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Measures of central tendency and dispersion

Introduction.
An average or a central value of a statistical series in the value of the variable which describes the

characteristics of the entire distribution.

The following are the five measures of central tendency.

(1) Arithmetic mean (2) Geometric mean (3) Harmonic mean (4) Median (5) Mode

Arithmetic Mean.
Arithmetic mean is the most important among the mathematical mean.

According to Horace Secrist,

“The arithmetic mean is the amount secured by dividing the sum of values of the items in a series by their number.”

(1) Simple arithmetic mean in individual series (Ungrouped data)
(i) Direct method : If the series in this case be x1, x 2, x3 ,......, xn then the arithmetic mean x is given by

Sum of the series , i.e., x= x1 + x2 + x3 + .... + xn 1 n xi
Number of terms n n i =1
∑x= =

(ii) Short cut method

Arithmetic mean (x) = A + ∑d ,
n

where, A = assumed mean, d = deviation from assumed mean = x – A, where x is the individual item,

Σd = sum of deviations and n = number of items.

(2) Simple arithmetic mean in continuous series (Grouped data)

(i) Direct method : If the terms of the given series be x1, x 2,...., xn and the corresponding frequencies be

n fi xi
∑∑f1, f2,.... fn , then the arithmetic mean x is given by, fi
x = f1x1 + f2x2 + .... + fnxn = i =1 .
f1 + f2 + .... + fn n

i =1

(ii) Short cut method : Arithmetic mean (x) = A + ∑ f(x − A)
∑ f

Where A = assumed mean, f = frequency and x – A = deviation of each item from the assumed mean.

(3) Properties of arithmetic mean
(i) Algebraic sum of the deviations of a set of values from their arithmetic mean is zero. If xi / fi , i = 1, 2, …,
n is the frequency distribution, then

n

∑ fi (xi − x) = 0 , x being the mean of the distribution.

i=1

(ii) The sum of the squares of the deviations of a set of values is minimum when taken about mean.

(iii) Mean of the composite series : If xi ,(i = 1, 2,....., k) are the means of k-component series of sizes

ni ,(i = 1, 2,...., k) respectively, then the mean x of the composite series obtained on combining the component

n1 x1 + n2 x 2 + .... + nk xk n n
n1 + n2 + .... + nk
ni xi ni .

i=1 i=1
∑ ∑series is given by the formulax= =

Geometric Mean.

If x1, x 2, x3 ,......, xn are n values of a variate x, none of them being zero, then geometric mean (G.M.) is

given by G.M. = (x1.x 2 .x 3 ......xn )1/ n ⇒ log(G.M.) = 1 (log x1 + log x 2 + ..... + log xn ) .
n

In case of frequency distribution, G.M. of n values x1, x 2,.....xn of a variate x occurring with frequency

f1, f2 ,....., fn is given by G.M. = (x f1 .x f2 ..... x nfn )1 / N , where N = f1 + f2 + ..... + fn .
1 2

Harmonic Mean.

The harmonic mean of n items x1, x 2 ,......, xn is defined as H.M. = 11 n 1 .
x1 + x2 + ..... + xn

If the frequency distribution is f1, f2 , f3 ,......, fn respectively, then H.M. = f1 + f2 + f3 + ..... + fn
..... +
 f1 + f2 + fn 
x1 x2 xn

Note :  A.M. gives more weightage to larger values whereas G.M. and H.M. give more weightage to smaller

values.

Example: 1 If the mean of the distribution is 2.6, then the value of y is [Kurukshetra CEE 2001]

Solution: (c) Variate x 12345
Example: 2
Solution: (b) Frequency f of x 4 5 y 1 2
Example: 3
Solution: (b) (a) 24 (b) 13 (c) 8 (d) 3

n
∑ fi xi
i=1
We know that, Mean = n

∑ fi
i=1

i.e. 2.6 = 1× 4 + 2×5 + 3×y + 4×1+ 5× 2 or 31.2 + 2.6y = 28 + 3y or 0.4y = 3.2 ⇒ y=8
4+5 + y +1+ 2

In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete

class are 72, then what are the average marks of the girls [AIEEE 2002]

(a) 73 (b) 65 (c) 68 (d) 74

Let the average marks of the girls students be x, then

72 = 70 × 75 + 30 × x (Number of girls = 100 – 70 = 30)
100

i.e., 7200 − 5250 = x,∴x = 65.
30

If the mean of the set of numbers x1, x2, x3, ....., xn is x , then the mean of the numbers xi + 2i , 1 ≤ i ≤ n is

[Pb. CET 1988]

(a) x + 2n (b) x + n + 1 (c) x + 2 (d) x + n

n
∑ xi
n
∑We know that
x= i=1 i.e., xi = nx

n i =1

n nn n(n + 1)
∑ ∑ ∑∴ 2
(xi + 2i) xi + 2 i nx + 2(1 + 2 + ...n) nx + 2
n n
i =1 = i =1 i =1 = = = x + (n + 1)

n n

Example: 4 The harmonic mean of 4, 8, 16 is [AMU 1995]

(a) 6.4 (b) 6.7 (c) 6.85 (d) 7.8

Solution: (c) H.M. of 4, 8, 16 = 1 3 = 48 = 6.85
4 11 7
+ 8 + 16

Example: 5 The average of n numbers x1, x2, x3, ......, xn is M. If xn is replaced by x′ , then new average is [DCE 2000]

(a) M − xn + x′ (b) nM − xn + x′ (c) (n − 1)M + x′ (d) M − xn + x′
n n n

Solution: (b) M = x1 + x2 + x3......xn i.e. nM = x1 + x2 + x3 + .....xn−1 + xn
n nM − xn = x1 + x2 + x3 + .....xn−1

nM − xn + x′ = x1 + x2 + x3 + .....xn−1 + x′

nn

∴ New average = nM − xn + x′
n

Example: 6 Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40,

20, 50 respectively. The correct mean is [Kurukshetra CEE 1994]

(a) 48 (b) 82 1 (c) 50 (d) 80
2

Solution: (c) Sum of 100 items = 49 ×100 = 4900

Sum of items added = 60 + 70 + 80 = 210

Sum of items replaced = 40 + 20 + 50 = 110

New sum = 4900 + 210 − 110 = 5000

∴ Correct mean = 5000 = 50
100

Median.

Median is defined as the value of an item or observation above or below which lies on an equal number of

observations i.e., the median is the central value of the set of observations provided all the observations are

arranged in the ascending or descending orders.

(1) Calculation of median
(i) Individual series : If the data is raw, arrange in ascending or descending order. Let n be the number of

observations.

 n + 1  th
 2 
If n is odd, Median = value of item.

1   n  th  n + 1 th item
2   2   2  
If n is even, Median =  value of item + value of

(ii) Discrete series : In this case, we first find the cumulative frequencies of the variables arranged in

ascending or descending order and the median is given by

Median =  n + 1  th observation, where n is the cumulative frequency.
 2 

(iii) For grouped or continuous distributions : In this case, following formula can be used

(a) For series in ascending order, Median = l +  N − C
 2 f ×i

Where l = Lower limit of the median class

f = Frequency of the median class

N = The sum of all frequencies

i = The width of the median class

C = The cumulative frequency of the class preceding to median class.

(b) For series in descending order

Median = u  N − C  , where u = upper limit of the median class
 2 f 
−   × i

n

∑N = fi
i=1

As median divides a distribution into two equal parts, similarly the quartiles, quantiles, deciles and percentiles
divide the distribution respectively into 4, 5, 10 and 100 equal parts. The jth quartile is given by

 j N − C 
+ 4 f 
Qj =l  i; j = 1, 2, 3 . Q1 is the lower quartile, Q2 is the median and Q3 is called the upper quartile.

 

(2) Lower quartile

 n + 1  th
 4 
(i) Discrete series : Q1 = size of item

(ii) Continuous series : Q1 = l +  N − C ×i
 4 

f

(3) Upper quartile

 3(n + 1) th
 4 
(i) Discrete series : Q3 = size of item

 3N − C 
 4 
(ii) Continuous series : Q3 =l + ×i
f

(4) Decile : Decile divide total frequencies N into ten equal parts.

Dj =l + N× j −C [j = 1, 2, 3, 4, 5, 6, 7, 8, 9]
×i
10
f

If j = 5, then D5 =l + N −C Hence D5 is also known as median.
2 f ×i.

(5) Percentile : Percentile divide total frequencies N into hundred equal parts

Pk =l + N×k −C
×i
100
f

where k = 1, 2, 3, 4, 5,.......,99.

Example: 7 The following data gives the distribution of height of students
Solution: (b)
Height (in cm) 160 150 152 161 156 154 155
3 37
Number of students 12 8 4 4
(d) 161
The median of the distribution is [AMU 1994]

(a) 154 (b) 155 (c) 160

Arranging the data in ascending order of magnitude, we obtain

Height (in cm) 150 152 154 155 156 160 161

Number of students 8 4 3 7 3 12 4

Cumulative frequency 8 12 15 22 25 37 41

Here, total number of items is 41, i.e. an odd number. Hence, the median is 41 + 1 th i.e. 21st item.
2

From cumulative frequency table, we find that median i.e. 21st item is 155.

(All items from 16 to 22nd are equal, each = 155)

Example: 8 The median of a set of 9 distinct observation is 20.5. If each of the largest 4 observation of the set is increased by 2, then
Solution: (d)
Example: 9 the median of the new set [AIEEE 2003]

Solution: (a) (a) Is increased by 2 (b) Is decreased by 2

(c) Is two times the original median (d) Remains the same as that of the original set

n = 9, then median term =  9 + 1 th = 5th term . Since last four observation are increased by 2.
 2 

∵ The median is 5th observation which is remaining unchanged.
∴ There will be no change in median.
Compute the median from the following table

Marks obtained No. of students

(a) 36.55 0-10 2 (d) None of these
10-20 18
20-30 30
30-40 45
40-50 35
50-60 20
60-70 6
70-80 3
(b) 35.55 (c) 40.05

Marks obtained No. of students Cumulative frequency
0-10 2 2
10-20 18 20
20-30 30 50
30-40 45 95
40-50 35
50-60 20 130
60-70 6 150
70-80 3 156
159

n = ∑ f = 159

Here n = 159, which is odd.

Median number = 1 (n + 1) = 1 (159 + 1) = 80 , which is in the class 30-40 (see the row of cumulative frequency
2 2

95, which contains 80).

Hence median class is 30-40.

∴ We have l = Lower limit of median class = 30

f = Frequency of median class = 45

C = Total of all frequencies preceding median class = 50

i = Width of class interval of median class = 10

∴ Required median = l + N −C = 30 + 159 50 10 30 + 295 36.55 .
2 f ×i 2− 45
45 × = =

Mode.

Mode : The mode or model value of a distribution is that value of the variable for which the frequency is

maximum. For continuous series, mode is calculated as, Mode = l1 +  f1 − f0  × i
 
 2 f1 − f0 − f2 

Where, l1 = The lower limit of the model class

f1 = The frequency of the model class

f0 = The frequency of the class preceding the model class

f2 = The frequency of the class succeeding the model class
i = The size of the model class.

Symmetric distribution : A symmetric is a symmetric distribution if the values of mean, mode and median

coincide. In a symmetric distribution frequencies are symmetrically distributed on both sides of the centre point of

the frequency curve.

Mean = Median = Mode

Mean Mode Mean
Median Mode Median

A distribution which is not symmetric is called a skewed-distribution. In a moderately asymmetric the interval
between the mean and the median is approximately one-third of the interval between the mean and the mode i.e. we
have the following empirical relation between them

Mean – Mode = 3(Mean – Median) ⇒ Mode = 3 Median – 2 Mean. It is known as Empirical relation.

Example: 10 The mode of the distribution [AMU 1988]
Solution: (b)
Marks 4 567 8
6 7 10 8 3
No. of students
(c) 8 (d) 10
(a) 5 (b) 6

Since frequency is maximum for 6

∴ Mode = 6

Example: 11 Consider the following statements [AIEEE 2004]
Solution: (d)
(1) Mode can be computed from histogram

(2) Median is not independent of change of scale

(3) Variance is independent of change of origin and scale

Which of these is/are correct

(a) (1), (2) and (3) (b) Only (2) (c) Only (1) and (2) (d) Only (1)

It is obvious.

Important Tips

 Some points about arithmetic mean
• Of all types of averages the arithmetic mean is most commonly used average.
• It is based upon all observations.
• If the number of observations is very large, it is more accurate and more reliable basis for comparison.

 Some points about geometric mean
• It is based on all items of the series.
• It is most suitable for constructing index number, average ratios, percentages etc.
• G.M. cannot be calculated if the size of any of the items is zero or negative.

• It is based on all item of the series.
• This is useful in problems related with rates, ratios, time etc.
• A.M. ≥ G.M. ≥ H.M. and also (G.M.)2 = (A.M.)(H.M.)

• It is an appropriate average in dealing with qualitative data, like intelligence, wealth etc.
• The sum of the deviations of the items from median, ignoring algebraic signs, is less than the sum from any other point.

• It is not based on all items of the series.
• As compared to other averages mode is affected to a large extent by fluctuations of sampling,.
• It is not suitable in a case where the relative importance of items have to be considered.

Pie Chart (Pie Diagram).

Here a circle is divided into a number of segments equal to the number of components in the corresponding

table. Here the entire diagram looks like a pie and the components appear like slices cut from the pie. In this

diagram each item has a sector whose area has the same percentage of the total area of the circle as this item has of

the total of such items. For example if N be the total and n1 is one of the components of the figure corresponding to

a particular item, then the angle of the sector for this item =  n1  × 360° , as the total number of degree in the angle
 N 

subtended by the whole circular arc at its centre is 360°.

Example: 12 If for a slightly assymetric distribution, mean and median are 5 and 6 respectively. What is its mode [DCE 1998]
Solution: (d)
(a) 5 (b) 6 (c) 7 (d) 8

We know that

Mode = 3Median – 2Mean

= 3(6) – 2(5) = 8

Example: 13 A pie chart is to be drawn for representing the following data

Items of expenditure Number of families

Education 150

Food and clothing 400

House rent 40

Electricity 250

Miscellaneous 160

The value of the central angle for food and clothing would be [NDA 1998]

(a) 90° (b) 2.8° (c) 150° (d) 144°

Solution: (d) Required angle for food and clothing = 400 × 360° = 144°
1000

Measure of Dispersion.
The degree to which numerical data tend to spread about an average value is called the dispersion of the

data. The four measure of dispersion are

(1) Range (2) Mean deviation (3) Standard deviation (4) Square deviation

(1) Range : It is the difference between the values of extreme items in a series. Range = Xmax – Xmin

The coefficient of range (scatter) = x max − x min .
x max + x min

Range is not the measure of central tendency. Range is widely used in statistical series relating to quality
control in production.

