Example 01:

Find the radius of curvature of the curve y = x³ at (2, 8)

Solution:

Equation of curve is y = x³ ………. (1)

Differentiating both sides w.r.t. x

 (dy/dx) = 3x² ………………. (2)

(dy/dx) _{at P(2, 8)} = 3(2)² = 12

Differentiating equation (2) again w.r.t. x

d²y/dx² = 6x

(d²y/dx²) _{at P(2, 8)} = 6(2) = 12

Ans: Radius of curvature of the given curve at (2, 8) is 145.5 units

Example 02:

Find the radius of curvature of the curve y = x³ at (1, 1)

Solution:

Equation of curve is y = x³  ………. (1)

Differentiating both sides w.r.t. x

 (dy/dx) = 3x²  ………………. (2)

(dy/dx) _{at P(1,1)} = 3(1)² = 3

Differentiating equation (2) again w.r.t. x

(d²y/dx²) = 6x  ………………. (2)

(d²y/dx²) _{at P(1,1)} = 6(1) = 6

Ans: Radius of curvature of the given curve at (1, 1) is 5√10/3 units

Example 03:

Find the radius of curvature of the curve y = e^x at (0, 1)

Solution:

Equation of curve is y = e^x   ………. (1)

Differentiating both sides w.r.t. x

 (dy/dx) = e^x  ………………. (2)

(dy/dx) _{at P(1,1)} = e¹ = e

Differentiating equation (2) again w.r.t. x

(d²y/dx²) = e^x  ………………. (2)

(d²y/dx²) _{at P(1,1)} = e¹ = e

Ans: Radius of curvature of the given curve at (0, 1) is 2√2 units

Example 04:

Find the radius of curvature of the curve xy = c²  at (c, c)

Solution:

Equation of curve is xy = c²    ………. (1)

Differentiating both sides w.r.t. x

X(dy/dx) + y = 0

(dy/dx) = -y/x

(dy/dx) _{at P(c, c)} = – c/c =  – 1

Differentiating equation (2) again w.r.t. x

Ans: Radius of curvature of the given curve at (0, 1) is c√2 units

Example 05:

Find the radius of curvature of the curve y² = 4x at (2, 2√2)

Solution:

Equation of curve is y² = 4x  ………. (1)

Differentiating both sides w.r.t. x

2y (dy/dx) = 4

(dy/dx) = 2/y ………………. (2)

(dy/dx) _{at P(2, 2√2)} = 2/2√2 = 1/√2

Differentiating equation (2) again w.r.t. x

(dy²/dx²) = -2/y² (dy/dx)

(dy²/dx²) = -2/y² (2/y) = -4/y³ ………………. (3)

(d²y/dx²) _{at P(2, 2√2)} = – 4/(2√2)³ = – 4/16√2 = – 1/4√2

Ans: Radius of curvature of the given curve at (2, 2√2) is – 6√ 3 units.

Example 06:

Find the radius of curvature at (3, – 4) to the curve x² + y² = 25

Solution:

Equation of curve is x² + y² = 25 ………. (1)

Differentiating both sides w.r.t. x

2x + 2y (dy/dx) = 0

2y (dy/dx) = – 2x

(dy/dx) = – x/y ………………. (2)

(dy/dx) _{at P(3, -4)} = – (3/-4) = 3/4

Differentiating equation (2) again w.r.t. x

Radius of curvature of the given curve at (3, -4) is 5 units

Example – 07:

Find the radius of curvature of the curve √x + √y = 1 at (1/4, 1/4)

Solution:

Equation of curve is √x + √y = 1 ………. (1)

Differentiating both sides w.r.t. x

Differentiating equation (2) again w.r.t. x

Ans: Radius of curvature of the given curve is 1/√2 units.

Example 08:

Find the radius of curvature of the curve √x + √y = root a at (a/4, a/4)

Solution:

Equation of curve is √x + √y = 1 ………. (1)

Differentiating both sides w.r.t. x

Differentiating equation (2) again w.r.t. x

Ans: Radius of curvature of the given curve is a/√2 units.

Example 09:

Find the radius of curvature of the curve x^2/3 + y^2/3  = 2 at (1, 1)

Solution:

Equation of curve is x^2/3 + y^2/3  = 2  ………. (1)

Differentiating both sides w.r.t. x

Differentiating equation (2) again w.r.t. x

Ans: Radius of curvature of the given curve is 3√2 units.

Example 10:

Find the radius of curvature of the curve x⁴ + y⁴ = 2 at (1, 1)

Solution:

Equation of curve is x⁴ + y⁴ = 2 ………. (1)

Differentiating both sides w.r.t. x

4x³ + 4y³ (dy/dx) = 0

4y³ (dy/dx) = – 4x³

4y³ (dy/dx) = – x³/y³    ………………. (2)

(dy/dx) _{at P(1,1)} = – 1³/1³=  – 1

Differentiating equation (2) again w.r.t. x

Ans: Radius of curvature of the given curve at (1,1) is – √2/3 units

The post Applications of Derivatives Radius of Curvature appeared first on The Fact Factor.

Example – 01:

xy = c

Solution:

Given xy = c ……….. (1)

Differentiating both sides w.r.t. x

x + y(1) = 0

∴ x + y = 0

This is the required differential Equation

Example – 02:

xy² = c²

Solution:

xy² = c²  ……….. (1)

Differentiating both sides w.r.t. x

x . 2y + y² (1) = 0

∴ 2x + y = 0

This is the required differential equation

Example – 03:

y = ce^-x

Solution:

y = ce^-x

∴  ye^x = c ………. (1)

Differentiating both sides w.r.t. x

y.e^x + e^x = 0

∴  + y = 0

This is the required differential Equation

Example – 04:

x² + y² = a²

Solution:

x² + y² = a²     ……………… (1)

Differentiating both sides w.r.t. x

2x + 2y= 0

∴ x + y= 0

This is the required differential Equation

Example – 05:

y = ax + 2

Solution:

y = ax + 2 ………… (1)

Differentiating both sides w.r.t. x

= a(1) = a

Substituting in equation (1)

y = x. + 2

∴ x. – y + 2 = 0

This is the required differential equation

Example – 06:

y = ax + a² + 5

Solution:

y = ax + a² + 5 ……….. (1)

Differentiating both sides w.r.t. x

= a(1) + 0 + 0 = a

Substituting in equation (1)

y = x. + ()² + 5

∴ ()² + x. – y + 5 = 0

This is the required differential equation

Example – 07:

y = ax + 6a² + a³

Solution:

y = ax + 6a² + a³……….. (1)

Differentiating both sides w.r.t. x

= a(1) + 0 + 0 = a

Substituting in equation (1)

y = x. + 6()² + ()³

∴ ()³+ 6()² + x. – y = 0

This is the required differential equation

Example – 08:

y = cx + x²

Solution:

y = cx + x² ……………… (1)

Differentiating both sides w.r.t. x

= c + 2x

∴  c = – 2x

Substituting in equation (1)

y = x( – 2x) + x²

∴  y = x – 2x² + x²

∴  x – x² – y = 0

This is the required differential equation

Example – 09:

(x – a) ² + y² = a²

Solution:

(x – a) ² + y² = a²

∴  x² – 2ax + a² + y² = a²

∴  x² – 2ax + y² = 0

∴  – 2ax + a² + y² = a²

∴  x² + y² = 2ax  ………… (1)

Differentiating both sides w.r.t. x

2x + 2y= 2a

x + y= a

Substituting in equation (1)

∴  x² + y² = 2(x + y)x

∴  x² + y² = 2x² + 2xy

∴ 2xy + x² – y² = 0

This is the required differential equation

Example – 10:

y² = 4ax

Solution:

y² = 4ax ……….. (1)

Differentiating both sides w.r.t. x

2y= 4a

Substituting in equation (1)

y² = 2xy

y = 2x

∴ 2x – y = 0

This is the required differential equation

Example – 11:

x² + y² = 2ax

Solution:

x² + y² = 2ax    ……………. (1)

Differentiating both sides w.r.t. x

2x + 2y= 2a

x + y= a

Substituting in equation (1)

∴  x² + y² = 2(x + y)x

∴  x² + y² = 2x² + 2xy

∴ 2xy + x² – y² = 0

This is the required differential equation

Example – 12:

x²  = 4ay

Solution:

x²  = 4ay  …………. (1)

Differentiating both sides w.r.t. x

∴ 2x = 4a

Substituting in equation (1)

∴ x²  = 2xy

∴ x  = 2y

∴ x  – 2y = 0

This is the required differential equation

Example – 13:

(y – b)² + x² = b²

Solution:

(y – b) ² + x² = b²

∴  y² – 2by + b² + x² = b²

∴  y² – 2by + x² = 0

∴  x² + y² = 2by  ………… (1)

Differentiating both sides w.r.t. x

2x + 2y= 2b

Substituting in equation (1)

∴  x² + y² = 2xy + 2y²

∴  x² – y² – 2xy = 0

∴  (x² – y² ) – 2xy = 0

This is the required differential equation

Example – 14:

y = c² + c/x

Solution:

y = c² + c/x  ………… (1)

Differentiating both sides w.r.t. x

= 0 + c(-1/x²) = -c/x²

c = – x²

Substituting in equation (1)

∴ y = x⁴()² – x.

