DIFFERENTIAL EQUATION AND AREA UNDER THE CURVE · 2020. 2. 1. · DIFFERENTIAL EQUATION & AREA...

DIFFERENTIAL EQUATIONAND

AREA UNDER THE CURVE

JEE (MAIN+ADVANCED)

DIFFERENTIAL EQUATIONAND


S.No Pages

1. Theory 01 – 13

2. Exercise-1 (Special DPP) 14 – 22

3. Exercise-2 22 – 25

4. Exercise-3 (Section-A) 26 – 32

[Previous years JEE-Advanced problems]

5. Exercise-3 (Section-B) 32 – 35

[Previous years JEE-Main problems]

6. Exercise-4 (Section-A) 36 – 41

[Previous years CBSE problems]

7. Exercise-4 (Section-B) 41 – 43

[Potential Problems for Board Preparations]

8. Exercise-5 (Rank Booster) 43 – 45

9. Answer Key 46 – 51

CONTENT

DIFFERENTIAL EQUATION & AREA UNDER THE CURVE

BANSAL CLASSES Private Ltd. ‘Bansal Tower’, A-10, Road No.-1, I.P.I.A., Kota-05 Page # 1

DIFFERENTIAL EQUATION

1. INTRODUCTION :

1.0 Anequationthat involves independentanddependentvariablesandat leastonederivativeof thedependentvariable w.r.t independent variable is called a differential equation.

For example : xdx

dy+ y log x = xex

xlog21

x , (x > 0);

4/16

2

2

dx

dyy

dx

yd

1.1 Adifferential equation is said to be ordinary, if the differential coefficients have reference to a singleindependent variable onlyand it is said to be Partial if there are two or more independent variables. Weareconcernedwith ordinarydifferential equationsonly.Whileanordinarydifferential equationcontainingtwo or more dependent variables with theirdifferential coefficients w.r.t. to a single independent variableis called a total differential equation.

eg. 0y2dx

dy3

dx

yd2

2

is an ordinarydifferential equation

0z

u

y

u

x

u

; yx

y

u

x

u 22

2

2

2

are partial differential equation.

1.2 Order and Degree of Differential Equation :

The order of a differential equation is the order of the highest differential coefficient occurring in it.The degree of a differential equation which is expressed or can be expressed as a polynomial in thederivatives is the degree of the highest order derivative occurring in it, after it has been expressedin a form free from radicals & fractions so far as derivatives are concerned, thus the differentialequation :

f(x , y)

p

m

m

xd

yd

+ (x , y)

q

1m

1m

xd

)y(d

+ ....... = 0 is order m & degree p .

2. FORMATION OF DIFFERENTIAL EQUATIONS :

Consider a familyof curvesf (x, y, c1, c2 , ……cn) = 0 …(i)

where c1, c2, ……, cn are n independent parameters.Equation (i) is known as an n parameter family of curves e.g. y = mx is 1-parameter familyof straightlines x2 + y2 + ax + by = 0 is a two-parameters family of circles.If we differentiate equation (i) n times w.r.t x, we will get n more relations between x, y, c1, c2, …… cnand derivatives of y with respect to x. By eliminating c1, c2, ……, cn from these n relations andequation (i), we get a differential equation.Clearlyorder of this differential equation will be n, i.e., equal to the number of independent parametersin the familyof curves.



3. LENGTH OF TANGENT, NORMAL, SUB-TANGENT, SUB-NORMAL:

y

xT

y = f(x)

P(x, y)

M NO

(i) Length of Tangent :

PT is defined as length of the tangent.In PMT, PT = | y cosec |

)cot1(y 2

2

dy

dx1y

Length of tangent =

2

dy

dx1y

(ii) Length of Normal :PN is defined as length of the normal.In PMN, PN = |y cosec (90º – )|

= | y sec |

2

dx

dy1y

Length of normal =

2

dx

dy1y

(iii) Length of Sub-tangent :TM is defined as sub-tangent.

In PTM, TM = |y cot | =dy

dxy

tan

y

Length of sub-tangent =dy

dxy

(iv) Length of Sub-normal :MN is defined as sub-normal.

In PMN, MN = |y cot (90º – )| = |y tan | =dx

dyy

Length of sub-normal =dx

dyy



4. SOLUTION OF A DIFFERENTIAL EQUATION :

Elementary types of first order & first degree differential equations.

Elementry Type of first order and first degree

Variable Separable Homogeneous Linear

T-1 T-2normal

Reducibleto Homogeneous

Reducibleto linear

(Bernouli's D.E.))zbyax(fdx

dy

4.1 Variables Separable :

If the differential equation can be expressed as; f(x)dx + g(y)dy = 0 then this is said to bevariable -separable type.

A general solution of this is given by f(x) dx + g(y) dy = c; where c is the arbitrary constant.

4.2 Differential Equation Reducible to the Separable Variable Type:

xd

yd= f (ax + by + c), a, b 0

To solve this , substitute t = ax + by + c. Then the equation reduces to separable type in thevariable t and x which can be solved.

4.3 Differential Equation of the Form :

222

111

cybxa

cybxa

dx

dy

where b1 + a2 = 0

4.4 Polar Coordinates :

Sometimes transformation to the polar coordinates facilitates separation of variables. In thisconnection it is convenient to remember the following differentials. If x = r cos ; y= r sin where rand both are variable.

(a) (i) x dx + y dy = r dr (ii) x dy y dx = r2 dProof : x = r cos ; y = r sin x2 + y2 = r2

x dx + y dy = rdrAlso tan= y/x xdy – y dx = x2 sec2 d

xdy – ydx = r2 d

(b) If x = r sec & y = r tan then(i) x dx y dy = r dr and (ii) x dy y dx = r2 secd.

P(x, y)r

O

x

y

Proof : x = r sec and y = r tan x2 – y2 = r2 xdx – ydy = rdry/x = sin xdy – ydx = x2 cos d = r2 sec d



4.5 Homogeneous Equations :

The function f(x, y) is said to be a homogeneous function of degree n if for any real number t (0), wehave f(tx, ty) = tn (x, y). For example, f(x, y) = ax2/3 + hx1/3 × y1/3 + by2/3 is a homogeneous function ofdegree 2/3.

Homogeneous Differential Equation :

A differential equation of the form)y,x(

)y,x(f

dx

dy

, where f(x, y) and(x, y) are homogeneous function of

x and y, and of the same degree, is called Homogeneous. This equation mayalso be reduced to the form

dx

dy= g

y

xand is solved by putting y = vx so that the dependent variable y is changed to another

variable v,where v is someunknown function, the differential equation is transformed toan equation withvariables separable.

4.6 Equations Reducible to the Homogenous Form :

Equation of the formdx

dy=

CByAx

cbyax

(aBAb andA+ b 0) can be reduced to a homogeneous

form bychanging the variable x, y, to X,Yby writing x = X + h and y=Y+ k; where h, k are constantto be chosen so as to make the given equation homogeneous. We have

dx

dy=

)hX(d

)kY(d

=

dX

dY

Hence the given equation becomes,

dX

dY=

)CBkAh(BkAh

)cbkAh(bYaX

Let h and k be chosen to satisfy the relation ah + bk + c = 0 and Ah + Bk + C = 0.