(i) Inter-quartile range : We know that quartiles are the magnitudes of the items which divide the
distribution into four equal parts. The inter-quartile range is found by taking the difference between third and first
quartiles and is given by the formula

Inter-quartile range = Q3 − Q1

Where Q1 = First quartile or lower quartile and Q3 = Third quartile or upper quartile.
(ii) Percentile range : This is measured by the following formula

Percentile range = P90 − P10

Where P90 = 90th percentile and P10 = 10th percentile.
Percentile range is considered better than range as well as inter-quartile range.

(iii) Quartile deviation or semi inter-quartile range : It is one-half of the difference between the third

quartile and first quartile i.e., Q.D. = Q3 − Q1 and coefficient of quartile deviation = Q3 − Q1 .
2 Q3 + Q1

Where, Q3 is the third or upper quartile and Q1 is the lowest or first quartile.
(2) Mean deviation : The arithmetic average of the deviations (all taking positive) from the mean, median or
mode is known as mean deviation.

(i) Mean deviation from ungrouped data (or individual series)

Mean deviation = ∑|x − M|
n

Where |x – M| means the modulus of the deviation of the variate from the mean (mean, median or mode).
M and n is the number of terms.

(ii) Mean deviation from continuous series : Here first of all we find the mean from which deviation is to

be taken. Then we find the deviation dM =| x − M | of each variate from the mean M so obtained.

Next we multiply these deviations by the corresponding frequency and find the product f.dM and then the

sum ∑ f dM of these products.

Lastly we use the formula, mean deviation = ∑ f |x − M | = ∑ f dM , where n = Σf.
n n

Important Tips

 Mean coefficient of dispersion = Mean deviation from the mean
Mean

 Median coefficient of dispersion = Mean deviation from the median
Median

 Mode coefficient of dispersion = Mean deviation from the mode
Mode

 In general, mean deviation (M.D.) always stands for mean deviation about median.

(3) Standard deviation : Standard deviation (or S.D.) is the square root of the arithmetic mean of the
square of deviations of various values from their arithmetic mean and is generally denoted by σ read as sigma.

(i) Coefficient of standard deviation : To compare the dispersion of two frequency distributions the
relative measure of standard deviation is computed which is known as coefficient of standard deviation and is given by

Coefficient of S.D. = σ , where x is the A.M.
x

(ii) Standard deviation from individual series

σ= ∑(x − x)2
N

where, x = The arithmetic mean of series

N = The total frequency.

(iii) Standard deviation from continuous series

σ= ∑ fi (xi − x)2
N

where, x = Arithmetic mean of series

xi = Mid value of the class

fi = Frequency of the corresponding xi

N = Σf = The total frequency

Short cut method

∑ fd 2  ∑ fd  2 ∑d2  ∑d  2
N  N  N  N 
(i) σ = − (ii) σ = −

where, d = x – A = Deviation from the assumed mean A

f = Frequency of the item

N = Σf = Sum of frequencies

(4) Square deviation
(i) Root mean square deviation

∑S = 1 n fi (xi − A)2
N i=1

where A is any arbitrary number and S is called mean square deviation.

(ii) Relation between S.D. and root mean square deviation : If σ be the standard deviation and S be
the root mean square deviation.

Then S 2 = σ 2 + d 2 .

Obviously, S 2 will be least when d = 0 i.e. x = A
Hence, mean square deviation and consequently root mean square deviation is least, if the deviations are

taken from the mean.

Variance.
The square of standard deviation is called the variance.

Coefficient of standard deviation and variance : The coefficient of standard deviation is the ratio of the

S.D. to A.M. i.e., σ . Coefficient of variance = coefficient of S.D.× 100 = σ × 100 .
x x

Variance of the combined series : If n1;n2 are the sizes, x1; x 2 the means and σ 1;σ 2 the standard

deviation of two series, then σ 2 = 1 [n1 (σ 2 + d12 ) + n2 (σ 2 + d 2 )]
+ n2 1 2 2
n1

Where, d1 = x1 −x, d2 = x2 −x and x = n1 x1 + n2 x 2 .
n1 + n2

Important Tips

 Range is widely used in statistical series relating to quality control in production.

 Standard deviation ≤ Range i.e., variance ≤ (Range)2.

 Empirical relations between measures of dispersion

• Mean deviation = 4 (standard deviation)
5

• Semi interquartile range = 2 (standard deviation)
3

 Semi interquartile range = 5 (mean deviation)
6

 For a symmetrical distribution, the following area relationship holds good

X ± σ covers 68.27% items

X ± 2σ covers 95.45% items

X ± 3σ covers 99.74% items

 S.D. of first n natural numbers is n2 − 1 .
12

 Range is not the measure of central tendency.

Skewness.

“Skewness” measures the lack of symmetry. It is measured by γ1 = ∑(xi − µ)3 2 and is denoted by γ 1 .
{∑(xi − µ 2)}3 /

The distribution is skewed if,

(i) Mean ≠ Median ≠ Mode

(ii) Quartiles are not equidistant from the median and

(iii) The frequency curve is stretched more to one side than to the other.
(1) Distribution : There are three types of distributions
(i) Normal distribution : When γ 1 = 0 , the distribution is said to be normal. In this case

Mean = Median = Mode
(ii) Positively skewed distribution : When γ 1 > 0 , the distribution is said to be positively skewed. In this case

Mean > Median > Mode
(iii) Negative skewed distribution : When γ 1 < 0 , the distribution is said to be negatively skewed. In this case

Mean < Median < Mode

(2) Measures of skewness
(i) Absolute measures of skewness : Various measures of skewness are

(a) SK = M − M d (b) SK = M − M o (c) Sk = Q3 + Q1 − 2M d

where, Md = median, Mo = mode, M = mean

Absolute measures of skewness are not useful to compare two series, therefore relative measure of dispersion

are used, as they are pure numbers.

(3) Relative measures of skewness M − Mo = 3 (M − Md ) ,
σ σ
(i) Karl Pearson’s coefficient of skewness : Sk = − 3 ≤ Sk ≤ 3 , where σ is

standard deviation.

(ii) Bowley’s coefficient of skewness : Sk = Q3 + Q1 − 2M d
Q3 − Q1

Bowley’s coefficient of skewness lies between –1 and 1.

(iii) Kelly’s coefficient of skewness : SK = P10 + P90 − 2Md = D1 + D9 − 2Md
P90 − P10 D9 − D1

Example: 14 A batsman scores runs in 10 innings 38, 70, 48, 34, 42, 55, 63, 46, 54, 44, then the mean deviation is [Kerala Engg. 2002]
Solution: (a)
(a) 8.6 (b) 6.4 (c) 10.6 (d) 9.6
Example: 15
Solution: (c) Arranging the given data in ascending order, we have
Example: 16
34, 38, 42, 44, 46, 48, 54, 55, 63, 70,

Here median M = 46 + 48 = 47 ( n = 10, median is the mean of 5th and 6th items)
2

∴ Mean deviation = Σ| xi − M | = Σ| xi − 47| = 13 + 9 + 5 + 3 + 1 +1 + 7 + 8 + 16 + 23 = 8.6
n 10 10

S.D. of data is 6 when each observation is increased by 1, then the S.D. of new data is [Pb. CET 1994]

(a) 5 (b) 7 (c) 6 (d) 8

S.D. and variance of data is not changed, when each observation is increased (OR decreased) by the same constant.

In a series of 2n observations, half of them equal a and remaining half equal –a. If the standard deviation of the

observations is 2, then |a| equals [AIEEE 2004]

(a) 2 (b) 2 (c) 2 (d) 1
n n

Solution: (c) Let a, a, ..........n times – a, – a, – a, – a, ----- n time i.e. mean = 0 and S.D. = n(a − 0)2 + n(−a − 0)2
2n

2= na2 + na2 = a2 = ±a . Hence |a| = 2
2n

Example: 17 If µ is the mean of distribution (yi, fi ) , then ∑ fi(yi − µ) = [Kerala PET 2001]

(a) M.D. (b) S.D. (c) 0 (d) Relative frequency

Solution: (c) We have, ∑ fi (yi − µ) = ∑ fi yi − µ ∑ fi = µ ∑ fi − µ ∑ fi = 0  µ = ∑ fi yi 
Example: 18  ∑ fi 
Solution: (c) 

Example: 19 What is the standard deviation of the following series [DCE 1996]
Solution: (a)
Example: 20 Measurements 0-10 10-20 20-30 30-40
Solution: (d) 42
Frequency 13
Example: 21
Solution: (b) (a) 81 (b) 7.6 (c) 9 (d) 2.26
Example: 22
Solution: (b) Class Frequency yi ui = yi − A , A = 25 fiui fiui2
10
0-10 1
10-20 3 5 –2 –2 4
20-30 4
30-40 2 15 – 1 – 3 3
10
25 0 0 0

35 1 2 2

–3 9

 ∑ fiui2  ∑ fiui2  2  9  −3  2 
 ∑ fi ∑ fi  10  10  
σ2 = c 2  −  = 10 2 −  = 90 − 9 = 81 ⇒ σ =9
 

In an experiment with 15 observations on x, the following results were available ∑ x2 = 2830 , ∑ x = 170 . On

observation that was 20 was found to be wrong and was replaced by the correct value 30. Then the corrected variance is
[AIEEE 2003]
(a) 78.00 (b) 188.66 (c) 177.33 (d) 8.33

∑ x = 170 , ∑ x2 = 2830
Increase in ∑ x = 10 , then ∑ x′ = 170 + 10 = 180

Increase in ∑ x2 = 900 − 400 = 500 , then ∑ x′ = 2830 + 500 = 3330

Variance = 1 ∑ x′2 −  ∑ x′ 2 = 3330 −  180 2 = 222 − 144 = 78
n  n  15  15 

The quartile deviation of daily wages (in Rs.) of 7 persons given below 12, 7, 15, 10, 17, 19, 25 is
[Pb. CET 1991, 96; Kurukshetra CEE 1997]

(a) 14.5 (b) 5 (c) 9 (d) 4.5
The given data in ascending order of magnitude is 7, 10, 12, 15, 17, 19, 25

Here Q1 = size of  n + 1 th item = size of 2nd item = 10
 4 

Q3 = size of  3(n + 1) th item = size of 6th item = 19
 4 

Then Q.D. = Q3 − Q1 = 19 − 10 = 4.5
2 2

Karl-Pearson’s coefficient of skewness of a distribution is 0.32. Its S.D. is 6.5 and mean 39.6. Then the median of the

distribution is given by [Kurukshetra CEE 1991]

(a) 28.61 (b) 38.81 (c) 29.13 (d) 28.31

We know that Sk = M − Mo , Where M = Mean, Mo = Mode, σ = S.D.
σ

i.e. 0.32 = 39.6 − Mo ⇒ Mo = 37.52 and also know that, Mo = 3median – 2mean
6.5

37.52 = 3(Median) – 2(39.6)

Median = 38.81 (approx.)

The S.D. of a variate x is σ. The S.D. of the variate ax + b where a, b, c are constant, is [Pb. CET 1996]
c

(a)  a  σ (b) a σ (c)  a2  σ (d) None of these
 c  c c2

Let y = ax + b i.e., y = a x + b i.e. y = Ax + B , where A= a , B = b
c c c c c

∴ y = Ax + B

∴ y − y = A(x − x) ⇒ (y − y)2 = A2(x − x)2 ⇒ ∑(y − y)2 = A2 ∑(x − x)2 ⇒ n.σ 2 = A2 .nσ 2 ⇒ σ 2 = A 2σ 2
y x y x

⇒ σ y =| A |σ x ⇒ σy = a σ x
c

Thus, new S.D. = a σ .
c

Funtion limit continuity & differentiability

If A and B are two non-empty sets, then a rule f which associated to each x ∈ A, a unique number y ∈ B, is
called a function from A to B and we write, f : A → B .

Some Important Definitions.

(1) Real numbers : Real numbers are those which are either rational or irrational. The set of real numbers is
denoted by R.

(i) Rational numbers : All numbers of the form p / q where p and q are integers and q ≠ 0, are called

rational numbers and their set is denoted by Q. e.g. 2 , − 5 , 4 as 4 = 4  are rational numbers.
3 2  1 

(ii) Irrational numbers : Those are numbers which can not be expressed in form of p / q are called

irrational numbers and their set is denoted by Qc (i.e., complementary set of Q) e.g. 2, 1 − 3, π are irrational
numbers.

(iii) Integers : The numbers …….– 3, – 2, – 1, 0, 1, 2, 3, …….. are called integers. The set of integers is
denoted by I or Z. Thus, I or Z ={…….,– 3, – 2, –1, 0, 1, 2, 3,……}

Number Chart:

Real Number (R)

Rational Numbers (Q) Irrational Numbers (QC)

Integers (I or Z) Non-Integers

Negative Integers (I –) Zero Integers (I0) Positive Integers (I+) or Natural number (N)

Whole Numbers (W)

Note :  Set of positive integers I+ = {1, 2, 3, ...}

 Set of negative integers I– = {– 1, – 2, – 3, ……}.
 Set of non negative integers ={0, 1, 2, 3, ..}
 Set of non positive integers = {0, – 1, – 2, – 3,…..}

 Positive real numbers: R+ = (0, ∞)  Negative real numbers: R − = (−∞, 0)

 R0 : all real numbers except 0 (Zero)  Imaginary numbers: C = {i,ω,....}
 Even numbers: E = {0, 2, 4, 6,......}  Odd numbers: 0 = {1, 3, 5,7,......}

 Prime numbers : The natural numbers greater than 1 which is divisible by 1 and itself only, called
prime numbers.

 In rational numbers the digits are repeated after decimal
 0 (zero) is a rational number
 In irrational numbers, digits are not repeated after decimal
 π and e are called special irrational quantities
 ∞ is neither a rational number nor an irrational number

(2) Related quantities : When two quantities are such that the change in one is accompanied by the
change in other, i.e., if the value of one quantity depends upon the other, then they are called related quantities.
e.g. the area of a circle (A = πr 2 ) depends upon its radius (r) as soon as the radius of the circle increases (or
decreases), its area also increases (or decreases). In the given example, A and r are related quantities.