∴  x⁴()² – x. – y = 0

This is the required differential equation

Example – 15:

e^x + c e^y = 1

Solution:

e^x + c e^y = 1 …… (1)

Differentiating both sides w.r.t. x

e^x + c e^y = 0

c e^y = – e^x

Substituting in equation (1)

This is the required differential equation

Example – 16:

y = ax³ + 4

Solution:

y = ax³ + 4 …………….. (1)

Differentiating both sides w.r.t. x

= l

Substituting in equation (1)

3y = x + 12

x – 3y + 12 = 0

This is the required differential equation

Example – 17:

e^x + e^y = k e^{x + y}

Solution:

e^x + e^y = k e^{x + y}

Differentiating both sides w.r.t. x

This is the required differential equation

Example – 18:

y  = e^cx

Solution:

y  = e^cx

∴  log y  = log e^cx

∴  log y  = cx log e = cx (1)

∴  log y = cx ……….. (1)

Differentiating both sides w.r.t. x

(1/y) = c

Substituting in equation (1)

This is the required differential equation

The post Formation of Differential Equations – 01 (Single Arbitrary Constant) appeared first on The Fact Factor.

Example – 01:

If x + y = 3 show that the maximum value of x² y is 4

Solution:

Given x + y = 3

hence y = 3 – x

∴  x² y = x²(3 – x) = 3x² – x³

Let ƒ(x) = 3x² – x³ ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 6x – 3x² ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 6 – 6x  ………….. (3)

For maximum or minimum value ƒ’(x) = 0

6x – 3x² = 0

∴   3x(2 – x) = 0

∴   x = 0 or/and 2- x = 0

∴   x = 0 or/and x = 2

Let us consider x = 0

ƒ’’(0) =6 – 6(0) = 6 – 0 = 6 > 0

Hence the function is increasing at x = 0 and has minimum value at x = 0

Let us consider x = 3

ƒ’’(2) = 6 –  6(2) = 6 – 12 = -6 < 0

Hence the function is decreasing at x = 2 and has maximum value at x = 2

Substituting x = 2 in equation (1)

Maximum value = ƒ(2) = 3(2)² – (2)³ = 12 – 8 = 4

Thus the maximum value is 4 (Proved as required)

Example – 02:

show that f(x) = x^x is minimum when x = 1/e

Solution:

Let ƒ(x) = x^x  ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = x^x(1 + logx)

Differentiating equation (2) w.r.t. x

ƒ’'(x) = x^x(0 + 1/x) + (1 + logx)x^x(1 + logx)

ƒ’'(x) = x^x/x + x^x(1 + logx)²

ƒ’'(x) = (1/x + 1(1 + logx)²)  ………….. (3)

For maximum or minimum value ƒ’(x) = 0

x^x(1 + logx) = 0

∴  x^x = 0 or/and (1 + logx) = 0

x^x = 0 implies x = 0 makes term logx resundant

Hence x = 0 is not possible

∴  (1 + logx) = 0

∴   log x = -1

∴   x = e^-1

∴   x = 1/e

ƒ’’(1/e) = (1/(1/e) + 1(1 + log(1/e))²)

ƒ’’(1/e) = (e + 1(1 – loge)²) = (e + 0) = e > 0

Hence the function is increasing at x = 1/e and has minimum value at x = 1/e (Proved as required)

Example – 03:

show that f(x) = (logx)/x (x ≠ 0) is maximum at x = e

Solution:

For maximum or minimum value ƒ’(x) = 0

(1 – logx)/x² = 0

∴  (1 – logx) = 0

∴   log x = 1

∴   x = e¹

∴   x = e

Hence the function is decreasing at x = e and has maximum value at x = e (Proved as required)

Example – 04:

Find the maximum and minimum values of x².e^x.

Solution:

Let ƒ(x) = x².e^x  ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = x².e^x + e^x. 2x

ƒ’(x) = x².e^x + 2 x e^x………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = x².e^x + e^x. 2x + 2(x e^x + e^x(1))

ƒ’’(x) = x².e^x + 2x e^x + 2x e^x + 2e^x

ƒ’’(x) = e^x(x² + 2x + 2x + 2)

ƒ’’(x) = e^x(x² + 4x + 2)………….. (3)

For maximum or minimum value ƒ’(x) = 0

x².e^x + 2 x e^x = 0

∴  x.e^x ( x + 2) = 0

e^x ≠ 0

∴   x = 0 or/and x + 2 = 0

∴   x = 0 or/and x = – 2

Let us consider x = 0

ƒ’’(0) = e⁰(0² + 4(0) + 2) = 1(0 + 0 + 2) = 2 > 0

Hence the function is increasing at x =0 and has minimum value at x = 0

Substituting x = 0 in equation (1)

Minimum value = ƒ(2) = 0².e⁰  = 0

Thus point of minimum is (0, 0)

Let us consider x = – 2

ƒ’’(-2) = e^-2((-2)² + 4(-2) + 2) = e^-2(4 – 8 + 2) = – 2/e² < 0

Hence the function is decreasing at x = – 2 and has maximum value at x = -2

Substituting x = – 2 in equation (1)

Maximum value = ƒ(-2) =  (-2)².e^-2  = 4/e²

Thus point of maximum is (- 2, 4/e²)

Ans: The maximum value is 4/e² at x = -2 and

minimum value is -0 at x = 0

Example – 05:

Find extreme values of ƒ(x) = a sin x + b cos x

Solution:

Let ƒ(x) = ƒ(x) = a sin x + b cos x ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = ƒ(x) = a cosx – b sin x  ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = – a sin x – b cos x = – ƒ(x)  ………….. (3)

For maximum or minimum value ƒ’(x) = 0

a cosx – b sin x = 0

a cosx = b sin x

Squaring both sides

a² cos²x = b² sin² x

a² (1 – sin²x) = b² sin² x

a²  – a²sin²x)= b² sin² x

a²  = a²sin²x + b² sin² x

a²  = (a² + b²)sin² x

Hence the function is increasing and has a minimum value

Hence the function is increasing and has a minimum value

The post Maximum and Minimum Value of Function appeared first on The Fact Factor.

Example – 01:

A particle is moving in such a way that is displacement ’s’ at any time ‘t’ is given by s = 2t² + 5t +20. Find the velocity and acceleration of the particle after 2 seconds.

Solution:

The displacement of the particle is given by

s = 2t² + 5t +20  ……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 4t + 5 ……………… (2)

Differentiating both sides of equation (2) w.r.t. t

Acceleration = a = dv/dt = 4 ……………… (3)

To find velocity after 2 seconds i.e. t = 2 s

Substituting in equation (2)

Velocity = v = 4(2) + 5 = 8 + 5 = 13 unit/s

To find acceleration after 2 seconds i.e. t = 2 s

Acceleration = a = 4 units/s²

Ans: The velocity and acceleration of the particle after 2 seconds are 13 units/s and 4 units/s² respectively.

Example – 02:

The displacement ‘s’ of a particle at a time ‘t’ is given by s = 5 + 20t – 2t². Find its acceleration when its velocity is zero.

Solution:

The displacement of a particle is given by

s = 5 + 20t – 2t²  ……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 20 – 4t  ……………… (2)

Differentiating both sides of equation (2) w.r.t. t

Acceleration = a = dv/dt = – 4 ……………… (3)

Given velocity is zero

20 – 4t = 0

4t = 20

T = 5 s

To find acceleration after 5 seconds i.e. t = 5 s

Acceleration = a = – 4 units/s²

Ans: The acceleration of the particle after 5 seconds is – 4 units/s²

Example – 03:

A particle is moving in such a way that is displacement’s’ at any time ‘t’ is given by s = t³ – 4t² – 5t. Find the velocity and acceleration of the particle after 2 seconds.

Solution:

The displacement of the particle is given by

s = t³ – 4t² – 5t  ……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 3t² -8t -5  ……………… (2)

Differentiating both sides of equation (2) w.r.t. t

Acceleration = a = dv/dt = 6t – 8   ……………… (3)

To find velocity after 2 seconds i.e. t = 2 s

Substituting in equation (2)

Velocity = v = 3(2)² -8(2) -5  = 12 – 16 – 5  = – 9 unit/s

To find acceleration after 2 seconds i.e. t = 2 s

Acceleration = a = 6(2) – 8 = 12 – 8 = 4 units/s²

Ans: The velocity and acceleration of the particle after 2 seconds are – 9 units/s and 4 units/s² respectively.

Example – 04:

A particle is moving in such a way that is displacement’s’ at any time ‘t’ is given by s = 2t³ – 5t² +  4t – 3. Find the time when acceleration is 14 ft/s². Also, find velocity and displacement at that time.

