Arihant CBSE Class 11 Book

CBSE Term II Applied Mathematics XII 35

ò\ = 27 3t3 = 27 ´ é t4 ù3 é æçè 3 ÷øö 2 ù 9/2
I dt ê 4 ú ê 2 ú
1 ë û1 = êë log t + t2 - ûú 5 / 2

= 27 [( 3)4 - (1)4 ] = log 9 + 81 - 9 - log 5 + 25 - 9
3 44

= 27 [ 81 - 1] = 27 ´ 80 2 44 2
44
= log 9 + 72 - log 5 + 16
= 540 24 24

ò(ii) Let I = 5 1 dx = log 9 + 6 2 - log 5 + 2

3 x2 -x - 2 22 2

ò= 5 1 dx 9+ 6 2

3 x2 - x + 1 - 2 - 1 = log 2 = log 9 + 6 2
44 9 9

ò= 5 dx dx 3 2 2
æçè x 1 ÷öø 9
- 2 - 4 ò(iii) Let I = 7 2 x4ex5 dx
1
Put x - 1 = t Put x5 = t Þ 5x4dx = dt
2
Lower limit When x = 1, then t =1
Þ dx = dt
Upper limit When x = 2, then t = 25 = 32
Lower limit When x = 3, then t = 3 - 1 = 5
22 ò\ I = 7 32 et dt

Upper limit When x = 5, then t = 5 - 1 = 9 15
22
= 7 t ]132 = 7[ e32 - e1 ]
ò\ I = 9/ 2 dt 5 [ e 5

5 / 2 t 2 - æç 3÷ö 2 = 7e ( e31 - 1)
è 2ø 5

Chapter Test ò(i) The value of 9 1 x dx is
0 x+

(a) 2log2 (b) 3log6

(c) 2log 4 (d) 2log 7

ò(ii) log5 1
Multiple Choice Questions The value of log 2 (ex + dx is
1)

ò1. 1 (x + 1) dx is equal to (a) log 4 (b) log 5
-1 5 4

(a) 3 (b) 2 (c) log 5 (d) log 2
2 5
(c) -3 (d) -2
ò(iii) Evaluate a x
ò2. The value of 1 x dx is 0 a2 + x 2 dx
0 x2 +
1

(a) log2 (b) 1 log2 (a) a( 2 + 1) (b) a( 2 - 1)
(c) 1 log2 2 (c) a 2 a
(d) 2 - 1)
3 (d) 2log2 (

3. If f (x) = (x + 1)2 + 1, then the integral of f (x) from limit ò(iv) The value of 4 e x dx is (b) e(e + 1)
x (d) 2e(e - 1)
12 x
1 to 5 is
(a) e(e - 1)
(a) 208 - log5 (b) 208 + log5 (c) e2
3 3
ò(v) 2 3 x4 + 4 dx is equal to

7x
0
(c) 208 - log5 (d) None of these
(a) 28 (5 5 + 1) (b) 28 (5 5 - 1)
ò4. The value of e6 dx is 5 3

e x log x (c) 28 (5 5 - 1) (d) None of these

(a) 3log2 (b) - log6 Short Answer Type Questions

(c) log6 (d) 2log3

5. If f (x) = 7x dx, then the value of integral from 7. Find the following integral by using partial fraction.
+ 74 - x
7x ò3 (2 + x + 5) dx.
1 x) (x
limit 0 to 4 is
(b) - 2 ò8. Evaluate 5 1 dx.
(a) 1 (d) 2
(c) - 1 0 x2 + 6x + 34

Case Based MCQs 9. òEvaluate 1 1 dx.
0 1+x - x
6. Suppose we have a definite integral of the form

òb (g(x)) g ¢ (x )dx. 10. òSolve 8 | x - 5 | dx.
2
a f

In this type of integral, we substitute g(x) = t Long Answer Type Questions
Þ g¢(x)dx = dt. And also the limits will change.
ò11. e2 log x
i.e. upper limit becomes g(b) and lower limit becomes Solve the integral e x2 dx.

òg(a). Then, the given integral will be g(b)f (t ) dt. 12. ò5 [| x - 2 | + | x - 3 | + | x - 5 | ] dx.
g(a)
Evaluate
On the basis of above information, solve the following
questions. 2

Answers

1. (b) 2. (b) 3. (b) 4. (c) 5. (d) 6. (i) (c) (ii) (b) (iii) (b) (iv) (a) (v) (b)
7. 2 log 3 + 5 log 4
8. log 8 + 89 9. 4 2 For Detailed Solutions
3 53 3 3 + 34 3 Scan the code
10. 9
11. 1 (2e - 3) 12. 23
e2 2

CBSE Term II Applied Mathematics XII 37

CHAPTER 03

Application of
Integrals

In this Chapter...

l Area Bounded by Curves
l Cost Function and Average Cost Function
l Revenue and Demand Function
l Consumer Surplus
l Producer Surplus
l Equilibrium Point

Area Bounded by Curves l If the curve y = f ( x) lies below the X-axis, then area
bounded by the curve y = f ( x), X-axis and the lines at x = a
l If the curve y = f ( x) lies above X-axis, then the area
bounded by the curve y = f ( x), the X-axis and the lines at òand x = b, is given by b y dx .
a
òx = a and x = b, is given by b y dx.
a Y

Y x=a x=b
X¢ X
A y = f(x) B
O

x=a y x=b y = f (x)
Y¢
X¢ O L dx M X

Y¢ l Generally, it may happen that some position of the curve is

l If the curve x = f ( y) lies to right of Y-axis, then the area above X-axis and some is below the X-axis which is shown
in the figure. The area A bounded by the curve y = f ( x),
bounded by the curve x = f ( y), the Y-axis and the lines at X-axis and the lines at x = a and x = b, is given by
A =| A2| + A1 .
òy = c and y = d, is given by d x dy.
c

Y Y A1 y = f(x)

B y=dC x=b
x X
dy x = f(y) X¢ x = a
O
A y=c D
X¢ X Y¢ A2

O
Y¢

38 CBSE Term II Applied Mathematics XII

l The area bounded by the two curves x = f ( y) and x = g( y) or dR = (MR)dx
such that 0 £ g( y) £ f ( y), " y Î[ c, d], the Y-axis and the
lines y = c and y = d is calculated as or ò dR = ò (MR) dx
or R = ò (MR)dx + k,
òArea = d [ f ( y) - g( y)] dy
c where k is a constant of integration.
Y When x = 0, then R = 0. Therefore, k = 0

y=d \ R = ò (MR) dx

x = f (y) If we divide the revenue function by x, we get the demand
x = g (y) function, which is represented by p.

y=c X i.e. p = R
X¢ x

O

Y¢ Consumer Surplus

l The area bounded by the two curves f ( x) and g( x) such that If the consumer purchase an item from the market at lower
f ( x) ³ g( x) in [ a, b] is calculated as
price instead of their expectation for paying it. i.e. some gain
òArea = b[ f ( x) - g( x)] dx
a by the consumer for expecting to pay the price, is said to be

consumer surplus.

If ‘p’ is the demand function of commodity (i.e. p = D( x)), then

Y y=f(x) the consumer surplus at the point ( x0 , p0 ) is defined as
B
òCS =x0 p dx - p0x0
A 0

y=g(x)

X¢ O x=a x=b X Producer Surplus
Y¢
If the producer sells an item from the market at higher price
Cost Function and Average Cost Function instead of their expectation for selling it i.e. some gain by the
producer for expecting to sell an item, is said to be producer
Suppose, C represents the total cost function, then the surplus.
marginal cost function is given by
If ‘p’ is the producer supply function of the commodity
MC = dC (i.e. P = S( x)), then the producer surplus at the point ( x0 , p0 )
dx is defined as

or dC = (MC) dx òPS = p0 x0 - x0 p dx

or ò dC = ò (MC) dx + k 0
or C = ò (MC)dx + k,
Equilibrium Point
where k is a constant of integration.
The equilibrium point ( x0 , p0 ) is the intersection point of the
If we divided the cost function by x, we get the average cost consumer’s demand curve and producer’s supply curve.
function. Hence, it is the point where the quantity demand is equal to
i.e. AC = C the quantity supplied.

x Price (P) Demand Supply curve
curve p=S (x)
p=D (x)

Revenue and Demand Functions Equilibrium
p0 point
Suppose, R represents the revenue function, then the
marginal revenue function is given by x0

MR = dR Quantity (x)
dx

where, x is the number of units sold

CBSE Term II Applied Mathematics XII 39

Solved Examples

Example 1. Find the area of the region bounded by the = é 0 - 4ù + é 9 - 0ùûú
êë 2 ûú êë 2
curve y = x + 3 and the lines x = 1 and x = 2.
=|- 2| + 9 = 2 + 9 = 13 sq units
Sol. Given, curve y = x + 3 is a straight line, which intersect the 2 22
coordinate axes at points A( - 3, 0) and B( 0, 3).

Y Example 3. Find the area under the curve y = 3 x

3 included between the lines x = 1 to x = 3.
2
1 Sol. Given, curve is y = 3 x.

y=x+3 X It is a positive square root curve, whose shape is upper half
123 of the horizontal parabola.
X¢
–3 –2 –1 Y
y=3 x

Y¢ x=1 x=2 X¢ X

2 y dx = 2
ò ò\ Area of region = + 3) dx
11 (x

= é x2 + ù2 = ( 2)2 + 3( 2) - 1 (1)2 - 3(1)
ê 2 3xú 2 2
ë x=1 x=3
û1 Y¢

= 2 + 6 - 1 - 3 = 8 - 7 = 9 sq units ò ò\Area of shaded region =3dx = 3( 3 x ) dx
2 22
y
11
Example 2. Find the area bounded by the line y = x, the
= é x 3/ 2 ù 3 = 2[( 3)3/ 2 - 1 3/ 2 ]
X-axis and the lines x = - 2 and x = 3. 3ê 3/ ú 1
ë 2 û
Sol. Given, line is y = x, which is passing through the origin and
= 2[ 3 3 - 1] = ( 6 3 - 2) sq units
making angle of 45° with the X-axis as shown in the given
figure. Example 4. Find the area of the region bounded by the

Y curve x - y = 4 and the lines x = 1 to x = 3.

D Sol. Given, curve is x - y = 4 is a straight line, which intersect
the coordinate axes at A( 4, 0) and B(0, 4).

Y

X¢ –2 –1 C X

B O 1 23 A(4,0)
X¢ X
A
x–y=4

y=x x=3 B(0,4)
x=–2 Y¢ x=1 x=3

\ Required Area = Area of region OAB Y¢

+ Area of region OCD ò\Area of shaded region = 3y dx
1
0 3 0 3

-2 y ydx = -2 x x dx
ò ò ò ò= dx + dx + ò= 3 ( x - 4) dx
0 1
0

= é x2 ù 0 + é x2 ù3 é x2 ù 3
ê 2 ú -2 ê 2 ú 4xú
ë û ë û0 = ê 2 - û1
ë

40 CBSE Term II Applied Mathematics XII

= é 9 - 12 - èæç 1 - 4ö÷ø ù It is given that, x = 1 and C = 15
êë 2 2 ûú
\ 15 = 60 [1 log1 - 1] + k

= é 9 - 1 - + 4ùúû 15 = 60( 0 - 1) + k
êë 2 2
12 Þ k = 15 + 60 = 75

=|[ 4 - 8]|= 4 sq units Put k = 75 in Eq. (i), we get

Example 5. Find the area of the region enclosed by the C = 60[ x logx - x] + 75

curves y = x 2 and y = x 3. Example 7. A manufacturer’s marginal revenue

Sol. We have, equation of the given curves as …(i) function is given by MR = 250 - x - 0.3x 2. If the
y = x2 …(ii) 2
y = x3
production is increased from 5 to 15 units, then find

the total increase in revenue by the manufacturer.

On equating Eqs.(i) and (ii), we get Sol. Given, MR = 250 - x - 0.3x2 Þ dR = 250 - x - 0.3x2
x3 = x2 2 dx 2

Þ x3 - x2 = 0 Þ dR = çæè 250 - x - 0.3x2 ÷øö dx
2
Þ x2( x - 1) = 0 Þ x = 0 or 1
On integrating both sides and taking limits from 5 to 15 is

Y y = x2 ò ò15 dR = 15 èçæ 250 - x - 0.3x2 ø÷ö dx
5 2
5

X¢ B X [ R]155 = é - x2 - 0.3 x3 ù 15
y = x3 CA ê 250x 4 3 ú 5
O ë û

é x2 ù 15
ê 250x 4 ú
= - - 0.1 x 3

Y¢ ë û5

Intersection points of given curves are (1, 1) and (0, 0). = 250 ´ 15 - 152 - 0.1 ´ (15)3
Area of shaded region = Area ( OCBAO) - Area ( OBAO) 4

ò\ Required area = 1( x2 - x3) dx - é 250 ´ 5 - 52 - 0.1( 5 ) 3 ù
0 ê 4 ú
ë û

= é x3 - x4 ù1 = 1 sq unit = 3750 - 225 - 337.5 - éêë1250 - 25 - 12.5ûúù
ê 3 4 ú 12 4 4
ë û0
= 3750 - 1250 - 225 + 25 - 337.5 + 12.5
Example 6. The marginal cost function of a firm is 44

given by MC = 60 log x. Find the total cost function = 2500 - 50 - 325

when the cost of producing one unit is ` 15. = ` 2125

Sol. Given, MC = 60logx Example 8. The supply function for a commodity is

p = 8 + x. If 10 units of goods are sold, then find the

Þ dC = 60logx producers surplus.
dx
Sol. We have, p = 8 + x …(i)
Þ dC = ( 60logx) dx
and x0 = 10

On integrating both sides, we get Put x0 = 10 and p = p0 in Eq. (i), we get

ò dC = ò( 60logx) dx + k p0 = 8 + 10 = 18

ò\ x0 pdx
Where, k is the constant of integration PS = p0x0 -
0

Þ C = 60ëêé logxò1 dx -ò îíì d (log x ) ò 1 dxýüþ dxùûú + k ò= 18 ´ 10 - 10 ( 8 + x) dx
dx 0

é èæç 1 xö÷ø ù = 180 - é + x2 ù 10
êë x logx ò x dxúû ê 8x ú
= 60 - ´ + k ë 2 û0

= 60[ x logx - ò1 dx] + k = 180 - é 80 + 100 - 0 - 0ùûú
êë 2

= 60 [ x logx - x] + k …(i) = 180 - 130 = 50

CBSE Term II Applied Mathematics XII 41

Chapter
Practice

PART 1 (b) 3 æç 1 ÷ö 4/ 3 - æç 3ö÷ 4/ 3 sq units
Objective Questions 4 è 2ø è 2ø

(c) 5 çæ 1 ö÷ 4/ 3 - æç 3÷ö 4/ 3 sq units
4 è 2ø è 2ø

l Multiple Choice Questions (d) 3 æç 1 ö÷ 4/ 3 + çæ 3ö÷ 4/ 3 sq units
4 è 2ø è 2ø
1. The area of the region bounded by the curve
y = x + 1 and the lines x = 2 and x = 3, is
7. Area bounded by the curve y = x 3, the X-axis and
(a) 7 sq units (b) 9 sq units the coordinates x = - 2 and x = 1 is
2 2

(c) 11 sq units (d) 13 sq units (a) - 9 sq unit (b) - 15 sq unit
2 2 (c) 15 sq units 4

2. The area of the region bounded by the curve 4 (d) 17 sq units
x = 2y + 3 and the lines y = 1 and y = - 1 is 4

(a) 4 sq units (b) 3 sq units 8. The area bounded by the curve y = x|x|, X-axis and
2
the coordinates x = - 1 and x = 1 is given by
(c) 6 sq units (d) 8 sq units
(a) 0 (b) 1 sq unit
3. The area of the region bounded by x 2 = 4y, y = 2, 3

y = 4 and the Y-axis in the first quadrant is (c) 2 sq unit (d) 4 sq units
3 3
(a) 8( 4 + 2) sq units (b) 8( 4 - 2) sq units
3 3 9. If we draw a rough sketch of the curve y = x - 1 in

(c) 8( 4 + 2) sq units (d) None of these the interval [1, 5], then the area under the curve

4. Area of the region bounded by the curve y2 = 4x, and between the lines x = 1 and x = 5 is

Y-axis and the line y = 3 is (a) 16 sq units (b) 8 sq units
9 3
(a) 2 sq units (b) 9 sq units
(c) 9 sq units 4 (c) 16 sq units (d) None of these
3
3 (d) 9 sq units
2 10. The marginal cost function for a product is given by
MC = 3 and the fixed cost is ` 20, then the
5. Area bounded by the curve y = log e x, x = 0, y £ 0 x2 + 9

and X-axis is

(a) 1 sq unit (b) 2 sq units average cost for 10 units of output is

(c) 3 sq units (d) 4 sq units (a) 3log|10 + 109| + 20 - log27
10
6. The area of region bounded by curve y = - x 3 and
(b) 3log|10 - 100| - 20 - log27
the lines y = - 1 and y = - 3 is 10
22
(c) 3log|10 + 109| - 20 - log7
(a) 4 çæ 1 ö÷ 4/ 3 - çæ 3÷ö 4/ 3 sq units
3 è 2ø è 2ø (d) None of the above

42 CBSE Term II Applied Mathematics XII

11. If the marginal revenue function of a commodity is l Case Based MCQs
MR = 8 - 5x 2 + 3x, then the demand function is 17. Consider the following equations of the parabola

(a) 8 - 5 x2 - 3 x y2 = 6x and the straight line y = 3x.
32
On the basis of the above information, solve the
(b) 8 - 5 x2 + 3 x following questions.
32 (i) The graph of given curve is

(c) 8 + 7 x2 + 3 x Y y=3x
32
y2=6x
(d) 8 + 5 x2 + 3 x
72

12. If the marginal revenue function for output is given (a) X¢ X

by MR = 7 + 10, then the demand function is x=1
(x + 3)2 Y¢

(a) -7 3) + 10 + 7 Y y2=6x
x(x + 3( x) y=3x

(b) 7 3) + 10 + 7
x(x + 3( x)

(c) 7 3) - 10 + 7 (b) X¢ X
x(x + 3( x)