(3) Variable: A variable is a symbol which can assume any value out of a given set of values. The quantities,
like height, weight, time, temperature, profit, sales etc, are examples of variables. The variables are usually denoted
by x, y, z, u, v, w, t etc. There are two types of variables mainly:

(i) Independent variable : A variable which can take any arbitrary value, is called independent variable.

(ii) Dependent variable : A variable whose value depends upon the independent variable is called
dependent variable. e.g. y = x 2, if x = 2 then y = 4 ⇒ so value of y depends on x. y is dependent and x is
independent variable here.

(4) Constant : A constant is a symbol which does not change its value, i.e., retains the same value
throughout a set of mathematical operation. These are generally denoted by a, b, c etc. There are two types of
constant.

(i) Absolute constant : A constant which remains the same throughout a set of mathematical operation is
known as absolute constant. All numerical numbers are absolute constants, i.e. 2, 3, π etc. are absolute constants.

(ii) Arbitrary constant : A constant which remains same in a particular operation, but changes with the
change of reference, is called arbitrary constant e.g. y = mx + c represents a line. Here m and c are constants, but

they are different for different lines. Therefore, m and c are arbitrary constants.

(5) Absolute value : The absolute value of a number x, denoted by |x|, is a number that satisfies the
conditions

− x if x < 0

| x |=  0 if x = 0. We also define |x| as follows, |x|= maximum {x, – x} or |x|= x2

 x if x > 0

The properties of absolute value are

(i) The inequality | x |≤ a means − a ≤ x ≤ a (ii) The inequality | x |≥ a means x ≥ a or x ≤ −a

(iii) | x ± y|≤| x | + | y| and | x ± y|≥| x | −| y| (iv) | xy |=| x || y | (v) x | x ||, y ≠ 0
y =| y

(6) Greatest integer: Let x ∈ R. Then [x] denotes the greatest integer less than or equal to x; e.g. [1.34]=1,
[– 4.57]= – 5, [0.69] = 0 etc.

(7) Fractional part : We know that x ≥ [x]. The difference between the number ‘x’ and it’s integral value
‘[x]’ is called the fractional part of x and is symbolically denoted as {x}. Thus, {x} = x − [x]

e.g., if x = 4.92 then [x] = 4 and {x}= 0.92.

Note :  Fractional part of any number is always non-negative and less than one.

Intervals.

If a variable x assumes any real value between two given numbers, say a and b (a<b) as its value, then x is
called a continuous variable. The set of real numbers which lie between two specific numbers, is called the interval.

There are four types of interval:

(1) Open interval : Let a and b be two real numbers (2) Closed interval : Let a and b be two real
such that a<b, then the set of all real numbers lying numbers such that a<b, then the set of all real
strictly between a and b is called an open interval and numbers lying between a and b including a and b is
is denoted by ]a, b[ or (a, b). Thus, ]a, b[ or called a closed interval and is denoted by [a, b].
(a, b) = {x ∈ R : a < x < b}
( a<x<b ) Thus, [a, b] = {x ∈ R : a ≤ x ≤ b}
ab
[ a ≤ x ≤b]
Open interval ab
Closed interval

(3) Open-Closed interval : It is denoted by ]a, b] or (a, (4) Closed-Open interval : It is denoted by [a, b[ or

b] and ]a, b] or (a, b] = {x ∈ R : a < x ≤ b} [a, b) and [a, b[ or [a, b) = {x ∈ R : a ≤ x < b}

( a<x≤b ] [ a≤x<b)
ab ab
Open closed interval Closed open interval

Definition of Function.

(1) Function can be easily defined with the help of the concept of mapping. Let X and Y be any two non-
empty sets. “A function from X to Y is a rule or correspondence that assigns to each element of set X, one and only
one element of set Y”. Let the correspondence be ‘f’ then mathematically we write f : X → Y where
y = f(x), x ∈ X and y ∈ Y. We say that ‘y’ is the image of ‘x’ under f (or x is the pre image of y).

Two things should always be kept in mind:

(i) A mapping f : X → Y is said to be a function if each element in the set X has it’s image in set Y. It is also
possible that there are few elements in set Y which are not the images of any element in set X.

(ii) Every element in set X should have one and only one image. That means it is impossible to have more
than one image for a specific element in set X. Functions can not be multi-valued (A mapping that is multi-valued is
called a relation from X and Y) e.g.

Set X Set Y Set X Set Y

a 1 a 1
b 2 b 2
c 3 c 3

Function Function

Set X Set Y Set X Set Y

a 1 a 1
b 2 b 2
c 3 c 3

Not function Not function

(2) Testing for a function by vertical line test : A relation f : A → B is a function or not it can be

checked by a graph of the relation. If it is possible to draw a vertical line which cuts the given curve at more than
one point then the given relation is not a function and when this vertical line means line parallel to Y-axis cuts the
curve at only one point then it is a function. Figure (iii) and (iv) represents a function.

Y YY Y

X′ O X X X′ O X X′ O X
Y′ (i) X′ O Y′ Y′
Y′ (ii) (iii) (iv)

(3) Number of functions : Let X and Y be two finite sets having m and n elements respectively. Then each
element of set X can be associated to any one of n elements of set Y. So, total number of functions from set X to set
Y is nm .

(4) Value of the function : If y = f(x) is a function then to find its values at some value of x, say x = a, we
directly substitute x = a in its given rule f(x) and it is denoted by f(a) .

e.g. If f(x) = x 2 + 1, then f(1) = 12 + 1 = 2, f(2) = 22 + 1 = 5, f(0) = 0 2 + 1 = 1etc.

Example: 1 If A contains 10 elements then total number of functions defined from A to A is [UPSEAT 1992]
Solution: (c)
Example: 2 (a) 10 (b) 210 (c) 1010 (d) 210 − 1

Solution: (b) According to formula, total number of functions = nn

Here, n = 10. So, total number of functions = 1010.

If f(x) = x−| x |, then f(−1) = [SCRA 1996]
|x |

(a) 1 (b) – 2 (c) 0 (d) 2

f (−1) = −1− | −1| = −1 − 1 = −2 .
| −1| 1

Example: 3 If f(y) = log y, then f(y) + f  1  is equal to [Rajasthan PET 1996]
y
(d) – 1
(a) 2 (b) 1 (c) 0

Solution: (c) Given f(y) = log y ⇒ f(1 / y) = log (1 / y) , then f(y) + f  1  = log y + log(1 / y) = log 1 = 0.
y

Example: 4 If f(x) = log 1 + x , then f  2x 2  is equal to [MP PET 1999; Rajasthan PET 1999; UPSEAT 2003]
1 − x 1 +x 

(a) [ f(x)]2 (b) [ f(x)]3 (c) 2 f(x) (d) 3 f(x)

Solution: (c) f (x) = log 1 + x 
 1 − x 

 2x   1 + 1 2x 2   x 2 + 1 + 2x  log  11 + x 2 1 + x 
 +x   − 1 +x 2   x 2 − x   1 − x 
∴ f = log  2x  = log  = = 2 log = 2 f (x)
1 2 +x  + 1 − 2x 
 1  


Example: 5 If f(x) = cos[π 2 ]x + cos [−π 2 ]x, then [Orissa JEE 2002]

(a) f  π  = 2 (b) f(−π ) = 2 (c) f(π ) = 1 (d) f  π  = −1
 4   2 

Solution: (d) f(x) = cos [π 2]x + cos [−π 2] x

Example: 6 f(x) = cos(9x) + cos(−10x) = cos(9x) + cos(10x) = 2 cos 19 x  cos x 
Solution: (d)  2  2 

Example: 7 f  π  = 2 cos 19π  cos π  ; f  π  = 2 × −1 × 1 = −1 .
Solution: (c)  2   4  4   2  2 2

n [AIEEE 2003]

∑If f : R → R satisfies f(x + y) = f(x) + f(y), for all x, y ∈ R and f(1) = 7, then f(r) is
r =1

(a) 7n (b) 7(n + 1) (c) 7n(n + 1) (d) 7n(n + 1)
2 2 2

f(x + y) = f(x) + f(y)

put x = 1, y = 0 ⇒ f(1) = f(1) + f(0) = 7

put x = 1, y = 1 ⇒ f(2) = 2.f(1) = 2.7 ; similarly f(3) = 3.7 and so on

∑n f(r) = 7 (1 + 2 + 3 + ..... + n) = 7n(n + 1) .
2

r =1

If f(x) = 1 + 1 for x > 2, then f(11) = [EAMCET 2003]
x + 2 2x − 4 x − 2 2x − 4

(a) 7 (b) 5 (c) 6 (d) 5
6 6 7 7

f(x) = 1+ 1
x + 2 2x − 4
x − 2 2x − 4

f(11) = 1 1 = 1 + 1 = 3− 2 + 3+ 2 = 6 .
+ 11 − 2 18 3+ 3− 7 7 7
2 2
11 + 2 18

Domain, Co-domain and Range of Function.

If a function f is defined from a set of A to set B then for f : A → B set A is called the domain of function f and set
B is called the co-domain of function f. The set of all f-images of the elements of A is called the range of function f.

In other words, we can say Domain = All possible values of x for which f(x) exists.

Range = For all values of x, all possible values of f(x).

A B Domain = {a, b, c, d} = A
p
af q Co-domain = {p, q, r, s} = B
Range Range = {p, q, r}
Co-domain b r

c s

d

Domain

(1) Methods for finding domain and range of function
(i) Domain

(a) Expression under even root (i.e., square root, fourth root etc.) ≥ 0

(b) Denominator ≠ 0.

(c) If domain of y = f(x) and y = g (x) are D1 and D2 respectively then the domain of f(x) ± g (x) or
f(x).g (x) is D1 ∩ D2.

(d) While domain of f(x) is D1 ∩ D2 − {g(x) = 0}.
g(x)

( )(e) Domain of f(x) = D1 ∩ {x : f(x) ≥ 0}

(ii) Range : Range of y = f(x) is collection of all outputs f(x) corresponding to each real number in the

domain.

(a) If domain ∈ finite number of points ⇒ range ∈ set of corresponding f(x) values.

(b) If domain ∈ R or R – [some finite points]. Then express x in terms of y. From this find y for x to be defined
(i.e., find the values of y for which x exists).

(c) If domain ∈ a finite interval, find the least and greatest value for range using monotonicity.

Important Tips

• If f(x) is a given function of x and if a is in its domain of definition, then by f(a) it means the number obtained by replacing x by a in f(x) or
the value assumed by f(x) when x = a.

• Range is always a subset of co-domain.

Example: 8 Domain of the function 1 is [Roorkee 1987; Rajasthan PET 2000]
Solution: (a) (a) (−∞, − 1) ∪ (1, ∞) x2 −1
Example: 9 (c) (−∞, − 1) ∪ [1, ∞) (d) None of these
Solution: (b) (b) (−∞, − 1] ∪ (1, ∞)
Example: 10
Solution: (c) For domain, x 2 − 1 > 0 ⇒(x − 1)(x + 1) > 0

Example: 11 ⇒ x < −1 or x > 1 ⇒ x ∈ (−∞,−1) ∪ (1, ∞) .
Solution: (b)
Example: 12 The domain of the function f(x) = 1 is [Roorkee 1998]
Solution: (b) | x|−x
Example: 13
Solution: (d) (a) R + (b) R − (c) R0 (d) R

Example: 14 For domain, | x | −x > 0 ⇒| x |> x . This is possible, only when x ∈ R − . [IIT 2001; UPSEAT 2001]

Solution: (c) Find the domain of definition of f(x) = log 2 (x + 3) (d) (−∞, ∞)
x2 + 3x + 2

(a) (−3, ∞) (b) {−1, − 2} (c) (−3, ∞) − {−1, − 2}

Here f(x) = log 2(x + 3) = log 2(x + 3) exists if,
x2 + 3x + 2 (x + 1)(x + 2)

Numerator x + 3 > 0 ⇒ x > −3 …... (i)

and denominator (x + 1)(x + 2) ≠ 0 ⇒ x ≠ −1, − 2 …… (ii)

Thus, from (i) and (ii); we have domain of f(x) is (−3, ∞) − {−1, − 2} .

The domain of the function f(x) = (2 − 2x − x 2 ) is [BIT Ranchi 1992]

(a) − 3 ≤ x ≤ 3 (b) − 1 − 3 ≤ x ≤ −1 + 3
(c) −2 ≤ x ≤ 2 (d) None of these

The quantity square root is positive, when − 1 − 3 ≤ x ≤ −1 + 3. [MP PET 1996]
If the domain of function f(x) = x 2 − 6x + 7 is (−∞, ∞) , then the range of function is

(a) (−∞, ∞) (b) [ − 2,∞) (c) (−2, 3) (d) (−∞, − 2)

x 2 − 6x + 7 = (x − 3)2 − 2 Obviously, minimum value is –2 and maximum ∞ .

The domain of the function f(x) = x − x 2 + 4 + x + 4 − x is [AMU 1999]

(a) [−4, ∞) (b) [−4, 4] (c) [0, 4] (d) [0,1]

f(x) = x − x 2 + 4 + x + 4 − x
clearly f(x) is defined if

4 + x ≥ 0 ⇒ x ≥ −4
4−x≥0 ⇒ x≤4
x(1 − x) ≥ 0 ⇒ x ≥ 0 and x ≤ 1
∴ Domain of f = (−∞, 4] ∩ [−4, ∞) ∩ [0,1] = [0,1] .

The domain of the function log(x 2 − 6x + 6) is [Roorkee 1999; MP PET 2002]

(a) (−∞, ∞) (b) (−∞,3 − 3) ∪ (3 + 3, ∞)
(c) (−∞,1] ∪[5, ∞) (d) [0, ∞)

The function f(x) = log(x 2 − 6x + 6) is defined when log(x 2 − 6x + 6) ≥ 0
⇒ x 2 − 6x + 6 ≥ 1 ⇒ (x − 5)(x − 1) ≥ 0
This inequality hold if x ≤ 1 or x ≥ 5 . Hence, the domain of the function will be (−∞,1] ∪ [5, ∞) .

