Question 1:( 3 marks each)
Solve the equations:
(x^2- 169)/x^2 =0

2x^2-7x+6=0 (use the quadratic formula)

x^2-12x+5 =7 (Apply the square root property)
Solve the following inequalities
ii)    5 |(2x+3)/3|  ≤ 10

iii)    |2x+1|≥ 13

..
Question 2 : ( 4+7+4marks)
Given f(x)= x^3-2x^2-4x +5, show that f(x) has a zero on the interval               [-1, 2].

Write the equation of the line with the following conditions:
Through the point  A(2, 9), and parallel  to the x-axis.
Through the points (1, -2), and (2, -3)

Find the center and the radius of the circle with the equation
x^2+y^(2 )-10x+2y+1=0

Question 3: (3+3+4+5 marks)
Given f(x) = x^3 and g(x)= x+1 Find (f o g) (4)?

Determine whether the function f(x) = 5-3x^2is one to one function.

Write an equation for the inverse of the one to one function f(x ) =2x-17.

Write the domain in interval notation for:  〖ln(〗⁡〖x^2+10x+16〗).

Question 4 : ( 6 +4 +5  marks )

Solve the equation:
ln x - lnx^2 +ln3=0
e^(x-3) = 4/5
b.Find the vertex  and axis of symmetry of the function f(x)=2 x^2-7x +6

Let's solve each problem step-by-step:

### Question 1

#### 1. Solve the equations:

1. **\((x^2 - 169) / x^2 = 0\)**

To solve the equation \(\frac{x^2 - 169}{x^2} = 0\), we start by setting the numerator equal to zero because a fraction equals zero if and only if its numerator is zero (and the denominator is non-zero).

\[ x^2 - 169 = 0 \]

This is a difference of squares, so we can factor it:

\[ (x - 13)(x + 13) = 0 \]

Setting each factor equal to zero gives us:

\[ x - 13 = 0 \quad \text{or} \quad x + 13 = 0 \]
\[ x = 13 \quad \text{or} \quad x = -13 \]

So, the solutions are \( x = 13 \) and \( x = -13 \).

2. **\(2x^2 - 7x + 6 = 0\)** (use the quadratic formula)

The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For the equation \(2x^2 - 7x + 6 = 0\), the coefficients are \( a = 2 \), \( b = -7 \), and \( c = 6 \).

First, we calculate the discriminant:

\[ b^2 - 4ac = (-7)^2 - 4(2)(6) = 49 - 48 = 1 \]

Now, apply the quadratic formula:

\[ x = \frac{-(-7) \pm \sqrt{1}}{2(2)} \]
\[ x = \frac{7 \pm 1}{4} \]

This gives us two solutions:

\[ x = \frac{7 + 1}{4} = \frac{8}{4} = 2 \]
\[ x = \frac{7 - 1}{4} = \frac{6}{4} = \frac{3}{2} \]

So, the solutions are \( x = 2 \) and \( x = \frac{3}{2} \).

3.

Explore detailed solutions to algebraic equations and inequalities, including quadratic equations, square root properties, and function compositions. This resource covers topics from Grades 10 to 12 mathematics curriculum.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation