Mathematics Past Papers PDF -

Maths-10 | Unit-1 (Quadratic Equations)

1- Circle the correct answer. (8×1=8)

اگلیں۔ ہر جواب کے گرد دائرہ لگائیں۔

Q1. Solution set of equation $5×2−125=05x^2 – 125 = 0$ is:
اسموات $5×2−125=05x^2 – 125 = 0$ کا حل سیٹ ہے:
Answer: {±5}

Q2. Standard form of $(x + 7) (x - 3) = - 7$ is:
$(x + 7) (x - 3) = - 7$ کی معیاری شکل ہے:
Answer: $x^2 + 4x – 14 = 0$

Q3. Equation $3^x + 3^{2-x} + 6 = 0$ is:
اسموات کس قسم کی ہے؟
Answer: Exponential Equation / قوت نمائی اسموات

Q4. $x+3=x+1\sqrt{x+3} = x+1$ is an example of:
یہ کس قسم کی اسموات ہے؟
Answer: Radical Equation / جذری اسموات

Q5. Extraneous root of $x+3=x+1\sqrt{x+3} = x+1$ is:
اسموات کا اضافی روٹ ہے:
Answer: $x = - 2$

Q6. Number of terms in $ax^2 + bx + c = 0$ :
معیاری دو درجی اسموات میں رقموں کی تعداد:
Answer: 3

Q7. Quadratic formula is:
دو درجی فارمولا ہے:
Answer: $\frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

Q8. Equation that remains unchanged when $x \to 1/ x$ :
وہ اسموات جو تبدیل نہ ہو:
Answer: Reciprocal Equation / معکوس اسموات

2- Short Questions (7×2=14)

درج ذیل سوالات کے مختصر جوابات لکھیں۔

Q1. Write in standard form:
$(x + 7) (x - 3) = - 7$
Answer: $x^2 + 4x – 14 = 0$

Q2. Solve by completing square:
$x^2 – 3x – 4 = 0$
Answer: $x = 4, - 1$

Q3. Solve using quadratic formula:
$1x−6−3x−5=4x\frac{1}{x-6} – \frac{3}{x-5} = \frac{4}{x}$
Answer: $x = 2, 3$

Q4. Solve using quadratic formula:
$6×2−7x−3=06x^2 – 7x – 3 = 0$
Answer: $\frac{3}{2}, -\frac{1}{3}$

Q5. Solve:
$\frac{1}{2} + \frac{1}{4}$
Answer: $\frac{15}{20} = \frac{3}{4}$

Q6. Solve:
$x+3=3x−1\sqrt{x+3} = 3x – 1$
Answer: $x = 1$

Q7. Solve by factorization:
$25×2−1=025x^2 – 1 = 0$
Answer: $\pm \frac{1}{5}$

3- Long Question (1×8=8)

درج ذیل سوال حل کریں۔

Q3(a)

Solve by completing square:
$7(x + 2a)^2 + 3a^2 = 5a(7x – 23a)$

Answer (Steps):
Expand both sides:
$7(x^2 + 4ax + 4a^2) + 3a^2 = 35ax – 115a^2$

Simplify:
$7×2+28ax+28a2+3a2=35ax−115a27x^2 + 28ax + 28a^2 + 3a^2 = 35ax – 115a^2$

$7×2+28ax+31a2=35ax−115a27x^2 + 28ax + 31a^2 = 35ax – 115a^2$

Bring all terms:
$7×2−7ax+146a2=07x^2 – 7ax + 146a^2 = 0$

Divide by 7:
$x2−ax+1467a2=0x^2 – ax + \frac{146}{7}a^2 = 0$

Solve ⇒
Answer: $\frac{a \pm \sqrt{a^2 – \frac{584}{7}a^2}}{2}$

Q3(b)

Solve:
$x2+x+1−x2+x−1=1\sqrt{x^2 + x + 1} – \sqrt{x^2 + x – 1} = 1$

Answer (Steps):
Let:
$\sqrt{x^2 + x + 1}$ ,
$\sqrt{x^2 + x – 1}$

Given:
$A - B = 1$

Square:
$A^2 + B^2 – 2AB = 1$

Substitute:
$x^2+x+1)+(x^2+x-1) – 2AB = 1$

$2×2+2x−2AB=12x^2 + 2x – 2AB = 1$

Solve further ⇒
Final Answer: $x = 1$

Maths-10 | Unit-2 (Theory of Quadratic Equations)

1- Circle the correct answer. (8×1=8)

اگلیں۔ ہر جواب کے گرد دائرہ لگائیں۔

Q1. Cube roots of 8 are:
8 کے جزر المکعب ہیں:
Answer: $2\omega, 2\omega^2$

Q2. $x^3 + y^3 = (x + y)\, ?$
Answer: $\omega y)(x + \omega^2 y)$

Q3. Sum of roots of $4×2−3x+6=04x^2 – 3x + 6 = 0$ :
روٹس کا مجموعہ ہے:
Answer: $3/4$

Q4. If $α,β\alpha, \beta$ are roots of $4×2−5x+6=04x^2 – 5x + 6 = 0$ , then $1/α+1/β=1/\alpha + 1/\beta =$
Answer: $5/6$

Q5. If $α,β\alpha, \beta$ are roots of $2×2−3x−5=02x^2 – 3x – 5 = 0$ , then $αβ=\alpha\beta =$
Answer: $- 5/2$

Q6. Remainder of $x^2 + 7x – 1) ÷ (x + 1)$ :
Answer: $- 7$

Q7. Two consecutive positive numbers whose product is 182:
Answer: 13, 14

Q8. Nature of roots is determined by:
Answer: Discriminant / فرق کنندہ

2- Short Questions (7×2=14)

درج ذیل سوالات کے مختصر جوابات لکھیں۔

Q1. Find discriminant of equation:
$2×2−7x+1=02x^2 – 7x + 1 = 0$
Answer:
$D = b^2 – 4ac = (-7)^2 – 4(2)(1) = 49 – 8 = 41$

Q2. Simplify:
$\sqrt{-3})^6 + (−1 − \sqrt{-3})^6$
Answer: $2$

Q3. Without solving, find sum and product of roots:
$7×2−5mx+9n=07x^2 – 5mx + 9n = 0$
Answer:
Sum = $5 m /7$
Product = $9 n /7$

Q4. If $α,β\alpha, \beta$ are roots of $lx^2 + mx + n = 0$ , then find:
Answer:
$α+β=−m/l\alpha + \beta = -m/l$
$αβ=n/l\alpha\beta = n/l$

Q5. Write quadratic equation with roots $\sqrt{2}, 3 – \sqrt{2}$ :
Answer:
Sum = 6, Product = 7
Equation: $x^2 – 6x + 7 = 0$

Q6. Define synthetic division:
Answer:
Synthetic division is a shortcut method to divide a polynomial by a linear factor.
مصنوعی تقسیم ایک مختصر طریقہ ہے جس سے کثیر رکنی کو خطی عامل سے تقسیم کرتے ہیں۔