Example: 15 The domain of definition of the function y(x) given by 2 x + 2y = 2 is [IIT Screening 2000; DCE 2001]
Solution: (d)
Example: 16 (a) (0, 1] (b) [0, 1] (c) (−∞, 0] (d) (−∞,1)
Solution: (a)
2y = 2 − 2x
Example: 17 y is real if 2 − 2 x ≥ 0 ⇒ 2 > 2 x ⇒ 1 > x
Solution: (c)
⇒ x ∈ (−∞, 1)
Example: 18
Solution: (d) The domain of the function f(x) = sin −1[log 2 (x / 2)] is [AIEEE 2002; Rajasthan PET 2002]

Example: 19 (a) [1, 4] (b) [– 4, 1] (c) [– 1, 4] (d) None of these

f(x) = sin −1[log 2 (x/2)]

Domain of sin −1 x is x ∈[−1,1]

⇒ −1 ≤ log 2 (x / 2) ≤ 1 ⇒ 1 ≤ x ≤ 2 ⇒ 1≤ x ≤4
2 2

∴ x ∈ [1, 4] .

The domain of the derivative of the function f (x)  tan −1 x , | x|≤ 1 [IIT Screening 2002]
 1 −1) , | x|> 1 is
=  2 (| x | (d) R − {−1, 1}

(a) R − {0} (b) R − {1} (c) R − {−1}

t12an(−−x1 − 1), x < −1 − 1 , x < −1
x, − 1 ≤ x ≤ 1  2 ,−1< x <1

f(x) = ⇒ f ′(x) = 1 1

 1 (x 1), x 1 + x2
 2
 + >  1 , x >1
 2

f ′(−1 − 0) = − 1 ; f ′(−1 + 0) = 1 = 1
2 1 + (−1 + 0)2 2

f ′(1 − 0) = 1 1 0) 2 = 1 ; f ′(1 + 0) = 1
(1 − 2 2
+

∴ f ′(−1) does not exist.

∴ domain of f ′(x) = R − {−1} .

Domain of definition of the function f (x) = 4 3 + log10 (x 3 − x), is [AIEEE 2003]
− x2
(d) (–1, 0) ∪ (1, 2) ∪ (2, ∞)
(a) (1, 2) (b) (–1, 0) ∪ (1, 2) (c) (1, 2) ∪ (2, ∞ )

f(x) = 3 + log10 (x 3 − x)
4 − x2

So, 4 − x 2 ≠ 0 ⇒ x ≠ ± 4 ⇒ x ≠ ± 2 –+ – +

and x 3 − x > 0 ⇒ x(x 2 − 1) > 0 ⇒ x > 0,| x | > 1 –1 0 1

∴ D = (−1,0) ∪ (1,∞) − {2}

D = (−1, 0) ∪ (1, 2) ∪ (2, ∞) .

The domain of the function f(x) = log3+x(x2 − 1) is [Orissa JEE 2003]
(a) (−3, − 1) ∪(1, ∞)
(c) (−3,−2) ∪ (−2, − 1) ∪ (1, ∞) (b) [−3, − 1) ∪[1, ∞)
(d) [−3, − 2) ∪ (−2, − 1) ∪[1, ∞]

Solution: (c) f(x) is to be defined when x 2 − 1 > 0
Example: 20
Solution: (a) ⇒ x 2 > 1, ⇒ x < −1 or x > 1 and 3 + x > 0
Example: 21 ∴ x > −3 and x ≠ −2
Solution: (d) ∴ Dr = (−3, − 2) ∪ (−2, − 1) ∪ (1, ∞) .

Example: 22 Domain of definition of the function f(x) = sin −1(2x) + π , for real value x, is [IIT Screening 2003]
Solution: (b) 6

Example: 23 (a) − 1 , 1 (b) − 1 , 1 (c)  − 1 , 1  (d) − 1 , 1
Solution: (a) 4 2  2 2   2 9  4 4 
Example 24
− π ≤ sin −1(2x) ≤ π ⇒ − 1 ≤ 2x ≤1 ⇒ x ∈ − 1 , 1 .
6 2 2 4 2 

The range of f(x) = cos x − sin x, is [MP PET 1995]

(a) (−1,1) (b) [−1,1) (c) − π , π  (d) [− 2, 2]
2 2 

Let, f(x) = cos x − sin x ⇒ f(x) = 2  1 cos x − 1 sin x  ⇒ f(x) = 2 cos x + π 
2 2  4 

Now since, −1≤ cos x + π  ≤ 1 ⇒ − 2 ≤ f(x) ≤ 2 ⇒ f(x)∈[− 2 , 2]
 4 

Trick :  Maximum value of cos x − sin x is 2 and minimum value of cos x − sin x is − 2 .

Hence, range of f(x) = [− 2, 2] .

The range of 1+ x2 is [Karnataka CET 1989]
x2
(d) [1, ∞)
(a) (0,1) (b) (1, ∞) (c) [0, 1]

Let y 1 + x 2 ⇒ x2y =1 + x2 ⇒ x 2 (y − 1) = 1 x2 1
x y −1
= 2 ⇒ =

Now since, x2 >0 ⇒ y 1 1 > 0 ⇒ (y − 1) > 0 ⇒ y >1 ⇒ y ∈ (1, ∞)

Trick : y 1 + x 2 =1+ 1 . Now since, 1 is always > 0 ⇒ y>1 ⇒ y ∈(1, ∞) .
x x2 x2
= 2

For real values of x, range of the function y= 1 is [DCE 1998]
2 − sin 3x

(a) 1 ≤ y ≤ 1 (b) − 1 ≤ y <1 (c) − 1 > y > −1 (d) 1 > y >1
3 3 3 3

 y= 1 , ∴ 2 − sin 3x = 1 ⇒ sin 3x = 2− 1
2 − sin 3x y y

Now since,

− 1 ≤ sin 3x ≤ 1 ⇒ −1 ≤ 2 − 1 ≤ 1 ⇒ − 3 ≤ − 1 ≤ −1 ⇒ 1 ≤ 1 ≤ 3 ⇒ 1 ≤ y ≤ 1 .
y y y 3

If f(x) = a cos ( bx + c ) + d, then range of f(x) is [UPSEAT 2001]

(a) [d + a, d + 2a] (b) [a − d, a + d] (c) [d + a, a − d] (d) [d − a, d + a]

Solution: (d) f(x) = a cos(bx + c) + d ….. (i)
(c) R
Example: 25 For minimum cos(bx + c) = −1
Solution: (b) from (i), f(x) = −a + d = (d − a) ,
for maximum cos(bx + c) = 1
from (i), f(x) = a + d = (d + a)
∴ Range of f(x) = [d − a, d + a] .

The range of the function f(x) = x + 2 is [Rajasthan PET 2002]
|x + 2|
(d) R − {−2}
(a) {0, 1} (b) {– 1, 1}

f(x) = x + 2 = −1, x < −2
|x + 2|  x > −2
 1,

∴ Range of f(x) is {−1,1} .

Example: 26 The range of f(x) = sec π cos 2 x  , −∞<x<∞ is [Orissa JEE 2002]
 4 
(d) (−∞, − 1] ∪[1, ∞)
(a) [1, 2] (b) [1, ∞) (c) [− 2, − 1] ∪[1, 2]

Solution: (a) f(x) = sec π cos 2 x 
 4 

We know that, 0 ≤ cos 2 x ≤ 1 at cos x = 0, f(x) = 1 and at cos x = 1, f(x) = 2

∴ 1 ≤ x ≤ 2 ⇒ x ∈[1, 2] .

Example: 27 Range of the function f(x) = x2 + x + 2 ; x ∈ R is [IIT Screening 2003]
Solution: (c) x2 + x +1
(d) (1,7 / 5]
(a) (1, ∞) (b) (1,11 / 7) (c) (1,7 / 3]

f(x) = 1 + 1 ⇒ Range = (1,7 / 3] .

 x 1  2 3
 2  4
+ +

Algebra of Functions.

Let f(x) and g (x) be two real and single-valued functions, with domains X f , Xg and ranges Yf and Yg
respectively. Let X = X f ∩ Xg ≠ φ. Then, the following operations are defined.

(1) Scalar multiplication of a function : (c f)(x) = c f(x), where c is a scalar. The new function c f(x)

has the domain X f .

(2) Addition/subtraction of functions :(f ± g)(x) = f(x) ± g(x). The new function has the domain X.

(3) Multiplication of functions : (fg)(x) = (g f)(x) = f(x)g (x). The product function has the domain X.

(4) Division of functions :

(i)  f  (x) = f(x) . The new function has the domain X, except for the values of x for which g (x) = 0.
g g(x)

(ii)  g  (x) = g(x) . The new function has the domain X, except for the values of x for which f(x) = 0.
f f(x)

(5) Equal functions : Two function f and g are said to be equal functions, if and only if
(i) Domain of f = domain of g
(ii) Co-domain of f = co-domain of g
(iii) f(x) = g(x)∀x ∈ their common domain
(6) Real valued function : If R, be the set of real numbers and A, B are subsets of R, then the function
f : A → B is called a real function or real –valued function.

Kinds of Function.

(1) One-one function (injection) : A function f : A → B is said to be a one-one function or an injection, if
different elements of A have different images in B. Thus, f : A → B is one-one.

⇔ a ≠ b ⇒ f(a) ≠ f(b) for all a, b ∈ A ⇔ f(a) = f(b) ⇒ a = b for all a, b ∈ A .

e.g. Let f : A → B and g : X → Y be two functions represented by the following diagrams.

A B X Y

a1 f b1 x1 g y1
a2 b2 x2 y2
b3 y3
a3 bb54 x3 yy54
a4 x4

Clearly, f : A → B is a one-one function. But g : X → Y is not one-one function because two distinct
elements x1 and x3 have the same image under function g.
(i) Method to check the injectivity of a function

Step I : Take two arbitrary elements x, y (say) in the domain of f.

Step II : Put f(x) = f(y).

Step III : Solve f(x) = f(y). If f(x) = f(y) gives x = y only, then f : A → B is a one-one function (or an
injection). Otherwise not.

Note :  If function is given in the form of ordered pairs and if two ordered pairs do not have same second
element then function is one-one.

 If the graph of the function y = f(x) is given and each line parallel to x-axis cuts the given curve at
maximum one point then function is one-one. e.g.

YY

X′ O X (0, 1)
Y′ f(x) = ax + b X′

OX
f (x) = ax (0 < a < 1)

Y′

(ii) Number of one-one functions (injections) : If A and B are finite sets having m and n elements

respectively, then number of one-one functions from A to B =  n Pm , if n ≥ m
 0 if n < m
 ,

(2) Many-one function : A function f : A → B is said to be a many-one function if two or more elements of
set A have the same image in B.

Thus, f : A → B is a many-one function if there exist x, y ∈ A such that x ≠ y but f(x) = f(y).

In other words, f : A → B is a many-one function if it is not a one-one function.

A f B X g Y
b1
a1 x1 y1
b2 x2 y2
a2 bb34 x3 y3
a3 bb65 x4 y4
a4 x5 y5
a5

Note :  If function is given in the form of set of ordered pairs and the second element of atleast two ordered
pairs are same then function is many-one.

 If the graph of y = f(x) is given and the line parallel to x-axis cuts the curve at more than one point
then function is many-one.

Y Y

Y Y

X′ O f (x) = x2 X X′ fO(x) = |x| X X

Y′ Y′

• If the domain of the function is in one quadrant then the trigonometrical functions are always one-one.

• If trigonometrical function changes its sign in two consecutive quadrants then it is one-one but if it does
not change the sign then it is many-one.

Y f : (0,π ), f(x) = sin x Y f : (0,π ), f(x) = cos x
Y Y one-one

many-one

+ π/2
X′
++ XX
O

X′ O π/2 π X

Y′ Y′

 In three consecutive quadrants trigonometrical functions are always many-one.

(3) Onto function (surjection) : A function f : A → B is onto if each element of B has its pre-image in A.
Therefore, if f −1(y) ∈ A, ∀y ∈ B then function is onto. In other words, Range of f = Co-domain of f.

e.g. The following arrow-diagram shows onto function.

A B XY
a1 f
b1 x1 g y1
a2 b2
b3 x2 y2
a3 x3
x4 y3

(i) Number of onto function (surjection) : If A and B are two sets having m and n elements respectively

n

∑such that 1 ≤ n ≤ m, then number of onto functions from A to B is (−1)n−r nCrrm.
r =1
(4) Into function : A function f : A → B is an into function if there exists an element in B having no pre-

image in A.
In other words, f : A → B is an into function if it is not an onto function.

e.g. The following arrow-diagram shows into function.

A B X Y
x1 g
a1 f b1 y1
a2 b2 x2 y2
y3
a3 b3 x3 y4

(i) Method to find onto or into function

(a) If range = co-domain, then f(x) is onto and if range is a proper subset of the co-domain, then f(x) is
into.

(b) Solve f(x) = y by taking x as a function of y i.e., g(y) (say).

(c) Now if g(y) is defined for each y ∈co-domain and g(y) ∈ domain for y ∈ co-domain, then f(x) is onto
and if any one of the above requirements is not fulfilled, then f(x) is into.

(5) One-one onto function (bijection) : A function f : A → B is a bijection if it is one-one as well as onto.

In other words, a function f : A → B is a bijection if A B
(i) It is one-one i.e., f(x) = f(y) ⇒ x = y for all x, y ∈ A.
(ii) It is onto i.e., for all y ∈ B , there exists x ∈ A such that f(x) = y. a1 f b1
a2 b2
Clearly, f is a bijection since it is both injective as well as surjective. b3
a3 b4
a4

Number of one-one onto function (bijection) : If A and B are finite sets and f : A → B is a bijection,

then A and B have the same number of elements. If A has n elements, then the number of bijection from A to B is

the total number of arrangements of n items taken all at a time i.e. n!.

(6) Algebraic functions : Functions consisting of finite number of terms involving powers and roots of the
independent variable and the four fundamental operations +, –, × and ÷ are called algebraic functions.

3 (ii) x +1 , x ≠ 1 (iii) 3x 4 − 5x + 7
x −1
e.g., (i) x 2 + 5x

The algebraic functions can be classified as follows:

(i) Polynomial or integral function : It is a function of the form a0xn + a1xn−1 + .... + an−1x + an,
where a0 ≠ 0 and a0, a1,........., an are constants and n ∈ N is called a polynomial function of degree n

e.g. f(x) = x3 − 2x2 + x + 3 is a polynomial function.

Note :  The polynomial of first degree is called a linear function and polynomial of zero degree is called a
constant function.

(ii) Rational function : The quotient of two polynomial functions is called the rational function. e.g.

f(x) = x2 −1 is a rational function.
2x3 + x2 + 1

(iii) Irrational function : An algebraic function which is not rational is called an irrational function. e.g.

f(x) = x + x + 6, g(x) = x3 − x are irrational functions.
1+ x1/4

(7) Transcendental function : A function which is not algebraic is called a transcendental function. e.g.,
trigonometric; inverse trigonometric , exponential and logarithmic functions are all transcendental functions.