4.7 Linear Differential Equations :

A differential equation is said to be linear if thedependent variable & all itsdifferential coefficientsoccur

in degree one onlyand are never multiplied together.The nth order linear differential equation is of the form ;

a0 (x)d y

dx

n

n+ a1(x)

d y

dx

n

n

1

1+ ...... + an (x) . y = (x) .

Where a0(x) , a1(x), ..... , an(x) are the coefficients of the differential equation.



Linear Differential Equations of First Order :

The most general form of a linear differential equations of first order isdy

dx+ Py= Q, where P &Q are

functions of x (Independent variable).

For solving such equations we multiplyboth sides by

Integrating factor = I.F. = dxPe

So we get dxPe

dxPQePy

dx

dy

dxPdxPdxPQeyPee

dx

dy

dxPdxP Qeye

dx

d

dxPdxP Pee

dx

dSince

dxeQdxye

dx

d dxPdxP

dxP

ye = dxP

Qe + C

which is the required solution of the given differential equation.

In some cases a linear differential equation may be of the formdy

dx+ P1x = Q1, where P1 and Q1 are

functions of yalone or constants. In such a case the integrating factor is dyP1e , and solutions is given by

CdyeQexdyP

1dyP 11

4.8 Equations Reducible To Linear Form (Bernoulli’s Equation):

The equationdy

dx+ py = Q . yn where P & Q functions of x , is reducible to the linear form by

dividing it by yn & then substituting yn+1 = Z. Its solution can be obtained as in the normal case.



Note : Following exact differentials must be remembered :

(i) xdy + y dx = d(xy) (ii)xdy ydx

xd

y

x

2

(iii)ydx xdy

yd

x

y

2 (iv)

xdy ydx

xyd xy

(ln )

(v)dx dy

x y

= d (ln (x + y)) (vi)

xdy ydx

xyd

y

x

ln

(vii)ydx xdy

xyd

x

y

ln (viii)

xdy ydx

x yd

y

x

2 2

1tan

(ix)ydx xdy

x yd

x

y

2 2

1tan (x)xdx ydy

x yd x y

2 2

2 2ln

(xi) dxy

xdy ydx

x y

12 2

(xii) de

y

ye dx e dy

y

x x x

2

(xiii) de

x

xe dy e dx

x

y y y

2

5. PHYSICAL APPLICATION OF DIFFERENTIAL EQUATION :

5.1 Mixture Problems :

A chemical in a liquid solution (or dispersed in a gas) runs into a container holding the liquid (or the gas)with, possibly, a specified amount of the chemical dissolved as well. The mixture is kept uniform bystirring and flows out of the container at a known rate. In this process it is often important to know theconcentration of the chemical in the container at anygiven time. The differential equation describing theprocess is based on the formula.

containerinamountof

changeofRate

=

arriveschemical

whichatrate

–

departschemical

whichatrate

....(1)

If y (t) is the amount of chemical in the container at time t and V (t) is the total volume of liquid in thecontainer at time t, then the departure rate of the chemical at time t is

Departure rate =)t(V

)t(y·(out flow rate)

=

ttimeatcontainer

inionconcentrat·(out flow rate)

Accordingly, Equation (1) becomes

dt

dy= (chemical's given arrival rate) –

)t(V

)t(y·(out flow rate) ....(2)

If y is measured in grams, V in liters, and t in minutes, then unit in equation (2) are

minute

grams=

minute

grams–

minute

litre

litre

grams·



5.2 Geometrical Applications of Differential Equation :

We alsouse differential equations for finding the familyofcurves for which somecondition involving thederivatives are given. For this we proceed in the following way

Equation of the tangent at a point (x, y) to the curve y = f(x) is given byY – y =dx

dy(X – x).

At the X axis, Y = 0, and X = x –dx/dy

y(intercept on X-axis)

At the Y axis, X = 0, and Y = y –dx

dy(intercept on Y-axis)

Similar information can be obtained for normals by writing equations as (Y – y)dx

dy+ (X – x) = 0.

5.3 Trajectories :

Suppose we are given the family of plane curves (x, y, a) = 0, depending on a single parameter a.

A curve makingat each of its points a fixed anglewith the curve of the familypassing through that pointis called an isogonal trajectory of that family ; if in particular =/2, then it is called an orthogonaltrajectory.

To find Orthogonal trajectories :We set up the differential equation of the given family of curves. Let it be of the form F (x, y, y') = 0

The differential equation of the orthogonal trajectories is of the form F

y

1,y,x = 0

The general integral of this equation 1 (x, y, C) = 0, gives the family of orthogonal trajectories.




DIFFERENT CASES OF BOUNDED AREA :

1. The area bounded by the continuous curve y= f (x), the axis ofx and the ordinates x = a and x = b (where b > a) is given by

b

a

b

aydxf(x)dxA

Ox'

y'

x = a

y = f(x)

x = b

x

y

2. The area bounded by the straight line x = a, x = b (a < b) andthe curves y = f (x) and y = g(x), provided f (x) < g(x) (wherea x b), is given by

dx)]x(f[g(x)Ab

a Ox'

y'

x=

a

y = g(x)

x = b

x

y

y = f (x)

3. When two curves y = f (x) and y = g(x) intersect, the boundedarea is

dx)]x(f[g(x)Ab

a

Ox'

y'

x=

a

y = g(x)

x = b

x

y

y = f (x)

where a and b are the roots of the equation f(x) = g(x)

4. If some part of a curve lies below the x-axis, then its areabecomes negative but area cannot be negative. Therefore, wetake its modulus.If the curves crosses the x-axis at c, then the area bounded bythe curve y = f(x) and ordinates x = a and x = b

(where b > a) is given by b

c

c

af(x)dxf(x)dxA

Ox'

y'

x=

a

x = b

x

y

y = f (x)

c

b

c

c

af(x)dxf(x)dxA

5. The area bounded by y = f(x) and y = g(x) (where a x b),when they intersect at x = c (a, b) is given by

b

adxg(x)f(x)A

or b

c

c

adx)x(f)x(gdx)x(g)x(f

Ox'

y'

x

y

x = a x = c x = b

y = f(x) y = g(x)



DIFFERENT CASES OF BOUNDED AREA :

1. The area bounded by the continuous curve x = f (y), the axis ofy and the abscissa y = a and y = b (where b > a) is given by

b

a

b

axdyf(y)dyA

Ox

y

y = b

y = a

x'

y'

2. The area bounded by the straight line y = a, y = b (a < b) andthe curves x = f (y) and x = g(y), provided f (y) < g(y)(where a y b), is given by

dy)]y(f[g(y)Ab

a O x

y

g(y)f(y)

y = a

y = b

x'

y'

3. When two curves x = f (y) and x = g(y) intersect, the boundedarea is

dy)]y(f[g(y)Ab

a

Ox

y

y = b

y = a

f (y)

g(y)

x'

y'

where a and b are the roots of the equation f (y) = g(y)

4. If some part of a curve lies left to y-axis, then its area becomesnegative but area cannot be negative. Therefore, we take itsmodulus.If the curves crosses the y-axis at c, then the area bounded bythe curve x = f(y) and abscissae y = a and y = b

(where b > a) is given by b

c

c

af(y)dydyf(y)A

Ox

y

x'

y'

a

c

b

x = f(y)

b

c

c

af(y)dyf(y)dyA

5. The area bounded by x = f(y) and x = g(y) (where a y b),when they intersect at y = c (a, b) is given by

b

adyg(y)f(y)A

or c

a

c

ady)y(f)y(gdy)y(g)y(f

Ox

y

g(y)

f(y)

y = b

y = c

y = a

x'

y'