Solution:

The displacement of the particle is given by

s = 2t³ – 5t² +  4t – 3……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 6t² – 10t + 4  ……………… (2)

Differentiating both sides of equation (2) w.r.t. t

Acceleration = a = dv/dt = 12t – 10   ……………… (3)

Given acceleration = a = 14 ft/s²

Substituting in equation (3)

12t – 10 = 14

12t = 24

T = 2 s

Substituting t = 2 in equation (2)

Velocity = v = 6t² – 10t + 4  = 6(2)² – 10(2) + 4 =  24 – 20 + 4 = 8 ft/s

Substituting t = 2 in equation (1)

Displacement = s = 2(2)³ – 5(2)² +  4(2) – 3 = 16 – 20 + 8 – 3 = 1 ft

Ans: After 4 s the acceleration will  be 14 ft/s²

The velocity is 8 ft/s and displacement is 1 ft

Example – 05:

A particle moves according to law s = t³ – 6t² + 9t + 15, find the velocity when t = 0

Solution:

The displacement of the particle is given by

s = t³ – 6t² + 9t + 15 ……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 3t² – 12t + 9  ……………… (2)

To find velocity at t = 0

Substituting t = 0 in equation (2)

Velocity = v = 3t(0)² – 12t(0) + 9  = 9 units/s

Ans: The velocity at t = 0 is 9 units/s

Example – 06:

A particle moves under the law s = t³ – 4t² – 5t. Find the displacement and velocity of particle when its acceleration is 4 units.

Solution:

The displacement of the particle is given by

s = t³ – 4t² – 5t ……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 3t² – 8t – 5  ……………… (2)

Differentiating both sides of equation (2) w.r.t. t

Acceleration = a = dv/dt = 6t – 8   ……………… (3)

Given acceleration = a = 4 units

Substituting in equation (3)

6t – 8 = 4

6t = 12

t = 2 s

Substituting t = 2 in equation (1)

Displacement = s = (2)³ – 4(2)² – 5(2) = 8 – 16 – 10 = – 18 units

Substituting t = 2 in equation (2)

Velocity = v = 3(2)² – 8(2) – 5 = 12 – 16 – 5 = – 9 units/s

Ans: When acceleration is 4 units displacement is -18 units and velocity is – 8 units/s

Example – 07:

A particle moves under the law s = t³/3 – t²/2 – t/2 +6. Find (i) its velocity at end of 4 s and (ii) acceleration and displacement when its velocity is 3/2 units.

Solution:

The displacement of the particle is given by

s = t³/3 – t²/2 – t/2 +6……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 3t²/3 – 2t/2 – 1/2 = t² – t – 1/2      ……………… (2)

Differentiating both sides of equation (2) w.r.t. t

Acceleration = a = dv/dt = 2t – 1   ……………… (3)

(i) To find its velocity at end of 4 s, t = 4 s

Substituting t = 4 in equation (2)

Velocity = v = (4)² – (4) – 1/2  = 16 – 4 – 1/2 = 11.5 units/s

(ii) To find acceleration and velocity when its velocity is 3/2 units.

t² – t – 1/2 = 3/2

t² – t – 2 = 0

(t – 2)(t + 1) = 0

t – 2 = 0 and t + 1 = 0

t = 2  and t = – 1

Time cannot be negative hence t = -1 not possible

t = 2 s

Substituting t = 2 in equation (3)

Acceleration =  2(2) – 1= 3 units/s²

Substituting t = 2 in equation (1)

Displacement = s = (2)³/3 – (2)²/2 – (2)/2 +6 = 8/3 – 2 – 1 + 6 = 17/3 units

Ans: Velocity at end of 4 s is 11.5 units/s

When velocity is 3/2 units, acceleration is 3 units/s² and displacement is 17/3 units

Example – 08:

The displacement x of a particle at time t is given by x = 160 t – 16t², show that its velocity at t = 1 and t =9 are equal in magnitude and opposite in direction.

Solution:

The displacement of the particle is given by

s = 160 t – 16t²……………… (1)

Differentiating both sides of equation (1) w.r.t. t

Velocity = v = ds/dt = 160 – 32t      ……………… (2)

Velocity at t = 1

Velocity = 160 – 32(1) = 128 units/s

Velocity at t = 9

Velocity = 160 – 32(9) = – 128 units/s

We can seet hat the velocities at t = 1 and t =9 are equal in magnitude and opposite in direction. (Proved)

Example – 09:

A particle is moving in a straight line and its displacement x from a fixed point O on the line at time t is given by . Show that the acceleration at time t is x^-3.

Solution:

The displacement of the particle is given by

Thus acceleration at time t is x^-3 (Proved as required)

The post Displacement, Velocity, and Acceleration appeared first on The Fact Factor.

Example – 01:

Find the maximum and minimum values of x³ – 12x – 5

Solution:

Let ƒ(x) = x³ – 12x – 5 ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 3x² – 12 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 6x ………….. (3)

For maximum or minimum value ƒ’(x) = 0

3x² – 12 = 0

∴   3(x² – 4) = 0

∴   3(x + 2)(x – 2) = 0

∴   x + 2 = 0 or/and x – 2 = 0

∴   x = – 2 or/and x = 2

Let us consider x = 2

ƒ’’(2) = 6 x 2 = 12 > 0

Hence the function is increasing at x = 2 and has minimum value at x = 2

Substituting x = 2 in equation (1)

Minimum value = ƒ(2) = (2)³ – 12(2) – 5 = 8 – 24 – 5 = – 21

Thus point of minimum is (2, -21)

Let us consider x = – 2

ƒ’’(-2) = 6 x (- 2) = – 12 < 0

Hence the function is decreasing at x = – 2 and has maximum value at x = -2

Substituting x = – 2 in equation (1)

Maximum value = ƒ(-2) = (-2)³ – 12(-2) – 5 = – 8 + 24 – 5 = 11

Thus point of maximum is (- 2, 11)

Ans: The maximum value is 11 at x = -2 and minimum value is – 21 at x = 2

Example – 02:

Find the maximum and minimum values of x³ – 9x² + 24 x

Solution:

Let ƒ(x) = x³ – 9x² + 24 x ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 3x² –  18x + 24 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 6x – 18 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

3x² –  18x + 24 = 0

∴   x² –  6x + 8 = 0

∴   (x – 4)(x – 2) = 0

∴   x – 4 = 0 or/and x – 2 = 0

∴   x = 4 or/and x = 2

Let us consider x = 4

ƒ’’(4) = 6 x 4 – 18 = 24 – 18 = 6 > 0

Hence the function is increasing at x = 4 and has minimum value at x = 4

Substituting x = 4 in equation (1)

Minimum value = ƒ(4) = (4)³ – 9(4)² + 24(4) = 64 – 144 + 96 = 16

Thus point of minimum is (2, -21)

Let us consider x = – 2

ƒ’’(-2) = 6 x (- 2) – 18 = -12 – 18 = –  30 < 0

Hence the function is decreasing at x = 2 and has maximum value at x = 2

Substituting x = 2 in equation (1)

Maximum value = ƒ(2) = (2)³ – 9(2)² + 24(2) = 8 – 36 + 48 = 20

Thus point of maximum is (2, 20)

Ans: The maximum value is 20 at x = 2 and minimum value is 16 at x = 4

Example – 03:

Find the maximum and minimum values of 2x³ – 3x² – 36x + 10

Solution:

Let ƒ(x) = 2x³ – 3x² – 36x + 10 ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 6x² –  6x – 36 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 12x – 6 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

6x² –  6x – 36 = 0

∴   x² –  x – 6 = 0

∴   (x – 3)(x + 2) = 0

∴   x – 3 = 0 or/and x + 2 = 0

∴   x = 3 or/and x = – 2

Let us consider x = 3

ƒ’’(3) = 12(3) – 6 = 36 – 6 = 30 > 0

Hence the function is increasing at x = 3 and has minimum value at x = 3

Substituting x = 3 in equation (1)

Minimum value = ƒ(3) = 2(3)³ – 3(3)² – 36(3) + 10 = 54 – 27 – 108 = -71

Thus point of minimum is (3, -71)

Let us consider x = – 2

ƒ’’(2) = 12(-2) – 6 = – 24 – 6 = – 30 < 0

Hence the function is decreasing at x = – 2 and has maximum value at x = – 2

Substituting x = – 2 in equation (1)

Maximum value = ƒ(-2) = 2(-2)³ – 3(-2)² – 36(-2) + 10 = – 16 – 12 + 72 + 10 = 54

Thus point of maximum is (-2, 54)

Ans: The maximum value is 54 at x = – 2 and minimum value is – 71 at x = 3

Example – 04

Find the maximum and minimum values of x³ + 3x² – 2

Solution:

Let ƒ(x) = x³ + 3x² – 2  ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 3x² +  6x  ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 6x + 6 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

3x² +  6x = 0

∴   3x(x  +  2) = 0

∴   x  = 0 or/and x + 2 = 0

∴   x = 0 or/and x = – 2

Let us consider x = 0

ƒ’’(-2) = 6(0) + 6 = 6  > 0

Hence the function is increasing at x = 0 and has minimum value at x = 0

Substituting x = 0 in equation (1)

Minimum value = ƒ(0) = (0)³ + 3(0)² – 2 = 0 + 0 – 2 = -2

Thus point of minimum is (0, -2)

Let us consider x = – 2

ƒ’’(-2) = 6(-2) + 6 = -12 + 6 = – 6 < 0

Hence the function is decreasing at x = – 2 and has maximum value at x = – 2

Substituting x = – 2 in equation (1)

Maximum value = ƒ(-2) = (-2)³ + 3(-2)² – 2 = – 8 + 12 – 2 = 2

Thus point of maximum is (-2, 2)

Ans: The maximum value is 2 at x = – 2 and minimum value is – 2 at x = 0

Example – 05:

Find the maximum and minimum values of 3x³ – 9x² – 27x + 15

Solution:

Let ƒ(x) = 3x³ – 9x² – 27x + 15  ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 9x² –  18x – 27 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 18x – 18 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