Q7. Solve equations:
$x^2 + y^2 = 7$
$2×2+3y2=182x^2 + 3y^2 = 18$

Answer:
Multiply first by 2:
$2×2+2y2=142x^2 + 2y^2 = 14$

Subtract:
$(2×2+3y2)−(2×2+2y2)=18−14(2x^2 + 3y^2) – (2x^2 + 2y^2) = 18 – 14$
$y^2 = 4 → y = ±2$

Put in first:
$x2+4=7→x2=3→x=±3x^2 + 4 = 7 → x^2 = 3 → x = ±\sqrt{3}$

3- Long Question (1×8=8)

درج ذیل سوال حل کریں۔

Q3(a)

Show that roots are rational:
$a(b – c)x^2 + b(c – a)x + c(a – b) = 0$

Answer (Steps):
Compare with $ax^2 + bx + c$

Discriminant:
$D = b^2(c-a)^2 – 4a(b-c)c(a-b)$

After simplification, $D$ becomes a perfect square
⇒ Roots are rational

Q3(b)

Prove that:
$x3+y3+z3−3xyz=(x+y+z)(x+ωy+ω2z)(x+ω2y+ωz)x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z)$

Answer (Proof idea):
Use identity:
$x^3 + y^3 + z^3 – 3xyz = (x+y+z)(x^2+y^2+z^2 – xy – yz – zx)$

Then factor second part using cube roots of unity:
$ω,ω2\omega, \omega^2$

Hence proved.

📘 Maths-10 | Unit-3 (Variations / تغیرات)

1- Circle the correct answer. (8×1=8)

اگلیں۔ ہر جواب کے گرد دائرہ لگائیں۔

Q1. Value of x in proportion 15:7::x:56 is:
تناسب 15:7::x:56 میں x کی قیمت ہے:
Answer: 120
(حل: $\times 56 ⇒ x = 120$ )

Q2. If $3 (4 x - 5 y) = 2 x - 7 y$ , then $x : y =$
اگر $3 (4 x - 5 y) = 2 x - 7 y$ ہو تو $x : y$ کیا ہے؟
Answer: 4:5
(حل: $12 x - 15 y = 2 x - 7 y \Rightarrow 10 x = 8 y \Rightarrow x / y = 4/5$ )

Q3. If $x /6 = 7/24$ , then value of $4 x$ is:
اگر $x /6 = 7/24$ ہو تو $4 x$ کی قیمت ہے:
Answer: 7
(حل: $x = 7/4 \Rightarrow 4 x = 7$ )

Q4. If $w ∝ u^3$ and $w = 81$ when $u = 3$ , then k is:
اگر $w ∝ u^3$ اور $w = 81$ , $u = 3$ ہو تو $k$ ہے:
Answer: 3
(حل: $81 = k (27) \Rightarrow k = 3$ )

Q5. If $A ∝ 1/r^2$ and $A = 2$ when $r = 3$ , find A when $r = 6$ :
Answer: 1/2
(حل: $k = 18 \Rightarrow A = 18/36 = 1/2$ )

Q6. If 12, p, 3 are in continued proportion, then p is:
اگر 12, p, 3 مسلسل تناسب میں ہوں تو p ہے:
Answer: ±6
(حل: $p^2=36 ⇒ p=±6$ )

Q7. Fourth proportional w in $x : y :: v : w$ is:
Answer: $v y / x$

Q8. If $a : b = x : y$ , then by alternendo theorem:
Answer: $a / x = b / y$

2- Short Questions (7×2=14)

درج ذیل سوالات کے مختصر جوابات لکھیں۔

Q1. In proportion $8 - x : 11 - x :: 16 - x : 25 - x$ , find x.
Answer: $x = 1$

Q2. If $y \propto x$ and $y = 7$ when $x = 3$ , find x when $y = 35$ .
Answer: $x = 15$

Q3. Find mean proportional of 20 and 45.
Answer: ±30

Q4. If $a : b = c : d$ , prove that:
$4a−5b4a+5b=4c−5d4c+5d\frac{4a-5b}{4a+5b} = \frac{4c-5d}{4c+5d}$
Answer:
Use componendo-dividendo theorem.

Q5. If $x ∝ y^3$ and $y = 3$ when $x = 81$ , find y when $x = 3$ .
Answer: $y = 1$

Q6. If $a : b = c : d$ , show that:
$p a + q b : ma - nb = p c + q d : m c - n d$
Answer:
Let $a / b = c / d = k$ , substitute and prove.

Q7. Surface area $S ∝ r^2$ . If $S = 16 π$ when $r = 2$ , find r when $S = 36 π$ .
Answer: $r = 3$

3- Long Question (1×8=8)

درج ذیل سوال حل کریں۔

Q3(a)

The intensity of light varies inversely as square of distance.
If $I = 20$ at $d = 12$ , find intensity at $d = 8$ .

روشنی کی شدت فاصلے کے مربع کے الٹ متناسب ہے۔

Answer: 45 candlepower

(حل:
$k = I × d^2 = 20 × 144 = 2880$
$I = 2880/64 = 45$ )

Q3(b)

Solve using componendo-dividendo theorem:

Find s if:

$\frac{6pq}{p – q}$

Answer:

$s−3ps+3p+s−3qs+3q=2\frac{s-3p}{s+3p} + \frac{s-3q}{s+3q} = 2$

(یہ ایک ثابت کرنے والا سوال ہے جس میں دی گئی ویلیو رکھ کر نتیجہ حاصل کیا جاتا ہے)

Maths-10: Test Paper (Unit-4)

Syllabus: Partial Fractions (جزوی کسریں)

1- Circle the correct answer. (8×1=8)

Q1. Which of the following is NOT an improper fraction?
ان میں سے کون سی غیر واجب کسر نہیں ہے؟
Answer: (A) $7x−9(x+1)(x−3)\frac{7x-9}{(x+1)(x-3)}$

Q2. Multiplication factors of $x^3+1$ are:
$x^3+1$ کے ضربی اجزاء ہیں:
Answer: (C) $x+1)(x^2 – x +1)$

Q3. Partial fractions of $x2(x+2)(x2+4)\frac{x^2}{(x+2)(x^2+4)}$ are of the form:
جزوی کسر کی شکل ہوگی:
Answer: (C) $Ax+2+Bx+Cx2+4\frac{A}{x+2} + \frac{Bx+C}{x^2+4}$

Q4. $x2+1×3+1\frac{x^2+1}{x^3+1}$ is a/an fraction.
یہ کسر ہے۔
Answer: Proper Fraction (واجب کسر)

Q5. $2x+1(x+1)(x−1)\frac{2x+1}{(x+1)(x-1)}$ is a/an .
یہ ایک ہے۔
Answer: Proper Fraction (واجب کسر)

Q6. $x+3)^2 = x^2+6x+9$ is a/an .
یہ ایک ہے۔
Answer: Identity (مماثلت)

Q7. $x3+1(x−1)(x+2)\frac{x^3+1}{(x-1)(x+2)}$ is a/an .
یہ ایک ہے۔
Answer: Improper Fraction (غیر واجب کسر)

Q8. Partial fraction of $x2+1(x+1)(x−1)\frac{x^2+1}{(x+1)(x-1)}$ is of the form:
جزوی کسر کی شکل ہوگی:
Answer: (D) $Ax+Bx+1+Cx−1\frac{Ax+B}{x+1} + \frac{C}{x-1}$

2- Answer the following questions. (7×2=14)

i. Define fraction and give an example.
کسر کی تعریف کریں اور مثال دیں۔

Answer:
A fraction is a ratio of two algebraic expressions or numbers.
Example: $x+1x\frac{x+1}{x}$

ii. Resolve $3x−25(x−4)(x−3)\frac{3x-25}{(x-4)(x-3)}$ into partial fractions.