STANDARD AREAS TO BE REMEMBERED :

(1) Area bounded by the curves y2 = 4ax ; x2 = 4by is equal to3

ab16:

At point of intersection

22

b4

x

= 4ax x4 = 64 ab2 x

x = 0, (64 ab2)1/3

Let k = 4 (ab2)1/3

A =

k

0

2

dxb4

xxa2

k

0

3

23 b12

xxa2

23

=3

a423

k –b12

k3=

b12

)ab(64ab8a

3

4 22 21

3

ab16ab

3

16ab

3

32

(2) Area bounded by the parabola y2 = 4ax and y = mx is equal to 3

2

3m

8a:

y2 = 4ax and y = mxAt point of intersection

m2x2 = 4ax x = 0, 2m

a4

Area = c

0

dxmxxa2 where c = 2m

a4

=

c

0

2

23 2

mxxa2

23

=

2

mcc

3

a4 223

3

2

3

2

3

2

4

2

3 m3

a8

m

a8

m3

a32

m

a16·

2

m

m

aa8·

3

a4



(3) Area enclosed by y2 = 4ax and its double ordinate at x = a :

(chord perpendicular to the axis of symmetry)Required area = OABO

a

023

a

0

23

xa4dxax2·2

3

a8)aa·(a

3

8 2

Area of rectangleABCD = 4a2

ABCD)(area3

2AOBofArea

(4) Whole area of ellipse 1b

y

a

x2

2

2

2

is equal to ab :

A = 4

a

02

2

dxa

x1b

Put x = a sin

2/

0

22/

0

2 dcosab4dcosab4A

4ab4d

2

2cos1ab4

2/

0

= ab

SHIFTING OF ORIGIN :

Since area remains invariant even if the coordinates axes are shifted, hence shifting of origin in manycases proves to be very convenient in computing the areas.

CURVE TRACING :

The approximate shape of a curve, the following procedure in order

(I) SYMMETRY:

(a) Symmetry about x-axisIf the equation of the curve remain unchanged by replacing yby–y then the curve is symmetrical aboutthe x-axis.e.g., y2 = 4ax.

(b) Symmetry about y-axisIf the equation of the curve remainunchanged byreplacing x by –x then the curve is symmetrical aboutthe y-axis.e.g., x2 = 4ay



(c) Symmetry about both axesIf the equation of the curve remain unchanged by replacing x by –x and y by –y then the curve issymmetrical about the axis of 'x' as well as 'y'.e.g., x2 + y2 = a2

(d) Symmetry about the line y = xIf the equation of curve remains unchanged on interchanging 'x' and 'y', then the curve is symmetricalabout the line y = xe.g., x3 + y3 = 3xy.

(II) Find the points where the curve crosses the x-axis and the y-axis.

(III) Finddx

dyand examine, ifpossible, the intervalswhen f (x) is increasingordecreasingandalso its stationary

points.

(IV) Examine y when x or x – .

GREATEST AND LEAST VALUE OF VARIABLE AREA :

An important concept :If y = f (x) is a monotonic function in (a, b) then the area bounded by the ordinates at x = a, x = b,

y = f (x) and y = f (c), [where c (a, b)] is minimum when c =2

ba .

Proof : A = c

a

b

c

dx)c(f)x(fdx)x(f)c(f

= c

a

b

c

)cb()c(fdx)x(fdx)x(f)ac()c(f

x

y

f(b)

f(c)

f(a)

y = f(x)

y = f(c)

x'

y'

O x = a c x = b

A = [2c – (a + b) f(c) + b

c

c

a

dx)x(fdx)x(f

Differentiating w.r.t. c,

dc

dA= bac2 f '(c) + 2 f(c) + 0 – f(c) – (f(c))

formaximaandminima 0dc

dA

f '(x)[2c – (a + b)] = 0 (as f '(c) 0)

hence c =2

ba

Also c <2

ba , 0

dc

dA and c > 0

dc

dA,

2

ba

.

HenceAis minimum when c =2

ba .

Note : Let f(x) be the bijective function and g(x) be the inverse of it then area bounded by y = g(x), and theordinate at x = a and x = b is same as area bounded by y = f(x) and the abscissa at y = a and y = b asf(x) and g(x) are mirror image with respect to line y = x.



AVERAGE VALUE OF A FUNCTION :

Average value of the function in y = f (x)w.r.t. x over an interval a x b is defined as

yavg = ab

1

b

a

dx)x(fyavg

a bx

y

O

Note :(i) Average value can be + ve, – ve or zero.(ii) If the function is defined in (0,) then

yavg =

b

0b

dx)x(fb

1Lim provided the limit exists.

Root mean square value (RMS) is defined as

2

1b

a

2 dx)x(fab

1

DETERMINATION OF FUNCTION :

The area functionA(x) satisfies the differential equationxd

)x(Ad= f (x) with initial conditionA(a)= 0

i.e. derivative of the area function is the function itself.

Note : If F(x) is any integral of f (x) then ,

A(x) = f (x) dx = [ F (x) + c ]A(a) = 0 = F (a) + c c = F (a)hence A(x) = F (x) F (a). Finally by taking x = b we get , A(b) = F (b) F (a).



EXERCISE-1 (SPECIAL DPP)

SECTION-A [DIFFERENTIAL EQUATION]

SPECIAL DPP-1

Q.1 Number of values of mN for which y = emx is a solution of the differential equationD3y – 3D2y – 4Dy + 12y = 0, is(A) 0 (B) 1 (C) 2 (D) more than 2

Q.2 Water is drained from a vertical cylindrical tank by opening a valve at the base of the tank. It is knownthat the rate at which the water level drops is proportional to the square root of water depth y, where theconstant of proportionality k > 0 depends on the acceleration due to gravity and the geometry of the

hole. If t is measured in minutes and k =15

1then the time to drain the tank if the water is 4 meter deep

to start with is(A) 30 min (B) 45 min (C) 60 min (D) 80 min

Q.3 A curve y= f (x) passing through the point

e

1,1 satisfies the differential equation

dx

dy+ 2

x2

ex

= 0.

Then which of the following does not hold good?(A) f (x) is differentiable at x = 0.(B) f (x) is symmetric w.r.t. the origin.(C) f (x) is increasing for x < 0 and decreasing for x > 0.(D) f (x) has two inflection points.

Q.4 Theorder &the degree of the differential equationwhose general solution is, y= c(xc)2, are respectively(A) 1, 1 (B) 1, 2 (C) 1, 3 (D) 2, 1

Q.5 If the x – intercept of normal to a curve at P(x,y) is twice the abscissa of P then the equation of curvepassing through M(2 , 4) is(A) x2 +y2 = 20 (B) x2 – y2 = – 12 (C) y2 = 8x (D) 2x2 + y2 = 24

Q.6 Let y = f (x) be a continuous function such thatdx

dy= | x – 1 |. If y (0) = 0 then y (3) equals

(A)2

3(B)

2

3(C)

2

5(D) 2

Q.7 Ifx

y2

dx

dy = 0, y(1) = 1, then y(2) is equal to

(A) 1 (B) 2 (C) 4 (D)4

1



Q.8 Let a solution y= y(x) of the differential equation ey dy– (2 + cos x) dx = 0 satisfyy(0) = 0 then the value

of

2f is equal to

(A) ln (B) ln (2 + ) (C) ln (1 + ) (D) does not exist

Q.9 The order of differential equation corresponding to y = c1 cos 2x + c2 cos2 x + c3 sin

2 x + c4 is(A) 2 (B) 4 (C) 3 (D) None of these

Q.10 Anormal is drawn at a point P(x, y) on a curve. It meets the x-axis and the y-axis atAand B respectivelysuch that (x-intercept)–1 + (y-intercept)–1 = 1, where O is origin, then find radius of the director circle ofthe curve passing through (3, 3).