9x² –  18x – 27 = 0

∴   x² –  2x – 3 = 0

∴   (x – 3)(x + 1) = 0

∴   x – 3 = 0 or/and x + 1 = 0

∴   x = 3 or/and x = – 1

Let us consider x = 3

ƒ’’(3) = 18(3) – 18 = 54 – 18 = 36 > 0

Hence the function is increasing at x = 3 and has minimum value at x = 3

Substituting x = 3 in equation (1)

Minimum value = ƒ(3) = 3(3)³ – 9(3)² – 27(3) + 15 = 81 – 81 -81 +15 = – 66

Thus point of minimum is (3, – 66)

Let us consider x = – 1

ƒ’’(2) = 18(-1) – 18 = – 18 – 18 = – 36 < 0

Hence the function is decreasing at x = – 1 and has maximum value at x = – 1

Substituting x = – 1 in equation (1)

Maximum value = ƒ(-1) = 3(-1)³ – 9(-1)² – 27(-1) + 15 = – 3 – 9 + 27 + 15 = 30

Thus point of maximum is (-1, 30)

Ans: The maximum value is 30 at x = – 1 and minimum value is – 66 at x = 3

Example – 06:

Find the maximum and minimum values of 2x³ – 21x² + 36x – 20

Solution:

Let ƒ(x) = 2x³ – 21x² + 36x – 20  ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 6x² –  42x + 36 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 12x – 42 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

6x² –  42x + 36 = 0

∴   x² –  7x + 6 = 0

∴   (x – 6)(x – 1) = 0

∴   x – 6 = 0 or/and x – 1 = 0

∴   x = 6 or/and x = 1

Let us consider x = 6

ƒ’’(6) = 12(6) – 42 =72 – 42 = 30 > 0

Hence the function is increasing at x = 6 and has minimum value at x = 6

Substituting x = 6 in equation (1)

Minimum value = ƒ(3) = 2(6)³ – 21(6)² + 36(6) – 20 = 432 – 756 = 216 – 20 = – 128

Thus point of minimum is (6, – 128)

Let us consider x = 1

ƒ’’(2) = 12(1) – 42 = 12 – 42 = – 30 < 0

Hence the function is decreasing at x = 1 and has maximum value at x = 1

Substituting x = 1 in equation (1)

Maximum value = ƒ(1) = 2(1)³ – 21(1)² + 36(1) – 20 = 2 – 21 +36 – 20 = – 3

Thus point of maximum is (1, -3)

Ans: The maximum value is -3 at x =  1 and minimum value is – 128 at x = 6

Example – 07:

Find the maximum and minimum values of x² + 16/x²

Solution:

Let ƒ(x) = x² + 16/x²  

ƒ(x) = x² + 16 x ^-2  ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 2x – 32x ^-3 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 2 + 96 x ^-4 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

2x – 32x ^-3 = 0

∴  2x = 32x ^-3

∴  2x = 32/x³

∴  x⁴ = 16

∴   x = ± 2

Let us consider x = 2

ƒ’’(2) = 2 + 96 x (2) ^-4 = 2 + 96/16 = 2 + 6 = 8 > 0

Hence the function is increasing at x = 2 and has minimum value at x = 2

Substituting x = 2 in equation (1)

Minimum value = ƒ(2) = 2² + 16/2² = 4 + 4 = 8

Thus point of minimum is (2, 8)

Let us consider x = – 2

ƒ’’(2) = – 2 + 96 x (- 2) ^-4 = – 2 + 96/16 = – 2 + 6 = 4 > 0

Hence the function is increasing at x = – 2 and has minimum value at x = – 2

Substituting x = – 2 in equation (1)

Minimumm value = ƒ(-2) =(-2)² + 16/(-2)² = 4 + 4 = 8

Thus point of minimum is (- 2, 8)

Ans: The minimum value is 8 at x = ±2

Example – 08:

Find the maximum and minimum values of x.logx

Solution:

Let ƒ(x) = x.logx  ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = x(1/x) + logx. (1)

ƒ’(x) = 1 + logx ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 1/x  ………….. (3)

For maximum or minimum value ƒ’(x) = 0

1 + logx = 0

∴  log x = -1

∴   x = e^-1 = 1/e

ƒ’’(1/e) = 1/(1/e) = e > 0

Hence the function is increasing at x = 1/e and has minimum value at x = 1/e

Substituting x = 1/e in equation (1)

Minimum value = ƒ(1/e) = (1/e). log(1/e) = – (1/e). log(e) =  – (1/e)

Thus point of minimum is (1/e, – 1/e)

Ans: The minimum value is – 1/e at x = 1/e

The post Concept of Maximum and Minimum Values of a Function appeared first on The Fact Factor.

Example – 12:

Find the equation of the tangent and normal to the curve x = sin θ and y = cos 2θ at θ = π/6

Solution:

The equation of the curve is x = sin θ and y = cos 2θ ………. (1)

Parameter θ = π/6

∴  x = sin π/6 = 1/2 and y = cos 2(π/6) = cos π/3 = ½

Therefore the point on the curve is P(1/2, 1/2)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(1/2, 1/2) is -2 and that of normal is 1/2.

Equation of tangent:

Its slope = m = -2

It passes through P(1/2, 1/2) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 1/2 = -2(x – 1/2)

∴  y – 1/2 = -2x + 1

∴  2y – 1 = -4x + 2

∴  4x + 2y – 3 = 0

Equation of normal:

By slope point form

y – y₁ = m(x – x₁)

∴  y – 1/2 = (1/2)(x – 1/2)

∴  2y – 1 = x – 1/2

∴  4y – 2 = 2x – 1

∴  2x – 4y + 1= 0

Ans: Equation of tangent is 4x + 2y – 3 = 0 and that of normal is 2x – 4y + 1= 0

Example – 13:

Find the equation of the tangent and normal to the curve x = acos³θ and y = asin³θ at θ = π/4

Solution:

Equation of curve is x = acos³θ and y = asin³θ ………. (1)

Parameter θ = π/4

∴  x = acos³ (π/4) = a(1/√2)³ = a (1/2√2) = a/2√2

∴  y = asin³ (π/4) = a(1/√2)³ = a (1/2√2) = a/2√2

Therefore the point on the curve is P(a/2√2, a/2√2)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(a/2√2, a/2√2) is – 1 and that of normal is 1.

Equation of tangent:

Its slope = m = – 1

It passes through P(a/2√2, a/2√2) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – a/2√2 = -1(x – a/2√2)

∴  2√2 y – a = – 2√2 x + a

∴  2√2 x + 2√2 y – 2a = 0

∴  √2 x + √2 y –a = 0

Equation of normal:

By slope point form

y – y₁ = m(x – x₁)

∴  y – a/2√2 = 1(x – a/2√2)

∴  y – a/2√2 = x – a/2√2

∴  2√2 y – a = 2√2 x – a

∴  2√2 x – 2√2 y = 0

∴  x – y = 0

Ans: Equation of tangent is √2 x + √2 y – a = 0 and that of normal is x – y = 0

Example – 14:

Find the equation of the tangent and normal to the curve x = a sec θ and y = a tan θ at θ = π/6

Solution:

Equation of curve is x = a sec θ and y = a tan θ ………. (1)

Parameter θ = π/6

∴  x = a sec (π/6) = 2a/√3 and y = a tan (π/6) = a/√3

Therefore the point on the curve is P(2a/√3, a/√3)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(2a/√3, a/√3) is 2 and that of normal is – 1/2.

Equation of tangent:

Its slope = m = 2

It passes through P(2a/√3, a/√3) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – a/√3 = 2(x – 2a/√3)

∴  y – a/√3 = 2x – 4a/√3

∴  √3 y – a  = 2 √3 x – 4a

∴  2√3 x – √3 y – 3a = 0

∴  2x – y –  √3a = 0

Equation of normal:

By slope point form

y – y₁ = m(x – x₁)

∴  y – a/√3 = (-1/2)(x – 2a/√3)

∴  2y – 2a/√3= – x + 2a/√3

∴  2√3 y – 2a  = – √3 x + 2a

∴  √3 x + 2√3 y – 4a = 0

Ans: Equation of tangent is 2x – y –  √3 a = 0 and that of normal is √3 x + 2√3 y – 4a = 0

Example – 15:

Find the equation of the tangent to the curve x = a(θ  + sin θ) and y = a(1  + cos θ) at θ = π/2

Solution:

Equation of curve is x = a(θ  + sin θ) and y = a(1  + cos θ) ………. (1)

Parameter θ = π/6

∴  x = a(θ  + sin θ) = a(π/2 + sin π/2)=a(π/2 + 1) and

∴  y = a(1  + cos θ) = a(1 + cos π/2) = a(1 + 0) = a

Therefore the point on the curve is P(a(π/2 + 1), a)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(a(π/2 + 1), a) is -1 and that of normal is 1.

Equation of tangent:

Its slope = m = -1

It passes through P(a(π/2 + 1), a) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – a = -1(x – a(π/2 + 1))

∴  y – a = – x + a(π/2) + a

∴  x + y – a – a(π/2) – a = 0

∴  x – y – a(π/2)  – 2a = 0

Ans: Equation of tangent is x – y – a(π/2)  – 2a = 0

Example – 16:

Find the equation of tangent and normal to the curve x = 1/t and y = t – 1/t at t = 2

Solution:

Equation of curve is x = 1/t and y = t – 1/t ………. (1)

Parameter t = 2

∴  x = 1/2 and y = 2 – 1/2 = 3/2

Therefore the point on the curve is P(1/2, 3/2)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(1/2, 3/2) is -5 and that of normal is 1/5.

