MTH101 Solved Practice Questions Lectures 1-22 Set 2

Lecture No. 1: Coordinates, Graphs, Lines

Q1. Solve the inequality x  2

x 1   1.

Solution:

x  2 x 1   1

 x  2  1  0

x  1  x  2  x  1  0 x  1  2x  1  0 x  1

Now there are two possibilities. Either 2x  1  0 and x  1  0 or 2x  1  0 and x  1  0

Consider, 2x  1  0 and x  1  0  x  1 and x  1

  1 ,   (1, )  2

Takingintersection of both intervals, we have  1 ,   ...............(1)

 2

Similarly, if we consider, 2x  1  0 and x  1  0  x  1 and x  1

  , 1   (, 1) 2 

Takingintersection of both intervals, we have (, 1) ................................................. (2)

Combining (1) and (2), we have the required solution set. That is:  1 ,   (, 1)  2

Q2. Solve the inequality and find the solution set of 3  1  1 .

x

Solution:

3  1  1

x   1  1  3   1   5  1  5  x  2

x x x So, solution set = 0  x  2 i.e.  0, 2  5 

Q3. List the elements in the following sets:

(i) {x : x2  4x  4  0}

Solution:

(ii) {x : x is an integer satisfying 1  x  5}

(i) Consider x2  4x  4  0  x2  2x  2x  4  0  x(x 2)  2(x  2)  0  (x 2)(x  2)  0

 (x 2)2  0  x 2  0  x  2 Solution set: 2

(ii) Solution set: 0,1, 2,3, 4

Q4. Find the solution set for the inequality: 9  x  2  3x

Solution:

9  x  2  3x  9  2  3x  x 11  2x  11  x or x  11

Hence, Solution set:  , 11  2 

Q5. Solve the inequality 2  1 3x  5 .

Solution:

2  1 3x  5  2 1 3x  5 1  3  3x  6  3  x  6 1 x  2 b4

b2 2

Q1. Solve for x ,

Solution:

Lecture No. 2: Absolute Value

 8.

x  7  8 or 4  x x  7  8 4  x

x  25 or x  39

Q2. Is the equality

reasoning.

Solution:

As we know that  b2 valid for all values of b ? Justify your answer with appropriate

 x if x is positive or zero i.e x  0 ,

 b2 ,

 b2 ,

but b2 is always positive , because if

So the given equality always holds. b  0 then b2 is always positive. x  7 4  x

b4

x2 x  7  84  x or x  7  84  x

x  7  32  8x or x  7  32  8x x  8x  32  7 or x  8x  32  7 9x  25 or  7x  39