(d) None of the above

13. A manufacturer’s marginal revenue function is Y¢
given by MR = 300 - 3x + 4 x 2. If the production is
3 y=3x Y

increased from 5 to 10 units, then increase in y2=6x

revenue is

(a) ` 1935 (b) ` 1776.38 (c) X¢ X

(c) ` 1940 (d) ` 1825

14. The demand function for a commodity is given by Y¢
p = 30 + 5x - x 2. When market price is ` 5, then
Y
consumer’s surplus (CS) is y=3x

(a) 17.5 (b) 42.5

(c) 20.83 (d) 39

15. The supply function of a producer is given by (d) X¢ X
p = 3 e3x , where x denotes thousands units. If sales
5

are 4000 units, then producer’s surplus (PS) is y2=6x
Y¢
(a) 11 e11 (b) 11 e12
7 6 (ii) The point of intersection of given curves are

(c) 11 e12 (d) None of these (a) (0, 0) and çæ - 1 , 2÷ö (b) (0, 0) and (2, 1)
5 è 3 3ø

16. The demand and supply functions for a commodity (c) (0, 0) and çæè 2, 2øö÷ (d) (0, 0) and çèæ 2, - 2÷öø
and p = x 2 - 6x + 16 and p = x 2 + 4 x + 4 3 3
33
(iii) The area of region bounded by the line between
respectively. The consumer’s surplus at the Açæè 2÷öø
equilibrium point, when x < 7, is points O( 0, 0), 2 , and X-axis is
3

(a) 17 (b) 20 (a) 2 sq unit (b) 3 sq unit
3 3 2

(c) 20 (d) 22 (c) 1 sq unit (d) None of these
3 3 3

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CBSE Term II Applied Mathematics XII 43

(iv) The area of region bounded by the parabola (iii) The area between the curves is
between points O( 0, 0), Açæ 2 , 2÷ö and X-axis is
è3 ø (a) 1 sq unit (b) 2 sq unit
3

(a) 8 sq units (b) 8 sq units (c) 3 sq unit (d) 3 sq unit
7 9 2

(c) 9 sq units (d) None of these (iv) The area of region bounded by the curve y = x
8
and the lines x = 1 and x = 2 is

(v) The area of region between parabola and the line is (a) 2( 2 2 - 1) sq units

(a) 9 sq unit (b) 8 sq unit (b) 1 ( 2 2 - 1) sq units
8 9 3

(c) 8 sq unit (d) 2 sq unit (c) 2( 2 2 - 1) sq units
7 9 3

18. Consider a square root curve y = 3x and the (d) None of the above
straight line 3x = 2y + 3.
(v) The area of region bounded by the lines
3x = 2y + 3, y = 1 and y = 3 is
On the basis of the above information, solve the
following questions. (a) 10 sq units (b) 5 sq units
3 3
(i) The intersection point of curve and line is
(c) 14 sq units (d) 11 sq units
(a) (- 3, 3) (b) (3, 3) 3 3

(c) (3, - 3) (d) ( - 3, - 3) 19. The marginal cost (MC) of producing x units of a
commodity in a day is given as MC = 14x - 1720.
(ii) The graph of given curve and line is

Y The selling price is fixed at ` 11 per unit and the
y =Ö3x fixed cost ` 1900 per day.

(a) X¢ (0,0) O 3x=2y+3 On the basis of the above information, solve the
following questions.
X

Y¢ (i) Cost function (C) is
Y 3x=2y+3 (a) 7x2 - 1720x + 1900
(b) 13x2 - 1000x + 1800
(c) 13x2 - 1000x + 1300
(d) 9x2 - 1000x + 1700

(ii) When x = 1, then value of cost function is

(b) X¢ OX (a) 311 (b) 187
(c) 410 (d) 176
y =Ö3x
Y¢ (iii) The revenue function (R) is
Y y =Ö3x 3x=2y+3
(a) 11x (b) 9x

(c) 17x (d) 20x

(c) X¢ O X (iv) The profit function (P) is
(1,0) (a) - 5x2 + 1731x - 1900
(b) - 6x2 + 1530x - 1300
(c) - 7x2 + 1731x - 1900
(d) - 5x2 + 1530x + 1200

(v) When x = 2 , then profit is

Y¢ (a) ` 1600 (b) ` 1534
(d) None of the above (c) ` 1320 (d) ` 1420

44 CBSE Term II Applied Mathematics XII

20. Consumer surplus and producer surplus. 3. Draw a rough sketch of the curve y =|x - 3|. Find
the area under the curve and lines x = 0 and x = 4.
Y
4. The area between x = y2 and x = 4 is divided into
Supply curve two equal parts by the line x = a, find the value of a.
p=S(x)
CS 5. Draw the rough sketch of the curve y = x 2 + 2 and
p0 Equilibrium point y = x. Find the area of the region bounded by the
lines, x = 0 and x = 3.
PS

Demand curve

p=D(x) 6. Using method of integration find the area bounded
by the curve|x|+|y|= 1.
O x0 X

The above graph showing the demand and supply 7. Using method of integration, find the area of region
curves of a tyre manufacturer company are linear. bounded by the curves y = ex and y = e-x and the
lines y = 1 and y = 3.
‘ABC’ tyre manufacturer sold 25 units every month
when the price of a tyre was ` 20000 per units and 8. The marginal cost function of producing x units of a
‘ABC’ tyre manufacturer sold 125 units every month x . Find the
when the prize dropped to ` 15000 per unit. When product is given by MC = x 2 + 2500
the price was ` 25000 per unit, 180 tyres were,
available per month for sale and when the price was total cost function, if the fixed cost is ` 1000.
only ` 15000 per unit, 80 tyres remained.
9. The marginal cost function of a firm is
On the basis of the above information, solve the MC = (log x)2. Find the total cost function when the
following questions.
cost of production one unit is ` 50.
(i) The demand function D(x) is
10. The marginal cost function MC for a product is
(a) - 40x + 21000 (b) - 50x + 21250 given by MC = 5 and the fixed cost is ` 20
2x + 9
(c) - 60x + 20000 (d) - 65x + 20000
find the average cost for 10 units of output.
(ii) The supply function S( x) is
11. The marginal revenue function of a commodity is
(a) 100x + 7000 (b) 110x + 6500 given as MR = 12 - 3x 2 + 4x. Find the total revenue
(c) 105x + 8000 (d) 90x + 6500 from the sale of 4 units.

(iii) The equilibrium point is 12. If demand for a consumer is given by the function
p = 27 - 3x - x 2 (where, x = quantity demand and
(a) (95, 16500) (b) (92, 16000) p = price). Find the consumer surplus at x 0 = 3.

(c) (90, 17000) (d) (97, 17200) 13. The demand function of a product is p = 10e-x . Find
the consumer’s surplus when the market price p = 1.
(iv) The consumer surplus (CS) is (given log10 e = 0.4343).

(a) ` 210720 (b) ` 227065

(c) ` 225625 (d) ` 2152729

(v) The producer surplus (PS) is

(a) ` 467230 (b) ` 451250

(c) ` 441623 (d) ` 468564

l Long Answer Type Questions

PART 2 14. Find the area of the region bounded by the
Subjective Questions parabola y = x 2 and y =|x|.

l Short Answer Type Questions 15. Sketch the graph of y =|x + 3|and evaluate
1. Find the area under given curves and given lines.
ò0 |x + 3|dx.
y = x 2; x = 1, x = 2 and X-axis -6
2. Find the area of the region bounded by y2 = 9x,
16. Find the area of the region
x = 2, x = 4 and the X-axis in the first quadrant. {(x, y) : 0 £ y £ x 2 + 1, 0 £ y £ x + 1, 0 £ x £ 2}.

17. Supply function of a producer is given by
40p = (x + 15)2. Find the producer’s surplus when
the market price is ` 40.

CBSE Term II Applied Mathematics XII 45

18. The demand and supply functions are p = 25 - x 2 On the basis of the above information solve the
and p = 2x + 1 respectively. Find each of the following questions.
following. (i) Find the point of intersection of the parabolas.
(i) The equilibrium point. (ii) Find the common region between parabolas.
(ii) The consumer’s surplus at the equilibrium point. (iii) Find the area of region between curve y 2 = x,
(iii) The producer’s surplus at the equilibrium point.
Y-axis and line y = 2 .
l Case Based Questions
19. The graph of barabolas y2 = x and x 2 = y is shown (iv) Find the area of region between curve x2 = y,
X-axis and line x = 3
below.
20. A long lasting company produces a parle biscuits,
Y x2=y whose marginal revenue function of a commodity is
y2=x given by MR = 6x 2 + 5x + 2.

X¢ X On the basis of the above information, solve the
following questions.
Y¢
(i) Find the total revenue function.
(ii) Find the demand function.
(iii) If the production increase from 10 to 20, find the

increase in their revenue.

SOLUTIONS

Objective Questions = é 2y2 + ù1 = [ y2 + 3y] 1
ê 2 3yú -1
3( é x2 ù3 ë û -1
ò1. (a)\Required + = ê 2 + xú
area = 2 x 1) dx ë û2 = [1 + 3 - 1 + 3] = 6 sq units

3. (b) The given curve x2 = 4y is a parabola which is

Y symmetrical about Y-axis (Q it contains even power of x) only
and passes through the origin.
The area of the region bounded by the curve x2 = 4y, y = 2

and y = 4 and the Y-axis is shown in the figure.

Y

y=x+1 X y=4
x=2 x=3 4

x2 = 4y
2 y=2

= é9 + 3 - 4 - 2úùû X¢ O X
êë 2 2

= é5 + 1 ûúù = 7 sq units Y¢
êë 2 2
y=b
ò2. (c)\Required area = 1 ( 2y + 3) dy ò\ Required area (shaded region) =
-1 y = a | x| dy

Y [here,| x| = 4y, a = 2 and b = 4]

y=1 ò= 4|x| dy [considering the elementary strip on Y-axis]
2

X ò= 4 y dy [Q x2 = 4y,\|x| = 2 y]

2
2

= 2 é y3/ 2 ù 4 = 4 [ 43/ 2 - 23/ 2 ] = 4 [8- 2 2]
y=–1 ê ú 3 3
ë 3/ 2 û
x=2y+3 2

= 8 [ 4 - 2] sq unit
3

46 CBSE Term II Applied Mathematics XII

4. (b) The area bounded by the curve, y2 = 4x, Y-axis and y = 3 7. (d) Given, curve is y = x3
is represented in the figure by shaded region.
Y
Y
y=3 X¢ (–2, 0)C O y = x3
B
3
A(1, 0)
X¢ O X X

Y¢ y 2 = 4x D Y¢

ò\ Required area = 3|x| dy On putting x = - x and y = - y, we get y = x3
0

ò= 3 y2 1 é y3 ù 3 Therefore, the curve is symmetrical in opposite quadrant
04 4 ê 3 ú 0
dy = ë û and passes through ( 0, 0).

\ Required area

= 1 ( 33 - 0) = 1 ( 27) ò ò ò=1 0 1|x3| dx
12 12 - |x3| dx = -2 |x 3| dx +
0
2

= 9 sq unit ò ò=0 ( - x3) dx + 1 x3 dx é x 3 | = íîì-xx33,, if x ³ 0üýúù
4 -2 êQ| if x < 0þúû
0 ëê

ò5. =½½ -¥ ey dy½½ [Q y = loge x Þ ey = x]
(a) \Required area 0 é x4 ù 0 é x4 ù 1
ê 4 ú -2 ê 4 ú 0
Y = - ë û + ë û

y = logex = - é - ( -2)4 ù + é1 - 0ûúù
ê0 4 ú êë 4
ë û

X¢ O X = - ( -4) + æèç 1 ÷øö = 17 sq units
4 4

Y¢ 8. (c) Given, y = x|x| = îïíìïx-2x,2x, ³0
x<
y -¥ 0
0
=|[ e ] | Y

= 1 sq unit

6. (b) The graph of a curve is shown below.

Y X¢ (–1, 0)C X

O A(1, 0)

X¢ X D Y¢

y =–12 \ Required area = 2
Area under the curve y = x2, between x = 0 and x = 1
y =–32 y = –x3
[Q the curve is symmetrical in opposite quadrant]
Y¢
1 1
ò ò= 2 = 2
x|x|dx 2 x dx
-1/ 2 --31//22( - y)1/ 3 dy 00

-3/ 2 x
ò ò\ Area of shaded region = dy = é x3 ù 1
ê 3 ú 0
ò= ( -1 )1/ 3 --31//22y1/ 3 dy = 2 ë û

= é y4/ 3 ù -1/ 2 = 2 (13 - 03)
ê 3 ûú 3
ë
4 / -3/ 2 = 2 sq unit
3
= 3 æç - 1 ÷ö 4/ 3 - çæ - 3ö÷ 4/ 3
4 è 2ø è 2ø 9. (c) We have, equation of the curve

y= x -1

3 æèç 1 öø÷ 4 / 3 çæè 3 öø÷ 4 / 3 On squaring both sides, we get
4 2 2 y2 = x - 1
= - sq units

CBSE Term II Applied Mathematics XII 47

Now, sketch the graph of given curve. When x = 0, then R = 0
\ 0 = 8( 0) - 5( 0) + 3( 0) + k Þ k = 0
Y x=5 Put k = 0 in Eq. (i), we get

(5, 2) y = Öx – 1 R = 8x - 5x3 + 3x2 + 0
32
X¢ O (1, 0) (5, 0) X
\ R = 8x - 5x3 + 3 x2
32

If p is the price per unit when x units are sold, then

Y¢ R = px

\ Area of the shaded region Þ p= R
x
ò= 5 x - 1 dx
1 = 8x - 5x3 + 3 x2 / x
32
= é ( x - 1 ) 3/ 2 ù 5 = 2 - 3/ 2 ] 15
ê 3 /2 ú 1 3
ë û [( x 1 ) = 8 - 5x2 + 3 x
32
= 2[( 4)3/ 2 - 0]
3 Hence, demand function is 8 - 5x2 + 3 x.
32
= 2 ´ 8 = 16 sq units
33 12. (a) Given, MR = 7 + 10
(x + 3)2
10. (a) Given that marginal cost function, dC = 3
dx x2 + 9 Þ dR = 7 + 10
dx (x + 3)2

On integrating both sides, we get On integrating both sides, we get

C=ò 3 dx = 3ò 1 dx ò dR = ò çèçæ ( x 7 + 10øö÷÷ dx
x2 + x2 + 32 + 3)2
9

Þ C = 3log|x + x2 + 32| + k …(i) R = 7 + 10x + k
3)1( -
It is given that fixed cost is ` 20 (x + 1)
When x = 0, then C = 20
\From Eq. (i), we get R = - 7 3) + 10x + k …(i)
(x +
20 = 3log|0 + 0 + 32| + k
where, k is a constant of integration.
20 = 3log3 + k When x = 0, then R = 0
Þ k = 20 - log27
Put k = 20 - log27 in Eq. (i), we get From Eq. (i),

C = 3log|x + x2 + 9| + 20 - log27 0=- 7 + 0+ kÞk= 7
33

Now, average cost, AC = C Put k = 7 in Eq. (i), we get
x 3

= 3log|x + x2 + 9| + 20 - log27 R = -7 + 10x + 7
x ( x + 3) 3

When x = 10 units, then R - 7 3) + 10x + 7
AC = 3log|10 + 100 + 9| + 20 - log27 (x + 3
10 \Demand function, p= =
= 3log|10 + 109| + 20 - log27 xx
10
= - x( x 7 3) + 10 + 7
11. (b) Given, MR = 8 - 5x2 + 3x + 3x

13. (b) We have, MR = 300 - 3x + 4 x2
3

Þ dR = 8 - 5x2 + 3x Þ dR = 300 - 3x + 4 x2
dx dx 3

On integrating both sides, we get \Required increase in revenue

R = ò( 8 - 5x2 + 3x) dx ò= 10 æèç 300 - 3x + 4 x2 ÷øö dx
5 3
R = 8x - 5x3 + 3x2 + k
32 …(i) = é - 3x2 + 4 x3 ù 10
ê 2
ë 300x 3 3 ú
û5
where, k is a constant of integration.

48 CBSE Term II Applied Mathematics XII

= 3000 - 3 ( 10) 2 + 4 (10)3 - é 300 ´5 - 3(5)2 + 4 (5 )3 ù Thus, x0 = 2 and p0 = 8
2 9 ëê 2 9 úû \ The consumer’s surplus (CS) at the equilibrium point

= 3000 - 300 + 4000 - æçè1500 - 75 + 500 øö÷ (2, 8) is given by
2 9 2 9
òCS = x0 D( x)dx - p0 x 0
0
= ( 3000 - 1500) + 75 - 300 + 4000 - 500
22 9 9 ò= 2( x2 - 6x + 16) dx - 8 ´ 2
0
= 1500 - 225 + 3500
29 é x3 6x2 ù 2
ê 3 2 16xú
= 27000 - 2025 + 7000 = 31975 = ` 1776.38 = ë - + û0 - 16
18 18
= ( 2)3 - 6( 2)2 + 16( 2) - ( 0 - 0 + 0) - 16
14. (c) We have, p = 30 + 5x - x2 32

when market demand x0 = 5, then = 8 - 12 + 32 - 16 = 8 + 4 = 20
p0 = 30 + 5(5) - 52 = 30 + 25 - 25 = 30 3 33

\Consumer’s surplus (CS) 17. (i) (a) Parabola y2 = 6x is standard right hand parabola

ò= x0 pdx - p0 x 0 with axis as X-axis and vertex as (0, 0). Straight line
0 y = 3x is a line passing through the origin.

