Chapter 1 Limit, Continuity and Derivative-I
1.1 Exercise 1 (i)
1.1.1 Question 1
Find the domain of the following functions:
- y=x2−4x
It is a polynomial function. Every real number R satisfies the equation. So the domain is D(f)=(−∞,∞).
- y=1x+2
Here x can be anything except a value that produces a zero in the denominator i.e x+2≠0 or x≠−2. So the domain is D(f)=(−∞,−2)∪(−2,∞).
- y=√3−x
Here
3−x≥0x≤3
Therefore, the domain is D(f)=(−∞,3].
- y=1√1−x2
Here 1−x2 should be >0 i.e.
1−x2>0x2<1−1<x<1
Also 1−x2 can’t be zero.
1−x2≠0x2≠1x≠±1
Thus, the domain is D(f)=(−1,1).
- y=√3−4x+x2
Here
3−4x+x2≥0(x−1)(x−3)≥0
Now, when x is ≤1 in the number line, LHS is satisfied.
When x lies between (1,3), LHS becomes negative and won’t satisfy the equation.
When x is ≥3, LHS is satisfied. Hence the domain of the function is D(f)=(−∞,1]∪[3,∞).
- y=xx−1
Here x−1 can’t equal zero i.e x−1≠0 or x≠1. The domain is thus D(f)=R−{1}.
Finding range of a function
A. Method 1
- Plot the graph (use all techniques of shifting, stretching, compression)
- Range will be the value along y-axis
B. Method 2
- Put y=f(x)
- Express x as a function of y
- Find possible values for y (just like domain)
- Eliminate values by looking at the definition to write the final range
1.1.2 Question 2
Find the domain and the range of the following functions:
- y=1x−2
Here x−2 can’t equal zero i.e x−2≠0 or x≠2. The domain is thus D(f)=R−{2}.
Plotting shows that the range of the function (the values along y-axis) is (−∞,0)∪(0,∞).
Figure 1.1: Plotting of y=1x−2
- √x−1
Domain: The function must be x−1≥0. So x≥1. Domain is thus D(f)=[1,∞).
Range: Plotting y=f(x)=√x shows the graph above x-axis. Plotting f(x−1)=√x−1 shifts the curve 1 unit to the right.
The range is thus R(f)=[0,∞).
Figure 1.2: Function y=√x−1
- y=√25−x2
Domain: For the given function 25−x2 should be ≥0.
25−x2≥0x2≤25−5≤x≤5
So the domain is D(f)=[−5,5].
Range: Plotting shows the function to be half circle above x-axis. The range is thus values along the y-axis i.e R(f)=[0,5].
Figure 1.3: y=√25−x2
- y=x2−25x−5
Domain: Here x−5≠0, so x≠5. Domain is thus D(f)=R−{5}.
Range: Plotting the equation gets a line y=x+5 except a hole at point (5,10). Range is therefore R(f)=(−∞,10)∪(10,∞) or R−{10}.
1.1.3 Question 3
Draw the graphs of the following functions:
- y=|x|+|x−1|
When x<0,
y=−x−(x−1)=−2x+1
When 0<x<1,
y=x−(x−1)=1
When x>1,
y=x+x−1=2x−1
The graph is thus plotted as below:
Figure 1.4: y=|x|+|x−1|
y=[x],x>0
y=x2−1x−1
For x≠1, y=x+1 which is a straight line not passing through the point (1,2) i.e hole at this point. The graph is given below:
Figure 1.5: y=x2−1x−1
- f(x)={x−if x≤0x2if 0<x<2
When x≤0, f(x) is a straight line with slope 1 passing through origin.
When 0<x<2, the f(x) is a parabola with concavity upwards. The graph is shown below:
1.2 Exercise 1(ii)
1.2.1 Question 1
Find the limits of the following:
limx→0+0|x|
limx→0+01x
limx→0−01x
limx→01+21/x3+21/x
limx→0(21/x+2x+12x)
limx→∞xx+1
limx→∞3x−3−x3x+3−x
limx→∞sinxx
limx→0tan−1xx
limx→0sin(1x)
1.2.2 Question 2
- A function f(x) is defined as follows:
f(x)={1−x2if x>0x+1if x≤0
−−Find limx→0f(x).
- A function f is defined as follows:
f(x)={x2when x<12.5when x=1x2+2when x>1
−−Does limx→1f(x) exist?
- A function f(x) is defined below:
f(x)={xif 0<x<12−1xif 1≤x<2x−18x2if x>2
−−Show the existence of the limits at x=1 and x=2. Find limx→1f(x) and limx→2f(x) if they exist.