Show that f: R → R defined as f(x) = x / sqrt(1 + x^2) is one-one but not onto.

Show that f: R → R defined as f(x) = x / √(1 + x²) is one-one but not onto.

 f: R → R,  f(x) = x / √(1 + x²)

 To check One-One / Injective

Let x₁, x₂ ∈ R 

such that f(x₁) = f(x₂)

x₁ / √(1 + x₁²) = x₂ / √(1 + x₂²)  --------[1]

Squaring both sides:

x₁² / (1 + x₁²) = x₂² / (1 + x₂²)

x₁²(1 + x₂²) = x₂²(1 + x₁²)

x₁² + x₁²x₂² = x₂² + x₁²x₂²

x₁² = x₂²

x₁ = ± x₂

[1] is possible only if x₁, x₂ have the same sign

x₁ = -x₂, is rejected

we have to conclude that x₁ =x₂,

Therefore f is one-one.

To check Onto / Surjective

Let y = x / √(1 + x²)

squaring

y² = x² / (1 + x²)

y²(1 + x²) = x²

y² + y²x² = x²

y² = x² - y²x² = x²(1 - y²)

x² = y² / (1 - y²)

For x to be real, RHS ≥ 0

Since y² ≥ 0, we need 1 - y² > 0

⇒ y² < 1

⇒ -1 < y < 1

Range of f = (-1, 1) ≠ R

Therefore f is not onto.

see this video for more explanation 

cbse 12th maths old board exam question paper 2025 2026 one to one injective function onto function

SAT Math Formula Cheat Sheet for Quick Reference

 

SAT Math Formulas

The Digital SAT includes a reference sheet with some geometry formulas, but it does not include everything you'll need. Knowing the most common formulas before test day helps you solve problems more quickly and reduces the chance of making simple mistakes.

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Exponent Rules

For any nonzero number a:

a⁰ = 1

a¹ = a

aᵐ × aⁿ = aᵐ⁺ⁿ

aᵐ ÷ aⁿ = aᵐ⁻ⁿ

(aᵐ)ⁿ = aᵐⁿ

(ab)ⁿ = aⁿbⁿ

(a/b)ⁿ = aⁿ/bⁿ

a⁻ⁿ = 1/aⁿ

Remember

• Multiply → add exponents.

• Divide → subtract exponents.

• A negative exponent means take the reciprocal.

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Radicals

√a × √b = √(ab)

√a ÷ √b = √(a/b)

Examples

√49 = 7

√81 = 9

∛125 = 5

Perfect squares worth memorizing:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100

121, 144, 169, 196, 225, 256, 289, 324, 361, 400

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Linear Equations

Slope

m = (y₂ − y₁)/(x₂ − x₁)

Slope-intercept form

y = mx + b

Point-slope form

y − y₁ = m(x − x₁)

Standard form

Ax + By = C

Parallel lines have the same slope.

Perpendicular lines have negative reciprocal slopes.

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Coordinate Geometry

Distance Formula

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Midpoint Formula

((x₁ + x₂)/2, (y₁ + y₂)/2)

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Quadratic Equations

Standard form

ax² + bx + c = 0

Quadratic Formula

x = (−b ± √(b² − 4ac))/2a

Discriminant

b² − 4ac

Positive → two real solutions

Zero → one real solution

Negative → no real solutions

Vertex

x = −b/(2a)

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Factoring Identities

Difference of Squares

a² − b² = (a + b)(a − b)

Perfect Square Trinomials

a² + 2ab + b² = (a + b)²

a² − 2ab + b² = (a − b)²

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Circle Formulas

Circumference

C = 2πr

Area

A = πr²

Arc Length

(θ/360) × 2πr

Sector Area

(θ/360) × πr²

Diameter = 2r

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Rectangles and Squares

Rectangle

Area = length × width

Perimeter = 2(length + width)

Square

Area = side²

Perimeter = 4 × side

Diagonal = side√2

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Triangles

Area

½ × base × height

Pythagorean Theorem

a² + b² = c²

The angles inside every triangle add up to 180°.

An exterior angle equals the sum of the two opposite interior angles.

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Special Right Triangles

45°–45°–90°

1 : 1 : √2

30°–60°–90°

1 : √3 : 2

These ratios are tested regularly.