ò= 5( 30 + 5x - x2 )dx - 30 ´ 5 Y y=3x
0
y2=6x
= é + 5 x2 - x3 ù 5 - 150 A
ê 30x 2 3 ú 0
ë û
X¢ X

= éëê150 + 125 - 125 - 0ûúù - 150
2 3

= 125 - 125 = 20.83 Y¢
23

15. (c) We have, p = 3 e3x (ii) (c) The point of intersection of line and parabola
5 i.e. y2 = 6x and y = 3x is

When sales are 4000 units i.e. x0 = 4, then we get ( 3x)2 = 6x Þ 9x2 = 6x

p0 = 3 e3 ´4 = 3 e12 Þ 3x( 3x - 2) = 0
5 5
Þ x = 0, 2
ò\ Producer surplus (PS) = p0x0 - x0 pdx 3

0 Þ y = 0, 2

ò= 3 e12 ´ 4 - 43 e3x dx \ The required point of intersection are O( 0, 0) and
Aèçæ 2÷øö.
5 05 2
3
= 12 e12 - 3 é e 3x ù 4 ,
5 5 ê 3 ú 0
ë û ò(iii) (a) Area of shaded region = 2/ 3

= 12 e12 - 1 ( e12 - e0 ) y dx
55 0

= 12 e12 - e12 = 11 e12 Y
555 y=3x

16. (c) Given, the demand and supply functions are X¢ (0,0) O A 2 ,2
p = D( x) = x2 - 6x + 16 and p = S( x) = 1 x2 + 4 x + 4. 3
33 X

For equilibrium point, D( x) = S( x) Y¢
Þ x2 - 6x + 16 = 1 x2 + 4 x + 4
2/ 3( é x2 ù2/ 3
33 0 3ê 2 ú
Þ 2 x2 - 22 x + 12 = 0 ò= 3x) dx = ë û0

33 = 3 é èæç 2÷ö 2 - ù = 3 é 4ù
Þ x2 - 11x + 18 = 0 2 ê 3ø 0ú 2 ëê 9úû
ëê ûú
Þ ( x - 2)( x - 9) = 0
Since, x < 7, therefore x = 2. = 2 sq unit …(i)
Put x = 2, we get 3

p = ( 2)2 - 6( 2) + 16 = 4 - 12 + 16 = 8

CBSE Term II Applied Mathematics XII 49

ò(iv) (b) Area of bounded region = 2/ 3 And straight line 3x = 2y + 3 line, which intersect the
coordinate axes at point (1,0) and æçè 0, - 32öø÷.
y dx The graph of given curve is shown below
0
Y y = Ö3x 3x=2y+3
ò= 2/ 3 6x dx
0

Y A 2 ,2
3

X¢ O(0,0) X X¢ X
(1,0)

y2=6x
Y¢

6 2/ 3x1/ 2dx = é x 3/ 2 ù 2 / 3 Y¢
ê 3 ú 0
ò= 0 6 ë / 2 û 3

= 2 6[( 2 / 3)3/ 2 - 0] ò(iii) (c) Area between the curves = ( x2 - x1) dy
3
0
=2 6 é æèç 2 ÷öø 2 ù
3 ê 3 3 ú Y y =Ö3x 3x=2y+3
ë û A (3,3)

= 4 ´2 X¢ X
9 (0,0)

= 8 sq unit …(ii)
9
Y¢
ò(v) 2/ 3
(d) Area of bounded region = - y1)dx 3 é çèæ 2y + 3 ø÷ö ççèæ y2 öø÷÷ ù
0 ( y2 0 ê 3 3 ú
ò= êë ûú
2/ 3 2/ 3 - dy
ò ò=
6x dx - 3xdx
00

Y y = 3x é 1 çèçæ 2y2 3y÷öø÷ y3 ù 3
ê 3 2 ú
y2=6x = ëê + - 3 ûú 0
A 2 ,2
X¢ (0,0) O = é y2 + y- y3 ù 3 = é 9 + 3- 27 - ( 0)úùû
3 ê 3 ú ëê 3 9
ë 9 û
X

0

= [ 3 + 3 - 3]

= 3 sq units

Y¢ ò(iv) (c) Area of bounded region = 2 y dx
1
=8-2
93 [from Eqs. (i) and (ii)] y =Öx
Y
= 8 - 6 = 2 sq unit
99

18. (i) (b) The intersection point of y = 3x and line X¢ X
x=1 x=2
3x = 2y + 3 is

y2 = 3x Þ y2 = 2y + 3

Þ y2 - 2y - 3 = 0

Þ ( y + 1) ( y - 3) = 0 Þ y = - 1, 3 Y¢

But y is positive, so we neglect y dx = 2 é x 3/ 2ù 2
y = -1 ê 3/ 1
1 ë
\ y = 3 Þ 3x = 2 ´ 3 + 3 Þ 3x = 9 Þ x = 3 ò ò= 2 x dx = 2 ú
\The point of intersection is A( 3, 3). 1 û
(ii) (c) Given, root curve is y = 3x or y2 = 3x, y ³ 0.
= 2[( 2)3/ 2 - (1)3/ 2 ]
It represents upper part of a right hand parabola with 3
vertex (0, 0).
= 2 [ 2 2 - 1] sq units
3

50 CBSE Term II Applied Mathematics XII

ò(v) (c) Area of bounded region = 3 and when x = 80, then p = 15000

1 x dy

Y From Eq. (i), we have … (iii)
15000 = 80c + d

y=3 3x=2y+3 On solving Eqs. (ii) and (iii), we get
c = 100 and d = 7000
y=1 X
X¢ \ Supply function S( x) = 100x + 7000
(iii) (a) For equilibrium point D( x) = S( x)

3x=2y+3 Þ - 50x0 + 21250 = 100x0 + 7000
Þ - 150x0 = - 14250
Y¢ Þ x0 = 95

ò=3æç 2y + 3 öø÷dy = 1 é 2y2 + ù3 On putting value of x0 in demand function or supply
1è 3 3 ê 2 3yú function, we get
ë
û1 p0 = 16500

= 1 [( 3)2 + 3( 3) - (1 + 3)] ò(iv) x0 -
3 (c) Consumer surplus (CS) = 0 D( x) dx p0 x 0

= 1 [ 9 + 9 - 4] = 14 sq units ò= 95 ( - 50x + 21250) dx - 16500 ´ 95
33 0

19. (i) (a) Cost function, C = ò(14x - 1720)dx + k = ççæè - 50 x2 + 21250x ÷÷öø 95 - 1567500
2 0
[Q C = ò( MC)dx + k]
= 1793125 - 1567500
= 14x2 - 1720x + k
2 = ` 225625

= 7x2 - 1720x + k ò(v) x0 S( x) dx
(b) The producer surplus (PS) = p0x0 -
When x = 0, then C = 1900 0
\ Cost function C = 7x2 - 1720x + 1900
ò= 16500 ´ 95 - 95 (100x + 7000) dx
0

(ii) (b) When x = 1, then C = 7 ´ 1 - 1720 + 1900 = 187 = 1567500 - é x2 + ù 95
ê100 2 7000xú
ë
(iii) (a) The selling price is fixed at ` 11 per unit. û0

So, the revenue function R is given by = 1567500 - [50 ´ 95 ´ 95 + 7000 ´ 95]
R = 11x
= 1567500 - 1116250
(iv) (c) The profit function P is given by
P=R -C = ` 451250
= 11x - ( 7x2 - 1720x + 1900)
Subjective Questions
= - 7x2 + 1731x - 1900
1. The given curve y = x2 represents an upward parabola with

vertex ( 0, 0) and axis along Y-axis.

(v) (b) When x = 2, then Y

P = - 7 ´ 4 + 1731 ´ 2 - 1900 = ` 1534 y = x2

20. (i) (b) Let us consider demand function be …(i) X¢ X
D( x) = ax + b …(ii) O 12
…(iii)
When x = 25, then p = 20000 Y¢
…(i)
From Eq. (i), we have …(ii) \Required area (shown in shaded region)
20000 = 25a + b = Area under the curve y= x2, X-axis and x = 1, x = 2

and when x = 125, then p = 15000 2 y dx = 2
ò ò= 2 dx
From Eq. (i), we have x
15000 = 125a + b 11

On solving Eqs. (ii) and (iii), we get = é x3 ù 2 = 1 [ 23 - 13]
a = - 50 and b = 21250 ê 3 ú 1 3
ë û
\ Demand function D( x) = - 50x + 21250
= 8 - 1 = 7 sq unit
(ii) (a) Let us consider supply function be 338
S( x) = cx + d
2. Since, the given curve y2 = 9x is a parabola which is
When x = 180, then p = 25000
symmetrical about X-axis (Q the power of y is even) and
From Eq. (i), we have passes through the origin.
25000 = 180c + d

CBSE Term II Applied Mathematics XII 51

The area of the region bounded by the curve, y2 = 9x, x = 2 4. Given, curve x = y2 is a parabola symmetrical about X-axis
and x = 4 and the X-axis is the area shown in the figure.
and passing through the origin.
Y
Y y2 = x
y 2 = 9x D C

X¢ x=2 x=4 X X¢ X
O E F

O

Y¢ A B
Y¢
\ Required area (shaded region)
x=a x=4
4 4

|y| 3
ò ò= dx = x dx [Q y2 = 9x Þ |y| = 3 x ] The line x = a, divides the area bounded by the parabola and
22 x = 4 into two equal parts.

é ù4 Area OAD = Area ABCD
ê x3/ 2 ú ´2
= 3 ê ú = 3 3 [ 43/ 2 - 23/ 2 ] = 2[4 4-2 2] \ Area OED = Area EFCD
êë ûú 2
3 òÞ a
2 Area OED =
y dx
0
= 2[8 - 2 2]
òand area of EFCD = 4 x dx [Q y2 = x Þ |y| = x]
= 4 [ 4 - 2] sq units a

3. Given, curve is y =|x - 3| a x dx = 4 é x 3/ 2 ù a é x 3/ 2 ù 4
0 ê 3/ ú 0 ê 3 ú a
a ë 2 û ë / 2 û
ò òÞ x dx Þ =

Since, it is an absolute function, therefore it make two

straight lines, which is define below. Þ 2 [ a3/ 2 - 0] = 2 [ 43/ 2 - a3/ 2 ]
33
y = ìx - 3, x³3
îí3 - x, x<3 Þ a3/ 2 = 43/ 2 - a3/ 2

The graph of an absolute function is defined below Þ 2a3/ 2 = 8 Þ a3/ 2 = 4 Þ a = ( 4)2/ 3.

Y y=x –3 Therefore, the value of a is ( 4)2/ 3.
y=3–x 5. Given, curve y= x2 + 2 represents a parabola and it is

symmetrical about Y-axis having vertex ( 0, 2).

X¢ 3 X The given region bounded by y= x2 + 2, y = x, x = 0 and
x = 3, is represented by the shaded area.

x=4 Y

Y¢ x2 + 2 B(3, 11)
y= C
\ The area of bounded region

ò ò= 3y dx + 4 y dx X¢ A x=3
03 y=x O DX

ò ò= 3 3 - x) dx + 4 x - 3) dx Y¢

( (
03

é x2 ù3 é x2 ù 4 The point of intersection of the curve y = x2 + 2 and the
ê 3x 3xú
= ë - 2 ú + ê 2 - û3 line x = 3 is (3, 11).
û0 ë
\ Required area (shown in shaded region)
é 32 ù é 42 æèç 9 3øö÷ ù
= ê 3 ´ 3 - 2 - ( 0)ú + ê - 3( 4) - 2 - 3 ´ ú = Area OABDO - Area OCDO
ë û ë 2 û
= [Area under y= x2 + 2 between x = 0, x = 3]
é 9ù é æçè 9 9ö÷ø ù
= ëê 9 - 2 ûú + êë 8 - 12 - 2 - úû – [Area under y = x between x = 0, x = 3]

3( x2 + 2) dx - 3x é x3 ù3 é x2 ù3
2xú ú
0 0 û0
= 9 + é - 4 - æçè - 9 öø÷ ù ò ò= dx = ê 3 + û0 - ê 2
2 êë 2 úû ë ë

= 9 + é - + 9ù = é 33 + 6 - ù - é 32 - ù
2 êë 4 2 ûú ê 0ú ê 2 0ú
ë 3 û ë û

= 9 + 1 = 5 sq units = 9 + 6 - 9 = 21 sq unit
22 22

52 CBSE Term II Applied Mathematics XII

6. The given curve is|x| + |y| = 1 So, C = ò tdt + k
t
Y
B(0, 1) Þ C = ò dt + k = t + k
y = x +1
x+y=1
(–1, 0) Þ C = x2 + 2500 + k

X¢ C O X Given, when x = 0, then C = ` 1000
A(1, 0) \ 1000 = 2500 + k

y = –x –1 y=x–1 Þ 1000 = 50 + k
Þ k = 1000 - 50 = 950
D(0, –1) \ Total cost function,
Y¢

In first quadrant, ( x > 0, y > 0) C = x2 + 2500 + 950
Then, the line AB is x + y = 1
9. We have, MC = (logx)2
In second quadrant, ( x < 0, y > 0)
Then, the line BC is -x + y = 1 \ C = ò( logx)2 . 1 dx + k [Q C = ò( MC) dx + k]
In third quadrant ( x < 0, y < 0) I II
Then, the line CD is - x - y = 1
Using integration by parts
In fourth quadrant ( x > 0, y < 0)
Then, the line DA is x - y = 1 = (log x ) 2 ò 1 dx - ò èçæ d (log x) 2 ò 1 dx÷öø dx + k
dx

Since, ABCD is a square. = (log x ) 2 x - ò 2 log x . x dx + k
x
\ Required area
= x(logx)2 - 2( x logx - x) + k
= 4 (Area of shaded region in the first quadrant)
= x(logx)2 - 2x logx + 2x + k …(i)
ò= 4 1 (1 - x) dx [Q x + y = 1 Þ y = 1 - x]
0 When, x = 1, then C = 50

= 4 é - x2 ù1 = 4 é èæç1 - 1 ÷øö - 0ùúû From Eq. (i), we get
êx 2 ú êë 2 50 = 2 + k
ë û0
Þ k = 48
= 2 sq unit \Total cost function, C = x(logx)2 - 2x logx + 2x + 48
7. Given, curves y = ex and y = e- x are exponential curve.
5
Y 10. Given that, MC = 2x + 9

y=ex Þ dC = 5 9 dx
y=3 dt 2x +

y=1 y=e–x On integrating both sides, we get
X¢
X ò dC = ò 5 9 dx + k
2x +

Þ C = 5( 2x + 9)-1/ 2 + 1 + k

Y¢ 2èæç - 1 + 1 øö÷
2
ò\Area of bounded region = 2 3
1 Þ C = 5 2x + 9 + k …(i)
x dy

ò= 2 13 It is given that x = 0 and C = 20
From Eq. (i), we get
ln y dy
20 = 5 2 ´ 0 + 9 + k
= 2[ y ln y - y] 3
1

= 2[( 3ln 3 - 3) - (|x|- 1)] Þ 20 = 15 + k
Þ k=5
= 2[ 3ln 3 - 2] Substituting k = 5 in Eq. (i), we get

= 6ln 3 - 4 sq units

8. We have, MC = x C = 5 2x + 9 + 5

x2 + 2500 Now, AC = C
x
\ C=ò x dx + k
x2 + 2500 Þ AC = 5 2x + 9 + 5
xx
On putting x2 + 2500 = t 2 Þ x dx = t dt

CBSE Term II Applied Mathematics XII 53

When x = 10, we get = - 10 é 1 - 1 ù - 2.30
AC = 5 2 ´ 10 + 9 + 5 ëê 10 ûú
10 10
= 5 20 + 9 + 1 = - 10 ´ ( - 0.9) - 2.30
10 2
= 9.0 - 2.30
= 29 + 1 = 5.39 + 1 = 6.70
2 22 2 14. Given, parabola y= x2 which is symmetrical about Y-axis and
passes through ( 0, 0) and the curve y =|x|.
= 2.695 + 0.5 On putting x = - x, we get y =|- x| =|x|
= ` 3.195 \Curve y =|x|is symmetrical about Y-axis and passes

Hence, average cost for 10 units of output is ` 3.195. through origin.
11. We have, MR = 12 - 3x2 + 4x The area bounded by the parabola, y = x2 and the line y=|x|
or y= ± x can be represented in the figure.
\ R = òMR dx + k

= ò(12 - 3x2 + 4x) dx + k Y

= 12x - x3 + 2x2 + k y = x2

We know that, when x = 0, then R = 0 B y = |x|
\ k=0 (–1, 1) A
Thus, R = 12x - x3 + 2x2
(1, 1)

Now, the total revenue from the sale of 4 units is given by X¢ D CX
Put x = 4, R = 48 - 64 + 32 = 16 O (0, 0) M
12. We have, demand function p = 27 - 3x - x2
Y¢

At x0 = 3, p0 = 27 - 3 ´ 3 - ( 3)2 = 27 - 9 - 9 = 9 The point of intersection of parabola, x2 = y and line, y= x in

\Consumer surplus, first quadrant is A (1, 1).

òCS =x0 D( x) dx - p0 x 0 The given area is symmetrical about Y-axis.
0
\Area OACO = Area ODBO

ò= 3 27 - 3x - x2 ) dx - 9 ´ 3 \ Required area = 2 (Area of shaded region in the first

( quadrant only)
0

= ççèæ 27x - 3x2 - x3 ÷÷øö 3 - 27 ò ò= 2 1 ( y2 - y1 ) dx = 2 1( x - x2 ) dx
2 3 0 0
0

æççè 27 øö÷÷ [Q the curve y =|x|lies above the curve y = x2 in [0, 1],

= ´ - 3( 3)2 - ( 3) 3 - so we take y2 = x and y1 = x2]
2 3
3 27 æç ö÷
ç ÷
= 81 - 27 - 9 - 27 = 45 - 13.5 = 31.5 ò ò=2 é 1 x dx - 1 x2 dx ù = 2 è é x2 ù 1 - é x3 ù 1 ø
2 êë 0 0 úû ê 2 ú 0 ê 3 ú 0
ë û ë û

13. Given, the demand function is = 2 íîìèçæ 1 - 0÷øö - çèæ 1 - 0øö÷ ýþü
p = 10e-x and p0 = 1 2 3

\ 1 = 10e-x0 = 1 sq unit
3
Þ e-x0 = 1
10 Therefore, required area is 1 sq unit.
3
Þ ex0 = 10
ì-( x + 3) for x < - 3
Þ x0 loge = loge 10 15. y =| x + 3| = îíx + 3 for x ³ - 3

Þ x0 = loge 10 When x < - 3, y = - x - 3
=1
Þ x0 log10 e

x -4 -5 -6

Þ x0 = 1 = 2.30 y 1 23
0.4343

\Consumer’s surplus (CS) When x ³ - 3, y = x + 3

ò= x0 10e-xdx - 1 ´ 2.30 x -1 -2 -3
0

= - 10[ e -x ] x0 - 2.30
0
y 2 10
= - 10[ e-x0 - e0 ]

54 CBSE Term II Applied Mathematics XII

Draw these points on the graph paper and we get the \ The region common to 0 £ y £ x 2 +1 ü is shown as
required figure. 0 £ y £ x +1 ï
ý
Y ï
y=x+3 0 £ x £2 þ

C(–6, 3) D(0, 3) shaded region OABCEFO.
y = –x – 3
Now, solving Eqs. (i) and (ii), we get
X¢ B A x2 + 1 = x + 1 Þ x2 - x= 0
–6 –5 –4 –3 –2 –1
O1 2 3 X Þ x( x - 1) = 0 Þ x = 0, 1
Y¢ When x = 0, then y = 0 + 1 = 1.
When x = 1, then y = 1 + 1 = 2. [using Eq. (ii)]
[using Eq. (ii)]

\Required area So, the points of intersection are A( 0, 1) and B( 1, 2).