Equation of tangent:

Its slope = m = – 5

It passes through P(1/2, 3/2) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 3/2   = – 5(x – 1/2)

∴  y – 3/2   = – 5x + 5/2

∴  5x + y – 3/2 – 5/2 = 0

∴  5x + y – 4 = 0

Equation of normal:

By slope point form

y – y₁ = m(x – x₁)

∴  y – 3/2   = 1/5(x – 1/2)

∴  5y – 15/2 = x – 1/2

∴  x – 5y – 1/2 + 15/2 = 0

∴  x – 5y + 7 = 0

Ans: Equation of tangent is 5x + y – 4 = 0 and that of normal is x – 5y + 7 = 0

Example – 17:

Show that the tangent to the curve 8y = (x -2)² at the point (-6, 8) is parallel to the tangent to the curve y = x + 3/x at point (1, 4)

Solution:

Consider the first curve 8y = (x -2)²

Differentiating both sides w.r.t. x

8 (dy/dx) = 2(x – 2) (1)

(dy/dx) = (x – 2)/4

∴  (dy/dx)_{at P(-6, 8)} = (- 6 – 2)/4 = -8/4 = -2

Thus the slope of the tangent to curve 8y = (x -2)²

at the point (-6, 8) is – 2 ………….. (1)

Consider the second curve y = x + 3/x

Differentiating both sides w.r.t. x

(dy/dx) = 1 – 3/x²

∴  (dy/dx)_{at Q(1, 4)} =  1- 3/1² = 1- 3 = -2

Thus slope of tangent to curve y = x + 3/x

at the point (1, 4) is – 2 ………….. (2)

From statements (1) and (2) we can see that the two slopes are equal.

Hence the two tangents are parallel to each other (proved as required)

Example – 18:

Show that the tangents to the curve x² = 2y and 6y = 5 – 2x³ at x = 1 are perpendicular to each other.

Solution:

Consider the first curve x² = 2y

Differentiating both sides w.r.t. x

2x = 2(dy/dx)

∴  (dy/dx) = x

∴  (dy/dx)_{at x = 1} = 1

Thus slope of tangent to curve x² = 2y at x = 1 is 1 ………….. (1)

Consider the second curve 6y = 5 – 2x³

Differentiating both sides w.r.t. x

6dy/dx = -6x²

∴  (dy/dx) = -x²

∴  (dy/dx)_{at x = 1} =  -(1)² = -1

Thus slope of tangent to curve 6y = 5 – 2x³ at x = 1 is -1 ………….. (2)

From statements (1) and (2) we can see that the product of the two slopes is -1.

Hence the two tangents are perpendicular to each other (proved as required)

Example – 19:

Find the points on the curve y = 3x² – 9x + 8 where the tangent is parallel to x- axis

Solution:

The tangent is parallel to x-axis hence its slope m = (dy/dx) = 0

The equation of the curve is y = 3x² – 9x + 8  …………… (1)

Differentiating both sides w.r.t. x

(dy/dx) = 6x – 9

Thus 6x – 9 = 0

∴  6x = 9

∴  x = 3/2

Substituting in equation (1)

y = 3(3/2)² – 9(3/2) + 8= 3(9/4) – 27/2 + 8

∴  y = 27/4 -27/2 + 8 = -27/4 + 8 = 5/4

The point on the curve is (3/2, 5/4)

Example – 20:

Find the points on the curve y = 3x² – 9x + 8 where the tangent makes an angle 45^o with the x-axis

Solution:

The tangent makes an angle 45^o with the x-axis

Slope of tangent = m = tan 45^o = 1 = (dy/dx)

The equation of the curve is y = 3x² – 9x + 8  …………… (1)

Differentiating both sides w.r.t. x

(dy/dx) = 6x – 9

Thus 6x – 9 = 1

∴  6x = 10

∴  x = 5/3

Substituting in equation (1)

y = 3(5/3)² – 9(5/3) + 8= 3(25/9) – 15 + 8

∴  y = 25/3 – 15 + 8 = 25/3 – 7 = 4/3

The point on the curve is (5/3, 4/3)

Example – 21:

Find the point on the curve y = √x – 3 where the tangent is perpendicular to the line 6x + 3y – 5 = 0

Solution:

Equation of given line 6x + 3y – 5 = 0

Slope of given line = – (coefficient of x/coefficient of y) = – (6/3) = – 2

As the tangent is perpendicular to given line

Slope of tangent = 1/2

Equation of curve is y = √x – 3

Differentiating both sides w.r.t. x

squaring both sides

1 = x – 3

∴  x = 4

Substituting in equation (1)

y = √4 – 3 = √1  = ±1

The points on the curve are (4, 1) and (4, -1)

Example – 22:

Find the coordinates of the point on the curve y = x – 4/x, where the tangent at that point is parallel to the line y = 2x.

Solution:

Equation of given line y = 2x

Slope of given line = 2

Now the tangent is parallel to the given line

Slope of tangent = m = 2 = (dy/dx)

Equation of the curve is y = x – 4/x

Differentiating both sides w.r.t.x

(dy/dx) = 1 – 4(- 1/x²) = 1 + 4/x²

∴  1 + 4/x² = 2

∴  4/x² = 1

∴  x² = 4

∴  x = ± 2

When x = 2

y = 2 – 4/2 = 2 – 2 = 0

Hence the point is (2, 0)

When x = -2

y = – 2 – 4/(-2) = – 2 + 2 = 0

Hence the point is (- 2, 0)

The required points are (2, 0) and (-2, 0)

Example – 23:

If the line x + y = 0 touches the curve 2y² = ax² + b at (1, -1) find a and b

Solution:

Equation of tangent is x + y = 0

Slope of tangent = – (coefficient of x/coefficient of y) = – (1/1) = -1

Equation of curve is 2y² = ax² + b

Differentiating both sides w.r.t. x

4y (dy/dx) = 2ax

∴  (dy/dx)=ax/2y

∴  (dy/dx)_{at P(1, -1)} = a(1)/2(-1) = – a/2

– a/2 = -1

∴  a = 2

Point (1, -1) lies on the curve 2y² = ax² + b

2(-1)² = 2(1)² + b

∴  2 = 2 + b

∴  b = 0

Ans: a = 2 and b = 0

Example – 24:

For the curve y = 3x – 2x², find the equation of the tangent whose slope is – 1.

Solution:

Slope of tangent = -1

Equation of curve is y = 3x – 2x²   ………. (1)

Differentiating both sides w.r.t. x

(dy/dx) = 3 – 4x

∴  3 – 4x = -1

∴  4 = 4x

∴  x = 1

Substituting in equation (1)

y = 3(1) – 2(1)² = 3 – 2 = 1

The tangent touches the curve at (1, 1) ≡ P(x₁, y₁) and its slope = m = -1

By slope point form

y – y₁ = m(x – x₁)

∴  y – 1 = – 1(x – 1)

∴  y – 1 = – x + 1

∴ x + y – 2 = 0

Ans: The equation of tangent is x + y – 2 = 0

Example – 25:

Find equations of tangents and normal to the curve y = 6 – x², where the normal is parallel to the cline x – 4y + 3 = 0

Solution:

Equation of given line x – 4y + 3 = 0

Slope of given line = – (coefficient of x/coefficient of y) = – (1/-4) = 1/4

As the normal is perpendicular to given line

Slope of normal = 1/4

Slope of tangent = – 4

The equation of curve isy = 6 – x²

Differentiating both sides w.r.t. x

(dy/dx) = – 2x = – 4

x = 2

Substituting in equation of curve

y = 6 – 2² = 6 – 4 = 2

The tangent touches the curve at (2, 2)

Equation of tangent:

Its slope = m = – 4

It passes through P(2, 2) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 2 = – 4(x – 2)

∴  y – 2 = – 4x + 8

∴  4x + y – 10 = 0

Equation of normal:

Its slope = m = 1/4

It passes through P(2, 2) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 2 = 1/4(x – 2)

∴  4y – 8 = x – 2

∴  x – 4y + 6 = 0

Ans: Equation of tangent is 4x + y – 10 = 0 and that of normal is x – 4y + 6 = 0

Example – 26: 

If the line y = 4x -5 touches the curve y² = ax³ + b at the point (2, 3) show that 7a + 2b = 0

Solution:

Equation of tangent is 4x – y -5 = 0

Slope of the tangent = – (coefficient of x/coefficient of y) = – (4/-1) = 4 = (dy/dx)

Equation of curve is

y² = ax³ + b

Differentiating both sides w.r.t. x

2y(dy/dx) = 3ax²

∴  (dy/dx) = 3ax²/2y

∴  (dy/dx) _{at(2, 3)} = 3a(2)²/ 2(3)= 2a

Thus slope of tangent to the curve at (2, 3) is 2a

2a = 4

∴  a = 2

Now point (2, 3) lies on the curve y² = ax³ + b

3² = 2(2)³ + b

∴  9 = 16 + b

∴   b = 9 – 16 = -7

To prove that 7a + 2b = 0

L.H.S. = 7a + 2b = 7(2) + 2(-7) = 14 – 14 = 0 = R.H.S.