(i) Trigonometric functions : A function is said to be a trigonometric function if it involves circular

functions (sine, cosine, tangent, cotangent, secant, Y
cosecant) of variable angles.

(a) Sine function : The function that associates to X’ X
each real numbers x to sin x is called the sine function.
Here x is the radian measure of the angle. The domain of Y’
the sine function is R and the range is [–1, 1].

(b) Cosine function: The function that associates to each real number x to cos x is called the cosine
function. Here x is the radian measure of the angle. The domain of the cosine function is R and the range is [– 1, 1].

Y
X’ X

Y’

(c) Tangent function : The function that associates a real number x to tan x is called the tangent function.

Clearly, the tangent function is not defined at odd Y
X
multiples of π i.e., ± π , ± 3π etc. So, the domain of the
2 2 2 Y’

π X’
2
tangent function is R − {(2n + 1) |n ∈ I} . Since it takes every

value between − ∞ and ∞ . So, the range is R. Graph of

f(x) = tan x is shown in figure.

(d) Cosecant function : The function that associates a real number x to cosecx is called the cosecant

function. Y
Clearly, cosec x is not defined at x = nπ , n ∈ I . i.e.,

0, ± π , ± 2π , ± 3π etc. So, its domain is R − {nπ |n ∈ I}. X’ X
Since cosec x ≥ 1 or cosec x ≤ −1 . Therefore, range is

(−∞, − 1] ∪[1, ∞) . Graph of f(x) = cosecx is shown in figure.

Y’

(e) Secant function : The function that associates a real number x to sec x is called the secant function.

Clearly, sec x is not defined at odd multiples of Y
X
π i.e., (2π + 1) π , where n∈ I. So, its domain is
2 2 Y′
X’

R − {(2n + 1) π | n ∈ I}. Also, |sec x |≥ 1, therefore its
2

range is (−∞, − 1] ∪ [1, ∞). Graph of f(x) = sec x is

shown in figure.

(f) Cotangent function : The function that associates a real number x to cot x is called the cotangent
function. Clearly, cot x is not defined at x = nπ , n ∈ I i.e., at n = 0, ± π , ± 2π etc. So, domain of cot x is
R − {nπ |n ∈ I}. The range of f(x) = cot x is R as is evident from its graph in figure.

Y

X’ X

Y’

(ii) Inverse trigonometric functions

Function Domain Range Definition of the function
sin−1 x [−1,1] [−π /2, π /2] y = sin−1 x ⇔ x = sin y
cos−1 x [–1, 1] y = cos−1 x ⇔ x = cos y
tan−1 x [0, π] y = tan−1 x ⇔ x = tan y
cot−1 x (–∞, ∞) or R (–π/2, π/2) y = cot−1 x ⇔ x = cot y
cosec −1 x (–∞, ∞) or R
sec−1 x R – (–1, 1) (0, π) y = cosec−1x ⇔ x = cosecy
R – (–1, 1) [−π /2, π /2] − {0} y = sec−1 x ⇔ x = sec y

[0, π ] − [π /2]

(iii) Exponential function : Let a ≠ 1 be a positive real number. Then f : R → (0, ∞) defined by f(x) = a x is
called exponential function. Its domain is R and range is (0, ∞) .

Y a>1 a<1 Y X
f(x) = ax f(x) = ax (0, 1)
(0, 1) O
X′ O X X′

Y′ Y′

graph of f(x) = ax, when a > 1 graph of f(x) = ax, when a < 1

(iv) Logarithmic function : Let a ≠ 1 be a positive real number. Then f : (0, ∞) → R defined by
f(x) = loga x is called logarithmic function. Its domain is (0, ∞)and range is R.

YY

f(x) = loga x X′ O (1, 0) X
X′ O (1, 0) X f(x) = loga x

Y′ Y′

graph of f(x) = loga x, when a > 1 graph of f(x) = loga x, when a < 1

(8) Explicit and implicit functions : A function is said to be explicit if it can be expressed directly in terms
of the independent variable. If the function can not be expressed directly in terms of the independent variable or
variables, then the function is said to be implicit. e.g. y = sin −1 x + log x is explicit function, while x 2 + y 2 = xy and

x 3 y 2 = (a − x)2 (b − y)2 are implicit functions.

(9) Constant function : Let k be a fixed real number. (10) Identity function : The function defined by
Then a function f(x) given by f(x) = k for all x ∈ R is f(x) = x for all x ∈ R , is called the identity function on
R. Clearly, the domain and range of the identity
called a constant function. The domain of the constant function is R.
function f(x) = k is the complete set of real numbers
The graph of the identity function is a straight line
and the range of f is the singleton set {k}. The graph of passing through the origin and inclined at an angle of
a constant function is a straight line parallel to x-axis as 45o with positive direction of x-axis.
shown in figure and it is above or below the x-axis
according as k is positive or negative. If k = 0, then the Y
straight line coincides with x-axis.

Y

k f (x) = k f (x) = x
X′ O X X′ X

Y′ Y′

(11) Modulus function : The function defined (12) Greatest integer function: Let f(x) = [x],

by f(x) =| x |= x, when x≥0 is called the modulus where [x] denotes the greatest integer less than or equal
-x, when x<0 to x. The domain is R and the range is I. e.g. [1.1] = 1,
[2.2] = 2, [– 0.9] = –1, [– 2.1] = – 3 etc. The function
function. The domain of the modulus function is the set
R of all real numbers and the range is the set of all non- f defined by f(x) = [x] for all x ∈ R , is called the

negative real numbers. Y greatest integer function. Y
3

f (x) =– x f (x) = x 2
X′ OX
1

X’ X

–3 –2 –1 1 2 3
–1

–2

Y′ Y’ –3

(13) Signum function : The function defined by (14) Reciprocal function: The function that
associates each non-zero real number x to be reciprocal
| x |, x 0  1, x > 0
 x ≠  1
f(x) = or f(x) =  0, x=0 is called the x is called the reciprocal function. The domain and

0 , x=0 − 1, x < 0 range of the reciprocal function are both equal to

signum function. The domain is R and the range is the R − {0} i.e., the set of all non-zero real numbers. The
set {–1, 0, 1}.
graph is as shown. Y

Y f (x) =1/ x
(0, 1)

X′ O O X X′ O X
Y′
(0, –1)
Y′

Domain and Range of Some Standard Functions

Function Domain Range
R
Polynomial function R R
R {K}
Identity function x R R0
R0
Constant function K R+ ∪{0}

Reciprocal function 1 R
x
{−1, 0, 1}
x2,| x | R R+ ∪{0}
R– ∪{0}
x3, x| x| R
I
Signum function R [0, 1)
R
x+| x | R R
R R+
x−| x | R R
[0, ∞) [−1, 1]
[x] [−1, 1]
R
x − [x]

x R
ax R+
log x R
sin x

cos x R

tan x R – ± π , ± 3π ,.........
2 2 

cot x R – {0, ± π , ± 2π ,..............} R
sec x R –(–1, 1)
R – ± π , ± 3π ,.............
 2 2 

cosec x R – {0, ± π, ± 2π,..............} R – ( – 1, 1)
sin−1 x
[−1, 1] −π , π
 2 2 

cos−1 x [−1, 1] [0, π ]
tan−1 x R
 −π , π 
 2 2 

cot−1 x R (0, π )
sec−1 x R– (−1,1)
[0, π ] − π 
 
 2 

cosec −1 x R– (−1,1) − π , π  − {0}
2 2 

Important Tips

• Any function, which is entirely increasing or decreasing in the whole of a domain, is one-one.
• Any continuous function f(x), which has at least one local maximum or local minimum, is many-one.

• If any line parallel to the x-axis cuts the graph of the function at most at one point, then the function is one-one and if there exists a
line which is parallel to the x-axis and cuts the graph of the function in at least two points, then the function is many-one.

• Any polynomial function f : R → R is onto if degree of f is odd and into if degree of f is even.

• An into function can be made onto by redefining the co-domain as the range of the original function.

Example: 28 Function f : N → N, f(x) = 2x + 3 is [IIT 1973; UPSEAT 1983]
Solution: (b)
(a) One-one onto (b) One-one into (c) Many-one onto (d) Many –one into
Example: 29
Solution: (b) f is one-one because f(x1) = f(x2) ⇒ 2x1 + 3 = 2x2 + 3 ⇒ x1 = x2
Example: 30
Solution: (b) Further f −1(x) = x − 3 ∉ N (domain) when x = 1, 2, 3 etc.
2
Example: 31
Solution: (d) ∴ f is into which shows that f is one-one into.
Example: 32
Solution: (d) The function f : R → R defined by f(x) = (x − 1)(x − 2)(x − 3) is [Roorkee 1999]

(a) One-one but not onto (b) Onto but not one-one
(c) Both one-one and onto (d) Neither one-one nor onto

We have f(x) = (x − 1)(x − 2)(x − 3) ⇒ f(1) = f(2) = f(3) = 0 ⇒ f(x) is not one-one

For each y ∈ R, there exists x ∈ R such that f(x) = y . Therefore f is onto.

Hence, f : R → R is onto but not one-one.

Find number of surjection from A to B where A = {1, 2, 3, 4}, B = {a,b} [IIT Screening 2001]

(a) 13 (b) 14 (c) 15 (d) 16

2

∑Number of surjection from A to B = (−1)2−r 2Cr (r)4
r =1

= (−1)2−1 2C1(1)4 + (−1)2−2 2C2(2)4 = −2 + 16 = 14
Therefore, number of surjection from A to B = 14.

Trick : Total number of functions from A to B is 24 of which two function f(x) = a for all x ∈ A and

g(x) = b for all x ∈ A are not surjective. Thus, total number of surjection from A to B

= 24 − 2 = 14.
If A = {a,b,c}, then total number of one-one onto functions which can be defined from A to A is

(a) 3 (b) 4 (c) 9 (d) 6

Total number of one-one onto functions = 3! [Rajasthan PET 2000]

If f : R → R, then f(x)=| x | is

(a) One-one but not onto (b) Onto but not one-one
(c) One-one and onto (d) None of these
f(−1) = f(1) = 1 ∴function is many-one function.

Obviously, f is not onto so f is neither one-one nor onto.

Example: 33 Let f :R→R be a function defined by f(x) = x −m , where m ≠ n . Then [UPSEAT 2001]
Solution: (b) x −n
(d) f is many one into
Example: 34 (a) f is one-one onto (b) f is one-one into (c) f is many one onto
Solution: (c)
Example: 35 For any x, y ∈ R, we have
Solution: (c)
f(x) = f(y) ⇒ x −m = y−m ⇒ x = y
x −n y−n

∴ f is one-one

Let α ∈ R such that f(x) = α ⇒ x−m =α ⇒ x = m − nα
x−n 1−α

Clearly x ∉ R for α = 1 . So, f is not onto.

The function f : R → R defined by f(x) = e x is [Karnataka CET 2002; UPSEAT 2002]

(a) Onto (b) Many-one (c) One-one and into (d) Many one and onto

Function f : R → R is defined by f(x) = e x . Let x1, x2 ∈ R and f(x1) = f(x2) or e x1 = e x2 or x1 = x2 . Therefore f is
one-one. Let f(x) = e x = y . Taking log on both sides, we get x = log y . We know that negative real numbers have no
pre-image or the function is not onto and zero is not the image of any real number. Therefore function f is into.

 n−1 , when nis odd
 when nis even
A function f from the set of natural numbers to integers defined by f (n) =  2 , is [AIEEE 2003]
− n
2 ,

(a) One-one but not onto (b) Onto but not one-one

(c) One-one and onto both (d) Neither one-one nor onto

f:N→I

f(1) = 0, f(2) = −1, f(3) = 1, f(4) = −2, f(5) = 2 and f(6) = −3 so on.

10
2 –1
31
4 –2
52
6 –3

In this type of function every element of set A has unique image in set B and there is no element left in set B. Hence f is
one-one and onto function.

Even and Odd function.

(1) Even function : If we put (–x) in place of x in the given function and if f(−x) = f(x) , ∀x ∈ domain then
function f(x) is called even function. e.g. f(x) = e x + e −x , f(x) = x 2 , f(x) = x sin x, f(x) = cos x, f(x) = x 2 cos x all
are even function.

(2) Odd function : If we put (–x) in place of x in the given function and if f(−x) = − f(x), ∀x ∈ domain then
f(x) is called odd function. e.g. f(x) = e x − e −x , f(x) = sin x, f(x) = x 3 , f(x) = x cos x, f(x) = x 2 sin x all are odd
function.

Important Tips

• The graph of even function is always symmetric with respect to y-axis.
• The graph of odd function is always symmetric with respect to origin.
• The product of two even functions is an even function.
• The sum and difference of two even functions is an even function.
• The sum and difference of two odd functions is an odd function.
• The product of two odd functions is an even function.
• The product of an even and an odd function is an odd function
• It is not essential that every function is even or odd. It is possible to have some functions which are neither even nor odd function. e.g.

f(x) = x2+ x3, f(x) = loge x, f(x) = ex.
• The sum of even and odd function is neither even nor odd function.

• Zero function f(x) = 0 is the only function which is even and odd both.

Example: 36 Which of the following is an even function [UPSEAT 1998]
Solution: (a)
(a) x ax − 1  (b) tan x (c) ax − a−x (d) ax +1
ax + 1 2 ax −1

We have : f(x) = x ax − 1 
ax + 1

− x a− x − 1   1 − 1  − x  1− ax  x ax −1 
a− x + 1 − x  ax  1+ ax ax +1
f (− x) = = 1  = = = f(x)
 ax 
+ 1

So, f(x) is an even function.

Example: 37 Let f(x) = x4 + 15, then the graph of the function y = f(x) is symmetrical about

(a) The x-axis (b) The y-axis (c) The origin (d) The line x = y

Solution: (b) f(x) = x4 + 15 ⇒ f(−x) = (−x)4 + 15 = x4 + 15 = f(x)
⇒ f(−x) = f(x) ⇒ f(x) is an even function ⇒ f(x) is symmetric about y-axis.