SPECIAL DPP-2

Q.1 A curve passes through the point 14

,

& its slope at any point is given by

y

x cos2

y

x

. Then the

curve has the equation

(A) y=x tan–1(lnx

e) (B) y=x tan–1(ln + 2)

(C) y =x

1tan–1(ln

x

e) (D) none

Q.2 If y = |xc|n

x

l (where c is an arbitrary constant) is the general solution of the differential equation

dx

dy =

x

y+

y

xthen the function

y

xis :

(A) 2

2

y

x(B) – 2

2

y

x(C) 2

2

x

y(D) – 2

2

x

y

Q.3 The real value of m for which the substitution, y= um will transform the differential equation,

2x4ydy

dx+ y4 = 4x6 into a homogeneous equation is :

(A) m = 0 (B) m = 1 (C) m = 3/2 (D) no value of m

Q.4 Let y' =2

22

x4

xxy4y4 and y (1) = 0, then

2ey equals

(A) 2e2

1

(B) 2e

(C) 2e4

1

(D)2



Q.5 Let y = f (x) satisfy the differential equationdx

dy=

x

yx , y (1) = 1, then

e

1y is equal to

(A) 2e (B)e

2(C) 0 (D)

e

1

Paragraph for question nos. 6 & 7

A differentiable function y = g(x) satisfies x

0

)1tx( g(t) dt = x4+ x2 for all x 0.

Q.6 y= g(x) satisfies the differential equation

(A)dx

dy– y = 12x2 + 2 (B)

dx

dy+ 2y = 12x2 + 2

(C)dx

dy+ y = 12x2 + 2 (D)

dx

dy+ y = 12x2 – 2

Q.7 The value of g(0) is equal to(A) 0 (B) 1 (C) e2 (D) data insufficient

Q.8 A curve is such that the ratio of the subnormal at any point to the sum of its co-ordinates is equal to theratio of the ordinate of this point to its abscissa. If the curve passes through M(1, 0), then possibleequation of the curve is(are)

(A) y = x ln x (B) y =x

xnl(C) y = 2x

)1x(2 (D) y =

x2

x1 2

Q.9 Identify the statement(s) which is/are True.

(A) f(x , y) = ey/x + tany

xis a homogeneous function of degree zero

(B) x . lny

xdx +

y

x

2

sin1y

xdy = 0 is a homogeneous differential equation of degree one

(C) f(x , y) = x2 + sin x . cosy is not homogeneous function.(D) (x2 + y2) dx - (xy2 y3) dy = 0 is a homegeneous differential equation.

Q.10 Let y = f (x) be a curve C1 passing through (2,2) and

2

1,8 and satisfying a differential equation

2

2

2

dx

dy2

dx

ydy

. Curve C2 is the director circle of the circle x

2 + y2 = 2. If the shortest distance

between the curves C1 and C2 is qp where p, qN, then find the value of (p2 – q).



SPECIAL DPP-3

Q.1 A function y = f (x) satisfies the differential equation f (x) · sin 2x – cos x + (1 + sin2x) f ' (x) = 0 with

initial condition y(0) = 0. The value of 6f is equal to(A) 1/5 (B) 3/5 (C) 4/5 (D) 2/5

Q.2 The solution of the differential equation, ex(x + 1)dx + (yey – xex)dy= 0 with initial condition f (0) = 0,is(A) xex + 2y2ey = 0 (B) 2xex + y2ey = 0 (C) xex – 2y2ey = 0 (D) 2xex – y2ey = 0

Q.3 If the differentiable equationdx

dy– y = y2(sin x + cos x) with y (0) = 1 then y () has the value equal to

(A) e (B) – e (C) e– (D) – e–

Q.4 The equation of curve passing through origin and satisfying the differential equation

(1 + x2)dx

dy+ 2xy = 4x2 , is

(A) 3(1 + x2) y = 2x3 (B) (1 + x2) y = x3 (C) (1 + x2) y = 3x3 (D) 3(1 + x2) y = 4x3

Q.5 Let y = f(x) be a real valued function satisfying xdx

dy= x2 + y – 2, f (1) = 1 then f (3) equals

(A) 0 (B) 3 (C) 5 (D) 8

Q.6 If y(t) satisfies the differential equation y'(t) + 2 y(t) = 2 e–2t, y(0) = 2 then y(1) equals

(A)e

3(B) 2e

3(C)

e

4(D) 2e

4

Q.7 Let y ' (x) + g ' x

g(x)· y(x) =

)x(g1

)x('g2

where f ' (x) denotesdf (x)

dxand g(x) is a given non-constant

differentiable function an R. If g(1) = y(1) = 1 and g(e) = 2e 1 then y(e) equals

(A)3

2g(e)(B)

1

2g(e)(C)

2

3g(e)(D)

1

3g(e)

Q.8 If the function y = f(x) satisfies f '(x) + f (x) cot x – 2 cos x = 0, f

2= 1, then f

3is equal to

(A) 0 (B)2

1(C)

2

3(D) 2



Q.9 If 2xy dy = (x2 + y2 + 1)dx, y(1) = 0 and y (x0) = 3 , then x0 can be

(A) 2 (B) – 2 (C) 3 (D) – 3

Q.10 Let function y = f(x) satisfies the differential equation x2dx

dy= y2 x

1

e (x 0) and 1)x(fLim0x

.

Identify the correct statement(s)?

(A) Range of f(x) is (0, 1) –

2

1. (B) f(x) is bounded

(C) 1)x(fLim0x

(D) 1

0

e

0

dx)x(fdx)x(f

SECTION-B [AREA UNDER THE CURVE]

SPECIAL DPP-1

Q.1 Suppose y= f (x) and y = g(x) are two continuous functions whose graphs intersect at the three points(0, 4), (2, 2) and (4, 0) with f (x) > g (x) for 0 < x < 2 and f (x) < g (x) for 2 < x < 4.