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Trigonometry

SOH CAH TOA

sin θ = opposite/hypotenuse

cos θ = adjacent/hypotenuse

tan θ = opposite/adjacent

tan θ = sin θ/cos θ

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Polygons

Interior Angle Sum

(n − 2) × 180°

Each Interior Angle of a Regular Polygon of n vertics

[(n − 2) × 180°]/n

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Volume

Cube

s³

Rectangular Prism

lwh

Cylinder

πr²h

Cone

⅓πr²h

Sphere

⁴⁄₃πr³

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Surface Area

Cube

6s²

Cylinder

2πr² + 2πrh

Sphere

4πr²

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Mean

Mean

Sum of all values ÷ Number of values

Weighted Mean

Σ(value × weight) ÷ Σ(weights)

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Probability

P(Event)

Number of Favorable Outcomes ÷ Total number of Outcomes

Complement Rule

P(complement event ) =1 − P(Event)

A probability is always between 0 and 1.

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Percents

Increase

Original × (1 + rate)

Decrease

Original × (1 − rate)

Percent Change

(New − Original)/Original × 100%

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Simple Interest

I = Prt

P = Principal

r = Interest Rate

t = Time

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Exponential Growth and Decay

Growth

A = P(1 + r)ᵗ

Decay

A = P(1 − r)ᵗ

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Functions

Example

f(x) = 2x + 3

f(5) = 13

Replace x with the given value.

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Useful Constants

π ≈ 3.14

√2 ≈ 1.414

√3 ≈ 1.732

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Find the mean of the following distribution :Class30 – 4040 – 5050 – 6060 – 7070 – 80Frequency61381211

 Find the mean of the following distribution :

Class               30 – 40    40 – 50   50 – 60      60 – 70      70 – 80

Frequency             6            13             8               12              11

calculate class mark, mid value or mid -x [x]   then the product f*x

Class    f     x     f*x

30-40    6     35    210

40-50    13    45    585

50-60    8     55    440

60-70    12    65    780

70-80    11    75    825

Total  Σf =50,  Σfx= 2840

Mean = Σfx / Σf = 2840 / 50 = 56.8

for more explanation watch this video 

statistics arithmetic mean by direct method 

cbse 10th maths previous years question papers 2025 2026 question in statistics

A boy has a collection of balls of different colours. He has a total of 35 balls in his basket out of which seven are black in colour and eight are yellow in colour. Out of remaining balls, some are white and the rest are red. Based on the above, answer the following questions: (a) If the probability of drawing a red ball at random from the basket is three times that of a white ball, then find the number of red balls in the basket. (b) Find the probability of drawing a ball at random from the basket which is either a black or a white ball.

 A boy  has a collection of balls of different colours. He has a total of 35 balls in his basket out of which seven are black in colour and eight are yellow in colour. Out of remaining balls, some are white and the rest are red.

Based on the above, answer the following questions:

(a) If the probability of drawing a red ball at random from the basket is three times that of a white ball, then find the number of red balls in the basket.

(b) Find the probability of drawing a ball at random from the basket which is either a black or a white ball

Total balls = 35

Number of Black balls = 7

Number of Yellow balls = 8

Remaining balls = 35 − 7 − 8 = 20  

Let number of white balls = w

Let number of red balls = r 

Given that 

Out of remaining balls, some are white and the rest are red.

Remaining balls =  20  

w + r  =  20  ---------[1]

Given 

P(red) = 3 × P(white) 

 P(red) = r/35

P(white) = w/35 

 r/35 = 3 × w/35

r = 3w  ---------(2) 

 Put (2) in (1)

w + r  =  20

w + 3w = 20

4w = 20

w = 5  

use r = 3w 

 r = 3 × 5 = 15 

 Number of red balls = 15

Number of  Black balls = 7

Numbe r of White balls [w] = 5  

 Number of  favourable outcomes = 7 + 5 = 12  

P(black or white) = 12/35

foe more details use the video 

probability, cbse 10th standard mathematics past years question papers 2025 2026

A survey was conducted on the patients who have undergone knee replacement surgeries. It was found that, Robotic Knee replacement surgeries have 90% success rate. On a particular day, robotic surgery was performed on three patients, A, B and C, one after the other. Assuming that the success and failure of each surgery is independent of each other, find the probability that : (i) exactly one surgery is successful, (ii) at most two surgeries are successful. probability of independent events cbse 12th maths old board exam question paper 2025 2026 independent events success failure type

 A survey was conducted on the patients who have undergone knee replacement surgeries. It was found that, Robotic Knee replacement surgeries have 90% success rate. On a particular day, robotic surgery was performed on three patients, A, B and C, one after the other. Assuming that the success and failure of each surgery is independent of each other, find the probability that : (i) exactly one surgery is successful, (ii) at most two surgeries are successful. probability of independent events cbse 12th maths old board exam question paper 2025 2026 independent events success failure type