= Area of region ABC + Area of region OAD \ Required area = Area of shaded region OABCEFO

ò ò= --63|x + 3| dx + -03|x + 3| dx = Area of region OABFO + Area of region FECBF

--63( - x - 3) dx + 0 ò ò= 1 y( parabola ) dx + 2
ò ò= -3 ( x + 3) dx 01
y(line) dx

é -x 2 ù - 3 é x2 ù0 ò ò= 1( x2 + 1) dx + 2( x + 1) dx
ê 2 3xú ê 2 3xú 01
ë ë
= - û-6 + + û-3 = é x3 + ù1 + é x2 + ù2
ê 3 xú ê 2 xú
é æèçç -( -3) 2 - 3 ´ ( - 3)÷ö÷ø ççæè -( -6)2 -6)÷÷øö ù ë û0 ë û1
ê 2 2 ú
= ëê - - 3 ´ ( úû = é èæç 1 + 1 öø÷ - 0úûù + é èçæ 4 + 2÷öø - çæè 1 + 1 ÷öø ù
êë 3 êë 2 2 ûú
é ççèæ -3)2 3)øö÷÷ ù
+ ê 0 - ( 2 + 3 ´(- ú = 4 + é - 3ù
êë ûú 3 ëê 2 úû
4

= é æèç -9 + 9ø÷ö - ( -18 + 18)ûúù + é 9ù =4+5
ëê 2 ëê 2ûú 32

=9+ 9 = 8 + 15
22 6

= 9 sq unit = 23 sq units
6
16. Given, equations of curve and lines are
y = x2 + 1 Hence, the required area is 23 sq units.
6
or x2 = y - 1 ...(i)

y=x+1 ...(ii) 17. Given, the supply function is
40p =( x + 15)2
and x = 2 ...(iii)

Here, Eq. (i) represents a parabola with vertex A( 0, 1) and At p0 = 40, we obtain
axis along the positive direction of Y-axis. 40 ´ 40 = ( x0 + 15)2

Eq. (ii) represents a line which intersects the coordinate Þ 1600 = ( x0 + 15)2
axes at( 0, 1) and ( -1, 0).
Þ ( x0 + 15) = 40 Þ x0 = 25
Eq. (iii) represents a line which is perpendicular to X-axis.
ò\Producer’s surplus (PS) = p0x0 - x0 S( x) dx
y = x2+1 Y y = x+1
(1, 2) C 0
B
ò= 40 ´ 25 - 25 ( x0 + 15)2 dx

0 40

(0, 1)A 1 é ( x + 15 )3 ù 25
40 ê 3 ú 0
= 1000 - 0 û

X¢ O FE X ë
(–1, 0) x=1 x=2
= 1000 - 1 [( 25 + 15)3 - (15)3]
Y¢ 120

As 0 £ y £ x2 + 1, therefore it represents the area below the = 1000 - 1 [ 64000 - 3375]
120
parabola and above the X-axis.
Similarly, 0 £ y £ x + 1 represents the area below the line = 1000 - 1 [ 60625]
AB and above X-axis and 0 £ x £ 2 represents the area 120
between the parallel lines x = 0 and x = 2.
= 1000 - 505.21

= 494.79

CBSE Term II Applied Mathematics XII 55

18. Given, the demand function and supply are Y
p = D( x) = 25 - x2 and p = S( x) = 2x + 1.
y=2
(i) At equilibrium point D( x) = S( x) X¢ y2=x
X
Þ 25 - x02 = 2x0 + 1

Þ x02 + 2x0 - 24 = 0

Þ ( x0 - 4) ( x0 + 6) = 0 [Q x0 + 6 ¹ 0]
Þ x0 = 4

Now, putting x0 = 4 in D( x) or S( x), we get p0 = 9. Y¢

Hence, the equilibrium point is ( x0, p0 ) i.e. (4, 9) é y3 ù 2
ê 3 ú 0
(ii) The consumer’s surplus (CS) at the equilibrium = ë û

point (4, 9) is given by

ò òx0 - = 4( 25 - x2 ) dx - 9 ´ 4 = ( 2)3 = 8 sq units
0 33
D( x)dx p0 x 0 0

= é - x3 ù4 - 36 ò ò(iv) \Required area of region = 3 = 3( x2 )dx
ê 25x ú dx
ë 3 û0 y
00

é ( 4)3 ù Y
ê 25
= ë ´4 - 3 ú - 36
û

= 100 - 64 - 36 X¢ X
3

= 128
3

(iii) The producer’s surplus (PS) at the equilibrium x=3
Y¢
point (4, 9) is given by
3
x0 S( x) dx = 9 ´ 4 - 4 é x3 ù 0 ( 3)3
ê 3 ú 3
0 ( ë û
ò òp0x0 - 0 2x + 1) dx = = = 9 sq units

= 36 - é 2x2 + ù4 20. (i) Given, marginal revenue function,
ê 2 xú MR = 6x2 - 5x + 2
ë û0

= 36 - [( 4)2 + 4] = 16 Þ dR = 6x2 - 5x + 2
dx
19. (i) The points of intersection of parabolas y2 = x and x2 = y
Þ dR = ( 6x2 - 5x + 2) dx
are
(x2 )2 = x On integrating both sides, we get

Þ x( x3 - 1) = 0 ò dR = ò( 6x2 - 5x + 2) dx

Þ x = 0, 1 R = 6x3 - 5x2 + 2x + k
Þ y = 0, 1 32

Hence, the points of intersection of two curves are [Q R = ò(MR) dx + k]

O( 0, 0), A(1, 1). R = 2x3 - 5x2 + 2x + k
2
ò(ii) 1 - y1) dx
The area of common region = …(i)
0( y2

ò= 1( x - x2 ) dx (ii) From Eq. (i), R = 2x3 - 5x2 + 2x + k
0 2

= é x3/ 2 - x3 ù 1 Initialy, when x = 0, then R = 0
ê 3/ 2 3 ú 0
ë û \ 0 = 2( 0) - 5 ( 0) + 2( 0) + k Þ k = 0
2
= é 2 1 3/ 2 - 1 (1)3 -(0 - 0)ùúû
ëê 3 3
Put k = 0 in Eq. (i), we get
é 2 1ù
= êë 3 - 3 úû R = 2x3 - 5 x2 + 2x + 0 = 2x3 - 5 x2 + 2x
22
= 1 sq units
3 (iii) The increase in revenue, when production increases
from 10 to 20 is

ò ò(iii) \Required area of region = 2 dy = 2y2 dy

x
00

56 CBSE Term II Applied Mathematics XII

òR( 20) - R(10) = 20 = é 80000 - 20000 + 400
êë 3
R dx
10

ò= 20 èæçç 2x 3 - 5x2 + 2x÷÷øö dx - íìî5000 - 2500 + 100üýþùûú
10 2 3

= é 2 x4 -5 ´ x3 + 2 x2 ù 20 = 80000 - 5000 - çèæ 20000 - 2500 ÷øö
ê 4 2 3 2 ú 10 3 3
ë û
+ 400 - 100
= é 1 20)4 - 5 ( 20)3 + ( 20)2
êë 2 ( 6 = 75000 - 17500 + 300
3
- îìí12 ( 10) 4 - 5 (10)3 + (10)2 üýþùûú
6 = 75300 - 17500
3
= é 160000 - 40000 + 400
êë 2 6 = 225900 - 17500
3
-ìíî100200 - 5000 + 100ýþüúûù
6 = 208400
3

= ` 69466.67

Chapter Test (a) 2 sq units (b) 4 sq units
(c) 3 sq units (d) None of these

Multiple Choice Questions (iii) The area of triangle formed by the curve and between
axes is
(a) 9 sq units (b) 8 sq units (c) 9 sq units (d) 7 sq units
98

1. The area of the region lying in the first quadrant and (iv) The area of region bounded by the curve f(x) and the
bounded by y = 4x2, x = 0, y = 1 and y = 4 is
lines y = 1 to y = 1 is
(a) 5 sq units (b) 7 sq units (c) 5 sq units (d) 8 sq units 4
(a) 32 sq units (b) 21 sq units
78 97 21 32

2. The area (in sq units) of the region bounded by the (c) 11 sq units (d) 9 sq units

curves 2x = y 2 - 1 and x = 0 is 32 32

(a) 1 (b) 2 (c) 1 (d) 2 (v) The area of region bounded by the curve f(x) and the
3 3
lines x = 1 to x = 3 is
3. The area (in sq units) of region bounded by the curve
y = xex and the lines x = 2 and x = 4 is ò(a) 3 - 3 dx sq units ò(b) 3 y - 3 dy sq units

(a) 3e2(e2 + 1) (b) e2(3e2 - 1) (c) 3e2(e2 - 1) (d) e2(3e2 + 1) x 12
12

2ò ò(c) èæç 3 ÷øö 23çèæ x 3 öø÷dx
1 - x - 2 + dx + - 2 sq units

4. The marginal cost function of a product is given by 2 çæè 3 ö÷ødy 23èçæ 3 öø÷dy
1 2 2
MC = 500 . If C is in `, then the costs involved to ò ò(d) - y - + y - sq units
2x + 25

increase production from 100 units to 300 units is Short Answer Type Questions

(a) ` 4500 (b) ` 4800 (c) ` 5000 (d) ` 4900 6. Find the area of the region bounded by the parabola
y 2 = 2x and the straight line x - y = 4.
Case Based MCQs
7. Find the area bounded by the curve y = log3x, X-axis
5. Suppose, f is an absolute function define from f : R ® R and the lines x = 1 and x = 3.
ïìx
such that f (x) = x - 3 , where x-3 = ïí3 -3, x³3 8. The marginal cost function of a product is given by
2 2 î2 2 2 MC = 2x . Find the total cost function and the
x2 + 900
- x, x<3
2 average cost function, if the fixed cost is ` 800.

On the basis of the above information, solve the 9. The demand function for a commodity is given by
p = 20 - 2x - x2. Find the consumer’s surplus when
following questions. p0 = 5.

(i) Identify the graph of the function f(x) = x - 3 10. The supply function for a commodity is p = 13 + x.
2
Determine producer’s surplus, if 22 units of the good
Y are sold.

f(x)=|x– 3
2

(a) X¢ X Long Answer Type Questions

(3/2,0) 11. Find the area of the region bounded by the curve
y = | x + 1 | + 1, x = - 3, x = 3 and y = 0.
Y¢
Y 12. The marginal cost MC of a product is given to be a
constant multiple of number of units (x) produced. Find
(b) X¢ X the total cost and the average cost function, if the
fixed cost is ` 2000 and the cost of producing 50 units
f(x)=|x– 3 is ` 3500.
2

Y¢ Answers

f(x)=|x– 3
2
Y 1. (b) 2. (b) 3. (b) 4. (c) 5. (i) (a) (ii) (c) (iii) (c) (iv) (b) (v) (a)

6. 18 sq units 7. [log 243 - 2]sq units 2 x 2 + 900 + 740
9. 27 10. 242 11. 16 sq units 8. xx

(c) X¢ X 12. 3x + 2000
5x

Y¢ For Detailed Solutions
Scan the code
(d) None of these
(ii) The area between the curve f(x) = x - 3 and the lines
2
x = 2 and x = 4 is

58 CBSE Term II Applied Mathematics XII

CHAPTER 04

Differential
Equations

In this Chapter...

l Ordinary Differential Equation
l Solution of a Differential Equation
l Applications of Differential Equation

An equation involving independent variable(s) dependent Particular Solution of a Differential Equation
variable and derivative(s) of dependent variable with respect
to independent variable(s), is called a differential equation. The solution obtained by giving particular values to the
arbitrary constants in general solution, is called a particular
Ordinary Differential Equation solution of the differential equation.
In other words, the solution free from arbitrary constants is
A differential equation involving derivative or derivatives of called particular solution.
the dependent variable with respect to only one
independent variable is called an ordinary differential Differential Equation in Variable Separable Form
equation.
Suppose a first order and first degree differential equation,
Order of a Differential Equation dy = f ( x, y), is such that f ( x, y) can be written as
dx g( x)× h( y)
The order of the highest order derivative of dependent variable ...(i)
with respect to independent variable occurring in the
differential equation, is called the order of differential equation. Then, expressed the given differential equation as ...(ii)
dy = h( y) × g( x)
Degree of a Differential Equation dx

The highest power (positive integral index) of the highest If h( y) ¹ 0, then separating the variables, Eq. (ii) can be written
order derivative involved in a differential equation, when it as 1 dy = g ( x) dx.
is written as a polynomial in derivatives, is called degree of a
differential equation. h( y)

Solution of a Differential Equation On integrating both sides, we get the required solution of
A function of the form y = f( x) + C, which satisfies given given differential equation.
differential equation, is called the solution of the differential
equation. Applications of Differential Equations
(Growth and Decay Model)
General Solution of a Differential Equation
Suppose the rate of change of the quantity of a substance or
A solution of a differential equation which contains as many population is proportional to the quantity present at any time.
arbitrary constants as the order of the differential equation, Then, we can say that substance or population is going through
is called the general solution or primitive solution of the either growth or decay, which depends upon the
differential equation. sign (+/ -) of the constant of proportionality.

CBSE Term II Applied Mathematics XII 59

Solved Examples

Example 1. Find the order and degree of the On integrating both sides, we get

æèçç d2y ö÷÷ø3 çæ dy ÷ö3 ò y + 1 dy = ò dx
dx 2 è dx ø y x
differential equation + + y = sin x.

Sol. Given, differential equation is Þ ò çæ1 + 1 ÷ö dy = ò dx
è y ø x

èççæ d2y ö÷ø÷ 3 çèæ dy ÷øö 3 Þ y + ln y = ln x + C
dx2 dx
+ + y = sin x Þ y = ln x - ln y + C

Q The highest order derivative is ddx2y2 . So, its order is 2. Example 5. Find the particular solution of differential

And the highest power of d2y is 3, so its degree is 3. equation dy = 4x - 2 at point (0, 1).
dx2 dx

Example 2. Find the order and degree of the Sol. Given, differential equation is
1 dy = 4x - 2
differential equation d3y + çæ = 1 dx
dy ö÷ 3
x3. Þ dy = ( 4x - 2) dx

dx 3 è dx ø On integrating both sides, we get

Sol. Given that, ò dy = ò( 4x - 2) dx
Þ
d3y çèæ 1 1 Þ y = 2x2 - 2x + C …(i)
dx3
+ dy÷ö 3 = x3 Put x = 0, y = 1 in Eq. (i), we get
dx ø 1 = 2( 0) - 2( 0) + C

1 1 d3y Þ C =1
dx3 \ From Eq. (i),
æçè dy öø÷ 3 = x3 -
dx y = 2x2 - 2x + 1

Taking cube on both sides, we get

dy = æç 1 - d3y ö÷ 3 Example 6. Find the general solution of
dx èç dx3 ø÷
x3 (x 4 + 2) dy - x 3dx = 0.

\ Order = 3 and degree = 3 Sol. Given differential equation is
( x4 + 2) dy - x3dx = 0
Example 3. Find the general solution of xdy - ydx = 0.
Þ ( x4 + 2) dy = x3dx

Sol. Given that, xdy - ydx = 0 Þ dy = x3 2
dx x4 +
Þ x dy = ydx
Þ dy = dx On separating the variables, we get

yx x3
x4 +
On integrating both sides, we get dy = dx
2
dy dx
ò y = ò x On integrating both sides, we get

Þ ln|y| = ln|x|+ ln C ò dy = ò x3 2 dx
x4 +
Þ ln|y| = ln|xC|
4x3
Þ y = xC Þ ò dy = 1 ò x4 + dx
4
Example 4. Find the general solution of 2

y dx - x (y + 1) dy = 0. é put x4 + 2 = t Þ 4x3dx = dt ù
êú
Sol. Given that, ydx - x( y + 1) dy = 0 òêëê\ 4x3dx dt 2|+ Cúúû
x4 + 2 = ò t = log|t|+ C = log|x4 +
Þ x( y + 1) dy = ydx
Þ y + 1 dy = dx \ y = 1 log|x4 + 2|+ C
4
yx

60 CBSE Term II Applied Mathematics XII

Chapter
Practice

PART 1 7. The number of arbitrary constants in the general
Objective Questions solution of a differential equation of fourth order is

(a) 0 (b) 2
(c) 3 (d) 4

l Multiple Choice Questions 8. The number of arbitrary constants in the particular

solution of a differential equation of third order is

1. The order and degree of a differential equation (a) 3 (b) 2 (c) 1 (d) 0
2
xçæèç d3 y ÷ø÷ö + æç dy ö÷4 9. The general solution of dy = 2x ex2 - y is
dx è dx ø + y2 = e-x is dx
3

(a) 2, 3 (b) 3, 2 (a) ex2 - y = C (b) e-y + ex2 = C
(c) 2, 2 (d) 3, 3
(c) ey = ex2 + C (d) ex2 + y = C

2. The sum of order and degree of differential 10. The general solution of dy - x = 0 is
÷øö÷3 dx x 2 + 1
equation 1 + æç dy ö÷4 = 7 æççè d2 y is
è dx ø dx (a) 3 log|x2 + 1| + C (b) 1 log|x2 + 1| + C
2 2 2

(a) 5 (b) 4 (c) 5 log|x2 + 1| + C (d) log|x2 + 1| + C
(c) 3 (d) 2 2

3. The order and degree of the differential equation 11. y = 2e2x - e-x is the solution of the differential
d 2y + çæ dy ö÷1/4 + x1/5 = 0 respectively, are
dx 2 è dx ø equation (b) y¢¢ - y¢ + 2y = 0
(d) y¢¢ - y¢ - 2y = 0
(a) 2 and 4 (b) 2 and 2 (a) y¢¢ + y¢ + 2y = 0
(c) 2 and 3 (d) 3 and 3 (c) y¢¢ + y¢ = 0

