Remainder Theorem)
(b) x3 - 21x + 20 (Remainder Theorem)
(c) x3 + 7x2 - x - 2 (Remainder Theorem)

(a) x3 + 8x2 + 19x + 12 (Remainder Theorem)
Solution:
Let P (x) = x3 + 8x2 + 19x + 12
The division of 12 are +1, +2, +3, +4, +6. +12
put x =1
= 13 + 8(1)2 + 19(1) + 12
= 1 + 8 + 19 + 12 = 40
∴ (x - 1) is not be the the factor of p (x)

put x = -1

= -13 + 8(-1)2 + 19(-1) + 12
= -1 +8 - 19 +12 = - 20 + 20 = 0
∴ x + 1 is one of the factor of p (x) now by dividing p (x) with (x + 1)


∴ x3 + 8x2 + 19x + 12 = (x + 1) (x2 + 7x +12)
By applying Mid term breaking formula
= (x + 1) (x2 + 4x + 3x +12)
= (x + 1) (x + 4) (x + 3) Ans.

(b) x3 - 21x + 20 (Remainder Theorem)
Solution:
Let P (x) = x3 - 21x + 20
The division of 20 are +1, +2, +4, +5, +20
put x = 1
= 13 - 21(1) + 20
= 1 - 21 + 20 = - 21 + 21
= 0
∴ x - 1 is one of the factor of p (x) now by dividing p (x) with (x - 1)


∴ x3 - 21x + 20 = (x - 1) (x2 + x - 20)
= (x - 1) (x2 + x - 20)
By applying Mid term breaking formula
= (x - 1) (x2 + x - 20)
= (x - 1) {x2 + 5x - 4x - 20}
= (x - 1) {x(x + 5) - 4(x + 5)}
= (x - 1) (x + 5) (x _ 4) Ans.

(c) x3 + 7x2 - x - 2 (Remainder Theorem)

Q.5: (a) x4 + 4x3 + 10x2 + ax + b, Find a, b (Perfect Square)
(b) 4x4 + 12x3 + 25x2 + 24x + q, Find q


(a) x4 + 4x3 + 10x2 + ax + b, Find a, b (Perfect Square)


(b) 4x4 + 12x3 + 25x2 + 24x + q, Find q.

Q.6: Solve graphically the following simultaneous linear equation.
  1. 3x - 11 = y; x - 3y = 9

1. 3x - 11 = y; x - 3y = 9
2.
3.





Q.7: (a) x2 - x - 56 = 0,
(b)
Solve it with the help of quadratic formula.



Q.8: Find the solution set of the following equation. And also verify the answer.

  1. 6/2x-5  - 4/x-3  =  0
  2. - 6 + |5x - 3| = 3



2. 

3. 





No comments:

Post a Comment