# Maxima and minima, inter maths 1b solutions

Inter maths 1b solutions for Maxima and minima are given for some problems.

These are very easy to understand. First you study the text book very well.

Observe the solutions and try them in your own method.

Text book exercise solutions links are also given.

You can see the solutions for junior inter maths 1b

1. Locus

Straight lines sa

11. Rate measure

You can see the solutions for junior inter maths 1A

3. Matrices

You can see the solutions for text book Maths 1A

Functions

Exercise 1(a)

Exercise 1(b)

Exercise 1(c)

Mathematical Induction

Exercise 2(a)

Matrices

Exercise 3(a)

Exercise 3(b)

Exercise 3(c)

Exercise 3(d)

Exercise 3(e)

Exercise 3(f)

Exercise 3(g)

Exercise 3(h)

Exercise 3(i)

Exercise 4(a)

Exercise 4(b)

Product of vectors

Exercise 5(a)

Exercise 5(b)

Exercise 5(c)

Trigonometric Ratios up to Transformations

Exercise 6(a)

Exercise 6(b)

Exercise 6(c)

Exercise 6(d)

Exercise 6(e)

Exercise 6(f)

Trigonometric Equations

Exercise 7(a)

Inverse Trigonometric Equations

Exercise 8(a)

Hyperbolic Functions

Exercise 9(a)

Properties of Triangles

Exercise 10(a)

Exercise 10(b)

Maths 1B solutions for

Locus

Exercise 1(a)

Transformation of Axes

Exercise 2(a)

The Straight Line

Three Dimensional Coordinates

Exercise 5(a)

Exercise 5(b)

Direction Cosines and Direction Ratios

Exercise 6(a)

Exercise 6(b)

The Plane

Exercise 7(a)

Limits and Continuity

Exercise 8(a)

Exercise 8(b)

Exercise 8(c)

Exercise 8(d)

Exercise 8(e)

Applications of Derivaties

## Intermediate mathematics 1b chapter 10.5 solutions

Maxima and Minima         Note – Observe the solutions and try them in your own methods.

Inter maths 1A solutions

Inter maths trigonometry solutions

Maths real numbers solutions class 10

You can see solutions for Inter Maths IIB

1. Circle

3. Parabola

4. Ellipse

You can also see solutions for Inter Maths IIA

For examination purpose you can see

Complex numbers

De Moivre’ s Theorem

### One Comment

1. Vaishali chadar September 24, 2019 at 1:32 pm - Reply

If x,y,z are length of the perpandicular dropped from a point inside the triangle of given area A,on the three sides of the triangle then the minimum value of x^2+y^2+Z^2 is 4A^2/a^2+b^2+c^2