If 4

0

dx)]x(g)x(f[ =10 and 4

2

dx)]x(f)x(g[ =5, the area between two curves for 0 < x < 2, is

(A) 5 (B) 10(C) 15 (D) 20

Q.2 Let 'a' be a positive constant number. Consider two curves C1: y = ex, C2 : y =e

a – x. Let S be the area

of the part surrounding by C1, C2 and the y-axis, then 20a a

SLim

equals

(A) 4 (B) 1/2 (C) 0 (D) 1/4

Q.3 Area enclosed by the curve y = (x2 + 2x)e–x and the positive x-axis is(A) 1 (B) 2 (C) 4 (D) 6

Q.4 The slope of the tangent to a curve y = f (x) at (x , f (x)) is 2x + 1 . If the curve passes through thepoint (1 , 2) then the area of the region bounded by the curve , the x-axis and the line x = 1 is

(A)5

6(B)

6

5(C)

1

6(D) 1

Q.5 The area bounded by the curves y = x and x = y where x, y 0

(A) cannot be determined(B) is 1/3(C) is 2/3

(D) is same as that of the figure bounded by the curves y = x ; x 0 and x = y ; y 0



Q.6 Area of region bounded by x = 0, y = 0, x = 2, y = 2, y ex & y ln x is(A) 6 – 4 ln 2 (B) 4 ln 2 – 2 (C) 2 ln 2 – 4 (D) 6 – 2 ln 2

Q.7 The area bounded by the curve y = | cos x – sin x |, 0 x 2

and above x-axis is

(A) 22 (B) 222 (C) 122 (D) 222

Q.8 The area bounded by the curve y =2

ee xx and the lines y = 0, x = ln 2 and x =

2

1nl is

(A)4

3(B)

4

5(C)

2

3(D)

2

5

Q.9 Find the area bounded by the curve y= 2 x x² & the straight line y = x.

Q.10 Find the area enclosed by the parabola y = 1 + x2 and a normal drawn to it with gradient – 1.

SPECIAL DPP-2

Q.1 The area bounded by the curve y = x e–x ; xy = 0 and x = c where c is the x-coordinate of the curve'sinflectionpoint, is(A) 1 – 3e–2 (B) 1 – 2e–2 (C) 1 – e–2 (D) 1

Q.2 The line y = mx bisects the area enclosed by the curve y = 1+ 4xx2 & the lines x = 0, x = 3

2

& y = 0 . Then the value of m is :

(A)13

6(B)

6

13(C)

3

2(D) 4

Q.3 If the line x – 1 = 0 divides the area bounded by the curves 2x + 1 = 1y4 , y = x and y = 2

in two regions of area A1 and A2 (A1 x – 1, x < 1 and x > 0, then area of R is

(A)6

11(B)

2

3(C)

6

5(D) 2



Q.6 Let S be the area enclosed by the curves f(x) = 4|x| – |x|3 and g(x) + 2x–4 = 0. The value of [S]

is equal to[Note : [k] denotes greatest integer less than or equal to k.](A) 10 (B) 12 (C) 14 (D) 16

Paragraph for question nos. 7 & 8

Let f be a twice differentiable function such that f " (x) = 1 – f (x) where f ' (0) = f (0) = 1.

Q.7 Area bounded by the curve y = f (x) and the x-axis from x = 0 to x = is(A) 2 + 4 (B) + 2 (C) + 4 (D) 2 + 2

Q.8 The value of dx1xcos

1)x(f2

02

is equal to

(A)4

(B)

4

(C)

2

(D)

4

3

Q.9 Consider the functions f (x) and g (x), both defined from R R and are defined asf (x) = 2x – x2 and g (x) = xn where nN. If the area between f (x) and g (x) in first quadrant is 1/2 thenn is a divisor of(A) 12 (B) 15 (C) 20 (D) 30

Q.10 Find the area bounded by y = xe|x| and lines |x| = 1, y = 0.

SPECIAL DPP-3

Q.1 Consider the following regions in the plane :R1 = {(x, y) : 0 x 1 and 0 y 1}R2 = {(x, y) : x

2 + y2 4/3}

The area of the region R1R2 can be expressed as 9

b3a , where a and b are integers. Then the

value of (a + b) equals(A) 2 (B) 3 (C) 4 (D) 5

Q.2 In the shown figure, half a period of sin x from 0 to is split into tworegions (light anddark shaded) of equalarea bya line through the origin.If the line and the sine function intersect at a point whose x co-ordinateis k, then k satisfies the equation(A) k cos k + 2 sin k = 0 (B) k sin k + 2 cos k = 0

x

y

O

(k, sin k)

k

(C) k sin k + 2 cos k – 2 = 0 (D) 2 cos k + k sin k + 2 = 0



Q.3 Let f be a positive continuous function on the interval [–2, 3] and A(t) is the area of the region bounded

by the graph of y = f (x) and the lines y= 0, x = – 2, and x = t where t (–2, 3). Ift3

)t(A)3(ALim

3t

is equal to 100 then the value of f (3) equals

(A) 3 (B) 100 (C)100

1(D) 300

Paragraph for Question no. 4 to 6

The graph of a polynomial f(x) of degree 3 is as shown in the figure and slope of tangent at Q (0, 5) is 3.

x

y

Q 5

4

3

2

1

1 2 3O– 1– 2

P R

Q.4 Number of solutions of the equation |)x(|f = 3, is

(A) 1 (B) 2 (C) 3 (D) 4

Q.5 The equation of normal at the point where curve crosses y-axis, is

(A) 3x + y = 15 (B) x + 3y = 15 (C) x + 3y = 5 (D) 3x + y = 5

Q.6 Area bounded by the curve y = f(x) with x axis and lines x + 1 = 0, x – 1 = 0 is

(A)2

13(B)

2

15(C)

2

17(D)

2

19

Q.7 Let T be the triangle with vertices (0, 0), (0, c2) and (c, c2) and let R be the region between y = cx andy = x2 where c > 0 then

(A) Area (R) =6

c3(B) Area of R =

3

c3

(C))R(Area

)T(AreaLim

0c = 3 (D)

)R(Area

)T(AreaLim

0c =

2

3



Q.8 Find the value of k > 0 so that the area of the bounded region enclosed betwen the parabolas

y = x – kx2 and y =k

x2is maximum.

Q.9 Let An be the area bounded by the curve y = xn (n 1) and the line x = 0, y = 0 and x =

2

1.

If3

1

n

A2n

1n

nn

then find the value of n.

Q.10 Let y = f (x) be a curve which satisfies the differential equation ex dy + (x – 1)dx = 0 and f(0) = 0.

If area enclosed by the curve, the x-axis and a line x = c where )c(f,c is the point of inflection on

the curve is

qe

p1 , p, qN then find the value of (p + q).

EXERCISE-2


(Formation & Variables Separable)]

Q.1 State the order and degree of the following differential equations:

(i)d x

d t

dx

d t

2

2

3 4

xt = 0 (ii)

d y

dx

dy

dx

2

2

23 2

1

/

Q.2(a) Form the differential equation of the family of curves represented by,c (y + c)2 = x3 ; where c is any arbitrary constant.

(b) Form a differential equation for the family of curves represented by ax2 + by2 = 1, where a & b arearbitaryconstants.

(c) Obtain the differential equation of the family of circles x2 + y2 + 2gx + 2fy + c = 0;where g , f & c are arbitary constants.

(d) Obtain the differential equation associated with the primitive,y = c1e

3x + c2e2x + c3 e

x, where c1, c2, c3 are arbitrary constants.

Q.3 Solve :dy

dx= sin (x + y) + cos (x + y)

Q.4 Solve :dy

dx=

x x

y y y

( ln )

sin cos

2 1

Q.5 Solve : sin x .dy

dx= y . lny if y = e, when x =

2



Q.6 The population P of a town decreases at a rate proportional to the number by which the populationexceeds 1000, proportionality constant being k > 0. Find

(a) Population at any time t, given initial population of the town being 2500.(b) If 10 years later the population has fallen to 1900, find the time when the population will be 1500.(c) Predict about the population of the town in the long run.