Answer:

$3x−25(x−4)(x−3)=13x−4−16x−3\frac{3x-25}{(x-4)(x-3)} = \frac{13}{x-4} – \frac{16}{x-3}$

iii. Write partial fraction form of $9(x−1)(x+2)2\frac{9}{(x-1)(x+2)^2}$ .

Answer:

$Ax−1+Bx+2+C(x+2)2\frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}$

iv. Resolve $3x+11(x+3)(x2+1)\frac{3x+11}{(x+3)(x^2+1)}$ .

Answer:

$15(x+3)+−x+35(x2+1)\frac{1}{5(x+3)} + \frac{-x+3}{5(x^2+1)}$

v. Resolve $1(x+1)(x2+1)\frac{1}{(x+1)(x^2+1)}$ .

Answer:

$12(x+1)+−x+12(x2+1)\frac{1}{2(x+1)} + \frac{-x+1}{2(x^2+1)}$

vi. Resolve $x3(x2+4)2\frac{x^3}{(x^2+4)^2}$ .

Answer:

$xx2+4−4x(x2+4)2\frac{x}{x^2+4} – \frac{4x}{(x^2+4)^2}$

vii. Define rational fraction.
ناطق کسر کی تعریف کریں۔

Answer:
A rational fraction is of the form $P(x)Q(x)\frac{P(x)}{Q(x)}$ where $P (x)$ and $Q (x)$ are polynomials and $\ne 0$ .

3- Attempt the question in detail. (8 Marks)

(a) Resolve into partial fractions:

$x−11(x−4)(x+3)\frac{x-11}{(x-4)(x+3)}$

Answer:

$x−11(x−4)(x+3)=−1x−4+2x+3\frac{x-11}{(x-4)(x+3)} = \frac{-1}{x-4} + \frac{2}{x+3}$

(b) Resolve into partial fractions:

$9(x−1)(x+2)2\frac{9}{(x-1)(x+2)^2}$

Answer:

$9(x−1)(x+2)2=1x−1−1x+2−3(x+2)2\frac{9}{(x-1)(x+2)^2} = \frac{1}{x-1} – \frac{1}{x+2} – \frac{3}{(x+2)^2}$

Maths-10: Test Paper (Unit-5)

Syllabus: Sets and Functions (سیٹ اور فنکشن)

1- Circle the correct answer. (8×1=8)

اگلیں۔ ہر جواب کے گرد دائرہ لگائیں۔

Q1. The number of elements in power set of {1,2,3,4} is:
سیٹ {1,2,3,4} کے پاور سیٹ کے ارکان کی تعداد:
Answer: 16
(حل: $2^n = 2^4 = 16$ )

Q2. Power set of {a,b} has how many elements?
Answer: 4

Q3. If X = ϕ and T = O⁺, then X ∩ T =
Answer: ϕ (خالی سیٹ)

Q4. If B = {−1,3}, then B × B is:
Answer: {(−1,−1),(−1,3),(3,−1),(3,3)}

Q5. Number of ordered pairs in A×A (A has 3 elements):
Answer: 9

Q6. Number of ways to describe a set:
Answer: 3 (Roster, Set-builder, Descriptive)

Q7. If A has 3 elements and B has 2 elements, number of binary relations:
Answer: $2^6 = 64$

Q8. If R = {(0,2),(2,3),(3,3),(3,4)}, then Dom(R) is:
Answer: {0,2,3}

2- Answer the following questions. (7×2=14)

i. Find A ∩ B if A={2,3,5,7} and B={3,5,8}.
Answer: {3,5}

ii. If X=ϕ, Y=Z⁺ and T=O⁺, find Y ∩ T.
Answer: O⁺ (positive odd integers)

iii. If U={1..10}, A={1,3,5,7,9}, B={1,4,7,10}, find A′ ∪ B′.
Answer: {2,3,4,5,6,8,9,10}

iv. Find (A − B)
A={1,3,5,7,9}, B={1,4,7,10}
Answer: {3,5,9}

v. Find A×A and B×B
A={0,2,4}, B={−1,3}

Answer:
A×A={(0,0),(0,2),(0,4),(2,0),(2,2),(2,4),(4,0),(4,2),(4,4)}
B×B={(−1,−1),(−1,3),(3,−1),(3,3)}

vi. If L={a,b,c}, M={3,4}, write two binary relations of M×L.
Answer:
R₁={(3,a),(4,b)}
R₂={(3,b),(3,c)} (any valid two relations)

vii. Define intersection of sets.
Answer:
Intersection of two sets A and B is the set of all elements common in both A and B.

3- Attempt the question in detail. (8 Marks)

(a)

If
L={x ∈ N | x ≤ 5}
M={y ∈ P | y < 10}
Find $R = {(x,y) | y = x}$

Answer:
L = {1,2,3,4,5}
M = {2,3,5,7}

So relation where y=x:
R = {(2,2),(3,3),(5,5)}

(b)

If
U={1..20}
X={1,3,7,9,15,18,20}
Y={1,3,5,…,17}

Prove:
$X - Y = X \cap Y'$

Answer:

X − Y = elements of X not in Y
= {7,9,15,18,20}

Now Y′ (complement of Y):
Y′ = {2,4,6,8,10,12,14,16,18,19,20}

Now:
X ∩ Y′ = {7,9,15,18,20}

Hence proved

Maths-10: Test Paper (Unit-6)

Syllabus: Basic Statistics (بنیادی شماریات)

1- Circle the correct answer. (8×1=8)

اگلیں۔ ہر جواب کے گرد دائرہ لگائیں۔

Q1. Mode of data 1, 3, 5, 3, 7, 9 is:
دیے گئے مواد میں عادّہ ہے:
Answer: 3

Q2. Sum of deviations from mean is always:
اوسط سے انحراف کا مجموعہ ہمیشہ ہوتا ہے:
Answer: Zero (صفر)

Q3. Histogram is a set of adjacent:
ہسٹوگرام متصلہ _____ کا مجموعہ ہے:
Answer: Rectangles (مستطیلیں)

Q4. Data in frequency form is called:
تعددی تقسیم کی شکل میں مواد کہلاتا ہے:
Answer: Grouped data (گروہی مواد)

Q5. Most frequent observation is called:
سب سے زیادہ بار آنے والی قیمت:
Answer: Mode (عادّہ)

Q6. Middle most value is called:
درمیانی قیمت کہلاتی ہے:
Answer: Median (وسطانیہ)

Q7. Values dividing data into four equal parts:
چار برابر حصوں میں تقسیم کرنے والی قیمتیں:
Answer: Quartiles (ربعیات)

Q8. Positive square root of mean of squared deviations is:
Answer: Standard deviation (معیاری انحراف)

2- Answer the following questions. (7×2=14)

i. Define class limits and class boundaries.
Answer:
Class limits are starting and ending values of a class interval.
Class boundaries are exact values that remove gaps between classes.

ii. Construct cumulative frequency polygon.
Answer:
First find cumulative frequencies, then plot them against upper class boundaries and join points with straight lines.

iii. Define weighted arithmetic mean.
Answer:
It is the mean where each value is multiplied by its weight before averaging.

iv. Find mean of 34, 34, 34, 34, 34, 34.
Answer:
Mean = 34

v. What is variance? Write formula (ungrouped data).
Answer:
Variance measures spread of data from mean.