Example: 38 The function f(x) = log(x + x2 + 1) is

(a) An even function (b) An odd function (c) Periodic function (d) None of these

Solution: (b) f(x) = log(x + x2 + 1) and f(−x) = − log(x + x2 + 1) = − f(x) , so f(x) is an odd function.
Example: 39
Which of the following is an even function [Rajasthan PET 2000]
Solution: (b)
(a) f(x) = ax +1 (b) f(x) = x ax − 1  (c) f(x) = ax − a−x (d) f(x) = sin x
ax −1 ax + 1 ax + a−x

In option (a), f (− x) = a−x +1 = 1+ ax = − ax +1 = − f(x) So, It is an odd function.
a−x −1 1− ax ax −1

In option (b), f (− x) = (− x) a− x −1 = −x (1 − ax) = x (a x − 1) = f(x) So, It is an even function.
a− x +1 1+ ax (a x + 1)

In option (c), f(−x) = a−x − ax = − f(x) So, It is an odd function.
a−x + ax

In option (d), f(−x) = sin(−x) = − sin x = − f(x) So, It is an odd function.

Example: 40 The function f(x) = sin  log(x + x2 + 1 ) is [Orissa JEE 2002]

(a) Even function (b) Odd function (c) Neither even nor odd (d) Periodic function

Solution: (b) f(x) = sin log(x + 1 + x2 )

⇒ f(−x) = sin[log(−x + 1 + x2 )] ⇒ f(−x) = sin log( 1+ x2 − x) ( 1+ x2 + x) 
 ( 1+ x2 
+ x) 

 1  ⇒ f(−x) = sinlog(x + 1 + x2 )−1 
⇒ f(−x) = sin log   
(x + 1 + x 2 )

⇒ f(−x) = sin− log(x + 1 + x 2 ) ⇒ f(−x) = − sinlog(x + 1 + x 2 ) ⇒ f(−x) = − f(x)
∴ f(x) is odd function.

Periodic Function.

A function is said to be periodic function if its each value is repeated after a definite interval. So a function f(x)
will be periodic if a positive real number T exist such that, f(x + T) = f(x), ∀x ∈ domain. Here the least positive value
of T is called the period of the function. Clearly f(x) = f(x + T) = f(x + 2T) = f(x + 3T) = ..... . e.g. sin x, cos x, tan x are
periodic functions with period 2π, 2π and π respectively.

Some standard results on periodic functions

Functions Periods
(1) sinn x, cosn x , secn x, cosecnx π ; if n is even
2π ; if n is odd or fraction
(2) tann x, cotn x
π ; n is even or odd.

(3) |sin x |,|cos x |,| tan x |, π
|cot x |, |sec x |, |cosec x |
1
(4) x − [x] Period does not exist

(5) Algebraic functions e.g., x, x2, x3 + 5,....etc

Important Tips

• If f(x) is periodic with period T, then c.f(x) is periodic with period T, f(x + c) is periodic with period T and f(x) ± c is periodic with period
T. where c is any constant.

• If a function f(x) has a period T, then the function f(ax+b) will have a period T.
|a|

• If f(x) is periodic with period T then 1 is also periodic with same period T.
f(x)

• If f(x) is periodic with period T, f(x) is also periodic with same period T.

• If f(x) is periodic with period T, then a f(x) + b, where a, b∈ R (a ≠ 0) is also a periodic function with period T.

• If f1(x), f2(x), f3(x) are periodic functions with periods T1, T2, T3 respectively then; we have
h(x) = af1(x) ± bf2(x) ± cf3(x) , has period as,

L.C.M.of {T1,T2,T3}; if h(x)is not an even function
 of {T1,T2,T3}; if h(x) is an even function
=  1 L.C.M.
 2

Example: 41 The period of the function f (x) = 2 cos 1 (x − π ) is [DCE 1998]
3 [EAMCET 1992]

(a) 6π (b) 4π (c) 2π (d) π

Solution: (a) f (x) = 2 cos 1 (x − π ) = 2 cos x − π 
3  3 3 

Now, since cos x has period 2π ⇒ cos x − π  has period 2π = 6π
 3 3  1

3

⇒ 2 cos x − π  has period = 6π .
 3 3 

Example: 42 The function f(x) = sin πx + 2 cos πx − tan πx is periodic with period
2 3 4

(a) 6 (b) 3 (c) 4 (d) 12

Solution: (d)  sin x has period = 2π ⇒ sin πx has period = 2π =4
2 π

2

 cos x has period = 2π ⇒ cos πx has period = 2π =6 ⇒ 2 cos πx has period = 6
3 π 3

3

 tan x has period = π ⇒ tan πx has period = π =4.
4 π

4

L.C.M. of 4, 6 and 4 =12, period of f(x) = 12.

Example: 43 The period of |sin 2x| is

(a) π (b) π (c) π (d) 2π
4 2

Solution: (b) Here | sin 2x |= sin2 2x = (1 − cos 4 x)
2
Example: 44
Solution: (a) Period of cos 4 x is π . Hence, period of | sin 2x | will be π
Example: 45 2 2
Solution: (b)
Example: 46 Trick :  sin x has period = 2π ⇒ sin 2x has period = 2π =π
Solution: (c) 2

Example: 47 Now, if f(x) has period p then | f(x)|has period p ⇒ | sin 2x | has period = π .
Solution: (c) 2 2

Example: 48 If f(x) is an odd periodic function with period 2, then f(4) equals [IIT 1991]
Solution: (b)
(a) 0 (b) 2 (c) 4 (d) – 4

Given, f(x) is an odd periodic function. We can take sin x, which is odd and periodic.

Now since, sin x has period = 2 and f(x) has period = 2.

So, f(x) = sin(π x) ⇒ f(4) = sin(4π ) = 0 .

The period of the function f(x) = sin2 x is [UPSEAT 1991, 2002; AIEEE 2002]

(a) π (b) π (c) 2π (d) None of these
2

sin 2 x = 1− cos 2x ⇒ Period = 2π = π.
2 2

The period of f(x) = x − [x], if it is periodic, is [AMU 2000]

(a) f(x) is not periodic (b) 1 (c) 1 (d) 2
2

Let f(x) be periodic with period T. Then,

f(x + T) = f(x) for all x ∈ R ⇒ x + T − [x + T] = x − [x] for all x ∈ R ⇒ x + T − x = [x + T] − [x]

⇒ [x + T] − [x] = T for all x ∈ R ⇒ T = 1, 2, 3, 4,........

The smallest value of T satisfying,
f(x + T) = f(x) for all x ∈ R is 1.

Hence f(x) = x − [x] has period 1.

The period of f (x) = sin  πx  + cos πx , n ∈ Z, n > 2 is [Orissa JEE 2002]
 n−1   n 
(d) None of these
(a) 2πn(n − 1) (b) 4n(n − 1) (c) 2n(n − 1)

f(x) = sin πx  + cos πx 
 n−1   n 

Period of sin πx  = 2π = 2(n − 1) and period of cos πx  = 2π = 2n
 n−1   n 
 π   π 
 n− 1   n 

Hence period of f(x) is LCM of 2n and 2(n − 1) ⇒ 2n(n − 1) .

1

If a,b be two fixed positive integers such that f(a + x) = b + [b3 + 1 − 3b2 f(x) + 3b{f(x)}2 − {f(x)}3]3 for all real x, then

f(x) is a periodic function with period [Orissa JEE 2003]

(a) a (b) 2 a (c) b (d) 2 b

f(a + x) = b + (1 + {b − f(x)}3)1 / 3 ⇒ f(a + x) − b = {1 − {f(x) − b}3}1 / 3

⇒ φ(a + x) = {1 − {φ(x)}3}1 / 3 [ φ(x) = f(x) − b ] ⇒ φ(x + 2a) = {1 − {φ(x + a)}3}1 / 3 = φ(x)

⇒ f(x + 2a) − b = f(x) − b ⇒ f(x + 2a) = f(x)

∴ f(x) is periodic with period 2a .

Composite Function.

If f : A → B and g : B → C are two function then the composite function of f and g,
gof A → C will be defined as gof(x) = g[ f(x)],∀x ∈ A
(1) Properties of composition of function :
(i) f is even, g is even ⇒ fog even function.
(ii) f is odd, g is odd ⇒ fog is odd function.
(iii) f is even, g is odd ⇒ fog is even function.
(iv) f is odd, g is even ⇒ fog is even function.
(v) Composite of functions is not commutative i.e., fog ≠ gof

(vi) Composite of functions is associative i.e., (fog)oh = fo(goh)
(vii) If f : A → B is bijection and g : B → A is inverse of f. Then fog = I B and gof = I A .

where, I A and I B are identity functions on the sets A and B respectively.
(viii) If f : A → B and g : B → C are two bijections, then gof : A → C is bijection and (gof)−1 = (f −1og −1).

(ix) fog ≠ gof but if , fog = gof then either f −1 = g or g −1 = f also, (fog)(x) = (gof)(x) = (x).

Important Tips

• gof(x) is simply the g-image of f(x), where f(x) is f-image of elements x∈ A.
• Function gof will exist only when range of f is the subset of domain of g.
• fog does not exist if range of g is not a subset of domain of f.
• fog and gof may not be always defined.
• If both f and g are one-one, then fog and gof are also one-one.
• If both f and g are onto, then gof is onto.

Example: 49 If f : R → R, f(x) = 2x − 1 and g : R → R, g(x) = x2 then (gof)(x) equals [Rajasthan PET 1987]

Solution: (b) (a) 2x2 − 1 (b) (2x − 1)2 (c) 4 x 2 − 2x + 1 (d) x2 + 2x − 1
Example: 50
Solution: (a) gof (x)= g{f(x)} = g (2x − 1) =(2x − 1)2 .
Example: 51
If f : R → R, f(x) = (x + 1)2 and g : R → R, g(x) = x2 + 1, then (fog)(−3) is equal to [Rajasthan PET 1999]
Solution: (a)
(a) 121 (b) 144 (c) 112 (d) 11

fog (x) = f {g (x)} = f(x 2 + 1) = (x 2 + 1 + 1)2 = (x 2 + 2)2 ⇒ fog (−3)=(9 + 2)2 =121 .

f(x) = sin 2 x + sin 2  x + π  + cos x cos x + π  and g 5  = 1, then (gof)(x) is equal to [IIT 1996]
 3   3   4 

(a) 1 (b) –1 (c) 2 (d) – 2

f(x) = sin2 x + sin2(x + π / 3) + cos x cos(x + π / 3) = 1 − cos 2x + 1 − cos(2x + 2π /3) + 1 {2 cos x cos(x + π / 3)}
2 2 2

= 1 [1 − cos 2x + 1 − cos(2x + 2π /3) + cos(2x + π / 3) + cos π / 3]
2

= 1 5 − {cos 2x + cos 2x + 2π } + cos 2x + π  = 1  5 − 2cos 2x + π  cos π + cos 2x + π  = 5 / 4 for all x.
2   3   3 2  2  3  3  3
 2 

∴ gof(x) = g(f(x)) = g(5 / 4) = 1 [ g(5/4) =1 (given)]

Hence, gof(x) = 1, for all x.

Example: 52 If g(x) = x2 + x − 2 and 1 (gof )(x) = 2x2 − 5x + 2, then f(x) is equal to [Roorkee 1998; MP PET 2002]
Solution: (a) 2
(d) 2x2 − 3x − 1
Example: 53 (a) 2x − 3 (b) 2x + 3 (c) 2x 2 + 3x + 1

g(x) = x2 + x − 2 ⇒ (gof)(x) = g[ f(x)] = [ f(x)]2 + f(x) − 2

Given, 1 (gof )(x) = 2x2 − 5x + 2 ∴ 1 [ f (x)]2 + 1 f(x) − 1 = 2x2 − 5x + 2
2 2 2

⇒ [ f(x)]2 + f(x) = 4 x2 − 10x + 6 ⇒ f(x)[ f(x) + 1] = (2x − 3)[(2x − 3) + 1] ⇒ f(x) = 2x − 3 .

If f(y) = y , g(y) = y , then ( fog)(y) is equal to [EAMCET 1996]
1 − y2 1 + y2

(a) y (b) y (c) y (d) 1 − y2
1 − y2 1 + y2 1 + y2

Solution: (c) f[g(y)] = y / 1 + y2 = y 1 + y2 = y
Example: 54 × 1 + y2 − y2
 y 2
Solution: (a)  1 + y2  1 + y2
Example: 55
Solution: (b) 1 −

Example: 56 If f(x) = 2x − 3 , then [ f{f(x)}] equals [Rajasthan PET 1996]
Solution: (b) x−2

(a) x (b) −x (c) x (d) 1
2 −x

2 2x − 3  − 3
 x−2 
f[ f(x)] = = x
 2x − 3 
 x−2  − 2

Suppose that g(x) = 1 + x and f(g(x)) = 3 + 2 x + x, then f(x) is [MP PET 2000; Karnataka CET 2002]

(a) 1 + 2x 2 (b) 2 + x 2 (c) 1 + x (d) 2 + x

g(x) = 1 + x and f(g(x)) = 3 + 2 x + x ….. (i)

⇒ f(1 + x ) = 3 + 2 x + x
Put 1 + x = y ⇒ x = (y − 1)2
then, f(y) = 3 + 2(y − 1) + (y − 1)2 = 2 + y2
therefore, f(x) = 2 + x2 .

− 1, x<0
 x = 0 , then for all x, f(g(x)) is equal to
Let g(x) = 1 + x − [x] and f(x) =  0, x>0 [IIT Screening 2001; UPSEAT 2001]

 1, (c) f(x) (d) g(x)

(a) x (b) 1

Here g(x) = 1 + n − n = 1, x = n ∈ Z

1 + n + k − n = 1 + k , x = n + k (where n ∈ Z, 0 < k < 1 )

− 1, g(x) < 0

Now f (g(x)) =  0, g(x) = 0

 1, g(x) > 0

Clearly, g(x) > 0 for all x. So, f(g(x)) = 1 for all x.

Example: 57 If f(x) = 2x + 1 , then (fof)(2) is equal to [Kerala (Engg.) 2002]
Solution: (d) 3x − 2

Example: 58 (a) 1 (b) 3 (c) 4 (d) 2
Solution: (d)
Here f(2) = 5
4

Hence ( fof )(2) = f ( f (2)) = f  5  = 2× 5 +1 = 2.
 4  3× 4 −2
5
4

If f : R → R and g : R → R are given by f(x)=| x | and g(x) = [x] for each x ∈ R, then {x ∈ R : g(f(x)) ≤ f(g(x))} =

(a) Z ∪(−∞,0) (b) (−∞,0) (c) Z (d) R

g(f(x)) ≤ f(g(x)) ⇒ g(| x |) ≤ f[x] ⇒ [| x |]≤|[x]|. This is true for x ∈ R.

Inverse Function.