Q.7 It is known that the decayrate of radium is directlyproportional to its quantityat each given instant. Findthe law of variation of a mass of radium as a function of time if at t = 0, the mass of the radius was m0and during time t0

% of the original mass of radium decay.

Q.8 A normal is drawn at a point P(x , y) of a curve. It meets the x axis at Q. If PQ is of constant length

k, then show that the differential equation describing such curves is, ydy

dx= ± k y2 2 . Find the

equation of such a curve passing through (0, k).

Q.9 Let C be a curve passing through M (2, 2) such that the slope of the tangent at any point to the curveis reciprocal of the ordinate of the point. If the area bounded by curve C and line x = 2 is expressed

as a rationalq

p(where p and q are in their lowest form), then find (p + q).

Q.10 A curve is such that the length of the polar radius of any point on the curve is equal to the length of thetangent drawn at this point. Form the differential equation and solve it to find the equation of the curve.

Q.11 Let f (x) is a continuous function which takes positive values for x 0 and satisfy x

0

dt)t(f = )x(fx

with f (1) =2

1. Find the value of 12 f .

Q.12 Solve :dy

dx=

x xy

x y

2

2 2

Q.13 Find the equation of a curve such that the projection of its ordinate upon the normal is equal to itsabscissa.

Q.14 The light rays emanating from a point source situated at origin when reflected from the mirror of a searchlight are reflected as beam parallel to the x-axis. Show that the surface is parabolic, byfirst forming thedifferential equation and then solving it.

Q.15 xy

xy

y

xcos sin

y = y

y

xx

y

xsin cos

x

dy

dx

Q.16dy

dx=

x y

x y

1

2 2 3



Q.17dy

dx=

2 2

1

2

2

( )

( )

y

x y

Q.18 Let y(x) be a real-valued differentiable function on the interval (0, ) such that y(1) = 0 and

satisfies y'(x) = ln x + 2 –nxx

)x(y

l. Find the value of [y(e) – y'(e)].

[Note: where [k] denotes largest integer less than or equal to k.]

Q.19 Find the differentiable function which satisfies the equation f (x) = – x

0

x

0

dt)xttan(dtttan)t(f

where x 2,2

Q.20 Atank contains 100 litres of fresh water.Asolution containing 1 gm/litre of soluble lawn fertilizer runsinto the tank at the rate of 1 lit/min, and the mixture is pumped out of the tank at the rate of 3 litres/min.Find the time when the amountof fertilizer in the tank is maximum.

Q.21dy

dxx y

dy

dxxy

2

0( )

Q.22 (1 xy + x2 y2) dx = x2 dy

Q.23dy

dx= y + y dx

0

1

given y = 1 , where x = 0

Q.24 Find the curve which passes through the point (2, 0) such that the segment of the tangent between thepoint of tangency & the yaxis has a constant length equal to 2 .

Q.25 Find the equationof the curvepassing through the orgin if the middle point of thesegment of its normalfrom any point of the curve to the x-axis lies on the parabola 2y2 = x.


Q.1 Find the area bounded on the right by the line x + y = 2, on the left by the parabola y = x2 and below bythe x-axis.

Q.2 Find the area of the region bounded by curves f (x) = (x – 4)2, g (x) = 16 – x2 and the x - axis.

Q.3 Afigure is bounded by the curves y = 24

sinx

, y = 0, x = 2 & x = 4.At what angles to the positive

xaxis straight lines must be drawn through (4 ,0) so that these lines partition the figure into three partsof the same size.

Q.4 Find the area bounded by the curves y = 1 2 x and y = x3 x.Also find the ratio in which the y-axis

divided this area.



Q.5 The line 3x + 2y = 13 divides the area enclosed by the curve,9x2 + 4y218x16y11 = 0 into two parts. Find the ratio of the larger area to the smaller area.

Q.6 Find the values of m (m > 0) for which the area bounded by the line y = mx + 2 andx = 2y – y2 is , (i) 9/2 square units & (ii) minimum.Also find the minimum area.

Q.7 Consider two curves C1 : y = x

1and C2 : y = ln x on the xyplane. Let D1denotes the region surrounded

by C1, C2 and the line x = 1 and D2 denotes the region surrounded by C1, C2 and the line x = a.If D1 = D2. Find the value of 'a'.

Q.8 Find the area enclosed between the curves : y = loge (x+ e) , x = loge (1/y) & the xaxis.

Q.9 For what value of 'a' is the area bounded by the curve y = a2x2 + ax + 1 and the straight line y = 0,x = 0 & x = 1 the least ?

Q.10 Let f (x) = 1 + cos x and g (x) =1cxbx

a2

. If f (0) = g (0), f ' (0) = g ' (0), f " (0) = g " (0).

and the area bounded by the graph of g (x) and x-axis is k then find the value of k.

Q.11 Find the area bounded by the curve y =2xex , the x-axis, and the line x = c where y(c) is maximum.

Q.12 The figure shows two regions in the first quadrant.

P(t, sin t )2 P(t, sin t )2y=sin x2

O Ot tX X

Y

A(t) B(t)X

Y

Y

XY

A(t) is the area under the curve y= sin x2 from 0 to t and B(t) is the area of the triangle with vertices O,

P and M(t, 0). Find)t(B

)t(ALim

0t.

Q.13 Consider the curve y = xn where n > 1 in the 1st quadrant. If the area bounded by the curve, the x-axisand the tangent line to the graph of y = xn at the point (1, 1) is maximum then find the value of n.

Q.14 Show that the area bounded by the curve y =x

cxn l, the x-axis and the vertical line through the

maximum point of the curve is independent of the constant c.

Q.15 LetAn be the area bounded by the curve y = (tan x)n & the lines x = 0, y = 0 & x = /4. Prove that for

n > 2 , An + An2 = 1/(n 1) & deduce that 1/(2n + 2) < An < 1/(2n 2).



EXERCISE-3

SECTION-A

(JEE-ADVANCED Previous Year's Questions)


Q.1 Let a solution y= y (x) of the differential equation, 0dx1yydy1xx 22 satisfyy (2) =3

2.

Statement-1 : y (x) = sec

6xsec 1

Statement-2 : y(x) is given by 2x

11

x

32

y

1

(A) Statement-1 is True, Statement-2 is True; statement-2 is a correct explanation for statement-1(B) Statement-1 is True, Statement-2 is True; statement-2 is NOT a correct explanation for statement-1(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True [JEE 2008, 3]

Q.2(i) Match the statements/expressions In Column I with the open intervals In Column II.