∴ 7a + 2b = 0 (Proved as required)

The post Equations of tangents and Normals – 02 appeared first on The Fact Factor.

Example – 01:

Find the equation of the tangent and normal to the curve y = 3x² – x + 1 at point P(1, 3)

Solution:

Equation of curve is y = 3x² – x + 1    ………… (1)

Differentiating equation (1) w.r.t. x

(dy/dx) = 6x – 1

∴  (dy/dx)_{at P(1, 3)} = 6(1) – 1 = 6 – 1 = 5

The slope of the tangent at P(1, 3) is 5 and that of normal is – 1/5.

Equation of tangent:

Its slope = m = 5

It passes through P(1, 3) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 3 = 5(x – 1)

∴  y – 3 = 5x – 5

∴  5x – y – 2 = 0

Equation of normal:

Its slope = m = – 1/5

It passes through P(1, 3) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 3 = (- 1/5)(x – 1)

∴  5y – 15 = -x + 1

∴  x + 5y -15 – 1 = 0

∴  x + 5y – 16 = 0

Ans: Equation of tangent is 5x – y – 2 = 0 and that of normal is x + 5y – 16 = 0

Example – 02:

Find the equation of the tangent and normal to the curve y = x² + 4x + 1 = 0 at point P(-1, -2)

Solution:

Equation of curve is y = x² + 4x + 1    ………… (1)

Differentiating equation (1) w.r.t. x

(dy/dx) = 2x + 4

∴  (dy/dx)_{at P(-1, -2)} = 2(-1) + 4 = -2 + 4 = 2

The slope of the tangent at P(-1, -2) is 2 and that of normal is – 1/2.

Equation of tangent:

Its slope = m = 2

It passes through P(-1, -2) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y + 2 = 2(x + 1)

∴  y + 2 = 2x + 2

∴  2x – y = 0

Equation of normal:

Its slope = m = -1/2

It passes through P(-1, -2) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y + 2 = (- 1/2)(x + 1)

∴  2y + 4  = -x – 1

∴  x + 2y + 5= 0

Ans: Equation of tangent is 2x – y = 0 and that of normal is x + 2y + 5= 0

Example – 03:

Find the equation of the tangent and normal to the curve 2x² + 3y² – 5 = 0 at point P(1, 1)

Solution:

Equation of curve is 2x² + 3y² – 5 = 0     ………… (1)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(1, 1) is – 2/3 and that of normal is 3/2.

Equation of tangent:

Its slope = m = – 2/3

It passes through P(1, 1) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 1 =(- 2/3)(x – 1)

∴  3y – 3 = -2x + 2

∴  2x + 3y – 5 = 0

Equation of normal:

Its slope = m = 3/2

It passes through P(1, 1) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 1 =(3/2)(x – 1)

∴  2y – 2 = 3x – 3

∴  3x – 2y – 1 = 0

Ans: Equation of tangent is 2x + 3y – 5 = 0 and that of normal is 3x – 2y – 1 = 0

Example – 04:

Find the equation of the tangent and normal to the curve x² + y³ + xy = 3 at point P(1, 1)

Solution:

Equation of curve is x² + y³ + xy = 3      ………… (1)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(1, 1) is – 3/4 and that of normal is 4/3.

Equation of tangent:

Its slope = m = – 4/3

It passes through P(1, 1) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 1 =(- 3/4)(x – 1)

∴  4y – 4 = -3x + 3

∴  3x + 4y – 7 = 0

Equation of normal:

Its slope = m = 3/4

It passes through P(1, 1) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 1 =(4/3)(x – 1)

∴  3y – 3 = 4x – 4

∴  4x – 3y – 1 = 0

Ans: Equation of tangent is 3x + 4y – 7 = 0 and that of normal is 4x – 3y – 1 = 0

Example – 05:

Find the equation of the tangent and normal to the curve xy = c² at point P(ct, c/t). Where t is a parameter

Solution:

Equation of curve is xy = c²     ………… (1)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(ct, c/t) is – 1/t² and that of normal is t².

Equation of tangent:

Its slope = m = – 1/t²

It passes through P(ct, c/t) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – c/t =(- 1/t²)(x – ct)

∴  t² (y – c/t) = – x + ct

∴  y t² – ct = – x + ct

∴  x + y t² – 2ct = 0

Equation of normal:

Its slope = m = t²

It passes through P(ct, c/t) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – c/t = t²(x – ct)

∴  yt – c  = t³(x – ct)

∴  yt – c  = x t³ – c t⁴

∴  t³x – yt + c – c t⁴ = 0

Ans: Equation of tangent is x + y t² – 2ct = 0 and that of normal is t³x – yt + c – c t⁴ = 0

Example – 06:

Find the equation of the tangent and normal to the curve √x – √y = 1 at P(9, 4)

Solution:

Equation of curve is √x – √y = 1     ………… (1)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(9, 4) is 2/3 and that of normal is -3/2.

Equation of tangent:

Its slope = m = 2/3

It passes through P(9, 4) ≡ P(x₁, y₁)\

By slope point form

y – y₁ = m(x – x₁)

∴  y – 4 =(2/3)(x – 9)

∴  3y – 12 = 2x – 18

∴  2x – 3y – 6 = 0

Equation of normal:

By slope point form

y – y₁ = m(x – x₁)

∴  y – 4 =(- 3/2)(x – 9)

∴  2y – 8 = – 3x + 27

∴  3x + 2y – 35 = 0

Ans: Equation of tangent is 2x – 3y – 6 = 0 and that of normal is 3x + 2y – 35 = 0

Example – 07:

Find the equation of the tangent and normal to the curve y = x³ – x² – 1 at a point whose abscissa is -2.

Solution:

Equation of curve is y = x³ – x² – 1   ………… (1)

Let P be the point whose abscissa (x-coordinate) is -2

Substituting in equation (1)

y = x³ – x² – 1 = (-2)³ – (-2)² – 1 = – 8 – 4 – 1 = -13

Hence coordinates of point P are (-2, -13)

Differentiating equation (1) w.r.t. x

(dy/dx) = 3x² – 2x

∴  (dy/dx)_{at P(-2, -13)} = 3(-2)² – 2(-2) = 12 + 4 = 16

The slope of the tangent at P(-2, -13) is 16 and that of normal is – 1/16.

Equation of tangent:

Its slope = m = 16

It passes through P(-2, -13) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y + 13 = 16 (x + 2)

∴  y + 13 = 16x + 32

∴  16x – y +32 -13 = 0

∴  16x – y + 19 = 0

Equation of normal:

Its slope = m = 1- 1/6

It passes through P(-2, -13) ≡ P(x₁, y₁)

slope point form

y – y₁ = m(x – x₁)

∴  y + 13 = (-1/16) (x + 2)

∴  16y + 208 = – x – 2

∴  x + 16y + 208 + 2 = 0

∴  x + 16y + 210 = 0

Ans: Equation of tangent is 16x – y + 19 = 0 and that of normal is x + 16y + 210 = 0

Example – 08:

Find the equation of the tangent and normal to the curve y = x² + 4x at a point whose ordinate is -3.

Solution:

Equation of curve is y = x² + 4x         ………… (1)

Let P be the point whose ordinate (y-coordinate) is -3

Substituting in equation (1)

y = x² + 4x = – 3

∴  x² + 4x + 3 = 0

∴  (x + 3)(x + 1) = 0

∴  x + 3 = 0 or/and x + 1 = 0

∴  x = -3 or/and x = -1

Hence the points are P(-3, -3) and Q (-1, -3)

Differentiating equation (1) w.r.t. x

(dy/dx) = 2x + 4

Let us consider point P(-3, -3)

(dy/dx)_{at P(-3, -3)} = 2(-3) + 4 = – 6 + 4 = – 2

The slopes of the f tangent at P(-3, -3) is -2 and that of normal is 1/2.

• Equation of tangent:

Its slope = m = -2

It passes through P(-3, -3) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y + 3 = -2 (x + 3)

∴  y + 3 = – 2x – 6

∴  2x + y + 9 = 0

• Equation of normal:

Its slope = m = 1/2 It passes through P(-3, -3) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y + 3 = (1/2)(x + 3)

∴  2y + 6 = x + 3

∴  x – 2y – 3= 0

Let us consider point P(-1, -3)

(dy/dx)_{at P(-1, -3)} = 2(-2) + 4

The slope of the tangent at P(-1, -3) is 2 and that of normal is – 1/2.

• Equation of tangent:

Its slope = m = 2

It passes through P(-1, -3) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y + 3 = 2 (x + 1)

∴  y + 3 =  2x + 2

∴  2x – y – 1 = 0

• Equation of normal:

Its slope = m = – 1/2

It passes through P(-1, -3) ≡ P(x₁, y₁)

By slope point form

∴  y – y₁ = m(x – x₁)

∴  y + 3 = (-1/2)(x + 1)

∴  2y + 6 = – x – 1

∴  x + 2y + 7 = 0

Ans: At point (-3, -3) equation of tangent is 2x + y + 9 = 0 and that of normal is x – 2y – 3= 0

At point (-1, -3) equation of tangent is 2x – y – 1 = 0 and that of normal is x + 2y + 7 = 0

Example – 09:

Find the equation of the tangent and normal to the curve y = x² – 5x at a point where the curve meets the x-axis.