If f : A → B be a one-one onto (bijection) function, then the mapping f −1 : B → A which associates each
element b ∈ B with element a ∈ A, such that f(a) = b, is called the inverse function of the function f : A → B

f −1 : B → A, f −1(b) = a ⇒ f(a) = b
In terms of ordered pairs inverse function is defined as f −1 = (b,a) if (a, b) ∈ f .

Note :  For the existence of inverse function, it should be one-one and onto.

Important Tips

• Inverse of a bijection is also a bijection function.
• Inverse of a bijection is unique.
• (f–1)–1= f
• If f and g are two bijections such that (gof) exists then (gof)–1=f–1og–1.
• If f : A→ B is a bijection then f–1: B → A is an inverse function of f. f–1of = IA and fof–1=IB. Here IA, is an identity function on set A, and

IB, is an identity function on set B.

Example: 59 If f : R → R is given by f(x) = 3x − 5, then f −1(x) [IIT Screening 1998]
Solution: (b)
Example: 60 (a) Is given by 1 (b) Is given by x+5
Solution: (c) 3x − 5 3

(c) Does not exist because f is not one-one (d) Does not exist because f is not onto

Clearly, f : R → R is a one-one onto function. So, it is invertible.

Let f(x) = y. then, 3x − 5 = y ⇒ x = y + 5 ⇒ f −1(y) = y + 5 . Hence, f −1(x) = x + 5 .
3 3 3

Let f : R → R be defined by f(x) = 3x − 4, then f −1(x) is [UPSEAT 1993]

(a) 3x + 4 (b) 1 x − 4 (c) 1 (x + 4) (d) 1 (x − 4)
3 3 3

f(x) = 3x − 4 = y ⇒ y = 3x − 4 ⇒ x = y+4 ⇒ f −1(y) = y+4 ⇒ f −1(x) = x +4 .
3 3 3

Example: 61 If the function f : R → R be such that f(x) = x − [x], where [y] denotes the greatest integer less than or equal to y, then
Solution: (c) f −1(x) is

Example: 62 (a) 1 (b) [x] − x (c) Not defined (d) None of these
x − [x]
Solution: (b)
f(x) = x − [x] Since, for x = 0 ⇒ f(x) = 0
Example: 63
Solution: (a) For x = 1 ⇒ f(x) = 0 .
Example: 64
Solution: (d) For every integer value of x, f(x) = 0

⇒ f(x) is not one-one ⇒ So f −1(x) is not defined. [IIT Screening 1999]
If f : [1, ∞) → [1, ∞) is defined as f(x) = 2x(x−1) then f −1(x) is equal to

 1  x( x −1) 1( )(b)
 2  2
(a) 1 + 1 + 4 log 2 x

( )(c)1 1− (d) Not defined
2 1 + 4 log 2 x

Given f(x) = 2x(x−1) ⇒ x(x − 1) = log 2 f(x)

⇒ x2 − x − log 2 f(x) = 0 ⇒ x 1± 1 + 4 log2 f(x)
= 2

Only x = 1 + 1 + 4 log 2 f(x) lies in the domain
2

∴ f −1(x) = 1 [1 + 1 + 4 log 2 x]
2

Which of the following function is invertible [AMU 2001]

(a) f(x) = 2x (b) f(x) = x3 − x (c) f(x) = x2 (d) None of these
(c) 3, 2
A function is invertible if it is one-one and onto. [UPSEAT 2003]

If f(x) = x 2 + 1, then f −1(17) and f −1(−3) will be (d) None of these

(a) 4, 1 (b) 4, 0

Let y = x 2 + 1 ⇒ x = ± y−1

⇒ f −1(y) = ± y − 1 ⇒ f −1(x) = ± x − 1

⇒ f −1(17) = ± 17 − 1 = ±4

and f −1(−3) = ± − 3 − 1 = ± − 4 , which is not possible.

Funtion limit continuity & differentiability

Limit of a Function.

Let y = f(x) be a function of x. If at x = a, f(x) takes indeterminate form, then we consider the values of the

function which are very near to ‘a’. If these values tend to a definite unique number as x tends to ‘a’, then the
unique number so obtained is called the limit of f(x) at x = a and we write it as lim f(x) .

x→a

(1) Meaning of ‘x → a’: Let x be a variable and a be the constant. If x assumes values nearer and nearer to
‘a’ then we say ’x tends to a’ and we write ' x → a' . It should be noted that as x → a , we have x ≠ a . By ' x tends
to a' we mean that

(i) x ≠ a (ii) x assumes values nearer and nearer to ‘a’ and

(iii) We are not specifying any manner in which x should approach to ‘a’. x may approach to a from left or
right as shown in figure.

xa ax

(2) Left hand and right hand limit : Consider the values of the functions at the points which are very near

to a on the left of a. If these values tend to a definite unique number as x tends to a, then the unique number so

obtained is called left-hand limit of f(x) at x = a and symbolically we write it as f(a − 0) = lim f(x) = lim f(a − h)
x→a− h→0

Similarly we can define right-hand limit of f(x) at x = a which is expressed as f(a + 0) = lim f(x)
x→a+

= lim f(a + h).
h→0

(3) Method for finding L.H.L. and R.H.L.

(i) For finding right hand limit (R.H.L.) of the function, we write x + h in place of x, while for left hand limit
(L.H.L.) we write x – h in place of x.

(ii) Then we replace x by ‘a’ in the function so obtained.

(iii) Lastly we find limit h → 0 .

(4) Existence of limit : lim f(x) exists when,
x→a

(i) lim f(x) and lim f(x) exist i.e. L.H.L. and R.H.L. both exists.
x→a− x→a+

(ii) lim f(x) = lim f(x) i.e. L.H.L. = R.H.L.
x→a− x→a+

Note :  If a function f(x) takes the form 0 or ∞ at x = a , then we say that f(x) is indeterminate
0 ∞

or meaningless at x = a . Other indeterminate forms are ∞ − ∞, ∞ × ∞, 0 × ∞, 1∞,00, ∞0

 In short, we write L.H.L. for left hand limit and R.H.L. for right hand limit.

 It is not necessary that if the value of a function at some point exists then its limit at that point must exist.

(5) Sandwich theorem : If f(x) , g(x) and h(x) are any three functions such that, f(x) ≤ g(x) ≤ h(x) ∀x ∈

neighborhood of x = a and lim f(x) = lim h(x) = l (say) , then lim g(x) = l. This theorem is normally applied when
x→a x→a x→a

the lim g(x) can't be obtained by using conventional methods as function f(x) and h(x) can be easily found.
x→a

Example: 1 If f (x) =  x, when x >1 , then lim f(x) = [MP PET 1987]
Solution: (d)  when x <1
 x 2 , x →1
Example: 2
Solution: (c) (a) x2 (b) x (c) – 1 (d) 1

Example: 3 To find L.H.L. at x = 1. i.e.,
Solution: (c)
Example: 4 lim f(x) = lim f(1 − h) = lim (1 − h)2 = lim (1 + h2 − 2h) = 1 i.e., lim f(x) = 1 ….(i)
Solution: (d) x →1− h→0 h→0 h→0 x →1− …..(ii)

Example: 5 Now find R.H.L. at x = 1 i.e., lim f(x) = lim f(1 + h) = 1 i.e., lim f(x) = 1
Solution: (d) x →1+ h→0 x →1+

From (i) and (ii), L.H.L. = R.H.L. ⇒ lim f(x) = 1 .

x →1

lim | x − 2| =
x − 2
x→2

(a) 1 (b) –1 (c) Does not exist (d) None of these
…..(i)
L.H.L.= lim | x − 2| = lim | 2 − h − 2 | = lim h = −1
x − 2 2 − h − 2 −h …..(ii)
x→2− h→0 h→0

and, R.H.L.= lim | x − 2| = lim | 2 + h − 2| = lim h =1
x − 2 2 + h − 2 h
x→2+ h→0 h→0

From (i) and (ii) L.H.L. ≠ R.H.L. i.e. lim | x − 2| does not exist.
x − 2
x→2

If f(x) = 55−−2xx, , when x<3, then
when x>3

(a) lim f(x) = 0 (b) lim f(x) = 0 (c) lim f(x) ≠ lim f(x) (d) None of these
x→3+ x→3−
x→3+ x→3− [SCRA 1996]

lim f(x) = 5 − 3 = 2 and lim f(x) = 2 =1 (d) lim f(x) does not exist
5−3
x→3+ x→3− x →1

Let the function f be defined by the equation f (x) =  3x, if 0 ≤ x ≤ 1 2 , then
5 − 3x, if 1 < x ≤

(a) lim f(x) = f(1) (b) lim f(x) = 3 (c) lim f(x) = 2

x →1 x →1 x →1

L.H.L. = lim f(x) = lim f(1 − h) = lim 3(1 − h) = lim (3 − 3h) = 3 − 3.0 = 3
x →1− 0 h→0 h→0 h→0

R.H.L. = lim f(x) = lim f(1 + h) = lim [5 − 3(1 + h)] = lim (2 − 3h) = 2 − 3.0 = 2
x →1+ 0 h→0 h→0 h→0

Hence lim f(x) does not exists.

x →1

lim | x | = [Roorkee 1982; UPSEAT 2001]
x
x→0 (d) Does not exist

(a) 1 (b) –1 (c) 0

 lim | x | = −1 and lim | x | = 1 , hence limit does not exists.
x x
x→0− x→0+

Fundamental Theorems on Limits.

The following theorems are very useful for evaluation of limits if lim f(x) = l and lim g(x) = m (l and m are
x→0 x→0

real numbers) then

(1) lim(f(x) + g(x)) = l + m (Sum rule) (2) lim(f(x) − g(x)) = l − m (Difference rule)
x→a x→a (Constant multiple rule)

(3) lim(f(x).g(x)) = l.m (Product rule) (4) lim k f(x) = k.l
x→a x→a

(5) lim f(x) = l , m ≠ 0 (Quotient rule) (6) If lim f(x) = +∞ or − ∞ , then lim 1 =0
g(x) m f(x)
x→a x→a x→a

(7) lim log{f(x)} = log {lim f(x)} (8) If f(x) ≤ g(x) for all x, then lim f(x) ≤ lim g(x)
x→a x→a x→a x→a

(9) lim[ f(x)]g(x) = {lim f ( x)} lim g ( x )

x→a

x→a x→a

(10) If p and q are integers, then lim(f(x))p / q = l p / q , provided (l)p / q is a real number.
x→a

(11) If lim f(g(x)) = f(lim g(x)) = f(m) provided ‘f’ is continuous at g(x) = m. e.g. lim ln[f(x)] = ln(l), only if l > 0.
x→a x→a x→a

Some Important Expansions.

In finding limits, use of expansions of following functions are useful :

(1) (1 + x)n = 1+ nx + n(n − 1) x2 + ..... (2) ax =1+ x log a + (x log a)2 + .....
2! 2!

(3) ex =1+ x+ x2 + x3 + ..... (4) log(1 + x) = x− x2 + x3 − x4 + .....,| x |< 1
2! 3! 2 3 4

(5) log(1 − x) = −x − x2 − x3 − x4 − ......., where | x |< 1
2 3 4

(6) (1 + 1 = e 1 log(1+ x) = e1− x + x2 ....... = e1 − x + 11 x2 − .......
x 2 3  2 24 
x)x

(7) sin x = x − x3 + x5 − ....... (8) cos x = 1 − x2 + x4 − x6 + ......
3! 5! 2! 4! 6!

(9) tan x = x+ x3 + 2x 5 + ..... (10) sinh x = x+ x3 + x5 + .....
3 15 3! 5!

(11) cosh x =1+ x2 + x4 + x6 + ..... (12) tanh x =x− x3 + 2x5 − .....
2! 4! 6! 3

(13) sin −1 x = x + 12. x3 + 3 2.12. x5 + ..... (14) cos −1 x =  π  − sin −1 x
3! 5!  2 

(15) tan −1 x = x − x3 + x5 x7 + .....
3 5 −7

Methods of Evaluation of Limits.
We shall divide the problems of evaluation of limits in five categories.

(1) Algebraic limits : Let f(x) be an algebraic function and ‘a’ be a real number. Then lim f(x) is known
x→a

as an algebraic limit.

(i) Direct substitution method : If by direct substitution of the point in the given expression we get a finite
number, then the number obtained is the limit of the given expression.

(ii) Factorisation method : In this method, numerator and denominator are factorised. The common
factors are cancelled and the rest outputs the results.

(iii) Rationalisation method : Rationalisation is followed when we have fractional powers (like 1 , 1 etc.)
2 3

on expressions in numerator or denominator or in both. After rationalisation the terms are factorised which on

cancellation gives the result.

(iv) Based on the form when x → ∞ : In this case expression should be expressed as a function 1/x and

then after removing indeterminate form, (if it is there) replace 1 by 0.
x

Step I : Write down the expression in the form of rational function, i.e., f(x) , if it is not so.
g(x)

Step II : If k is the highest power of x in numerator and denominator both, then divide each term of
numerator and denominator by x k .