Given:

 P(Success) = 90% = 0.9 = 9/10

 P(Failure) = 1 − 0.9 = 0.1 = 1/10

let S debite Success,

 F denote Failure

Probability that exactly one surgery is successful

possibilities SFF, FSF, FFS

P(SFF) = 0.9 × 0.1 × 0.1 = 0.009

P(FSF) = 0.1 × 0.9 × 0.1 = 0.009

P(FFS) = 0.1 × 0.1 × 0.9 = 0.009

  P(exactly one auccess) = 0.009 + 0.009 + 0.009 = 0.027 = 27/1000 

 Probability that at most two surgeries are successful

At most two successes means 0 or 1 or 2 successes

so

 Probability that at most two surgeries are successful 

= 1 − P(all three successful)

= 1 - P[ SSS]

= 1 - [0.9 × 0.9 × 0.9]

 = 1 -  0.729 

= 0.271 

= 271/1000

see this video for more explanation 

basic probability , independent events , cbse 12th maths old board exam question paper 20252026

From a solid cylinder whose height is 2.8 cm and radius 2.1 cm, a conical cavity of the same height and same radius is hollowed out. Find the volume and the total surface area of the remaining solid.

 From a solid cylinder whose height is 2.8 cm and radius 2.1 cm, a conical cavity of the same height and same radius is hollowed out. Find the volume and the total surface area of the remaining solid.

Cylinder and conical cavity have same radius and height

r = 2.1 cm, h = 2.8 cm

.Volume of remaining solid

 Volume of cylinder − Volume of cone 

Volume of cylinder = πr²h = π × (2.1)² × 2.8 = π × 4.41 × 2.8 = 12.348π cm³  

Volume of cone = (1/3)πr²h = (1/3) × 12.348π = 4.116π cm³

Remaining volume = 12.348π − 4.116π = 8.232π cm³  

Using π = 22/7

Remaining volume  = 8.232 × 22/7 = 25.872 cm³ ≈ 25.87 cm³

. Total surface area of remaining solid

Surfaces left after hollowing:  

Bottom circular base of cylinder = πr²  

Curved surface of cylinder = 2πrh  

Curved surface of cone = πrl  

Slant height of cone: l = √(r² + h²) = √(2.1² + 2.8²) = √(4.41 + 7.84) = √12.25 = 3.5 cm 

Required surface area = πr² + 2πrh + πrl

 = π(2.1)² + 2π(2.1)(2.8) + π(2.1)(3.5)

 = 4.41π + 11.76π + 7.35π = 23.52π cm²  Using π = 22/7:

 = 23.52 × 22/7 = 73.92 cm²  

for more details watch this video 

mensuration,   surface area and volume of solids , cbse th maths old board exam question paper 2025 2026

cbse 10th maths questions

 cbse 10th maths questions

Linear equations in two unknowns 

 The sum of numerator and denominator of a fraction is 4 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃. Find the fraction.

solution

Quadratic equations

Three consecutive positive integers are such that sum of square of the first and the product of the other two is 67, find the integers.

solution

Coordinate Geometry

Find a relation between x and y such that the point P(x, y) is equidistant from the

 points A(5, 3) and B(1, 7)

solution

The coordinates of the centre of a circle are (x − 7, 2x). Find the value(s) of ‘x’, if the circle passes through the point (−9, 11) and has radius 5√2 units.

solution

Trigonometry

 If cos A + sin A = (√2) cos A, prove that cos A − sin A = (√2) sin A

solution

Mensuration surface area and volume of solids

 From a solid cylinder whose height is 2.8 cm and radius 2.1 cm, a conical cavity of the same height and same radius is hollowed out. Find the volume and the total surface area of the remaining solid.

solution

Statistics

Find the mean of the following distribution :

Class               30 – 40    40 – 50   50 – 60      60 – 70      70 – 80

Frequency             6            13             8               12              11

solution

Probability

A boy  has a collection of balls of different colours. He has a total of 35 balls in his basket out of which seven are black in colour and eight are yellow in colour. Out of remaining balls, some are white and the rest are red.

Based on the above, answer the following questions:

(a) If the probability of drawing a red ball at random from the basket is three times that of a white ball, then find the number of red balls in the basket.

(b) Find the probability of drawing a ball at random from the basket which is either a black or a white ball

solution