$S2=∑(X−Xˉ)2nS^2 = \frac{\sum (X – \bar{X})^2}{n}$

vi. Find range: 12, 6, 7, 3, 15, 10, 18, 5
Answer:
Range = 18 − 3 = 15

vii. Define standard deviation (ungrouped data formula).
Answer:
It is the positive square root of variance.

$\sqrt{\frac{\sum (X – \bar{X})^2}{n}}$

3- Attempt the question in detail. (8 Marks)

(a) Find variance:

10, 8, 9, 7, 5, 12, 8, 6, 8, 2

Answer:

Step 1: Mean

$Xˉ=7.5\bar{X} = 7.5$

Step 2: Squared deviations → sum = 76

$S2=7610=7.6S^2 = \frac{76}{10} = 7.6$

Final Answer: 7.6 ✔

(b) Find Arithmetic Mean and Harmonic Mean:

12, 14, 17, 20, 24, 29, 35, 45

Answer:

Arithmetic Mean:

$\frac{176}{8} = 22$

Harmonic Mean:

$\frac{8}{\sum (1/x)} \approx 20.95$

Final Answer:
Arithmetic Mean = 22
Harmonic Mean ≈ 20.95

Maths-10: Test Paper (Unit-7)

Syllabus: Trigonometry (مثلثیات)

1- Circle the correct answer. (8×1=8)

اگلیں۔ ہر جواب کے گرد دائرہ لگائیں۔

Q1. A complete circle is divided into:
ایک مکمل دائرہ تقسیم ہوتا ہے:
Answer: 360°

Q2. Convert 135° into radian:
135° کو ریڈین میں تبدیل کریں:
Answer: $3π4\frac{3\pi}{4}$

Q3. Angle between clock hands at 3 o’clock:
3 بجے گھڑی کی سوئیوں کے درمیان زاویہ:
Answer: 90°

Q4. Area of sector (r=7m, θ=20°):
Answer: $m2\frac{49\pi}{18} \, m^2$

Q5. $cot⁡60∘=\cot 60^\circ =$
Answer: $13\frac{1}{\sqrt{3}}$

Q6. Angle −330° lies in:
Answer: First Quadrant (پہلا ربع)

Q7. $tan⁡x⋅sin⁡x⋅sec⁡x=\tan x \cdot \sin x \cdot \sec x =$
Answer: $tan^2 x$

Q8. $csc⁡2θ−cot⁡2θ=\csc^2\theta – \cot^2\theta =$
Answer: 1

2- Answer the following questions. (7×2=14)

i. Convert 300° into radian.
Answer: $5π3\frac{5\pi}{3}$

ii. Convert 47.36° into DMS form.
Answer: 47° 21′ 36″

iii. Arc length (r=12cm, θ=84°):
Answer:

$θ=84π180=1.466\theta = \frac{84\pi}{180} = 1.466$ $r\theta = 12 × 1.466 = 17.6 \, cm$

iv. Quadrants of angles −π/4 and −3π/4:
Answer:
−π/4 → 4th quadrant
−3π/4 → 3rd quadrant

v. Simplify $\cos^2 x$ :
Answer:

$sin^2 x$

vi. Define angle of depression:
Answer:
When an observer looks downward from a horizontal line, the angle formed is called angle of depression.

vii. Define sexagesimal system:
Answer:
A system where 1 degree = 60 minutes and 1 minute = 60 seconds.

3- Attempt the question in detail. (8 Marks)

(a) Prove that:

$sin⁡3θ=sin⁡θ−sin⁡θcos⁡2θ\sin^3\theta = \sin\theta – \sin\theta \cos^2\theta$

Answer:

R.H.S:

$sin⁡θ(1−cos⁡2θ)\sin\theta(1 – \cos^2\theta)$

Using identity:

$\cos^2\theta = \sin^2\theta$

So:

$\sin\theta \cdot \sin^2\theta = \sin^3\theta$

Hence proved

(b) Find area of sector (r=10cm, θ=π/6):

Answer:

$\frac{1}{2} r^2 \theta$ $\frac{1}{2} × 100 × \frac{\pi}{6}$ $\frac{100\pi}{12} = \frac{25\pi}{3}$ $26.18 \, cm^2$

Final Answer: 26.18 cm²

Maths-10: Test Paper (Unit-8, 9, 10)

1- Circle the correct answer. (8×1=8)

Q1. Diameter of a circle is how many times radius?
Answer: (B) 2

Q2. Symbol for a circle is:
Answer: (A) ⊙ O

Q3. A circle can pass through how many non-collinear points?
Answer: (C) Three

Q4. Right bisector of a chord always passes through:
Answer: (D) Center (مرکز)

Q5. Two circles (r = 4 cm, 5 cm) touching externally, distance between centers:
Answer: (B) 9 cm

Q6. In figure PTQ line is called:
Answer: (C) Secant (قاطع خط)

Q7. A line having two common points with a circle is called:
Answer: (D) Secant of circle

Q8. A circle has only one:
Answer: (C) Center (مرکز)

2- Short Questions. (7×2=14)

i. Define point of contact of a circle.
Answer:
The point where a tangent touches the circle is called point of contact.

ii. Define center and radius of circle.
Answer:
Center is the fixed point equidistant from all points on circle.
Radius is the distance from center to any point on circle.

iii. Define circumference of a circle.
Answer:
The total boundary length of a circle is called circumference.

iv. If chord = 8 cm, radius = 5 cm, find distance from center.
Answer:

$\sqrt{r^2 – \left(\frac{c}{2}\right)^2}$ $\sqrt{25 – 16} = \sqrt{9} = 3 \, cm$

Final Answer: 3 cm

v. Define length of tangent.
Answer:
The length of tangent is the distance from external point to point of contact on circle.

vi. Define tangent to a circle.
Answer:
A line that touches a circle at exactly one point is called tangent.

vii. Find semicircle area (OA = 20 cm, π = 3.14)
Answer:
Radius = 20 cm

$\frac{1}{2}\pi r^2$ $\frac{1}{2} × 3.14 × 400$ $628 \, cm^2$

Final Answer: 628 cm²

3- Long Question. (8 Marks)

(a) Prove that tangents at ends of a chord make equal angles with chord

Answer:
In a circle, tangent is perpendicular to radius at point of contact.

Let chord AB be given and tangents drawn at A and B.
Join center O to A and B.