Step III : Use the result lim 1 = 0 , where n > 0.
xn
x→∞

Note :  An important result : If m, n are positive integers and a0 ,b0 ≠ 0 are non-zero real numbers,

then lim a0 xm + a1 x m−1 + .... + am−1 x + am =  a0 , if m=n
b0 x n + b1 x n−1 + ..... + bn−1 x + bn  b0 if m<n
x→∞  0,

 ∞, if m > n

Example: 6 lim (3x2 + 4 x + 5) = (b) –1 (c) Does not exist (d) None of these
Solution: (a)
Example: 7 x →1

Solution: (c) (a) 12
Example: 8
lim (3x2 + 4 x + 5) = 3(1)2 + 4(1) + 5 = 12 .

x →1

The value of lim 3x / 2 − 3 is [MP PET 2000]
3x − 9
x→2

(a) 0 (b) 1 (c) 1 (d) ln 3
3 6

lim 3x / 2 − 3 = lim (3x / (3x / 2 − 3) + 3) = 1 .
(3x / 2)2 − (3)2 2 − 3)(3x / 2 6
x→2 x→2

The value of lim xn − an is [Rajasthan PET 1989, 92]
x −a
x→a (d) 1

(a) 0 (b) nan−1 (c) nan

Solution: (b) lim xn − an = lim (x − a) (x n−1 + xn−2a + .. + an−1) = lim (xn−1 + xn−2a + .... + an−1) = n . an−1 .
Example: 9 x − a (x − a)
Solution: (d) x→a x→a x→a
Example: 10
Solution: (b) lim 1  1 h − 1 equals [Rajasthan PET 1987]
Example: 11 h  x + x 
Solution: (d) h→0

Example: 12 (a) 1 (b) − 1 (c) 1 (d) 1
Solution: (c) 2x 2x x2 − x2

Example: 13 lim 1 1 h − 1 = lim 1  x − (x + h) = lim 1  −h  = − 1 .
h  x + x  h  (x + h)x  h (x + h)x  x2
Solution: (b) h→0 h→0 h→0
Example: 14
The value of lim 1− x2 − 1+ x2 is [MP PET 1999]
x2
x→0
(b) –1 (c) – 2 (d) 0
(a) 1

 1 − x2 − 1 + x2   1− x2 + 1 + x2  = lim (1 − x2) − (1 + x2) = −2 = −1 .
lim   1− x2 +  x→0 x2  1 − x2 + 1 + x2  2

x→0 x2  1 + x2 

 

lim x − 3 equals [UPSEAT 1991]
x→3 x − 2 − 4 − x

(a) 1 (b) 3 (c) 1 (d) None of these
2 4

( )lim − 3)
( ) ( )x→3 x−
x−3 = lim (x x−2 + 4−x
x−2− 4−x 4 −x 2
x→3 2 2

( )= lim (x − 3)
x→3
x−2+ 4−x = lim x−2+ 4− x = 1+1 =1.
(2x − 6) 2 2
x→3

lim ax 2 + bx + c =
dx 2 + ex + f
x→∞

(a) b (b) c (c) a (d) d
e f d a

Here the expression assumes the form ∞ . We note that the highest power of x in both the numerator and denominator is

2. So we divide each terms in both the numerator and denominator by x 2 .

lim ax 2 + bx + c = lim a + b + c = a +0+0 = a .
dx 2 + ex + f d + x + x2 d +0+0 d
x→∞ x→∞ e f
x x2

lim  x+ x+ x− x  is equal to
 
x→∞  

(a) 0 (b) 1 (c) log 2 (d) e4
2

lim  x+ x+ x− x  = lim x + x + x − x = lim x+ x = lim 1 + x−1 / 2 = 1 .
  x + x + x + x x→∞ x+ x + x x→∞ x−1 + x−3 / 2 2
x→∞   x→∞
x+ 1+ +1

The values of constants a and b so that lim  x2 +1 − ax − b  = 0 is
 x +1
x→∞

(a) a = 0, b = 0 (b) a = 1, b = −1 (c) a = −1, b = 1 (d) a = 2, b = −1

Solution: (b) We have lim  x2 +1 − ax − b  = 0 ⇒ lim x2(1 − a) − x(a + b) + 1 − b = 0
 x +1  x +1
Example: 15 x→∞ x→∞
Solution: (a)
Example: 16 Since the limit of the given expression is zero, therefore degree of the polynomial in numerator must be less than that of
Solution: (a) denominator. As the denominator is a first degree polynomial. So, numerator must be a constant i.e., a zero degree
Example: 17
Solution: (d) polynomial. ∴ 1 − a =0 and a + b = 0 ⇒ a = 1 and b = –1. Hence, a = 1 and b = – 1.
Example: 18
lim x x = (b) ∞ (c) Not defined (d) None of these
Solution: (a)
Example: 19 x →1

Solution: (c) (a) 1

Example: 20 lim x x =  lim x  lim x = 11 =1

Solution: (b) x →1 x →1 x →1

Example: 21 lim (1 + x)1 / x = (b) e (c) Not defined (d) None of these

x →1

(a) 2

lim (1 + x)1 / x =  lim (1 + x) lim  1  = 2
x →1 x 
x →1 x →1

The value of the limit of x3 − x2 − 18 as x tends to 3 is
x−3

(a) 3 (b) 9 (c) 18 (d) 21

Let y = lim x3 − x 2 − 18 = lim (x2 + 2x + 6) = 9 + 6 + 6 = 21
x−3
x→3 x→3

The value of the limit of x3 − 8 as x tends to 2 is
(x2 − 4)

(a) 3 (b) 3 (c) 1 (d) 0
2

xl→im2 x 3 −8 = xl→im2 (x 2 + 2x + 4)(x − 2) = xl→im2 x2 + 2x + 4 = 4+4+4 = 3.
x 2 −4 (x + 2)(x − 2) x+2 2+2

lim x is equal to [Rajasthan PET 1988]
x→0 1 + x − 1 − x

(a) 1 (b) 2 (c) 1 (d) 0
2

lim  x 1− x  = lim  x 1− x × 1+ x + 1 − x 
1+ x −  1+ x − 1+ x + 1 − x 
x→0 x→0

( ) ( )=  x   
lim  1+ x + 1− x  = lim  1+ x + 1− x  = 2 =1
1+ x −1+ x 2 2
x→0 x→0

lim a + 2x − 3x equals [IIT 1978; Kurukshetra CEE 1998]
x→a 3a + x − 2 x
(d) None of these
(a) 2a (b) 2 (c) 0
33 33

xl→ima a + 2x − 3x = xl→ima  a + 2x − 3x  ×  a + 2x + 3x  ×  3a + x + 2 x 
3a + x − 2 x 3a + x − 2x a + 2x + 3x 3a + x + 2 x

= xl→ima  3a + x + 2 x = 2 .
 a + 2x +  33
 3( 3 x ) 

lim 199 + 299 + 399 + .... + n99 = [EAMCET 1994]
n100
n→∞

(a) 99 (b) 1 (c) 1 (d) 1
100 100 99 101

199 + 299 + 399 + .... + n99 r∑=n1 r 99  1 n  r 99 1 x99 dx  x100 1 1
Solution: (b) ∑ ∫lim n100 = lim n100 = lim n =1  n  = = = 100 .
Example: 22 0  100 
Solution: (c) n→∞ n→∞ n→∞ r 
Example: 23 0

Solution: (c) The values of constants ‘a’ and ‘b’ so that lim  x2 −1 − ax − b  = 2 is
 x +1
Example: 24 x→∞
Solution: (b)
Example: 25 (a) a = 0, b = 0 (b) a = 1, b = −1 (c) a = 1, b = −3 (d) a = 2, b = −1
Solution: (c)
Example: 26 lim  x2 −1 − ax − b  = 2 ⇒ lim x − 1 − ax −b = 2 ⇒ lim x(1 − a) − (1 + b) = 2.
Solution: (b)  x +1 
x→∞ x→∞ x→∞

Comparing the coefficient of both sides, 1 − a = 0 and 1 + b = −2 ⇒ a = 1,b = −3

lim  ∑ n2  = [Rajasthan PET 1999, 2002]
 n3 
n→∞ 

(a) 1 (b) 1 (c) 1 (d) −1
−6 6 3 3

 n(n + 1)(2n + 1) 1 + 1   2 + 1  1
 6n3   n   n  3
lim = lim =
6
n→∞ n→∞

Note :  Students should remember that,

lim ∑n = 1 and lim ∑ n2 = 1.
n2 2 n3 3
n→∞ n→∞

nl→im∞ 1 1 2 + ...... + n is equal to [IIT 1984; DCE 2000]
− n2 + 1 − n2 1 − n2 
(d) None of these
(a) 0 (b) 1 (c) 1
−2 2 (d) 0

nl→im∞ 1 1 2 + ...... + 1 n  = lim 1 ∑ n = 1 lim n2 + n = − 1 . [Kurukshetra CEE 1998]
− n2 + 1 − n2 − n2  − n2 2 1 − n2 2
n→∞ n→∞ (d) 4
π
If f(x) = x 2 3 , g(x) = x−3 and h(x) = − 2(2x + 1) then lim [ f(x) + g(x) + h(x)] is
− x+4 x2 + x − 12
x→3

(a) – 2 (b) – 1 (c) − 2
7

We have f(x) + g(x) + h(x) = x2 − 4 x + 17 − 4 x − 2 = x 2 − 8x + 15 = (x − 3)(x − 5)
x2 + x − 12 x 2 + x − 12 (x − 3)(x + 4)

∴ lim [ f(x) + g(x) + h(x)] = lim (x − 3)(x − 5) = − 2 .
(x − 3)(x + 4) 7
x→3 x→3

 n! 1 / n
 nn 
If lim equal

n→∞

(a) e (b) 1 (c) π
e 4

lim  n! 1 / n lim  1 2 3 4 n 1 / n
n→∞ nn  n→∞ n n n n n 
Let P = ⇒ P = . . . ..........

1 lim  log 1 2 n  n 1 r
n n→∞ n n n  r =1 n n
∑∴log P log
= + log + ......... + ⇒ log P = lim log

n→∞

∫log P = 1 x dx = [x log x − x] 1 = (−1) ⇒ P = 1 .
0 e
log
0

Example: 27 If lim  x3 + 1 − (ax + b) = 2, then [Karnataka CET 2000]
Solution: (c)  x2 + 1 
x→∞ 

(a) a = 1 and b = 1 (b) a = 1 and b = −1 (c) a = 1 and b = −2 (d) a = 1 and b = 2

lim  x3 +1 − (ax + b) = 2 ⇒ lim  x 3 (1 − a) − bx2 − ax + (1 − b)  = 2 ⇒ lim [x3(1 − a) − bx 2 − ax + (1 − b)] = 2(x 2 + 1) .
 x2 +1  x2 +1 
x→∞ x→∞ x→∞

Comparing the coefficients of both sides, 1 − a = 0 and −b = 2 or a = 1,b = −2 .

Example: 28 lim (x + 1)10 + (x + 2)10 + ..... + (x + 100)10 is equal to [AMU 2000]
x10 + 1010
Solution: (d) x→∞
Example: 29
Solution: (c) (a) 0 (b) 1 (c) 10 (d) 100

1)10 + 2)10 + ...... + 100)10 x10 1 + 1 10 + 1 + 2 10 + ... + 1 + 100 10 
x10 + 1010  x   x   x 
 
lim (x + + (x + (x = lim = 100 .

x→∞ x→∞ x10 1 + 1010 
 x10 


Let f(x) = 4 and f ' (x) = 4 , then lim xf (2) − 2f(x) equals [Rajasthan 2000; AIEEE 2002]
x −2
x→2 (d) 3

(a) 2 (b) – 2 (c) – 4

y = lim xf(2) − 2 f(x) ⇒ y = lim xf(2) − 2 f(2) + 2 f(2) − 2 f(x)
x−2 x−2
x→2 x→2

⇒ y = lim −2 f(x) + 2 f(2) + xf(2) − 2 f(2) ⇒ y = lim − 2 [ f(x) − f (2)] + lim f(2).(x − 2)
(x − 2) x − 2 (x − 2)
x→2 x→2 x→2

⇒ y = −2 lim f(x) − f(2) + f (2) ⇒ y = −2 lim f ′(x) + f(2) = − 8 + 4 = − 4 .
x−2
x→2 x→2

(2) Trigonometric limits : To evaluate trigonometric limits the following results are very important.

(i) lim sin x =1= lim x (ii) lim tan x =1 = lim x
x sin x x tan x
x→0 x→0 x→0 x→0

(iii) lim sin −1 x =1 = lim x (iv) lim tan −1 x =1 = lim x
x sin −1 x tan −1
x→0 x→0 x x→0 x→0 x

(v) lim sin x 0 π (vi) lim cos x = 1
x 180 x→0
x→0 =

(vii) lim sin(x − a) = 1 (viii) lim tan(x − a) = 1
x−a x−a
x→a x→a

(ix) lim sin −1 x = sin −1 a, |a | ≤ 1 (x) lim cos −1 x = cos −1 a; |a | ≤ 1
x→a x→a

(xi) lim tan−1 x = tan−1 a; − ∞ < a < ∞ (xii) lim sin x = lim cos x = 0
x→a x x
x→∞ x→∞

(xiii) lim sin(1 / x) = 1
(1 / x)
x→∞

Example: 30 lim (1 − x) tan πx  = [IIT 1978, 84; Rajasthan PET 1997, 2001; UPSEAT 2003]
 2 
x →1

(a) π (b) π (c) 2 (d) 0
2 π
[IIT 1998; UPSEAT 2001]
Solution: (c) lim (1 − x) tan πx  , Put 1 − x = y ⇒ as x → 1, y → 0
Example: 31  2 
Solution: (d) x →1

Thus lim y tan π (1 − y) = lim 2 .  πy  = 2 ×1 = 2 .
2 π  2  ππ
y→0 y→0
tan  πy 
 2 

lim 1 − cos 2(x − 1)
x −1
x →1

(a) Exists and it equal 2

(b) Exists and it equals − 2
(c) Does not exist because x − 1 → 0
(d) Does not exist because left hand limit is not equal to right hand limit

f(1+) = lim f(1 + h) = = lim 1 − cos 2h = lim 2 sinh = 2
h→0 h→0 h h
h→0

f(1−) = lim f(1 − h) = lim 1 − cos(−2h) = lim 2 sinh = − 2.
h→0 h→0 −h −h
h→0

∴ limit does not exist because left hand limit is not equal to right hand limit.

Example: 32 lim (1 − cos 2x) sin 5x = [MP PET 2000; UPSEAT 2000; Karmataka CET 2002]
x2 sin 3x
Solution: (a) x→0
Example: 33
(a) 10 (b) 3 (c) 6 (d) 5
Solution: (a) 3 10 5 6
Example: 34
lim 2 sin2 x sin 5x 3x 5x = lim 2 sin2 x . 3x . sin 5x . 5x = 2. 5 = 10 .
Solution: (c) x2 sin 3x 3x 5x x2 sin 3x 5x 3x 3 3
Example: 35 x→0 x→0
Solution: (b)
lim x3 =
sin x2
x→0

(a) 0 (b) 1 (c) 3 (d) 1
3 2

lim x3 = lim x2 . x =  lim x2   lim x  = 1.0 = 0.
sin x2 sin x2  sin x2 
x→0 x→0 x→0 x→0

lim sin 3x + sin x =
x
x→0

(a) 1 (b) 3 (c) 4 (d) 1
3 4

lim sin 3x + sin x = lim sin 3x + lim sin x = lim sin 3x .3 + lim sin x = 1.3 + 1 = 4.
x x x 3x x
x→0 x→0 x→0 x→0 x→0

If f(x) = x sin 1 , x ≠ 0 , then lim f(x) = [IIT 1988; UPSEAT 1988; SCRA 1996]
 x
x→0 (d) None of these
 0 , x=0

(a) 1 (b) 0 (c) –1

lim x sin 1  =  lim x   lim sin 1  = 0 × (A number oscillating between – 1 and 1) = 0.
 x   x 
x→0 x→0 x→0

Aniprogramm 2019-08.10
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