Solution:

The equation of the curve is y = x² – 5x  …………….. (1)

Let P be the point at which the curve meets x-axis (y = 0).

Substituting y = 0 in equation (1)

0 = x² – 5x

∴  x(x – 5) = 0

∴  x = 0 or x = 5

Hence the curve cuts x-axis at P(0, 0) and Q(5, 0)

Differentiating equation (1) w.r.t. x

(dy/dx) = 2x – 5

Consider point P(0, 0)

(dy/dx)_{at P(0, 0)} = 2(0) – 5 = -5

The slope of the tangent at P(0, 0) is – 5 and that of normal is 1/5.

• Equation of tangent:

Its slope = m = – 5

It passes through P(0, 0) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 0 = -5 (x – 0)

∴  y = – 5x

∴  5x + y = 0

• Equation of normal:

Its slope = m = 1/5

It passes through P(0, 0) ≡ P(x₁, y₁)

By slope point form

∴  y – y₁ = m(x – x₁)

∴  y – 0 = (1/5) (x – 0)

∴  5y = x

∴  x – 5y = 0

Consider point P(5, 0)

(dy/dx)_{at P(0, 0)} = 2(5) – 5 = 5

The slope of the tangent at P(0, 0) is 5 and that of normal is – 1/5.

• Equation of tangent:

Its slope = m =  5

It passes through P(5, 0) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 0 = 5 (x – 5)

∴  y =  5x – 25

∴  5x – y – 25 = 0

• Equation of normal:

Its slope = m =  – 1/5

It passes through P(5, 0) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 0 = (-1/5) (x – 5)

∴  5y = – x + 5

∴  x + 5y – 5 = 0

Ans: At point (0, 0) equation of tangent is 5x + y = 0 and that of normal is x – 5y = 0

At point (5, 0) equation of tangent is 5x – y – 25 = 0 and that of normal is x + 5y – 5 = 0

Example – 10:

Find the equation of the tangent and normal to the curve y = x² + 4x at the point where it cuts the y-axis

Solution:

Equation of curve is y = x² + 4x    ………… (1)

Let P be the point at which the curve cuts y-axis (x = 0).

Substituting x = 0 in equation (1)

y = 0² – 5(0) = 0

Hence the curve cuts x-axis at P(0, 0)

Differentiating equation (1) w.r.t. x

(dy/dx) = 2x + 4

(dy/dx)_{at P(0, 0)} = 2(0) + 4 = 0 + 4 = 4

The slope of the tangent at P(0, 0) is 4 and that of normal is – 1/4.

Equation of tangent:

Its Slope = m = 4

It passes through P(0, 0) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 0 = 4(x – 0)

∴  y = 4x

∴  4x – y = 0

Equation of normal:

Its Slope = m = -1/4

It passes through P(0, 0) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 0 = (-1/4) (x – 0)

∴  4y = – x

∴  x + 4y = 0

Example – 11:

Find the equation of the tangent and normal to the curve y = √2 sin (2x + π/4) at x = π/4

Solution:

Equation of curve is y = √2 sin (2x + π/4)         ………… (1)

Substituting x = π/4 in equation (1)

y = √2 sin (2(π/4) + π/4) = √2 sin (π/2 + π/4) =√2 cos π/4 = √2 x (1/√2 )= 1

Therefore the coordinates of point P are (π/4, 1)

Differentiating equation (1) w.r.t. x

(dy/dx) = √2 cos (2x + π/4). 2

∴  (dy/dx) = 2√2 cos (2x + π/4)

∴  (dy/dx)_{at P(π/4, 1)} = 2√2 cos (2(π/4) + π/4)

∴  (dy/dx)_{at P(π/4, 1)} = 2√2 cos (π/2 + π/4)

∴  (dy/dx)_{at P(π/4, 1)} = – 2√2 sin (π/4)

∴  (dy/dx)_{at P(π/4, 1)} = – 2√2 x (1/√2 ) = – 2

The slope of the tangent at P(9, 4) is -2 and that of normal is 1/2.

Equation of tangent:

Its slope = m = -2

It passes through P(π/4, 1) ≡ P(x₁, y₁)

By slope point form

y – y₁ = m(x – x₁)

∴  y – 1 = -2(x – π/4)

∴  y – 1 = -2x + π/2

∴  2x + Y – 1 – π/2 = 0

Equation of normal:

Its slope = m = 1/2

It passes through P(π/4, 1) ≡ P(x₁, y₁)

By slope point form

 y  – y₁ = m(x – x₁)

∴  y – 1 = (1/2)(x – π/4)

∴  2y – 2 = x – π/4

∴  x – 2y + 2 – π/4 = 0

Ans: Equation of tangent is 2x + Y – 1 – π/2 = 0 and that of normal is x – 2y + 2 – π/4 = 0

The post Equations of Tangents and Normals appeared first on The Fact Factor.

Real Function:

A function whose domain and co-domain are the set or subset of real numbers R, then the function is called a real function.
Thus if ƒ: R → R then ƒ is a real function.

Example:

Consider function y = ƒ(x) = x² + 3x + 2

For every x ∈ R, y = ƒ(x) = x² + 3x + 2 ∈ R,

Thus the function y = ƒ(x) = x² + 3x + 2 is a real function.

Constant Function:

If a real function ƒ is defined as ƒ(x) = k, k is constant for all x ∈ R Then is called a constant function.

Examples: 

ƒ(x) = 3, ƒ(x) = – 4 etc.

Note:

• The domain for the constant function is a set of real number R, i.e. D_ƒ = R, while its range is {k}
• The range contains only one element. i.e. R_ƒ = {k}
• Constant function is many-one function
• The graph for a constant function is as follows.

Zero Function:

For constant function k = 0 then the function is called zero function.

Example:

ƒ(x) = 0, g(x) = 0 etc

Note:

• The domain for zero function is a set of real number R i.e. D_ƒ = R, while its range is {0}.
• The range contains only one element i.e. zero. i.e. R_ƒ = {0}
• Zero function is many-one function.
• The graph for the zero function is as follows.

The graph is x-axis

Identity Function:

If real function ƒ is defined as ƒ(x) = x, for all x ∈ R Then ƒ is called identity function.

Example:

y = x

Note:

The domain and range for identity function is a set of real number R i.e. D_ƒ = R.

The graph for the zero function is as follows.

Absolute value function:

A function ƒ is defined by ƒ(x) = |x|, Where

is called an absolute value function.

Note:

The domain for absolute value function is a set of real number R i.e. D_ƒ = R.

The graph for the absolute value function is as follows.

Signum function:

A function is defined by

is called signum function.

Note:

• The domain for signum function is a set of real number R i.e. D_ƒ = R.
• The range of signum function contains three elements only. = {-1, 0, 1}
• The graph for the signum function is as follows.

Greatest integer function:

The greatest integer function ƒ is defined as [x], the greatest integer ≤ x, for each x ∈ R. Thus, [x] = x if x is integer and [x] = an integer immediately on the left side of x if x is not an integer.

Examples:

[5] = 5, [-6.9] = -7, [0] = 0, [2.3] = 2, [17/3] = 5

Note:

• The domain for greatest integer function is a set of real number R i.e. D_ƒ = R.
• The range of greatest integer function is a set of integers. R_ƒ =  I
• The graph for the greatest integer function is as follows.

Fractional part function:

A functionƒ defined by ƒ(x) = x – [x], is called fractional part function.

Examples: 

(3.9) =3.9 -3 = 0.9 and (-6.9) = -6.9 – (-7) = 0.1

Note:

• The domain for fractional part function is a set of real number R i.e. D_ƒ =  R.
• Range of fractional part function is R_ƒ =  [0, 1) i.e. 0 ≤ f(x) < 0
• The graph for the fractional part function is as follows.

Linear function:

A function defined by ƒ(x) = mx + c, where m, c ∈ R and m ≠ 0 is called a linear function.

Example:

ƒ(x) = y = 3x + 5

Note:

• The domain for a linear function is a set of real number R i.e. D_ƒ =  R.
• The range of a linear function is a set of real numbers. R_ƒ = R
• The graph of a linear function is a straight line.
• If c = 0 then the graph passes through the origin.

Polynomial Function:

If real function ƒ is defined as ƒ(x) = a₀ + a₁x + a₂x² + a₃x³ + ……… +a_nxⁿ. Where a₀, a₁, a₂, a₃, …,a_n ∈ R and n is a whole number. Then ƒ is called as a polynomial function.

Example :

ƒ(x) = x² + 3x + 2

Note:

• The domain and range for a polynomial function is a set of real number R. Thus, D_ƒ = R. and R_ƒ = = R

Reciprocal function:

A function ƒ defined by ƒ(x) = 1/x, Where x ∈ R and x ≠ 0. is called reciprocal function.

Note:

• The domain for reciprocal function is a set of real number R except x ≠ 0 i.e. D_ƒ = R – {0}.
• The graph for the reciprocal function is as follows.

Exponential function:

A function ƒ defined by ƒ(x) = e^x is called exponential function.

Note:

• The domain for the exponential function is a set of real number R i.e. D_ƒ = R
• The graph for the exponential function is as follows.

Logarithmic function:

A function ƒ defined by ƒ(x) =log x, x > 0 is called logarithmic function.