12. The equation of the curve passing through the point

4. The degree of the differential equation (1, 1), whose differential equation is
x dy = (2x 2 + 1) dx (x ¹ 0) is
æççè d2y ÷öø÷2 + æç dy ö÷2 = x sin çæ dy ÷ö is
dx 2 è dx ø è dx ø (a) x2 = y + log| x| (b) y = x2 + log| x|

(a) 1 (b) 2 (c) y2 = x + log| x| (d) y = x + log| x|

(c) 3 (d) not defined 13. The number of solutions of dy = y + 1,
dx x - 1
ççæè 1 d2y öø÷÷
5. If log e + dx 2 = x, then the order and degree of when y(1) = 2 is

the given differential equation are (a) none (b) one
(c) two (d) infinite
(a) 1, 1 (b) 1, 2
14. The particular solution of differential equation
(c) 2, 2 (d) 2, 1
dy = - 3x 2y at point (0, 1) is
6. The order and degree of the differential equation dx
y¢ ¢ ¢ + y3 + 3ey¢ = 0 are
(a) logy = x3 (b) logy = - x3
(a) 1, 3 (b) 3,1
(c) 3, 2 (d) 3, not defined (c) logy = 3x2 (d) None of these

CBSE Term II Applied Mathematics XII 61

15. Suppose a initial population is p0 grows at the rate (ii) The degree of the differential equation
y = 1 + æç dy ÷ö2 is
of 8% per year. After 3 yr, the population becomes. è dx ø

66

(a) 2P0e 25 (b) P0e 25

6 -6 (a) 2 (b) 3
(c) 1 (d) does not exist
(c) 3P0e 25 (d) P0e 25

16. A population grows at the rate of 5% per year. The (iii) The degree of the differential equation

time taken to triple the population is ççèæ d2 y ö÷÷ø 2 + çæ dy ö÷3 = sinçæ dy ÷ö is
dx è dx ø è dx ø
(a) 20log5 (b) 20log3 2

(c) 10log3 (d) 15 log3

17. In a bank, principal increases continuously at the (a) 0 (b) 1
(c) 2 (d) not determined
rate of 8% per year. The number of years in which
(iv) The sum of the order and degree of a differential
` 2000 double itself, is equation 4 d 2 y = 3 dy is
dx2 dx
(a) 25 log2 yr (b) 21 log3 yr
2 2

(c) 25 log2 yr (d) None of these (a) 4 (b) 2

18. The decay rate of radium at any time is proportional (c) 3 (d) 6

to its mass at that time. If the mass will be one third (v) The solution of the differential equation dy = x2 y 2
dx
of its initial mass, then time taken is (where k is

constant of proportionality) is

(a) - klog4 (b) klog3 (a) 3x3 + y + c = 0 (b) x3 + 1 + c = 0
3y
(c) log3 (d) 1 log3
k
(d) x3 - 1 + c = 0
19. If the interest is compounded continuously at 10% (c) 3x3 - y + c = 0 3y
per annum, then the number of years in which the
worth ` 500 will become ` 1500, is 22. A Veterinary doctor was examining a sick cat
(given log 3 = 1.0986) brought by a pet lover. When it was brought to the
hospital, it was already dead. The pet lover wanted
(a) 10.560 yr to find its time of death. He took the temperature of

(b) 10.986 yr the Cat at 11:30 pm which was 94.6°F. He took the
temperature again after 1 h; the temperature was
(c) 10.870 yr lower than the first observation. It was 93.4°F. The
room in which the Cat was put is always at 70°F.
(d) None of the above The normal temperature of the cat is taken as
98.6°F when it was alive.
20. The volume of a spherical balloon being inflated

changes at a constant rate. If initially its radius is

5 units and after 6 s its radius becomes 8 units, then

the value of constant is The doctor estimated the time of death using

(a) 265 p (b) 260 p Newton law of cooling which is governed by the
differential equation: dT µ (T - 70), where 70°F is
(c) 258p (d) 270p
dt
l Case Based MCQs
the room temperature and T is the temperature of
21. The order of a differential equation is the order of
the highest order derivative appearing in the the object at time t.
equation and the power of the highest order
derivative is a degree. Substituting the two different observations of T and

On the basis of the above information, solve the t made, in the solution of the differential equation
following questions. dT = k(T - 70) where k is a constant of
dt
(i) The order of the differential equation
proportionality, time of death is calculated.

çèæç d3 y ÷øö÷ 3 5æççè d2 y ÷øö÷5 Answer the following questions using the above
dx dx information.
3 + 2 - 8y = 0 is

(a) 4 (b) 3 (i) State the degree of the given differential
(c) 2 (d) 1 equation.

(a) 1 (b) 2 (c) 3 (d) 4

62 CBSE Term II Applied Mathematics XII

(ii) If the temperature was measured 2 h after PART 2
Subjective Questions
11:30 pm, will the time of death change.

(a) Yes (b) No

(c) Can’t say (d) None of these

(iii) The solution of the differential equation l Short Answer Type Questions
dT = k(T - 70) is given by,
dt 1. Write the sum of the order and degree of the

(a) log|T - 70| = kt + C differential equation d íîïïìçèæ dy ÷ö3 ýïü = 0.
(b) log|T - 70| = log| kt | + C dx dx ø ïþ
(c) T - 70 = kt + C
(d) T - 70 = kt + C 2. Find the order and degree of the differential equation

(iv) If t = 0 when T is 72, then the value of C is x3 + y = log ½½ dy + 3½½- log 4 èæçç d2y + 2 dy + 4 ÷ö÷ø.
2½ dx ½ dx 2 dx
(a) - 2 (b) 0 (c) 2 (d) log 2

(v) If C = 10, T = 90 and t = 5, then the value of 3. Find the general solution of y log y dx - x dy = 0.

constant K is

(a) log20 -10 (b) log20 + 10 4. Show that the function xy = log y + C is a solution of
5 5
the differential equation y¢ = y2 , ( ¹ 1).
(c) log20 - 10 (d) None of these 1 - xy xy

23. It is known that, if the interest is compounded 5. Show that y = Aemx + Benx is a solution of the
continuously, then the principal changes at the rate differential equation d 2y - (m + n) dy + mny = 0.
equal to the product of the rate of bank interest per dx 2 dx
annum. Consider P be the principal amount at any
time t with rate of interest r and P0 be the initial 6. Verify that y = em sin -1 x is a solution of the
principal amount. differential equation (1 - x 2 ) d 2y - x dy - m 2y = 0.
dx 2 dx
On the basis of the above information, solve the
following questions. 7. Verify that the function x 2 = 2y2 log y is a solution
of the differential equation (x 2 + y2 ) dy - xy = 0.
(i) From the given statement, the principal equation dx

in terms of initial principal amount is 8. Verify that y = A cos x - B sin x is a solution of the
differential equation d 2y + y = 0.
(a) logèçæç 2P ÷ö÷ø = rt (b) logççèæ P øö÷÷ = rt dx 2
P0 100 P0 100
9. Verify that the function y = e-3x is a solution of the
(c) logçèæ P0 ÷öø = rt (d) logæèçç P øö÷÷ = rt differential equation d 2y + dy - 6y = 0.
P 100 P0 50 dx 2 dx

(ii) If the interest is compounded continuously at 5%

per annum, in how many years will ` 100 double
itself? (given log 2 = 0.6931)

(a) 13.700 yr (b) 14.520 yr

(c) 13.8620 yr (d) None of these

(iii) The rate of interest at which ` 100 double itself in 10. Find the general solution of the differential
10 yr is (given, log e 2 = 0.6931) equation dy =1 - x + y - xy.
dx
(a) 6.725% (b) 6.5%

(c) 6.931% (d) None of these

(iv) The rate of interest at which ` 100 triple itself in 11. Solve the differential equation
5 yr is (given, log e 3 = 1.0986) dy = (1 + x 2 ) (1 + y2 ).
dx
(a) 22.365% (b) 21.972%
12. Verify that the function y = a2 - x 2 , x Î(- a, a) is a
(c) 21.820% (d) None of these solution of differential equation x + y dy = 0 (y ¹ 0).
dx
(v) If rate of interest is 5% and amount is ` 1000, then
after 10 yr, the worth of ` 1000 will be 13. Find the general solution of logæç dy ÷ö = 3x + 4y.
(given, e0.5 = 1.648) è dx ø

(a) ` 1700 (b) ` 1648
(c) ` 1800 (d) ` 1850

CBSE Term II Applied Mathematics XII 63

14. Find the general solution of the differential 27. Find a particular solution of the differential
equation (x + 2) dy = x 2 + 5x - 3, x ¹ - 2. equation (x - y) (dx + dy) = dx - dy, given that
dx y = - 1, when x = 0.

15. Solve the differential equation dy = ex - y + x 2e- y . 28. At any point (x, y) of a curve, the slope of the
dx tangent is twice the slope of the line segment
joining the point of contact to the point (–4, – 3).
16. Solve the differential equation dy - x = 0. Find the equation of the curve given that it passes
dx x 2 + 1 through (–2, 1).

17. Find the equation of a curve passing through the 29. It is given that helium decomposes at a rate
point (-2, 3) having the slope of tangent to the curve proportional to the amount present. If q% of the
at any point as 2x. original amount of helium disappears in n years,
y2 what percentage of it will remain after 2n years?

18. Assume that a spherical rain drop evaporates at a 30. Experiments show that radium disintegrates at a
rate proportional to its surface area. If its radius
originally is 5 mm and 2 h later has been reduced to rate proportional to the amount of radium present
3 mm. Find an expression for the radius of the rain
drop at any time. at the moment. Its half-life is 1590 yr. What

19. A population grows at the rate of 10% per year. percentage will disappear in one year?
How long does it take for the population to double?
Use differential equation for it. é - log 2 = 0.9996 ù
êUse ú
20. Suppose it is given that the rate at which some e 1590
bacteria multiply is proportional to the êë ûú
instantaneous number present. If the original
number of bacteria triples in three hours. In how l Case Based Questions
many hours will it be four times?
31. Suppose a function f (x, y) can be expressed as the
21. If the interest is compounded continuously at 6%
per annum, how much worth ` 1000 will be 10 product of two functions x and y
years? (given, e0.6 = 1.822 ) i.e. F(x, y) = g(x) × g(y), such that dy = F(x, y)

l Long Answer Type Questions dx

22. Solve (x - 1 ) dy = 2x 3y. = g(x) × g(y), g(y) ¹ 0
dx
In which, we separate the variables and integrate it,
23. Find the particular solution of the differential to get the solution of differential equation.
equation dy = 1 + x + y + xy, given that y = 0, when
dx On the basis of the above information, solve the
x = 1.
following questions.
24. Find the particular solution of the differential
equation (1 - y2 ) ( 1 + log|x|) dx + 2xy dy = 0, given (i) Find the general solution of dy = ex +y .
that y = 0 when x = 1. dx

25. Find the particular solution of the following (ii) Suppose that dy = x+1 and y = 2, when x = 1. Find
differential equation ex 1 - y2 dx + y dy = 0, given dx y+1
x
that y = 1, when x = 0. the positive value of x, when y = 5.

26. Find the particular solution of the differential (iii) Find the solution of dy = xy + y.
equation x (1 + y2 ) dx - y(1 + x 2 ) dy = 0, given that dx xy + x
y = 1, when x = 0.
32. Suppose the height and radius of cylindrical pot be
10 cm and ‘r’ cm. The volume of a cylindrical pot
changes at a constant rate. At initial time, the radius
of the pot is 3 cm.

(i) Write the solution of differential equation formed
by the given statement.

(ii) After 5 s, the radius of pot becomes 6 cm. Then,
find the value of constant.

(iii) Find the radius of the balloon after 1 s.

64 CBSE Term II Applied Mathematics XII

SOLUTIONS

Objective Questions 7. (d) We know that the number of constants in the general
solution of a differential equation of order n is equal to its
1. (b) Given differential equation is order. Therefore, the number of constants in the general
equation of fourth order differential equation is four.
èçæç d3y ÷÷øö 2 çæè dy ö÷ø 4
dx3 dx 8. (d) In the particular solution of a differential equation of
x + + y2 = e-x. third order, there is no arbitrary constant because in the
is ddx3y3. particular solution of any differential equation we remove all
Since, the highest order derivative So, its order is 3. the arbitrary constants by substituting some particular
values.
It is a polynomial equation in dy, d 3y and the highest power
dx dx3 9. (c) Given that, dy = 2x ex2 - y = 2x ex2 × e-y
d3y dx
of dx3 is 2. So, its degree is 2.
Þ ey dy = 2x ex2 dx
æèç dy ÷öø 4 7èæçç d2y ÷÷öø 3
dx dx2
2. (a) Given differential equation is 1 + = .

On integrating both sides, we get

Here, highest order derivative is d2 y2 , whose highest power ò òy dy = 2 x e x2 d x
dx
e

is 3. Put x2 = t in RHS integral, we get

So, order = 2 and degree = 3 2x dx = dt

Sum of order and degree = 2 + 3 = 5 ò ey dy = ò et dt

d2y èæç dy ÷öø1/ 4
dx2 dx
3. (a) Given that, + = - x1/ 5 Þ ey = et + C
Þ ey = ex2 + C
Þ d2y + çæ dy÷ö1/ 4 = - x1/ 5
dx2 è dx ø 10. (b) Given differential equation is

çèæ dy öø÷1/ 4 = - ççèæ x1/5 d2y ö÷÷ø dy - x = 0 Þ dy = x
dx dx2 dx x2 + dx +
Þ + 1 x2 1

On squaring both sides, we get On separating the variables, we get

èçæ dy öø÷1/ 2 = çæçè x1/5 d2 y øö÷÷ 2 dy = x dx
dx dx x2 +
+ 2 1

On integrating both sides, we get

Again, on squaring both sides, we have ò dy = ò x dx Þ ò dy = 1 ò 2x dx
x2 + 2 x2 + 1
dy = ççèæ x1/5 d2y øö÷÷ 4 1
dx dx2
+ putòé x2 +1=t Þ 2x dx = dt ù
x ú
order = 2, degree = 4 êëê\ê 2x dx =ò 1 dt = log|t| + C = log|x2 + 1| + C úûú
2+ t
4. (d) The degree of given differential equation is not defined 1
because when we expand sin èæç ddxy÷øö, we get an infinite series
\ y = 1 log|x2 + 1| + C
2

in the increasing powers of dy. Therefore, its degree is not which is the required solution.
dx
11. (d) Given, y = 2e2x - e-x
defined.
Þ y¢ = 4e2x + e-x
èçæç1 d2y ÷÷öø
5. (d) Given differential equation is loge + dx2 = x, which Þ y¢¢ = 8e2x - e-x

can be rewritten as 1 + d2y = ex. Þ y¢¢ = 4e2x + e-x + 4e2x - 2e-x
dx2
Þ y ¢¢ = y¢ + 2( 2e2x - e-x )

Clearly, the order of differential equation is 2 and its degree Þ y¢¢ = y¢ + 2y
is 1.
Þ y¢¢ - y¢ - 2y = 0
6. (d) Given differential equation is
y¢¢¢ + y3 + 3ey¢ = 0 12. (b) The given differential equation can be expressed as

The highest order derivative present in the given dy = 2x2 + 1 dx
differential equation is y¢¢¢, so its order is 3. x
In the given equation, ey¢ is not polynomial in y¢, so the
or dy = èçæ 2x + 1 ÷øö dx ...(i)
degree of the given differential equation is not defined. x

CBSE Term II Applied Mathematics XII 65

On integrating both sides of Eq. (i), we get when t = 3, then

ò dy = ò çèæ 2x + 1 ÷øö dx 2 ´ 3 = log P
x 25 P0

Þ y = x2 + log| x| + C ...(ii) Þ P 6 Þ 6

Eq. (ii) represents the family of solution curves of the given P0 = e 25 P = P0e 25
differential equation but we are interested in finding the
equation of a particular member of the family which passes 16. (b) Let P0 be the initial population and the population after
through the point (1, 1).
Therefore, substituting x = 1, y = 1 in Eq. (ii), we get t yr be P.