Column I Column II

(A) Interval contained in the domain of definition of non-zero (p)

2,

2

solutions of the differential equation (x - 3)2 y+y= 0

(B) Interval containing the value of the integral (q)

2,0

5

1

dx)5x()4x()3x()2x()1x(

(C) Interval in which at least one of the points of local maximum (r)

4

5,

8of cos2x + sin x lies

(D) Interval in which tan–1 (sin x + cos x) is increasing (s)

8,0

(t) ( – , )



(ii) Match the statements/expressions given in Column I with the values given in Column II.Column I Column II

(A) The number of solutions of the equation (p) 1

0xcosxe xsin in the interval

2,0

(B) Value(s) of k for which the planes (q) 2kx + 4y + z = 0, 4x + ky + 2z = 0 and 2x + 2y + z = 0intersect in a straight line

(C) Value(s) of k for which |x – 1| + |x – 2| + |x + 1| + |x + 2| = 4k (r) 3has integer solution(s)

(D) If y = y + 1 and y(0) = 1, then value(s) of y (ln 2) (s) 4(t) 5

[JEE 2009, (2+2+2+2)×2]

Q.3 Let f be a real valued differentiable function on R (the set of all real numbers) such that f (1) = 1. If they-intercept of the tangent at any point P(x, y) on the curve y = f (x) is equal to the cube of the abscissaof P, then the value of f (–3) is equal to [JEE 2010, 3]

Q.4

(a) Let f : [1,) [2,) be a differentiable function such that f (1) = 2. If x

1

dt)t(f6 = 3x f (x) – x3 – 5

for all x 1, then the value of f (2) is

(b) Let y'(x) + y(x) g'(x) = g(x) g'(x), y (0) = 0, xR, where f '(x) denotes)x(d

)x(fdand g (x) is a given

non-constant differentiable function on R with g (0) = g (2) = 0. Then the value of y(2) is[JEE 2011, 4+4]

Q.5 If y (x) satisfies the differential equation y' – y tan x = 2x sec x and y(0) = 0, then

(A)284

y2

(B)

184'y

2

(C)93

'y2

(D)

33

2

3

4

3'y

2

[JEE 2012, 4]

Q.6 A curve passes through the point

6,1 . Let the slope of the curve at each point (x, y) be

x

ysec

x

y, x > 0. Then the equation of the curve is

(A)

x

ysin = log x +

2

1(B)

x

ycosec = log x + 2

(C)

x

y2sec = log x + 2 (D)

x

y2cos = log x +

2

1

[JEE (Advanced) 2013, 2]



Q.7 The function y = f (x) is the solution of differential equation2

4

2x1

x2x

1x

xy

dx

dy

in (–1, 1)

satisfying f (0) = 0. Then

2

3

2

3

dx)x(f is

(A)2

3

3

(B)

4

3

3

(C)

4

3

6

(D)

2

3

6


Q.8 Let y (x) be a solution of the differential equation (1 + ex)y' + yex = 1. If y (0) = 2, then which of thefollowing statements is(are) true?(A) y (– 4) = 0(B) y (– 2) = 0(C) y (x) has a critical point in the interval (– 1, 0)(D) y (x) has no critical point in the interval (– 1, 0)


Q.9 Consider the familyof all circles whose centers lie on the straight line y = x. If this familyof circles isrepresented by the differential equation Py'' + Qy' + 1 = 0, where P, Q are functions of x, y and y'

(here y' =dx

dy, y'' = 2

2

dx

yd), then which of the following statement(s) is(are) true?

(A) P = y + x (B) P = y – x(C) P + Q = 1 – x + y + y' + (y')2 (D) P – Q = x + y – y' – (y')2.


Q.10 Let f : (0, ) R be a differentiable function such that f '(x) = 2 –x

)x(ffor all x (0, ) and

f (1) 1. Then

(A)

x

1'fLim

0x= 1 (B)

x

1xfLim

0x= 2

(C) )x('fxLim 20x

= 0 (D) | f (x) | 2 for all x (0, 2)


Q.11 A solution curve of the differential equation (x2 + xy + 4x + 2y + 4)dx

dy– y2 = 0, x > 0, passes through

the point (1, 3). Then the solution curve(A) intersects y = x + 2 exactly at one point (B) intersects y = x + 2 exactly at two points(C) intersects y = (x + 2)2 (D) does NOT intersect y = (x + 3)2




Q.12 If y= y(x) satisfies the differential equation

dyx9x8

= dxx94

1

, x > 0

and y(0) = 7 , then y(256) =

(A) 16 (B) 80 (C) 3 (D) 9[JEE (Advanced) 2017, 3]

Q.13 Let f : (0, )R be a twice differentiable function such that

xt

xsin)t(ftsin)x(fLim

xt

= sin2x for all x (0, ).

If

6f = –

12

, then which of the following statement(s) is(are) TRUE ?

(A)

4f =

24

(B) f (x) <6

x4– x2 for all x (0, )

(C) There exists (0, ) such that f '() = 0

(D)

2f

2"f = 0 [JEE (Advanced) 2018, 4]

Q.14 Let f : RR be a differentiable function with f (0) = 0. If y = f (x) satisfies the differential equation

dx

dy= (2 + 5y) (5y – 2),

then the value of )x(fLimx

is______. [JEE (Advanced) 2018, 3]


Q.1

(a) The area of the region between the curves y =xcos

xsin1and y =

xcos

xsin1bounded by the lines

x = 0 and x =4

is

(A)

12

022

dtt1)t1(

t(B)

12

022

dtt1)t1(

t4

(C)

12

022

dtt1)t1(

t4(D)

12

022

dtt1)t1(

t



(b) Comprehension (3 questions together):

Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in thereal line.If x (–, –2)(2, ), the equation implicitly defines a unique real valued differentiablefunction y = f (x).If x (–2, 2), the equation implicitly defines a unique real valued differentiable function y=g(x)satisfyingg(0)=0.

(i) If f (–10 2 ) = 22 , then f '' (–10 2 ) =

(A) 2337

24(B) 2337

24 (C)

37

243 (D) 37

243

(ii) The area of the region bounded by the curve y = f (x), the x-axis, and the lines x = a and x = b,where – < a < b < –2, is

(A) b

a 1))x(f(3x

2 dx + b f (b) – a f (a) (B) –

b

a 1))x(f(3x

2 dx + b f (b) – a f (a)

(C) b

a 1))x(f(3x

2 dx – b f (b) + a f (a) (D) –

b

a 1))x(f(3x

2 dx – b f (b) + a f (a)

(iii) dx)x('g1

1

=

(A) 2g(–1) (B) 0 (C) – 2 g(1) (D) 2 g(1)[JEE 2008, 3 + 4 + 4 + 4]

Q.2 Area of the region bounded by the curve y = ex and lines x = 0 and y = e is

(A) e – 1 (B) e

1

dy)y1e(nl (C) 1

0

xdxee (D) e

1

dyynl

[JEE 2009, 4]

Q.3 Consider the polynomial f(x) = l + 2x + 3x2 + 4x3. Let s be the sum of all distinct real roots of f(x)and let t = | s |.

(i) The real number s lies in the interval

(A)

0,

4

1(B)

4

3,11 (C)

2

1,

4

3(D)

4

1,0

(ii) The area bounded by the curve y = f(x) and the lines x = 0, y = 0 and x = t, lies in the interval

(A)

3,

4

3(B)

16

11,

64

21(C) (9, 10) (D)

64

21,0



(iii) The function f (x) is

(A) increasing in

4

1,t and decreasing in

t,

4

1

(B) decreasing in

4

1,t and increasing in

t,

4

1

(C) increasing in (– t, t)(D) decreasing in (– t, t) [JEE 2010, 3+3+3]

Q.4(a) Let the straight line x = b divide the area enclosed by y = (1 – x)2, y = 0 and x = 0 into two parts

R1 (0 x b) and R2 (b x 1) such that R1 – R2 = 4

1. Then b equals

(A)4

3(B)

2

1(C)

3

1(D)

4

1

(b) Let f : [–1, 2] [0, ] be a continuous function such that f (x) = f (1 – x) for all x [–1, 2].