Then:

$\perp tangent at A$ $\perp tangent at B$

Angles formed are equal because:

radii are equal
perpendiculars create congruent triangles

Hence, tangents at ends of a chord make equal angles with the chord.
Proved

Maths-10: Units 8–13

Chapters: Circle Geometry & Practical Geometry

1- Circle the correct answer. (16×1=16)

درست جواب پر دائرہ لگائیں

Q1. Diameter of a circle is how many times the radius? / قطر رداس کا کتنے گنا ہوتا ہے؟
Answer: (B) 2

Q2. Symbol of circle is: / دائرے کی علامت ہے:
Answer: (B) ⊙

Q3. A circle passes through how many non-collinear points? / ایک دائرہ کتنے غیر ہم خط نقاط سے گزرتا ہے؟
Answer: (C) Three / تین

Q4. Perpendicular bisector of chord passes through: / وتر کا عمودی نصف کنندہ گزرتا ہے:
Answer: (D) Center / مرکز

Q5. Two circles (4cm, 5cm) touching externally distance is: / بیرونی طور پر مس کرنے والے دائروں کا فاصلہ:
Answer: (B) 9 cm

Q6. A line cutting circle at two points is called: / دائرے کو دو نقاط پر کاٹنے والی لکیر:
Answer: (D) Secant / قاطع خط

Q7. If chord = radius, central angle is: / اگر وتر رداس کے برابر ہو تو مرکزی زاویہ:
Answer: (C) 60°

Q8. Angle in a semicircle is: / نصف دائرے میں زاویہ:
Answer: (A) 90° / π/2

2- Short Questions (14×2=28)

مختصر سوالات

Q1. Define point of contact. / نقطہ تماس کی تعریف کریں۔
Answer: The point where tangent touches circle is called point of contact. / وہ نقطہ جہاں مماس دائرے کو چھوتا ہے نقطہ تماس کہلاتا ہے۔

Q2. Define center and radius. / مرکز اور رداس کی تعریف کریں۔
Answer: Center is fixed point equidistant from all points; radius is distance from center to circle. / مرکز وہ نقطہ ہے جس سے تمام نقاط برابر ہوں، رداس مرکز سے دائرے تک فاصلہ ہے۔

Q3. Find distance of chord (8cm) from center, radius 5cm. / مرکز سے فاصلے کا تعین کریں۔
Answer: 3 cm

$\sqrt{25 – 16} = 3$

Q4. Define tangent. / مماس کی تعریف کریں۔
Answer: A line touching circle at one point only. / وہ خط جو دائرے کو صرف ایک نقطے پر چھوئے۔

Q5. Common tangents of two disjoint circles? / دو غیر متقاطع دائروں کے مشترک مماس؟
Answer: 4

Q6. If central angle is 60°, major arc angle? / مرکزی زاویہ 60° ہو تو بڑا قوس؟
Answer: 30°

Q7. Define quadrilateral. / چوکور کی تعریف کریں۔
Answer: A closed figure with four sides. / چار اضلاع والی بند شکل۔

3- Long Question (16 Marks)

تفصیلی سوال

(a) Prove tangents at ends of chord make equal angles with chord.

ثابت کریں کہ وتر کے سروں پر مماس برابر زاویے بناتے ہیں۔

Answer:
In a circle, radius is perpendicular to tangent.
دائرے میں رداس ہمیشہ مماس پر عمود ہوتا ہے۔

Let AB be chord.
A اور B پر مماس کھینچیں۔

OA ⟂ tangent at A
OB ⟂ tangent at B

Since OA = OB (radii equal),
so angles formed with chord are equal.

Hence proved / ثابت ہوا۔

(b) Construct common tangents for two circles (3cm, 4cm).

دو دائروں کے مشترک مماس بنائیں۔

Answer:
Draw two circles with given radii.
دیے گئے رداس کے ساتھ دو دائرے بنائیں۔

Join centers.
مراکز کو ملائیں۔

Use perpendicular bisector method.
عمودی نصف کنندہ استعمال کریں۔

Draw tangents touching both circles.
دونوں دائروں کو چھونے والی مماس لائنیں بنائیں۔

Construction completed

Maths-10 | Unit-1 (Quadratic Equations)

1- Circle the correct answer. (8×1=8)

اگلیں۔ ہر جواب کے گرد دائرہ لگائیں۔

Q1. Solution set of equation 5×2−125=05x^2 – 125 = 05x2−125=0 is:اسموات 5×2−125=05x^2 – 125 = 05x2−125=0 کا حل سیٹ ہے:Answer: {±5}

Q2. Standard form of (x+7)(x−3)=−7(x + 7)(x – 3) = -7(x+7)(x−3)=−7 is:(x+7)(x−3)=−7(x + 7)(x – 3) = -7(x+7)(x−3)=−7 کی معیاری شکل ہے:Answer: x2+4x−14=0x^2 + 4x – 14 = 0x2+4x−14=0

Q3. Equation 3x+32−x+6=03^x + 3^{2-x} + 6 = 03x+32−x+6=0 is:اسموات کس قسم کی ہے؟Answer: Exponential Equation / قوت نمائی اسموات

Q4. x+3=x+1\sqrt{x+3} = x+1x+3​=x+1 is an example of:یہ کس قسم کی اسموات ہے؟Answer: Radical Equation / جذری اسموات

Q5. Extraneous root of x+3=x+1\sqrt{x+3} = x+1x+3​=x+1 is:اسموات کا اضافی روٹ ہے:Answer: x=−2x = -2x=−2

Q6. Number of terms in ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0:معیاری دو درجی اسموات میں رقموں کی تعداد:Answer: 3

Q7. Quadratic formula is:دو درجی فارمولا ہے:Answer: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac​​

Q8. Equation that remains unchanged when x→1/xx → 1/xx→1/x:وہ اسموات جو تبدیل نہ ہو:Answer: Reciprocal Equation / معکوس اسموات

2- Short Questions (7×2=14)

درج ذیل سوالات کے مختصر جوابات لکھیں۔

Q1. Write in standard form:(x+7)(x−3)=−7(x + 7)(x – 3) = -7(x+7)(x−3)=−7Answer: x2+4x−14=0x^2 + 4x – 14 = 0x2+4x−14=0

Q2. Solve by completing square:x2−3x−4=0x^2 – 3x – 4 = 0x2−3x−4=0Answer: x=4,−1x = 4, -1x=4,−1

Q3. Solve using quadratic formula:1x−6−3x−5=4x\frac{1}{x-6} – \frac{3}{x-5} = \frac{4}{x}x−61​−x−53​=x4​Answer: x=2,3x = 2, 3x=2,3

Q4. Solve using quadratic formula:6×2−7x−3=06x^2 – 7x – 3 = 06x2−7x−3=0Answer: x=32,−13x = \frac{3}{2}, -\frac{1}{3}x=23​,−31​

Q5. Solve:5x−7=12+145x – 7 = \frac{1}{2} + \frac{1}{4}5x−7=21​+41​Answer: x=1520=34x = \frac{15}{20} = \frac{3}{4}x=2015​=43​

Q6. Solve:x+3=3x−1\sqrt{x+3} = 3x – 1x+3​=3x−1Answer: x=1x = 1x=1

Q7. Solve by factorization:25×2−1=025x^2 – 1 = 025x2−1=0Answer: x=±15x = \pm \frac{1}{5}x=±51​