C=0 Then, dP = 5P
Put C = 0 in Eq. (ii), we get dt 100

y = x2 + log| x| Þ dP = P
dt 20

13. (b) Given that, dy = y +1 Þ dP = 1 dt
dx x -1 P 20

dy dx On integrating both sides, we get
y+1 x -1
Þ = logP = 1 t + C
20

On integrating both sides, we get Initially, at t = 0, P = P0

log( y + 1) = log( x - 1) - logC log P0 = 1 ´ 0 +C
20
Þ C( y + 1) = ( x - 1)

Þ C = x - 1 Þ C = logP0
y + 1 1t
\ log P = 20 + log P0
When x = 1 and y = 2, then C = 0
So, the required solution is x - 1 = 0. Þ t = 20 (logP - logP0 )

Hence, only one solution exist. Þ t = 20 logæççè P ÷öø÷
P0
14. (b) Given differential equation is
dy = - 3x2y Þ dy = - 3x2dx When, P = 3P0, then 20 logçæèç 3P0 ø÷÷ö
dx y P0
t = = 20 log 3
On integrating both sides, we get
logy = - 3 x3 + C = logy = - x3 + C 17. (a) Let P be the principal at any time t, then according to the
3 given condition, dP = 8% of P
dt
Put ( x, y) = ( 0, 1) Þ x = 0, y = 1
\ log(1) = ( 0) + C Þ dP = 8 P
Þ 0= 0+ CÞ C = 0 dt 100
\ logy = - x3
Þ dP = 2 P
15. (b) Let P be the population after t yr. dt 25

Then, dP = 8P Þ dP = 2 dt
dt 100 P 25

Þ dP = 2 dt On integrating both sides, we get
P 25
logP = 2 t + C …(i)
25
On integrating both sides, we get
Initially at t = 0, P = ` 2000
ò dP = 2 ò dt
P 25 \ log2000 = 2 ´ 0 + C
25
logP = 2 t + C …(i)
25 Þ C = log2000
Put C = log2000 in Eq. (i), we get
Initially, at t = 0, P = P0, then
2 ( 0)
log P0 = 25 + C Þ C = log P0 logP = 2 t + log2000
25

Put C = logP0 in Eq. (i), we get Þ t = 25 (logP - log2000) = 25 logèçæ 20P00øö÷
2 2 2
log P = 25 t + log P0

Þ 2 t = log P When, P = 2 ´ 2000 = ` 4000,
25 P0
t = 25 logæèç 24000000 ö÷ø = 25 log2 yr
2 2

66 CBSE Term II Applied Mathematics XII

18. (d) Let M be the mass of radium at any time t and M0 be the 20. (c) Let at any time t, the radius of the balloon be r units.
mass of radium initially at t = 0.
Then, according to the question,

According to the given condition, d çæè 4 pr 3ø÷ö = k, where k is some positive constant.
dM = - kM, dt 3
dt
On integrating both sides, we get

where k is constant of proportionality. ò d èçæ 4 pr 3 ø÷ö = ò kdt
Þ dM = - kdt 3

M 4 pr 3 = kt + C
3
On integrating both sides, we get

logM = - kt + C …(i) When, t = 0, r = 5; then

Initially, when t = 0, M = M0 4 p(5)3 = 0 + C Þ C = 500 p
Þ logM0 = - k( 0) + C 33
Þ C = logM0
Put C = logM0 in Eq. (i), we get \ 4 pr 3 = kt + 500 p
33
logM = - kt + logM0
Þ kt = (logM0 - logM) When, t = 6 s, r = 8,
Þ kt = log M0
4 p( 8)3 = k6 + 500 p
M 33

When the mass of radium is one-third i.e. M = 1 M0, then Þ k6 = 2048 p - 500 p
3 33

kt = log M0 Þ k = 1548p = 258p
6

1 M0 21. (i) (b) Given, differential equation is
3
æççè d3y ÷øö÷ 3 æèçç d2y ÷ö÷ø5
dx3 dx2
Þ t = 1 log3 + 5 - 8y = 0
k
The highest order derivative present in the given
19. (b) If P denotes the principal at any time t and the rate of differential equation is 3.
interest be r % per annum compounded continuously.
(ii) (a) Given, differential equation is
Then, according to the given condition,
dP = Pr æçè dy ø÷ö 2
dt 100 dx
y= 1 +
Þ dP = r dt
P 100 On squaring both sides, we get
y2 = 1 + æç dy÷ö 2
On integrating both sides, we get è dx ø

ò dP = ò r dt The highest order derivative present in the given
P 100 differential equation is 1, whose degree is 2.

logP = r t + C …(i) (iii) (d) Given, differential equation is
100
2
Let P0 be the initial principal i.e. at t = 0, P = P0 èççæ d2y ÷øö÷ çæè dy ö÷ø 3 sinèæç dy ö÷ø
dx2 dx dx
r ( 0) + =
100
Then, log P0 = + C

Þ C = logP0 This equation is not polynomial in its derivative. So,
degree cannot be determined.
Put the value of C in Eq. (i), we get
(iv) (c) Given differential equation is
rt
log P = 100 + log P0 4 d2y = 3 dy
dx2 dx
Þ rt = log P
100 P0 èççæ d2y ø÷÷ö1/ 4 èçæ 3dy ø÷ö1/ 2
dx2 dx
When, r = 10%, P0 = ` 500 and P = ` 1500, =

\ 10 ´ t = logèæç 1550000öø÷ Raising power 4 both sides, we get
100
d2y = çèæ 3 dy÷ö 2
Þ t = log3 dx2 dx ø
10
Here, the highest order derivative present in the given
Þ t = 10 ´1.0986 differential equation is 2, whose degree is 1.
\Required sum of order and degree = 2 + 1 i.e. 3.
= 10.986 yr

CBSE Term II Applied Mathematics XII 67

(v) (b) Given differential equation is Þ rt = logæèçç P ö÷ø÷
100 P0
dy = x2y2
dx logèæçç P ÷ø÷ö rt
P0 100
Þ dy = x 2 dx Þ =
y2
(ii) (c) Given, r = 5, P0 = ` 100 and P = ` 200 = 2P0
On integrating both sides, we get
logççèæ ÷÷øö
ò dy = ò x2 dx Q P = rt
y2 P0 100

Þ - 1 = x3 + C Þ log æççè 2P0 øö÷÷ = 5 ´t
y3 P0 100

Þ x3 + 1 + C = 0 Þ log( 2) = t
3y 20

22. Given differential equation is Þ t = 20log2

dT µ (T - 70) = 20 ´ 0.6931 = 13.8620 yr
dt
(iii) (c) Given, P0 = ` 100, P = ` 200 = 2P0 and t = 10 yr
dT
T - 70 = kdt Q logçæçè P ø÷÷ö = rt
P0 100

(i) (a) Degree of the given differential equation is 1. logæçèç 2P0 ÷ö÷ø r ´ 10
P0 100
(ii) (b) No; The time of death of Cat is 11:30. \ =

Time of death does not depend upon temperature Þ r = 10log2

\Time of death does not change. = 10 ´ 0.6931 = 6.931% p.a.

(iii) (a) We have, dT = k(T - 70) (iv) (b) Given, P0 = ` 100, P = ` 300 = 3P0 and t = 5 yr
dt

T dT = kdt Q logæççè P ÷öø÷ = rt
- 70 P0 100

On integration, we get \ logæççè 3P0 ÷öø÷ = r ´5
log|T - 70| = kt + C P0 100

(iv) (d) Now, log|T - 70|= kt + C, Þ r = 20log3

when t = 0, T = 72, we get = 20 ´ 1.0986 = 21.972%
log( 72 - 70) = 0 + C
(v) (b) Given, P0 = ` 1000, r = 5% and t = 10 yr
Þ C = log 2
(v) (a) Given, C = 10, T = 90, t = 5, Q logçæèç P øö÷÷ = rt
P0 100
Then, log|90 - 70| = k(5) + 10
Þ log20 = 5k + 10 Þ k = log20 - 10 \ logæèç P ø÷ö = 5 ´ 10
1000 100
5
logæèç 10P00÷øö
23. (i) (b) According to the given statement, Þ = 1
dP = Pr 2
dt 100
Þ P = e0.5 = P = 1000 ´ 1.648 = ` 1648
Þ dP = r dt 1000
P 100
Subjective Questions

On integrating both sides, we get d îìíïïçèæ dy ø÷ö 3 üïýþï
dx dx
ò dP = r ò dt 1. Given equation is = 0
P 100
dy÷ö 2
logP = r t + C …(i) Þ 3 æçè dx ø d2y = 0
100 dx2

When t = 0, P = P0; then Clearly, the highest order derivative occurring in the
differential equation is d2y/ dx2.
log P0 = r ( 0) + C …(ii)
100
So, its order is 2. Also, it is a polynomial equation in its
Þ C = logP0 derivative and highest power raised to d2y/ dx2 is 1, so its

Put the value of C in Eq. (i), we get degree is 1.
Hence, the sum of the order and degree of the above
log P = rt + log P0 differential equation is 2 + 1 = 3.
100

68 CBSE Term II Applied Mathematics XII

2. The given differential equation is Þ xy¢ + y = y¢
y
x3 + y = log2 ½ dy + 3 ½½- log4 çèçæ d2 y + 2 dy + 4öø÷÷
½ dx dx dx Þ xy× y¢ + y2 = y¢
2

Þ x3 + = log2 ½ dy + 3½½- log2 2èçæç d2y + 2dy + 4ø÷÷ö Þ y2 = y¢ - xy× y¢
½ dx dx2 dx
y Þ y2 = y¢(1 - xy)

= log2 ½ dy + 3 ½½ - 1 log 2 èçæç d2y + 2 dy + 4ø÷÷ö Þ y¢ = -y2xy, xy ¹ 1
½ dx 2 dx2 dx 1

é logab = 1 loga ù 5. Given, y = Aemx + Benx ...(i)
êëQ b úû
x x On differentiating both sides of Eq. (i) w.r.t. x, we get

= log2 ½ dy + 3 ½- log2 èæçç d2y + 2 dy + 4÷÷øö1/ 2 dy = A × emx × m + Benx × n ...(ii)
½ dx ½ dx2 dx dx

Again, on differentiating both sides of Eq. (ii) w.r.t. x,

[Q log xa = a log x] we get

½ dy + 3 ½ d2y = m× Ae mx ×m + n × Benx ×n
½ dx ½ dx2
Þ x3 + y = log2
ççæè d2y 4öø÷÷1/ 2 d2y
dx2 + 2 dy + Þ dx2 = m2 Aemx + n 2 Be nx
dx
d2y dy
é log a - log b = log aù Now, consider LHS = dx2 - (m + n) dx + mn y
ëêQ b úû
= m2 Aemx + n2Benx - ( m + n) ( m Aemx + nBenx )
½ dy ½
x3 ½ dx + 3 ½ + mn( Aemx + Benx )

Þ 2y + = èæçç d2y 4ø÷ö÷1/ 2 [Q logk a = b Þ kb = a] = m2 Aemx + n2Benx - m2 Aemx - mnBenx
dx2
+ 2 dy + - nm Aemx - n2Benx + mn Aemx + mnBenx
dx

Þ ( 2y + x3 )2 æèçç d2y + 2 dy + 4÷øö÷ = æèç dy + 3÷øö 2 = 0 = RHS
dx2 dx dx Thus, LHS = RHS
Hence, y = Aemx + Benx is a solution of given equation.

Hence, its order is 2 and degree is 1. 6. Given, y = em sin -1 x

3. Given differential equation is On differentiating w.r.t. x, we get
ylogydx - x dy = 0
dy = my Þ (1 - x2 ) æç dy÷ö 2 = m2y2
Þ dy = dx dx 1 - x2 è dx ø
ylogy x

On integrating both sides, we get Again, differentiating w.r.t. x, we get

ò dy = ò 1 dx x2 ) 2èçæ dy ÷øö d2y çæè dy ö÷ø 2 dy
y log y x dx dx2 dx dx
Þ (1 - + ( -2x) = m2 2y

Put logy = t Þ 1 dy = dt On dividing both sides by 2èæç dy ÷öø, we get
y dx

\ ò 1 dt = ò 1 dx (1 - x2 ) d2y - xdy - m2y = 0
t x dx2 dx

Þ logt = logx + logc 7. Given function is
x2 = 2y2 logy
Þ logc = log t
x

Þ c=t On differentiating w.r.t. x, we get
x
2x = 4ylogy dy + 2y2 1 dy
Þ c = logy dx y dx
x
Þ 2x = dy( 4ylogy + 2y)
Þ logy = cx dx

4. Given, xy = log y + C Þ x = dy( 2ylogy + y)
dx
On differentiating both sides w.r.t. x, we get
xy¢ + y×1 = 1 × y¢ + 0 [using product rule of derivative] Þ dy = y(1 + x log y)
y dx 2

CBSE Term II Applied Mathematics XII 69

8. Given, y = A cos x - B sin x …(i) 12. Given, y = a2 - x2

Here, order of the given differential equation is 2. So, we On differentiating w.r.t. x, we get
differentiate the Eq. (i) two times. dy = 1 ( - 2x)
dx 2 a2 - x2
On differentiating both sides of Eq. (i) w.r.t. x two times, we
Þ dy = -x
get dx a2 - x2
dy = - A sin x - B cos x
dx

Þ d2y = - A cos x + B sin x Þ a2 - x2 dy = - x
dx2 dx

Þ d2y = - ( A cos x - B sin x) Þ y dy + x = 0 Hence proved.
dx2 dx

Þ d2y = - y [from Eq. (i)] 13. Given differential equation is log çèæ dy ÷øö = 3x + 4y
dx2 dx
d2y
Þ dx2 + y = 0, Þ dy = e3x + 4y Þ dy = e3xe4y
dx dx

which is the given differential equation. On separating the variables, we get
Hence, y = A cos x - Bsin x is a solution of given equation.
1 dy = e3xdx
9. Given differential equation is e4y

d2y + dy - 6y = 0 ...(i) On integrating both sides, we get
dx2 dx ...(ii)
ò òe-4ydy = e3xdx Þ e -4 y = e 3x +C …(i)
and given function is y = e-3x. -4 3

Here, order of the given differential equation is 2. which is the required general solution of given differential
equation.
So, we differentiate Eq. (ii) two times.
14. Given differential equation is ( x + 2) dy = x2 + 5x - 3
On differentiating Eq. (ii), w.r.t. x two times, we get dx

dy = - 3e-3x and d2y = 9e -3x Þ dy = x2 + 5x - 3 dx [separating the variables]
dx dx2 (x + 2)

On putting the values of y, dy and d2y in LHS of Eq. (i), On integrating both sides, we get
dx dx2
x2
we get LHS = 9e-3x + ( -3e-3x ) - 6e-3x ò dy = ò ( + 5 x- 3 dx
x + 2)
= 9e-3x - 3e-3x - 6e-3x = 0 = RHS
Þ = ò ì + - 9 ü
Hence, the given function is a solution of given differential y í( x 3) ( x + 2) ýdx
equation. î þ

10. Given, differential equation is [dividing (x2 + 5x - 3) by ( x + 2)]

dy = 1 - x + y - xy = (1 - x) + y(1 - x) Þ y = x2 + 3x - 9log|x + 2| + C
dx 2

Þ dy = (1 - x) (1 + y) Þ dy = (1 - x) dx which is the required general solution.
dx 1+y
15. Given differential equation is

On integrating both sides, we get dy = ex - y + x2e-y
dx
log (1 + y) = - (1 - x)2 + C
2 Þ dy = e-y ( ex + x2 )
dx
11. Given differential equation is dy = (1 + x2 ) (1 + y2 )
dx Þ 1 dy = ( ex + x2 ) dx Þ eydy = ( ex + x2 )dx
e-y
On separating the variables, we get
[separating the variables]
ççèæ 1 ø÷÷ö
1 + y2 dy = (1 + x2 ) dx On integrating both sides, we get

On integrating both sides, we get ò eydy = ò ( ex + x2 ) dx

ò 1 dy = ò (1 + x2 ) dx Þ ey = ex + x3 + C
1 + y2 3

Þ tan-1 y = x + x3 + C ò òé xn x n+ 1 ù
3 n 1ú
êQ + 1
ë
ex dx = exand dx = , where n ¹ - û

where, C is a constant of integration. which is the required solution.

70 CBSE Term II Applied Mathematics XII

16. Given differential equation is ò dr = ò k dt

dy - x2 x 1 = 0 Þ r = kt + C …(i)
dx + When, t = 0, then r = 5

Þ dy = x 5 =0+C Þ C =5
dx x2 + 1 When, r = 3, then t = 2

On separating the variables, we get 3 = k2 + C
Þ 3 = k2 + 5
dy = x 1 dx Þ 2k = - 2
x2 + Þ k=-1
Put k = - 1 and c = 5 in Eq. (i), we get
On integrating both sides, we get

ò dy = ò x2 x 1 dx Þ ò dy = 1 ò 2x dx
+ 2 x2 + 1
r = - 1t + 5 Þ r = - t + 5,

òé put x2 +1 =t Þ 2x dx = dt ù where 0 £ t £ 5.
x dx ú
êëêê\ 2x =ò 1 ú 19. Let P0 be the initial population and P be the population after
2 +1 t dt = log|t| + C = log|x2 + 1| + C ûú t years.

\ y = 1 log|x2 + 1| + C Then, dP = 10P
2 dt 100

which is the required solution. Þ dP = P
dt 10
17. We know that, the slope of the tangent to a curve is given
by dy.
dx Þ dP = 1 dt
P 10
dy = 2x
So, dx y2 ...(i) On integrating both sides, we get

On separating the variables, Eq. (ii) can be written as logP = 1 t + C …(i)
10

y2dy = 2x dx ...(ii) Initially at t = 0, P = P0, then

On integrating both sides of Eq. (ii), we get log P0 = 1 ( 0) + CÞC = log P0
10
ò y2dy = ò 2x dx
Put C = logP0 in Eq. (i), we get
Þ y3 = x2 + C 1t
3 ...(iii) log P = 10 + log P0

On substituting x = - 2, y = 3 in Eq. (iii), we get Þ 1t = log P - log P0
C =5 10

On substituting the value of C in Eq. (iii), we get Þ 1 t = log P
10 P0
the equation of the required curve as
y3 = x2 + 5 When, P = 2P0, then
3 1 t = log 2P0
10 P0
1
Þ t = 10log2
or y = ( 3x2 + 15) 3

18. Let V be the volume, S be the surface area and r be the 20. Let the original count of bacteria be N0 and at any time, the
radius of rain drop at any time t. count of bacteria be N.
Then, V = 4 pr 3 and S = 4pr 2
3 Then, dN µ N
Þ dt
According to the given condition, dN = kN ,
(Rate of change of volume of the rain drop) µ (surface area) dt
\ dV µ S Þ dV = kS,
where k is a constant of proportionality
dt dt Þ dN = k dt

where k is a constant of proportionality. N
d æç 4 pr 3ö÷ = k( 4pr 2 )
dt è 3 ø On integrating both sides, we get

Þ 4 p 3r 2 dr = k( 4pr 2 ) ò 1 dN = kò dt + C
3 dt N

Þ dr = k Þ dr = k dt Þ logN = kt + C …(i)
dt

On integrating both sides, we get

CBSE Term II Applied Mathematics XII 71

Initially at t = 0, N = N0, therefore Þ 0.6 = logçæè P ÷øö
1000
logN0 = k( 0) + C

Þ C = logN0 Þ P = e0.6 Þ P = 1000 ´ 1.822 = ` 1822
1000
Put C = logN0 in Eq. (i) we get

logN = kt + logN0 Hence, in 10 yr, the worth of ` 1000 will become ` 1822.

Þ logèæçç N ö÷ø÷ = kt …(ii) 22. Given differential equation is ( x - 1) dy = 2x3y.
N0 dx

But it is given that the original number of bacteria triples On separating the variables, we get

in 3 h. dy = 2x3 dx
y (x - 1)
i.e. t = 3h, N = 3N0.