Note:

• The domain for the logarithmic function is a set = {x| x ∈ R and x > 0}
• The graph for the logarithmic function is as follows.

Trigonometric functions:

Graphs of Trigonometric Functions:

Inverse trigonometric functions:

Some Important Results of Inverse Functions:

SET – I

1. sin(sin^-1x) = x, for |x|<1
2. sin^-1(sin x) = x, for |x| ≤ π/2
3. cos(cos^-1x) = x, for |x|<1
4. cos^-1(cos x) = x, for x = [0, π]
5. tan(tan^-1x) = x, ∀ x ∈ R
6. tan^-1(tan x) = x, for x ∈ (- π/2, π/2)
7. cot(cot^-1x) = x, x ∈ R
8. cot^-1(cot x) = x, for x ∈ (0, π)
9. sec(sec^-1x) = x, |x| ≥ 1
10. sec^-1(sec x) = x, for x ∈ [0, – π/2) ∪ (π/2, π]
11. cosec(cosec^-1x) = x, |x| ≥ 1
12. cosec^-1(cosec x) = x, for x ∈ [- π/2, 0) ∪ (0, π/2]

SET- II

1. cosec^-1x = sin^-1(1/x)
2. sec^-1x = cos^-1(1/x)
3. cot^-1x = tan^-1(1/x)

SET – III

1. sin^-1x + cos^-1x = π/2
2. tan^-1x + cot^-1x = π/2
3. sec^-1x + cosec^-1x = π/2

SET – IV

SET – V

Rational functions:

A function ƒ of the form p(x)/q(x) = 0, q(x) ≠ 0 is called rational function. Its domain is R except for q(x) ≠ 0

The post Types of Functions appeared first on The Fact Factor.

Let A and B two non-empty set and If each and every element x of set A is related by the relation ‘ƒ’ to one and only one element y of the set B such that y = ƒ(x) then such a relation ƒ from set A to set B is called a function.

In short, this is written as ƒ: A → B

Example:

Consider function y = ƒ(x) = x² + 3x + 2

Here we should know that y is an element of set B and x is an element of set A. They are related to each other by the given function ƒ(x)

Characteristics of Function:

• For each element x ∈ A, there exists a unique element y ∈ B.
• The element y ∈ B is called the image of x under the function ‘ƒ’.
• If there is an element in A which has more than one image in B then ƒ: A → B is not a function.
• If there is an element in A which has no image in B then ƒ: A → B is not a function.

Value of Function:

The value ƒ(x) = y is called the value of the function at x. Sometimes the value of the function is also referred as the image of x under function ƒ. x is known as preimage or inverse image of y.

Example:

Consider function,y = ƒ(x) = x² + 3x + 2, then

ƒ(2) = (2)² + 3(2) + 2

∴ ƒ(2) = 4 + 6 + 2

∴ ƒ(2) = 12

Thus the value of function at x = 2 is 12 or y = 12 when x = 2

Domain and Co-domain of a Function:

The set A is called the domain of the function ‘ƒ’. The set B is called the co-domain of the function.

Example:

Consider a function y = ƒ(x) = x²

Consider set A = {1,2,3} and set B = {1,4,9,16.25}

Here we can see each and every element of set A is related to only one element in set B. hence there is a function ƒ. Then the set A is called the domain and set B is called the co-domain.

Range of Function:

The set of all values of ƒ at x i.e. {ƒ(x) = y | x ∈ A} is called the range of ƒ and it may or may not be whole of B.

Example:

Consider a function y = ƒ(x) = x²

Consider set A = {1,2,3}, then

Range = {1, 4, 9}

Range of function is a subset of co-domain of function

Onto Function:

If the range of ƒ = Co-domain of ƒ, then the function ƒ is called onto function.

Example:

Consider a function y = ƒ(x) = x²

Consider set A = {1,2,3} and set B = {1,4,9}

Set B is co-domain of function,

Range = {1, 4, 9}

Here Range of ƒ = Co-domain of function ƒ, hence the function is onto function.

Notes:

• On to function is also called as a surjective function or surjection.
• If ƒ: A → B then every element in B is an image of at least one element in X. Diagrammatically,

Into Function:

If the range of ƒ ≠ Co-domain of ƒ, then the function is called into function

Example:

Consider a function y = (x) = x²

Consider set A = {1,2,3} and set B = {1,4,9,16, 25}

Set B is co-domain of function,

Range = {1, 4, 9}

Here Range of ƒ ≠ Co-domain of function ƒ, hence the function is into function.

Note:

• If the function is not onto then it is into function ƒ: A →B. Diagramatically,

Method for checking whether the function ƒ: A → B is onto or into:

• Find the range of function ƒ.
• If the range of ƒ = B then the function ƒ is onto, otherwise into.

One-One Function:

When ƒ relates one element of set A to one element of set B, then the function ƒ is called one-one function,

Example:

Consider a function y = ƒ(x) = x²

Consider set A = {1,2,3} and set B = {1,4,9,16, 25}

Here function ƒ is one-one function.

Notes:

• One-one function is also called as injective function.
• For one – one function, x₁ ≠ x₂, ƒ(x₁) ≠ ƒ(x₂) for all x₁ , x₂ ∈ A
• For one – one function, x₁= x₂, ƒ(x₁) = ƒ(x₂) for all x₁ , x₂ ∈ A

One-one Correspondence:

A function ƒ from A to B is said to be one-one correspondence (bijective) iff ƒ is both one-one (injective) and onto (surjective)

Example:

Let A = {a, b, c, d} and B = {p, q, r, s}

The function ƒ: A → B defined by ƒ(a) = p, ƒ(b) = q, ƒ(c) r and ƒ(d) = s is one-one onto.

Therefore their exist a one-one correspondence. Diagrammatically

One – one onto function

One – one not onto function

The function ƒ is one-one but not onto because the element t has no preimage in A.

Neither one-one nor onto function:

The function ƒ is not one-one because the distinct elements a and c have the same image p and the function is not onto because the element r has no preimage in A.

Many – one onto function:

When ƒ relates many elements of set A to one element of set B, then the function ƒ is called many-one function,

Example:

Consider a function y = ƒ(x) = x²

Consider set A = {-1, 1, -2, 2, -3, 3}

And set B = {1,4,9,16, 25}

Here function ƒ is a many-one function.

Method for checking whether the function ƒ: A → B is one – one or many-one:

Test – 01:

• Consider any two points x, y ∈ A.
• Put ƒ(x) = ƒ(y) and solve the equation.
• If we get x = y only, then ƒ is one-one, otherwise, it is many-one.

Test – 02:

• If function ƒ is either strictly increasing or strictly decreasing in the whole domain i.e. ƒ’(x) > 0 or ƒ’(x) < 0, for all x ∈ A respectively, then it is one-one otherwise it is many-one.

Test – 03:

• If any straight line parallel to x-axis intersects the graph of the function atmost at one point, then the function is one – one, otherwise, it is many-one. In case of many – one function the line cuts the graph atleast in two distinct points.

Other Tests:

• Any continuous function having local maxima or local minima is many – one.
• All even functions are many – one.
• All polynomial of even degree defined on R have atleast one local maxima or minima and hence many – one on the domain R. Polynomial of odd degree may be one – one or may be many – one.
• If A and B are any two finite sets having m and n elements respectively, then the number of one-one function from A to B

Concept of Intervals :

Closed Interval:

Let a and b be the two real numbers such that a < b. Then, the set of all real numbers between a and b and including a and b is called a closed interval. It is denoted as [a,b], Mathematically, [a,b] = {x ∈ R, a ≤ x ≤ b }

Open Interval:

Let a and b be the two real numbers such that a < b. Then, the set of all real numbers between a and b is called open interval.
It is denoted as (a,b), Mathematically, (a,b) = {x ∈ R, a < x < b }

Semi-open Interval:

Let a and b be the two real numbers such that a < b. Then, following sets represent semi-open interval It Mathematically
[a,b)= {x R, a ≤ x < b } and (a,b]= {x R, a < x ≤ b }

Note:

• The real numbers a and b are called the endpoints of interval. ‘a’ is called the left end point and ‘b’ is called right end point.

Neighbourhoods:

Let ‘a’ be any real number and ‘δ’ be any positive real number, then the open interval (a – δ, a + δ) is called the δ-neighbourhood of a and written as” δ-nbd of a”

If x δ-nbd of a

∴ x ∈ (a – δ, a + δ)

∴ a – δ < x < a + δ

∴ – δ < x – a < a

∴  | x – a| < δ

Thus a is center and 2δ is the length of δ-neighbourhood of a

Deleted δ-neighborhood of a:

If δ-neighbourhood of ‘a’ is defined by deleting the center a, then the neighbourhood obtained is called deleted δ-neighbourhood of a.
Mathematically, The deleted δ-neighbourhood of a is given by

{x ∈ (a – δ < x < a + δ), x ≠ a} or

{x | x (a – δ, a + δ) , x ≠ a}

In the modulus form, it is written as

0 < |x – a| < δ

One-sided neighborhoods:

If a, b, c R such that a < c < b then

• (a, c] is called left neighbourhood of c.
• [c, b) is called right neighbourhood of c.
• (a, c) is called left deleted neighbourhood of c.
• (c, b) is called right deleted neighbourhood of c.

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