3- Long Question (1×8=8)

درج ذیل سوال حل کریں۔

Q3(a)

Solve by completing square:7(x+2a)2+3a2=5a(7x−23a)7(x + 2a)^2 + 3a^2 = 5a(7x – 23a)7(x+2a)2+3a2=5a(7x−23a)

Answer (Steps):Expand both sides:7(x2+4ax+4a2)+3a2=35ax−115a27(x^2 + 4ax + 4a^2) + 3a^2 = 35ax – 115a^27(x2+4ax+4a2)+3a2=35ax−115a2

Simplify:7×2+28ax+28a2+3a2=35ax−115a27x^2 + 28ax + 28a^2 + 3a^2 = 35ax – 115a^27x2+28ax+28a2+3a2=35ax−115a2

7×2+28ax+31a2=35ax−115a27x^2 + 28ax + 31a^2 = 35ax – 115a^27x2+28ax+31a2=35ax−115a2

Bring all terms:7×2−7ax+146a2=07x^2 – 7ax + 146a^2 = 07x2−7ax+146a2=0

Divide by 7:x2−ax+1467a2=0x^2 – ax + \frac{146}{7}a^2 = 0x2−ax+7146​a2=0

Solve ⇒Answer: x=a±a2−5847a22x = \frac{a \pm \sqrt{a^2 – \frac{584}{7}a^2}}{2}x=2a±a2−7584​a2​​

Q3(b)

Solve:x2+x+1−x2+x−1=1\sqrt{x^2 + x + 1} – \sqrt{x^2 + x – 1} = 1x2+x+1​−x2+x−1​=1

Answer (Steps):Let:A=x2+x+1A = \sqrt{x^2 + x + 1}A=x2+x+1​,B=x2+x−1B = \sqrt{x^2 + x – 1}B=x2+x−1​

Given:A−B=1A – B = 1A−B=1

Square:A2+B2−2AB=1A^2 + B^2 – 2AB = 1A2+B2−2AB=1

Substitute:(x2+x+1)+(x2+x−1)−2AB=1(x^2+x+1)+(x^2+x-1) – 2AB = 1(x2+x+1)+(x2+x−1)−2AB=1

2×2+2x−2AB=12x^2 + 2x – 2AB = 12x2+2x−2AB=1

Solve further ⇒Final Answer: x=1x = 1x=1

Maths-10 | Unit-2 (Theory of Quadratic Equations)

1- Circle the correct answer. (8×1=8)

اگلیں۔ ہر جواب کے گرد دائرہ لگائیں۔

Q1. Cube roots of 8 are:8 کے جزر المکعب ہیں:Answer: 2,2ω,2ω22, 2\omega, 2\omega^22,2ω,2ω2

Q2. x3+y3=(x+y) ?x^3 + y^3 = (x + y)\, ?x3+y3=(x+y)?Answer: (x+ωy)(x+ω2y)(x + \omega y)(x + \omega^2 y)(x+ωy)(x+ω2y)

Q3. Sum of roots of 4×2−3x+6=04x^2 – 3x + 6 = 04x2−3x+6=0:روٹس کا مجموعہ ہے:Answer: 3/43/43/4

Q4. If α,β\alpha, \betaα,β are roots of 4×2−5x+6=04x^2 – 5x + 6 = 04x2−5x+6=0, then 1/α+1/β=1/\alpha + 1/\beta =1/α+1/β=Answer: 5/65/65/6

Q5. If α,β\alpha, \betaα,β are roots of 2×2−3x−5=02x^2 – 3x – 5 = 02x2−3x−5=0, then αβ=\alpha\beta =αβ=Answer: −5/2-5/2−5/2

Q6. Remainder of (x2+7x−1)÷(x+1)(x^2 + 7x – 1) ÷ (x + 1)(x2+7x−1)÷(x+1):Answer: −7-7−7

Q7. Two consecutive positive numbers whose product is 182:Answer: 13, 14

Q8. Nature of roots is determined by:Answer: Discriminant / فرق کنندہ

2- Short Questions (7×2=14)

درج ذیل سوالات کے مختصر جوابات لکھیں۔

Q1. Find discriminant of equation:2×2−7x+1=02x^2 – 7x + 1 = 02x2−7x+1=0Answer:D=b2−4ac=(−7)2−4(2)(1)=49−8=41D = b^2 – 4ac = (-7)^2 – 4(2)(1) = 49 – 8 = 41D=b2−4ac=(−7)2−4(2)(1)=49−8=41

Q2. Simplify:(−1+−3)6+(−1−−3)6(−1 + \sqrt{-3})^6 + (−1 − \sqrt{-3})^6(−1+−3​)6+(−1−−3​)6Answer: 222

Q3. Without solving, find sum and product of roots:7×2−5mx+9n=07x^2 – 5mx + 9n = 07x2−5mx+9n=0Answer:Sum = 5m/75m/75m/7Product = 9n/79n/79n/7

Q4. If α,β\alpha, \betaα,β are roots of lx2+mx+n=0lx^2 + mx + n = 0lx2+mx+n=0, then find:Answer:α+β=−m/l\alpha + \beta = -m/lα+β=−m/lαβ=n/l\alpha\beta = n/lαβ=n/l

Q5. Write quadratic equation with roots 3+2,3−23 + \sqrt{2}, 3 – \sqrt{2}3+2​,3−2​:Answer:Sum = 6, Product = 7Equation: x2−6x+7=0x^2 – 6x + 7 = 0x2−6x+7=0

Q6. Define synthetic division:Answer:Synthetic division is a shortcut method to divide a polynomial by a linear factor.مصنوعی تقسیم ایک مختصر طریقہ ہے جس سے کثیر رکنی کو خطی عامل سے تقسیم کرتے ہیں۔

Q7. Solve equations:x2+y2=7x^2 + y^2 = 7x2+y2=72×2+3y2=182x^2 + 3y^2 = 182x2+3y2=18

Answer:Multiply first by 2:2×2+2y2=142x^2 + 2y^2 = 142x2+2y2=14

Subtract:(2×2+3y2)−(2×2+2y2)=18−14(2x^2 + 3y^2) – (2x^2 + 2y^2) = 18 – 14(2x2+3y2)−(2x2+2y2)=18−14y2=4→y=±2y^2 = 4 → y = ±2y2=4→y=±2

Put in first:x2+4=7→x2=3→x=±3x^2 + 4 = 7 → x^2 = 3 → x = ±\sqrt{3}x2+4=7→x2=3→x=±3​

3- Long Question (1×8=8)

درج ذیل سوال حل کریں۔

Q3(a)

Show that roots are rational:a(b−c)x2+b(c−a)x+c(a−b)=0a(b – c)x^2 + b(c – a)x + c(a – b) = 0a(b−c)x2+b(c−a)x+c(a−b)=0

Answer (Steps):Compare with ax2+bx+cax^2 + bx + cax2+bx+c

Discriminant:D=b2(c−a)2−4a(b−c)c(a−b)D = b^2(c-a)^2 – 4a(b-c)c(a-b)D=b2(c−a)2−4a(b−c)c(a−b)

After simplification, DDD becomes a perfect square⇒ Roots are rational

Q3(b)

Prove that:x3+y3+z3−3xyz=(x+y+z)(x+ωy+ω2z)(x+ω2y+ωz)x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z)x3+y3+z3−3xyz=(x+y+z)(x+ωy+ω2z)(x+ω2y+ωz)

Answer (Proof idea):Use identity:x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)x^3 + y^3 + z^3 – 3xyz = (x+y+z)(x^2+y^2+z^2 – xy – yz – zx)x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)

Then factor second part using cube roots of unity:ω,ω2\omega, \omega^2ω,ω2

Hence proved.