\ logçèæç 3N 0 ø÷÷ö = k3 Þ dy = 2 ççèæ x3 ÷÷öø dx
N0 y -
x 1
Þ log3 = 3k Þ k = 1 log3
3 Þ dy = é ( x 3 - 1) + 1 ù
y ê x -1 ú
2 ë û dx

Put k = 1 log3 in Eq. (i), we get = é ( x - 1) (x2 +x + 1) + 1 ù dx
3 2ê (x - 1) - ú
û
logèæçç N ÷öø÷ = 1 log3t ë (x 1)
N0 3
[Q a3 - b3 = ( a - b) ( a2 + ab + b2 )]
æççè log ÷øö÷
Þ t = 3 N Þ dy = 2 æççè x2 + x + 1 + 1 øö÷÷ dx
log 3 N0 y -
x 1

Suppose the count of bacteria becomes 4 times. i.e. 4N0 in On integrating both sides, we get
t1 h.
ò dy 2ò ççæè x2 1 ÷öø÷
\ t1 = 3 ççèæ log 4N 0 ÷ö÷ø y = + x + 1 + x - 1 dx
log 3 N0
èæçç x3 x2 - 1|÷ø÷ö
= 3 log4 h Þ log|y| = 2 3 + 2 + x + log| x + C,
log 3

21. Let P be the principal amount at any time t and the rate of which is the required solution.

interest be r % per annum compounded continuously. 23. Given, dy = 1 + x + y + x y
dx
Then, dP = Pr
dt 100 Þ dy = 1(1 + x) + y(1 + x)
dx
Þ dP = r dt
P 100 Þ dy = (1 + x)(1 + y) …(i)
dx …(ii)
On integrating both sides, we get

ò 1 dP = r ò dt On separating the variables, we get
P 100
1 dy = (1 + x) dx
Þ logP = rt + C (1 + y)
100 …(i)

Initially at t = 0, P = P0, then On integrating both sides of Eq. (ii), we get

r ´ ( 0) ò 1 1 y dy = ò(1 + x) dx
100 +
log P0 = + C

Þ C = logP0 Þ log|1 + y| = x + x2 + C …(iii)
2
Put C = logP0 in Eq. (i), we get
On substituting x = 1, y = 0 in Eq. (iii), we get
log P = rt + log P0 log|1 + 0| = 1 + 1 + C
100 2

Þ rt = log P - log P0 Þ C=-3 [Q log1 = 0]
100 2

Þ rt = logæççè P ÷øö÷ Now, substituting the value of C in Eq. (iii), we get
100 P0 log |1 + y| = x + x2 - 3
22
when r = 6, P0 = ` 1000 and t = 10,
\ 6 ´ 10 = logæç P ÷ö which is the required particular solution of differential
equation.
100 è 1000ø

72 CBSE Term II Applied Mathematics XII

24. Given differential equation is On integrating both sides, we get
(1 - y2 )(1 + log|x|) dx + 2xy dy = 0.
1 ò 2x dx - 1 ò 2y dy = 0
On separating the variables, we get 2 1 + x2 2 (1 + y2 )

(1 + log|x|) dx + 2y dy = 0 Þ 1 log(1 + x2 ) - 1 log(1 + y2 ) = C
x - y2 22
1

On integrating, we get Þ logçæçè 1 + x2 ö÷ø÷ =C …(i)
1 + y2
ò æèç 1 + logx|x|öø÷ dx + ò1 2y dy = 0
x - y2
Put x = 0, y = 1, then
Þ log|x| + (log|x|)2 - log|1 - y2| = log C
2 ...(i) logççèæ 1 +0 øö÷÷ =C Þ logèçæ 1 ø÷ö =C
1 + 12 2
Also, given y = 0 and x = 1
\ log 1 + (log 1)2 - log|1 - 0| = log C \ From Eq. (i), we get

2 logççèæ 1 + x2 ÷ø÷ö = logèçæ 1 öø÷
Þ 0 + 0 - 0 = log C 1 + y2 2
Þ log C = 0
Put log C = 0 in Eq. (i), we get Þ 1 + x2 = 1
1 + y2 2
log|x| + (log|x|)2 - log|1 - y2| = 0
2 Þ 2 + 2x2 = 1 + y2

25. Given differential equation is Þ y2 - 2x2 - 1 = 0

ex 1 - y2 dx + y dy = 0 27. Given differential equation is
x
( x - y) ( dx + dy) = dx - dy
Þ ex 1 - y2 dx = -y dy
x Þ dx + dy = dx - dy
x - y

On separating the variables, we get On integrating both sides, we get
-y dy = xexdx
ò( dx + dy) = ò dx - dy …(i)
1 - y2 x -y

On integrating both sides, we get Put x - y = t, then
dx - dy = dt
ò ò-y dy = x exdx
From Eq. (i), we get
1 - y2

On putting 1 - y2 = t, then -ydy = dt in LHS, we get x+y=ò dt = log|t|+ C
2 t

ò 1 dt = ò xexdx Þ x + y = log| x - y| + C [ put t = x - y] ...(ii)
2
t It is given that, when x = 0, then y = -1
\ 0 + ( -1) = log| 0 + 1| + C
Þ 1[2 t ] = xò exdx - ò é d ( x)ò exdx ù dx Þ C =-1
2 êë dx ûú
[Q log1 = 0]
òÞ
1 - y2 = xex - exdx [ put t = 1 - y2 ] On substituting the value of C in Eq. (ii), we get the
required particular solution
Þ 1 - y2 = xex - ex + C …(i)
x + y = log| x - y| - 1
On putting y = 1 and x = 0 in Eq. (i), we get Þ log| x - y| = x + y + 1
1 - 1 = 0 - e0 + C
[Q e0 = 1] 28. It is given that ( x, y) is the point of contact of tangent to the
Þ C =1
curve.
On substituting the value of C in Eq. (i), we get
The slope of the line segment joining the points
1 - y2 = xex - ex + 1
( x2, y2 ) º ( x, y) and ( x1, y1) º (– 4, – 3)
which is the required particular solution of given differential
equation. = y – (–3) = y+ 3 é of tangent = y2 – y1 ù
x – (–4) x+ 4 êëQ slope x2 – x1 ûú
26. Given differential equation is
x(1 + y2 ) dx - y(1 + x2 )dy = 0 According to the question, slope of tangent is twice the

It can be rewritten as slope of the line segment.

x dx - y dy = 0 \ dy = 2 çèæ y + 3 ÷öø
+ x2 + y2 dx x + 4
1 (1 )

CBSE Term II Applied Mathematics XII 73

Now, separating the variables, we get æç ÷ö
èçæ 4 ÷øö logç ÷
dy = 2 dx Þ k = 1 1
y+ 3 + n çèç 1
x - q ÷÷ø
100
On integrating both sides, we get

ò dy = ò èçæ 2 4ø÷ö dx Put the value of k in Eq. (ii), we get
y+ 3 +
x çæ ÷ö logçèæ ö÷ø
logç ÷
Þ log|y + 3|= 2log|x + 4| + log|C| t 1 = A0 …(iii)
n èçç 1 A
Þ log|y + 3|= log|x + 4|2 + log|C| - q ø÷÷
100
Þ log||xy++43||2 = log|C| é m ù
êëQ log m – log n = log n ûú Let A be the amount of helium available after 2n years.

( y + 3) Put t = 2n in Eq. (iii), we get
( x + 4)2
Þ =C …(i) çæ ö÷
logç ÷
2n 1 = logçèæ A0 øö÷
n çèç 1 A
Since, the curve passes through the point (–2, 1), therefore - q ÷ø÷
100
we have

( 1 + 3) =C Þ C =1 æç ÷ö 2
(–2 + 4)2 logç ÷
Þ 1 = logæèç A0 ÷øö
Þ çèç 1 çæ A
On putting C = 1 in Eq. (i), we get - q ÷÷ø
100 =ç 1
(y+ 3) =1 ÷ö 2
( x + 4)2 ÷
A0
Þ y + 3 = ( x + 4)2
A ççè 1 - q ø÷÷
which is the required equation of curve. 100

29. Let A0 be the original amount of helium and A be the Þ A ´ 100 = æç1 - q ö÷ 2 ´ 100
amount of helium at any time t. Then, the rate of A0 è 100ø
decomposition of helium is dA.
dt [multiply both sides by 100]

According to the given condition, Hence, required percentage is çèæ10 - q ÷øö 2
dA µ A 10
dt .

Þ dA = - kA, 30. Let A denote the amount of the radium substance present at
dt any instant t and let A0 be the initial amount of the
substance.
where k is a positive constant. Then, dA µ A Þ dA = - lA
Þ dA = - k dt dt dt
Where, l is a positive constant of proportionality and we
A take ‘-’ sign because A decreases with the increase in t.
Þ dA = - ldt
On integrating both sides, we get …(i) A
…(ii)
log A = - kt + C On integrating both sides, we get

When t = 0, A = A0, then log A = - lt + C …(i)
log A0 = - k( 0) + C Þ C = log A0 …(ii)
Initially at t = 0, A = A0, then
Put C = log A0 in Eq. (i), we get
log A = - kt + log A0 log A0 = - l( 0) + C Þ C = log A0

Þ kt = logæç A0 ö÷ Put C = log A0 in Eq. (i), we get
è Aø
log A = - lt + log A0
But it is given that q% of the original amount of helium
disintegrates in ‘n’ years. Þ lt = log A0 - log A

i.e. at t = n, A = çæè A0 - q A0 ö÷ø Þ lt = logçèæ A0 ÷öø
100 A

= A0 æèç1 - 1q00÷öø When A = 1 A0 at t = 1590 , then
2
çæ ö÷
\From Eq. (ii), we get 1590l = logç A0 ÷
ø÷÷
æç ÷ö ççè 1 A0
logçç A0 ÷ 2
k( n) = èç ÷
èçæ1 q ÷öø ÷ø l = log( 2)
A0 - 100 1590

74 CBSE Term II Applied Mathematics XII

Put l = 1590log( 2) in Eq. (ii), we get Þ x = -2 ± 124
2
èæç log 2 ÷öø t = logèçæ A0 ÷øö
1590 A Þ x = 4.65, -6.65

Þ A0 log 2t Hence, positive value of x is 4.65.

A = e 1590 (iii) Given, dy = xy + y
dx xy + x
- log 2 t
Þ A = 1590 +
A0 e Þ dy = y( x + 1)
dx x( y 1)
In t = 1 yr, the amount of the radium substance present is
given by Þ çæ y + 1 ÷ö dy = ( x + 1) dx

A = A0 ´ 0.9996 é - log 2 = 0. 9996 ù è yø x
êQ 1590 ú
e
ëê ûú
On integrating both sides, we get

\ The amount that disintegrates in one year = A0 - A ò y + 1 dy = ò x + 1 dx
= A0 - 0.9996 A0 y x

= 0.0004 A0 Þ ò æç1 + 1 ÷ö dy = ò æèç1 + 1 öø÷ dx
è yø x
\ The percentage of the amount disintegrated in 1 year
Þ y + logy = x + logx + C
= 0.0004 A0 ´ 100 = 0.04 Þ logy - logx + y - x = C
A0 Þ log y + y - x = C

31. (i) We have, dy = ex + y x
dx
32. (i) According to the given statement,
Þ dy = exey d (volume of cylinder) = k,
dx dt

Þ 1 dy = exdx where k is some positive constant.
ey
d ( pr 2h) = k
Þ e-ydy = exdx dt

On integrating both sides, we get Þ d ( pr 2(10)) = k [Q h = 10 cm given]
- e-y = ex + C dt

Þ ex + e-y + C = 0 Þ 10p dr 2 = k dt

(ii) We have, dy = x + 1 On integrating both sides, we get
dx y + 1
10pò d( r 2 ) = kò dt
Þ ( y + 1) dy = ( x + 1)dx
Þ 10p( r 2 ) = kt + C …(i)
On integrating both sides, we get
Initially at t = 0, r = 3, then
ò( y + 1)dy = ò( x + 1)dx 10p( 3)2 = k( 0) + C

Þ y2 + y = x2 + x + C …(i) Þ C = 90p
22 Put C = 90p in Eq. (i), we get

When x = 1, then y = 2 10pr 2 = kt + 90p
\ 22 + 2 = (1)2 + 1 + C
(ii) When t = 5 s, r = 6 cm, then [Q10pr 2 = kt + 90p]
22 10p( 6)2 = k(5) + 90p
Þ 4= 3+C
Þ 5k = 360p - 90p
2 Þ 5k = 270p
Þ C=4- 3=5 Þ k = 54p
(iii) Q10pr 2 = 54pt + 90p
22
Put C = 5 in Eq. (i), we get Þ r 2 = 5.4t + 9

2 When t = 1,
y2 + y = x2 + x + 5 r 2 = 5.4(1) + 9
2 22
Þ r 2 = 5.4 + 9
When y = 5, then
25 + 5 = x2 + x + 5 Þ r = 3.8 cm
2 22

Þ x2 + 2x - 30 = 0

Chapter Test (a) log |50 - y | = kx + C
(b) - log |50 - y | = kx + C
(c) log |50 - y | = log | kx | + C
(d) 50 - y = kx + C

Multiple Choice Questions (iv) The value of c in the particular solution given that

y(0) = 0 and k = 0.049 is (b) log 1

1. The order of the differential equation (a) log 50 50

2x2 d2y -3 dy + y = 0 is (c) 50 (d) - 50
dx2 dx
(v) Which of the following solutions may be used to find
(a) 2 (b) 1 (c) 0 (d) not defined
the number of children who have been given the polio

2. The degree of the differential equation drops?
(a) y = 50 - ekx
y = x dy + a æçè dy öø÷ 2 (c) y = 50(1 - e- kx ) (b) y = 50 - ekx
dx dx (d) y = 50(ekx - 1)
1 + is

(a) 2 (b) 1 (c) 0 (d) not defined Short Answer Type Questions

3. The solution of dy - y = 1, y(0) = 1 is given by 7. Find the sum of order and degree of the differential
dx
çæ èæç dy ø÷ö 2 ÷ö 3 / 4 çæçè d2 y ÷öø÷ 1 / 3
èç dx ÷ø dx
(a) xy = - ex (b) y = 2ex - 1 (c) xy = - 1 (d) y = 2ex - 1 equation 1 + = 2 .

4. In a bank, principal increases continuously at the rate 8. Find the solution of the equation
of 5% per year. In how many years ` 1000 double itself? (2 y - 1) dx - (2x + 3) dy = 0.

(a) 2 (b) 20 9. Solve log y dy - x2 sin x dx = 0.

(c) 20 loge 2 (d) 2 loge 20 10. Solve 2 ( y + 3) - xy dy = 0, given that y (1) = - 2.
dx
5. Suppose the growth of a population is proportional to
11. The surface area of a balloon being inflated changes at
the number present (say P). And the population of a a rate proportional to time. If initially its radius is 2 units
colony double in 40 days. The number of days in which and after 2 s it is 4 units, find its radius after time t.

the population become triple is

(a) 40 æç log3 ÷ö yr (b) 20 çæ log3 ö÷ yr
è log2 ø è log2 ø

(c) 10 çæ log2 ÷ö yr (d) None of these Long Answer Type Questions
è log3 ø
12. We have very well known that, if the interest is
Case Based MCQs compounded continuously, the principal changes at
the rate equal to the product of the rate of bank
6. Polio drops are delivered to 50K children in a district. interest per annum.
The rate at which polio drops are given is directly
proportional to the number of children who have not (i) If the interest is compounded continuously at 5% per
been administered the drops. By the end of 2nd week annum, find the number of years in which the amount
half the children have been given the polio drops. How of ` 100 will be double of itself.
many will have been given the drops by the end of 3rd
week can be estimated using the solution to the (ii) At what interest rate will ` 100 double itself in
differential equation dy = k(50 - y ) where x denotes 10 years? (given, loge 2 = 0.6931).
dx
the number of weeks and y the number of children (iii) In after 10 years, find the worth of the amount of
who have been given the drops. ` 1000, when rate of interest is 5%. (given, e0.5 = 1.648).

Answer the following questions using the above 13. Find the equation of the curve passing through the
information.
point (1, 0), if the sxlyo2 p-+e1xo.f the tangent to the curve at
(i) State the order of the above given differential any point (x, y ) is
equation.
(a) 1 (b) 2 (c) 3 (d) 4 Answers

(ii) Find the method to solve the differential equation 1. (a) 2. (a) 3. (b) 4. (c) 5. (a) 8. 2x + 3 = k
dy = k(50 - y ). 6. (i) (a) (ii) (a) (iii) (b) (iv) (b) (v) (c) 7. 6 2y - 1
dx
(a) Variable separable method 9. y (log y - 1) = cos x(2 - x 2) + 2x sin x + C 10. x 2 ( y + 3)3 = e y + 2
(b) Solving Homogeneous differential equation
(c) Solving Linear differential equation 11. 3t 2 + 4 12. (i) 13.862 yr (ii) 6.931% p.a. (iii) ` 1648
(d) All of the above
13. ( y - 1)(x + 1) + 2x = 0
(iii) The solution of the differential equation
dy = k(50 - y ) is given by, For Detailed Solutions
dx Scan the code

76 CBSE Term II Applied Mathematics XII

CHAPTER 05

Inferential
Statistics

In this Chapter...

l Population and Sample
l Hypothesis Testing
l t-Test

Population (ii) Non-probability sampling When selection of objects
from the population is non-random, so object of the
In statistics, the term population refers to group of objects, population has not equal chance of selection.
creatures, experiments and observations etc. under study.
e.g. Height of all females aged below 40 yr in Delhi and all Simple Random Sampling
students of class-XIIth in India etc.
In simple random sampling each and every object of the
Types of Population population has an equal chance of being included in the
sample.
According to number of objects in population, it may be of
two types e.g. If we want to take sample of 20 persons out of a
population of 200, write the names of all the 200 persons on
(i) Finite population If number of objects in population is separate slips of paper, fold these slips, mix them thoroughly
countable, then it is finite population. and then make a blindfold selection of 20 slips.

(ii) Infinite population If number of objects in population is Systematic Sampling
uncountable, then it is infinite population.
Systematic sampling is relatively a simple technique and may
Sample be more efficient than simple random, provided the lists are
arranged wholly as random.
A finite subset (smaller group) of objects (individuals) in a
population is called a sample. Process of selecting samples e.g. If a complete list of 1500 students of a university is
from a population is called sampling and the number of available and, if we want to draw a sample of 500 students by
objects in a sample is called the sample size. using this method, select every kth item ( k = 3) from the list,
where ‘k’ refers to the sample interval. The first item between
Types of Sampling one and three will be selected at random.

Sampling are of two types Purpose of Sampling

(i) Probability Sampling When selection of objects from the Sampling is only a tool which helps us to know the
population is random, so objects of the population has an characteristics of the population by examining only a small
equal chance of selection. part of it.