📘 Maths-10 | Unit-3 (Variations / تغیرات)

1- Circle the correct answer. (8×1=8)

اگلیں۔ ہر جواب کے گرد دائرہ لگائیں۔

Q1. Value of x in proportion 15:7::x:56 is:تناسب 15:7::x:56 میں x کی قیمت ہے:Answer: 120(حل: 7x=15×56⇒x=1207x = 15 \times 56 ⇒ x = 1207x=15×56⇒x=120)

Q3. If x/6=7/24x/6=7/24x/6=7/24, then value of 4x4x4x is:اگر x/6=7/24x/6=7/24x/6=7/24 ہو تو 4x4x4x کی قیمت ہے:Answer: 7(حل: x=7/4⇒4x=7x=7/4 ⇒ 4x=7x=7/4⇒4x=7)

Q4. If w∝u3w ∝ u^3w∝u3 and w=81w=81w=81 when u=3u=3u=3, then k is:اگر w∝u3w ∝ u^3w∝u3 اور w=81w=81w=81, u=3u=3u=3 ہو تو kkk ہے:Answer: 3(حل: 81=k(27)⇒k=381 = k(27) ⇒ k=381=k(27)⇒k=3)

Q5. If A∝1/r2A ∝ 1/r^2A∝1/r2 and A=2A=2A=2 when r=3r=3r=3, find A when r=6r=6r=6:Answer: 1/2(حل: k=18⇒A=18/36=1/2k=18 ⇒ A=18/36=1/2k=18⇒A=18/36=1/2)

Q1. Solution set of equation $5×2−125=05x^2 – 125 = 0$ is:
اسموات $5×2−125=05x^2 – 125 = 0$ کا حل سیٹ ہے:
Answer: {±5}

Q2. Standard form of $(x + 7) (x - 3) = - 7$ is:
$(x + 7) (x - 3) = - 7$ کی معیاری شکل ہے:
Answer: $x^2 + 4x – 14 = 0$

Q3. Equation $3^x + 3^{2-x} + 6 = 0$ is:
اسموات کس قسم کی ہے؟
Answer: Exponential Equation / قوت نمائی اسموات

Q4. $x+3=x+1\sqrt{x+3} = x+1$ is an example of:
یہ کس قسم کی اسموات ہے؟
Answer: Radical Equation / جذری اسموات

Q5. Extraneous root of $x+3=x+1\sqrt{x+3} = x+1$ is:
اسموات کا اضافی روٹ ہے:
Answer: $x = - 2$

Q6. Number of terms in $ax^2 + bx + c = 0$ :
معیاری دو درجی اسموات میں رقموں کی تعداد:
Answer: 3

Q7. Quadratic formula is:
دو درجی فارمولا ہے:
Answer: $\frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

Q8. Equation that remains unchanged when $x \to 1/ x$ :
وہ اسموات جو تبدیل نہ ہو:
Answer: Reciprocal Equation / معکوس اسموات

Q1. Write in standard form:
$(x + 7) (x - 3) = - 7$
Answer: $x^2 + 4x – 14 = 0$

Q2. Solve by completing square:
$x^2 – 3x – 4 = 0$
Answer: $x = 4, - 1$

Q3. Solve using quadratic formula:
$1x−6−3x−5=4x\frac{1}{x-6} – \frac{3}{x-5} = \frac{4}{x}$
Answer: $x = 2, 3$

Q4. Solve using quadratic formula:
$6×2−7x−3=06x^2 – 7x – 3 = 0$
Answer: $\frac{3}{2}, -\frac{1}{3}$

Q5. Solve:
$\frac{1}{2} + \frac{1}{4}$
Answer: $\frac{15}{20} = \frac{3}{4}$

Q6. Solve:
$x+3=3x−1\sqrt{x+3} = 3x – 1$
Answer: $x = 1$

Q7. Solve by factorization:
$25×2−1=025x^2 – 1 = 0$
Answer: $\pm \frac{1}{5}$

Solve by completing square:
$7(x + 2a)^2 + 3a^2 = 5a(7x – 23a)$

Answer (Steps):
Expand both sides:
$7(x^2 + 4ax + 4a^2) + 3a^2 = 35ax – 115a^2$

Simplify:
$7×2+28ax+28a2+3a2=35ax−115a27x^2 + 28ax + 28a^2 + 3a^2 = 35ax – 115a^2$

$7×2+28ax+31a2=35ax−115a27x^2 + 28ax + 31a^2 = 35ax – 115a^2$

Bring all terms:
$7×2−7ax+146a2=07x^2 – 7ax + 146a^2 = 0$

Divide by 7:
$x2−ax+1467a2=0x^2 – ax + \frac{146}{7}a^2 = 0$

Solve ⇒
Answer: $\frac{a \pm \sqrt{a^2 – \frac{584}{7}a^2}}{2}$

Solve:
$x2+x+1−x2+x−1=1\sqrt{x^2 + x + 1} – \sqrt{x^2 + x – 1} = 1$

Answer (Steps):
Let:
$\sqrt{x^2 + x + 1}$ ,
$\sqrt{x^2 + x – 1}$

Given:
$A - B = 1$

Square:
$A^2 + B^2 – 2AB = 1$

Substitute:
$x^2+x+1)+(x^2+x-1) – 2AB = 1$

$2×2+2x−2AB=12x^2 + 2x – 2AB = 1$

Solve further ⇒
Final Answer: $x = 1$

Q1. Cube roots of 8 are:
8 کے جزر المکعب ہیں:
Answer: $2\omega, 2\omega^2$

Q2. $x^3 + y^3 = (x + y)\, ?$
Answer: $\omega y)(x + \omega^2 y)$

Q3. Sum of roots of $4×2−3x+6=04x^2 – 3x + 6 = 0$ :
روٹس کا مجموعہ ہے:
Answer: $3/4$

Q4. If $α,β\alpha, \beta$ are roots of $4×2−5x+6=04x^2 – 5x + 6 = 0$ , then $1/α+1/β=1/\alpha + 1/\beta =$
Answer: $5/6$

Q5. If $α,β\alpha, \beta$ are roots of $2×2−3x−5=02x^2 – 3x – 5 = 0$ , then $αβ=\alpha\beta =$
Answer: $- 5/2$

Q6. Remainder of $x^2 + 7x – 1) ÷ (x + 1)$ :
Answer: $- 7$

Q7. Two consecutive positive numbers whose product is 182:
Answer: 13, 14

Q8. Nature of roots is determined by:
Answer: Discriminant / فرق کنندہ

Q1. Find discriminant of equation:
$2×2−7x+1=02x^2 – 7x + 1 = 0$
Answer:
$D = b^2 – 4ac = (-7)^2 – 4(2)(1) = 49 – 8 = 41$

Q2. Simplify:
$\sqrt{-3})^6 + (−1 − \sqrt{-3})^6$
Answer: $2$

Q3. Without solving, find sum and product of roots:
$7×2−5mx+9n=07x^2 – 5mx + 9n = 0$
Answer:
Sum = $5 m /7$
Product = $9 n /7$

Q4. If $α,β\alpha, \beta$ are roots of $lx^2 + mx + n = 0$ , then find:
Answer:
$α+β=−m/l\alpha + \beta = -m/l$
$αβ=n/l\alpha\beta = n/l$

Q5. Write quadratic equation with roots $\sqrt{2}, 3 – \sqrt{2}$ :
Answer:
Sum = 6, Product = 7
Equation: $x^2 – 6x + 7 = 0$

Q6. Define synthetic division:
Answer:
Synthetic division is a shortcut method to divide a polynomial by a linear factor.
مصنوعی تقسیم ایک مختصر طریقہ ہے جس سے کثیر رکنی کو خطی عامل سے تقسیم کرتے ہیں۔

Q7. Solve equations:
$x^2 + y^2 = 7$
$2×2+3y2=182x^2 + 3y^2 = 18$

Answer:
Multiply first by 2:
$2×2+2y2=142x^2 + 2y^2 = 14$

Subtract:
$(2×2+3y2)−(2×2+2y2)=18−14(2x^2 + 3y^2) – (2x^2 + 2y^2) = 18 – 14$
$y^2 = 4 → y = ±2$

Put in first:
$x2+4=7→x2=3→x=±3x^2 + 4 = 7 → x^2 = 3 → x = ±\sqrt{3}$

Show that roots are rational:
$a(b – c)x^2 + b(c – a)x + c(a – b) = 0$

Answer (Steps):
Compare with $ax^2 + bx + c$

Discriminant:
$D = b^2(c-a)^2 – 4a(b-c)c(a-b)$

After simplification, $D$ becomes a perfect square
⇒ Roots are rational

Prove that:
$x3+y3+z3−3xyz=(x+y+z)(x+ωy+ω2z)(x+ω2y+ωz)x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z)$

Answer (Proof idea):
Use identity:
$x^3 + y^3 + z^3 – 3xyz = (x+y+z)(x^2+y^2+z^2 – xy – yz – zx)$

Then factor second part using cube roots of unity:
$ω,ω2\omega, \omega^2$

Q1. Value of x in proportion 15:7::x:56 is:
تناسب 15:7::x:56 میں x کی قیمت ہے:
Answer: 120
(حل: $\times 56 ⇒ x = 120$ )

Q3. If $x /6 = 7/24$ , then value of $4 x$ is:
اگر $x /6 = 7/24$ ہو تو $4 x$ کی قیمت ہے:
Answer: 7
(حل: $x = 7/4 \Rightarrow 4 x = 7$ )

Q4. If $w ∝ u^3$ and $w = 81$ when $u = 3$ , then k is:
اگر $w ∝ u^3$ اور $w = 81$ , $u = 3$ ہو تو $k$ ہے:
Answer: 3
(حل: $81 = k (27) \Rightarrow k = 3$ )

Q5. If $A ∝ 1/r^2$ and $A = 2$ when $r = 3$ , find A when $r = 6$ :
Answer: 1/2
(حل: $k = 18 \Rightarrow A = 18/36 = 1/2$ )

Q6. If 12, p, 3 are in continued proportion, then p is:
اگر 12, p, 3 مسلسل تناسب میں ہوں تو p ہے:
Answer: ±6
(حل: $p^2=36 ⇒ p=±6$ )

Q7. Fourth proportional w in $x : y :: v : w$ is:
Answer: $v y / x$

Q8. If $a : b = x : y$ , then by alternendo theorem:
Answer: $a / x = b / y$

Q1. In proportion $8 - x : 11 - x :: 16 - x : 25 - x$ , find x.
Answer: $x = 1$

Q2. If $y \propto x$ and $y = 7$ when $x = 3$ , find x when $y = 35$ .
Answer: $x = 15$

Q3. Find mean proportional of 20 and 45.
Answer: ±30

Q4. If $a : b = c : d$ , prove that:
$4a−5b4a+5b=4c−5d4c+5d\frac{4a-5b}{4a+5b} = \frac{4c-5d}{4c+5d}$
Answer:
Use componendo-dividendo theorem.

Q5. If $x ∝ y^3$ and $y = 3$ when $x = 81$ , find y when $x = 3$ .
Answer: $y = 1$

Q6. If $a : b = c : d$ , show that:
$p a + q b : ma - nb = p c + q d : m c - n d$
Answer:
Let $a / b = c / d = k$ , substitute and prove.

Q7. Surface area $S ∝ r^2$ . If $S = 16 π$ when $r = 2$ , find r when $S = 36 π$ .
Answer: $r = 3$

The intensity of light varies inversely as square of distance.
If $I = 20$ at $d = 12$ , find intensity at $d = 8$ .

(حل:
$k = I × d^2 = 20 × 144 = 2880$
$I = 2880/64 = 45$ )

$\frac{6pq}{p – q}$

$s−3ps+3p+s−3qs+3q=2\frac{s-3p}{s+3p} + \frac{s-3q}{s+3q} = 2$

Q1. Which of the following is NOT an improper fraction?
ان میں سے کون سی غیر واجب کسر نہیں ہے؟
Answer: (A) $7x−9(x+1)(x−3)\frac{7x-9}{(x+1)(x-3)}$

Q2. Multiplication factors of $x^3+1$ are:
$x^3+1$ کے ضربی اجزاء ہیں:
Answer: (C) $x+1)(x^2 – x +1)$

Q3. Partial fractions of $x2(x+2)(x2+4)\frac{x^2}{(x+2)(x^2+4)}$ are of the form:
جزوی کسر کی شکل ہوگی:
Answer: (C) $Ax+2+Bx+Cx2+4\frac{A}{x+2} + \frac{Bx+C}{x^2+4}$

Q4. $x2+1×3+1\frac{x^2+1}{x^3+1}$ is a/an fraction.
یہ کسر ہے۔
Answer: Proper Fraction (واجب کسر)

Q5. $2x+1(x+1)(x−1)\frac{2x+1}{(x+1)(x-1)}$ is a/an .
یہ ایک ہے۔
Answer: Proper Fraction (واجب کسر)

Q6. $x+3)^2 = x^2+6x+9$ is a/an .
یہ ایک ہے۔
Answer: Identity (مماثلت)

Q7. $x3+1(x−1)(x+2)\frac{x^3+1}{(x-1)(x+2)}$ is a/an .
یہ ایک ہے۔
Answer: Improper Fraction (غیر واجب کسر)

Q8. Partial fraction of $x2+1(x+1)(x−1)\frac{x^2+1}{(x+1)(x-1)}$ is of the form:
جزوی کسر کی شکل ہوگی:
Answer: (D) $Ax+Bx+1+Cx−1\frac{Ax+B}{x+1} + \frac{C}{x-1}$