CBSE Term II Applied Mathematics XII 77

Parameter and Statistics Then, sample mean x = Sxi
n
Parameter
Sample proportion (probability), p = x
The values (measurable characteristics) obtained from the n
study of population such as the population mean (m),
population variance ( s2 ), population standard deviation ( s) Sample standard deviation S = S( xi - x) 2
n -1
etc. are called parameters. Parameter is used to describe the
entire population. Interval Estimation

Statistics The interval estimation of a population parameter is given
by two numbers between which the parameter may be
The values (measurable characteristics) of a sample is called considered to lie.
statistics. Such as sample mean ( x), sample variance (S2 ),
sample standard deviation (S) etc. are called statistics. The sample distribution of sample mean is normally
distributed with m and standard deviation s .
Sample Errors
n
Sampling errors arise due to the fact that samples are of used
and to the particular method used in selecting the items from The sampling distribution of x can be transformed into the
the population. There are mainly two types of sampling standard normal distribution by
errors.
Z = x -m .
(i) Biased (ii) Unbiased s/ n

Sampling Distribution Interval Estimation of Population Proportion

If S1 , S2 , S3 , ......, Sn are values of statistics S(like mean, Consider p is sample proportion and p is population
variance, etc.) obtained from n independent random samples
proportion, then the interval estimation of population
of a definite size selected from a given population, then S1 ,
S2 , S3 , …, Sn form a sampling distribution of statistics. proportion is p = p ± margin of error

Then, Mean ( S ) = 1 n Si and margin of error = Z a/ 2 p(1 - p), where Za/2 is the Z
n n
i S value providing an area of a in the upper tail of the standard

=1

åand = 1 n - S)2 2
Var (S) n
(Si normal probability distribution.

i=1

Statistical Inference Hypothesis Testing

The process through which inference about the population Hypothesis is an assumption about the population
are drawn which is based on population parameter is called parameter to be tested based on sample information,
statistical inference. assumptions, which may or may not be true are called
statistical hypothesis.
There are mainly three types of statistical inferences
In hypothesis testing, hypothesis are normally referred to as
(i) Point Estimation
(i) H0 : null hypothesis
(ii) Interval Estimation
(ii) H1 : alternative hypothesis
(iii) Hypothesis testing
Here, null hypothesis asserts that there is no true difference
Point Estimation in sample statistics and population parameter under
consideration i.e. there is no difference between assumed and
It is used in sample data to calculate a single value of actual value of the parameter and the hypothesis that is
unknown population parameter. This single value of sample different from the null hypothesis is alternative hypothesis.
data is known as point of estimation. The rejection of the null hypothesis ( H0 ) indicates that there
e.g. Sample mean x is a point estimate of population mean m. is enough evidence to show that alternate hypothesis ( H1 ) is
sample standard deviation S, is a point estimate of population true.
standard deviation s.
Not rejection of the null hypothesis ( H0 ) indicates that there
Let us consider, a sample consists of n objects and the is not enough evidence to show that alternate hypothesis ( H1 )
population consists of N objects and every possible samples is true.
of n objects are equally likely to occur.

78 CBSE Term II Applied Mathematics XII

Hypothesis Test for Population Mean There are two types of errors that can be made in the
hypothesis testing.
There are three types of possibilities

H0 :m ³m0, H0 :m £m0, H0 :m =m0 Decision H0 true H0 False
Accept H0 Type II error
H1 : m < m 0, H1 : m > m 0, H1 : m ¹ m 0 Correct
conclusion
mWahnedreitmm0adyebneotneottheed hypothesized value of population mean
that the equality sign ( =) should always Reject H0 Type I error Correct conclusion

appear in the null hypothesis. In most cases of hypothesis testing Type-I error can be
controlled.
Level of Significance
One-Tailed and Two-Tailed Test
The confidence with which an experimenter rejects or retains
null hypothesis depends on the significance level adopted Mainly, there are two types of problems of test of hypothesis
and it is usually denoted by “a”. It is generally specified
before any sample is drawn. (i) One-tailed test
For hypothesis test about a population mean ( s), a test
statistic is standard normal random variable Z and value of (ii) Two-tailed tests
test statistic Z is given by
One-tailed Test
Z = x -m0,
s/ n A one-tailed test is a statistical hypothesis test set up to show
that the sample mean would be higher or lower than the
where x : sample mean population mean, but not both i.e. hypothesis about
m 0 : hypothesized value of population mean population mean is rejected only for value of falling into one
s : population standard deviation of the tails of the sampling distribution.

n : number of sample points Right-tailed test
e.g. If we take s = 5% level of significance, then there are
about 5 chance out of 100 that we would reject the null Acceptance Rejection
hypothesis when it should be accepted means that 95% Region Region
confident that we have made the right decision.
If a = 0.5, the test result is said to be “significant” and when Za
the null hypothesis is rejected at a = 0.01, the test result is H0 : m ³ m0
said to be highly significant. Ha : m < m0

Determination of Test Statistics Rejection Left-tailed test
Region
To determine a suitable test statistics, generally test statistics Acceptance
will encounter in the following form – Za Region
(1 – a)
Test statistics
= Sample statistics - Hypothesized population parameter
Standard error of the sample statistics

Critical Value and Critical Region Two-Tailed Test

Critical value is a cut off value that define the boundaries In two-tailed test the critical area of a distribution is two
beyond which less than 5% of sample means can be obtained, sided and tests whether a sample is greater or less than a
if the null hypothesis is true sample means obtained beyond a range of values i.e. the hypothesis about the population mean
critical value will result in a decision to reject the null is rejected for values falling into either tail of the sampling
hypothesis. distribution.

The values of the test statistic which leads to a rejection of H0 Rejection Acceptance Rejection
and which lead to acceptance of H0 . The former is called the Region Region Region
critical region.
H0 : m = m0
Type I and Type II Errors H1 : m ¹ m0

If we reject a hypothesis when it should be accepted, then we
say that it is a Type I error and, if we accept a hypothesis
when it should be rejected, then we say that it is a Type II
error.

CBSE Term II Applied Mathematics XII 79

t-Test t-distribution

t-test is a statistical test that is used to compare the means of It is type of normal distribution used for smaller sample sizes,
two groups. It is often used in hypothesis testing to (say 30 or less) where the variance in the data is unknown.
determine whether two groups are different from one The t-distribution also known as student’s t-distribution is a
another. way of describing data that follow a bell curve when plotted
The t-test assumes, data are on a graph, with the maximum number of observation close to
the mean and fewer observation in the tails.
(i) independent
Standard normal
(ii) approximately normally distributed. distribution

(iii) have a similar amount of variance within each group t-distribution
being compared. (20-degree of freedom)

There are mainly three types of t-tests t-distibution
(10-degree of freedom)
(i) One Sample t-test If the groups comes from a single
population (e.g. measuring before and after an –4 –3 –2 –1 0 12 34
experimental treatment), perform one sample t-test.
Consider x1 , x2 , x3 , .... xn be random sample of size n from a
If there is one group being compared against a standard normal population with mean m and variance s2 .
value (e.g. comparing the acidity of a liquid with neutral
pH of 7), perform a one-sample t-test. Then, t-distribution is defined by the statistics,
t =x -m0
(ii) Two sample t-Test If the groups come from two different S/ n
populations (e.g. two different species or people from
two separate cities), perform a two-sample t-Test (or where, x : Sample mean
independent t-test). m 0 : Hypothesized value of population mean

(iii) Pair t-Test If the group come from a single population S: Sample standard deviation
(e.g. measuring before and after an experimental n : Sample size
treatment) perform a paired t-test.
Properties of t-distribution
Degrees of Freedom
(i) t-distribution range from - ¥ to ¥ just as does a normal
The maximum number of logically independent values, which distribution.
are values that have the freedom to vary, in the sample.
The number of degrees of freedom (ii) t-distribution with more degrees of freedom exhibits less
variability and more closely resembles the standard
= Total number of observation ( n) deviation.
- number of independent constraints imposed
on the observation ( k) (iii) Like the standard normal distribution, it is bell-shaped
and symmetrical around mean zero.
e.g. If S xi = x1 + x2 + x3 , has 3 degrees of freedom. Since, if
we assign values to any three of these we get value of the
fourth observation automatically determined.

80 CBSE Term II Applied Mathematics XII

Solved Examples

Example 1. A random sample of size 16 has 53 mean. Sol. Given, m 0 = 30, n = 50, x = 32.6 and s = 8

The sum of the squares of the deviations taken from \Test statistic, Z = x- m0
s/ n
mean is 150. Compute the value of the test
statistics. = 32.6 - 30

Sol. Given, n = 16, X = 53, m = 56 8 / 50

16 - x)2 = 2.6 ´ 50
and S ( = 150 8
xi
i =1

S2 = 1 16 - x)2 = 18.38 = 2.3
\ -1 S ( 8
n xi
i =1

= 1 ´ 150 = 10 Example 4. Suppose a student measuring the boiling
(16 - 1)
temperature of a certain liquid observes the reading
\ Test statistics, t = X -m (in degree celsius) 102.5, 101.7, 103.1, 100.9, 100.5
S/ n and 102.2 on 6 different samples of the liquid. If he
knows that the standard deviation for this
= 53 - 56 = - 3´4 = -12 = - 3.794 procedure is 1.2°C, what is the interval estimation
10 10 10 for the population mean at a 95% confidence level?

16 Sol. We have, n = 6 and s = 1.2

Example 2. Consider the following hypothesis test Now, Sample mean ( x)
= 102.5 + 101.7 + 103.1 + 100.9 + 100.5 + 102.2
H 0 : m = 15 6
H a : m ¹ 15

A sample of 50 provided a sample mean of 14.15, = 610.9 = 101.82
the population standard deviation is 3, if a = 0.05 6

then what is the p-value? Given, confidence level = 95%

Sol. Given, x = 14.15, n = 50, s = 3 and m 0 = 15 Þ (1 - a ) = 0.95

Now, Z = x -m0 = 14.15 - 15 Þ a = 1 - 0.95 = 0.05
s/ n 3/ 50 Þ a = 0.025

= ( - 0.85) ´ 50 2
3
\ Za/ 2 = Z0.025 s
n
= - 2.003 Now, margin of error = Za/ 2 .

\ Z=-2 = ´ 1.2
6
\ p-value = 2 (Area under the standard normal curve to the Z0. 025
left of Z)
= 1.96 ´ 1.2 = 0.96
= 2 ´ ( 0.0228) = 0.0456 2.45

\ p-value = 0.0456 \Interval estimation for the population mean,

Example 3. Consider the following hypothesis test m = x ± margin of error

H 0 : m £ 30; H a : m > 30 = 101.82 ± 0.96

A sample of 50 provides a sample mean of 32.6 and = (101.82 - 0.96, 101.82 + 0.96)
the population standard deviation is 8. Compute the
value of the test statistic. = (100.86, 102.78)

Thus, confidence interval = (100.86, 102.78)

CBSE Term II Applied Mathematics XII 81

Chapter
Practice

PART 1 9. A hypothesis which defines the population
Objective Questions
distribution is called

(a) statistical hypothesis (b) null hypothesis

(c) composite hypothesis (d) simple hypothesis

l Multiple Choice Questions 10. The probability of Type I error is referred as

1. The process of using sample data to estimate the (a) b (b) 1 - a
(c) a (d) 1 - b

values of unknown population parameters is called 11. The point estimate of the population mean from a

(a) estimation (b) population simple random sample 3, 6, 8, 10, 12, 15 is

(c) sampling (d) interval estimation (a) 6 (b) 9

2. The process of making estimates about the (c) 4 (d) 12

population parameter from a sample is called 12. In a statistical hypothesis test, then hypothesis is

statistical true but our test rejects it, it is

(a) decision (b) inference (a) Type II error (b) Type I error

(c) hypothesis (d) independence (c) Both Type I and II error (d) None of these

3. A member of the population is called 13. If 700 throws of six-faced die, odd points appeared

(a) data (b) element 400 times, the die is fair at 6% level of significance,

(c) family (d) group then hypothesis is

4. Random sampling is useful as it is (a) rejected (b) accept

(a) reasonably more accurate as compared to other methods (c) error (d) None of these
(b) economical in nature
(c) free from personal biases of the investigator 14. Consider the following hypothesis test
(d) All of the above
H 0 : m £ 20
5. A statement made about a population for testing H a : m > 20

purpose is called A sample of 40 provided a sample mean of 24.3 and

(a) testing statistics (b) level of significance population standard deviation is 5. The value of the

(c) statistics (d) hypothesis test statistics is

6. If the null hypothesis is false, then which of the (a) 4.24 (b) 5.63

following is accepted? (c) 6.84 (d) 2.36

(a) Alternate hypothesis (b) Null hypothesis 15. If a coin is tossed 20 times and the coin falls on

(c) Negative hypothesis (d) Positive hypothesis head after any toss, it is success. Suppose the

probability of success is 0.5, the probability that the

7. Population value is called number of success is less than or equal to 12 is

(a) variable (b) parameters (a) 0.542 (b) 0.749

(c) data (d) statistics (c) 0.8133 (d) 0.6431

8. Sampling which provides for a known non-zero 16. A 95% confidence interval for a population was

equal chance of selection is reported to be 152 to 162. If s = 15, then sample

(a) probability sampling (b) non-probability sampling size is

(c) snowball sampling (d) convenience sampling (a) 60 (b) 54 (c) 72 (d) 65

82 CBSE Term II Applied Mathematics XII

17. If the critical region is evenly distributed, then the PART 2
Subjective Questions
test is referred as?
l Short Answer Type Questions
(a) Zero tailed (b) One tailed
1. What is one of the distinct difference between a
(c) Two tailed (d) Three tailed
population parameter and a sample statistics?
18. The assumed hypothesis which is tested for
2. What is sampling Distribution?
rejection considering it to be true is called
3. What is systematic random sampling?
(a) true hypothesis (b) alternative hypothesis
4. Describe probability sampling.
(c) null hypothesis (d) simple hypothesis
5. What is the significance level?
19. Consider the following hypothesis test
6. How do you know, if a p-value is statistically
H 0 : m ³ 45
H a : m < 45 significant?
A sample of 36 provided a sample mean, x = 44 and
a sample standard deviation, S = 5.2. If a = 0.01, 7. Define t-distribution?

then 8. Give at least two important application of

(a) do not reject H0 (b) reject H0 t-distribution?
(c) t = 4.82 (d) t = 9.2
9. What is paired t-test?
20. Consider the following hypothesis test
10. Explain point estimation.
H 0 : p ³ 0.84
H a : p< 0.84 11. What is the two-sample t-test?

A sample of 400 provided a sample proportion of 12. State central limit theorem?

0.75, then the value of the test statistics 13. The following data : 6, 7, 9, 12, 16, 10 form a

(a) - 4.86 (b) - 2.51 (c) - 6.90 (d) - 5.42 random sample. What is the point estimate of
population standard deviation?
l Case Based MCQs
14. In a survey question for a sample of 500
21. A company producing electric bulbs may be
individuals, 180 persons gave response ‘Yes’ 230
interested in the average life of bulbs produced by gave response ‘No’ and 90 gave no response. What
them during a shift for this consider the following is the point estimate of the proportion in the
hypothesis test population who respond ‘No’?

H 0 : m £ 36 15. Consider the following hypothesis
H a : m > 36
H 0 : m = 295
A sample of 60 provided a sample mean of 38.6. H a : m ¹ 295
The population standard deviation is 7. A sample of 50 provided a sample mean of 297.6.
The population standard deviation is 12 and level of
On the basis of above information, answer the significance, a = 0.05. Check the hypothesis test
following questions. given above using interval estimation.

(i) The value of Sxi is (c) 2325 (d) 2340 16. Consider the following hypothesis test
(a) 2310 (b) 2316
H 0 : m £ 18
(ii) The value of the test statistics is H a : m > 18
A sample of 28 provided a sample mean, x = 20 and
(a) 2.88 (b) 3.10 (c) 3.05 (d) 2.67 a sample standard deviation, S = 5.68, use the
t-distribution table to compute a range for the
(iii) What is the p-value? p-value.

(a) 0.642 (b) 0.012

(c) 0.002 (d) 0.50

(iv) What is your conclusion, when a = 0.01?

(a) do not reject H0 (b) reject H0
(c) can not say (d) None of these

(v) What is the rejection rule using critical value?

(a) reject H0 (b) do not reject H0
(c) reject Ha (d) do not reject Ha

CBSE Term II Applied Mathematics XII 83

17. Consider the following hypothesis tests 22. Define test statistics and consider the prices of

H 0 : m = 22; shares (in `) of a company on the different days in a
H a : m ¹ 22 month were found to be : 66, 65, 69, 70, 69, 71, 70,
A sample of 65 provided a sample mean, x = 20 and 63, 64 and 68. Calculate value of t-statistics from
sample standard deviation, S = 6.4. Compute the given data whether the mean price of the shares in
value of the test statistic. the month of ` 65.

18. Consider the following hypothesis test 23. Consider a simple random sample of two household

H 0 : p ³ 0.55; H a : p< 0.55 from a population of five households having
A sample of 250 provided a sample proportion of monthly income (in `) as follows.
0.48. Compute the p-value.
Household 1 2 3 4 5

19. A drug manufacturer has installed a machine which Income (in `) 1560 1490 1660 1640 1550

fills automatically 5 gm of drug in each phial. A Enumerate all possible samples of size 2 and show
random sample of phials was taken and it was found that same mean gives an unbiased estimate of
to contain 5.02 gm on an average in a phial. The SD population mean.
of the sample was 0.002 gm. Test at 5% level of
significance, if the adjustment in the machine is in 24. Consider the following hypothesis test
order.
H 0 : p = 0.20
20. Consider the following hypothesis test H a : p ¹ 0.20
A sample of 400 provided a sample proportion
H0 :m1 - m2 = 0; Ha :m1 - m2 ¹ 0 p = 0.175
The following results are from independent sample
taken from two populations. (i) Compute the value of the test statistics

Sample 1 Sample 2 (ii) What is the p-value?
(iii) At a = 0.05, what is your conclusion?

n1 = 43 n2 = 48 l Case Based Questions
x1 = 16.2 x 2 = 14.5
S1 = 6.4 S2 = 9.6 25. Consider the following hypothesis test

Then, what is the value of test statistic? H 0 : m £ 16
H a : m > 16
l Long Answer Type Questions A sample of 30 provided mean x = 18 and a sample
standard deviation, S = 5.14.
21. Two types of batteries are tested for their length of
Based on the above information, answer the
life and the following results are obtained. following questions.

Battery Sample Size Mean Standard (i) Value of the test statistics.

ABC 100 (hours) deviation (hours) (ii) Range of the p-value.
XYZ 120
1000 10 (iii) Sum of the minimum value and maximum value
of p is.
1050 11
(iv) At a = 0.05, what is your conclusion.
Conclude that the two types of batteries are having
the same mean life? (v) Write the rejection rule using the critical value.