Let R1 =

2

1

dx)x(fx , and R2 be the area of the region bounded by y = f (x), x = – 1, x = 2 and the

x-axis. Then(A) R1 = 2R2 (B) R1 = 3R2 (C) 2R1 = R2 (D) 3R1 = R2

[JEE 2011, 3+3]

Q.5 The area enclosed by the curves y = sin x + cos x and y = | cos x – sin x | over the interval

2,0 is

(A) 4 ( 2 – 1) (B) 2 2 ( 2 – 1) (C) 2 ( 2 + 1) (D) 2 2 ( 2 + 1)[JEE (Advanced) 2013, 2]

Q.6 Let F (x) =

6x

x

2

2

dttcos2 for all x R and f :

2

1,0 [0, ) be a continuous function. For

a

2

1,0 , if F ' (a) + 2 is the area of the region bounded by x = 0, y = 0, y = f (x) and x = a, then

f (0) is [JEE (Advanced) 2015, 4]

Q.7 Suppose that F() denotes the area of the region bounded by x = 0, x = 2, y2 = 4x and

y = | x – 1 | + | x – 2 | + x, where {0, 1}.Then the value(s) of F() +3

28,

when = 0 and = 1, is (are)(A) 3 (B) 4 (C) 5 (D) 6

[JEE (Advanced) 2015, MTC, 2]



Q.8 Area of the region {(x, y)R2 : y |3x| , 5y x + 9 15} is equal to

(A)6

1(B)

3

4(C)

2

3(D)

3

5


Q.9 If the line x =divides the area of region R = {(x, y)R2 : x3 y x, 0 x 1} into two equal parts,then

(A) 24 – 42 + 1 = 0 (B) 4 + 42 – 1 = 0 (C) 0 < 2

1(D)

2

1< < 1


Q.10 Let f : [0,)R be a continuous function such that f(x) = 1 – 2x +

x

0

tx dt)t(fe for all x [0,).

Then, which of the following statement(s) is(are) TRUE?(A) The curve y = f(x) passes through the point (1, 2).(B) The curve y = f(x) passes through the point (2, – 1).

(C) The area of the region 2x1y)x(f:R]1,0[)y,x( is4

2.

(D) The area of the region 2x1y)x(f:R]1,0[)y,x( is4

1.


SECTION-B

(JEE-MAIN Previous Year's Questions)


Q.1 The differential equation of the familyof circles with fixed radius 5 units and centre on the line y= 2 is -(1) (y – 2) y2 = 25 – (y – 2)2 (2) (y – 2)2 y2 = 25 – (y – 2)2

(3) (x – 2)2 y2 = 25 – (y – 2)2 (4) (x – 2) y2 = 25 – (y – 2)2 [AIEEE 2008]

Q.2 The solution of the differential equationdx

dy=

x

yx satisfyingtheconditiony(1)=1 is -

(1) y = x n x + x2 (2) y = xe(x–1)

(3) y = x n x + x (4) y = n x + x [AIEEE 2008]

Q.3 The differential equation which represents the family of curves y = c1

xc2e where c1and c

2are arbitrary

constants, is - [AIEEE 2009](1) y= y2 (2) y= y y (3) yy = y (4) yy = (y)2

Q.4 Solution of the differential equation cos x dy = y (sin x – y) dx,2

x0

, is

(1) sec x = (tan x + c) y (2) y sec x = tan x + c(3) y tan x = sec x + c (4) tan x = (sec x + c)y [AIEEE 2010]



Q.5 Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years. The

value V(t) depreciates at a rate given by differential equationdt

)t(dV= –k(T – t), where k > 0 is a

constant and T is the total life in years of the equipment. Then the scrap value V(T) of the equipment is

(1) T2 –k

I(2) I –

2

kT2(3) I –

2

)tT(k 2(4) e – kT [AIEEE 2011]

Q.6 Ifdx

dy= y + 3 > 0 and y (0) = 2, then y (ln 2) is equal to

(1) 7 (2) 5 (3) 13 (4) –2 [AIEEE 2011]

Q.7 The population p(t) at time t of a certain mouse species satisfies the differential equation

dt

)t(dp= 0.5 p(t) – 450. If p(0) = 850, then the time at which the population becomes zero is

(1) 18n2

1l (2) ln 18 (3) 2 ln 18 (4) ln 9 [AIEEE 2012]

Q.8 At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P

w.r.t. additional number of workers x is given bydx

dP= 100 – 12 x . If the firm employs 25 more

workers, then the new level of production of items is(1) 3000 (2) 3500 (3) 4500 (4) 2500

[JEE (Main) 2013]

Q.9 Let the population of rabbits surviving at a time t be governed by the differential equation

dt

)t(dp=

2

1p(t) – 200.

If p(0) = 100, then p(t) equals

(1) 400 – 300 2t

e

(2) 400 – 300 2t

e

(3) 300 – 200 2t

e

(4) 600 – 500 2t

e [JEE (Main) 2014]

Q.10 Let y(x) be the solution of the differential equation (x log x)dx

dy+ y = 2x log x, (x 1). Then y(e)

is equal to(1) 2 (2) 2e (3) e (4) 0

[JEE (Main) 2015]



Q.11 If a curve y = f(x) passes through the point (1, – 1) and satisfies the differential equation

y (1 + xy) dx = x dy, then

2

1f is equal to

(1)5

4(2)

5

2(3)

5

4(4)

5

2[JEE (Main) 2016]

Q.12 If (2 + sin x)dx

dy+ (y + 1) cos x = 0 and y(0) = 1, then

2y is equal to

(1)3

1(2)

3

2(3)

3

1(4)

3

4[JEE (Main) 2017]

Q.13 Let y = y(x) be the solution of the differential equation ),0(x,x4xcosydx

dyxsin .

If 02

y

, then

6y is equal to: [JEE (Main) 2018]

(1)2

9

8 (2)

2

9

4 (3)

2

39

4 (4)

2

39

8


Q.1 The area of the plane region bounded by the curves x + 2y2 = 0 and x + 3y2 = 1 is equal to -

(1)3

1(2)

3

2(3)

3

4(4)

3

5[AIEEE 2008]

Q.2 The area of the region bounded by the parabola (y – 2)2 = x – 1, the tangent to the parabola at the point(2, 3) and the x – axis is -(1) 3 (2) 6 (3) 9 (4) 12 [AIEEE 2009]

Q.3 The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and x =2

3is

(1) 224 sq. unit (2) 224 sq. unit(3) 124 sq. unit (4) 124 sq. unit [AIEEE 2010]

Q.4 The area of the region enclosed by the curves y = x, x = e, y = 1/x and the positive x-axis is

(1)2

1square units (2) 1 square units

(3)2

3square units (4)

2

5square units [AIEEE - 2011]



Q.5 The area bounded between the parabolas x2 =4

yand x2 = 9y, and the straight line y = 2 is

(1)3

220(2) 210 (3) 220 (4) 3

210[AIEEE 2012]

Q.6 The area (in square units) bounded by the curves y = x , 2y – x + 3 = 0, x-axis and lying in the firstquadrant is

(1) 36 (2) 18 (3)4

27(4) 9 [JEE (Main) 2013]

Q.7 The area of the region described by A = {(x, y) : x2 + y2 1 and y2 1 – x} is

(1)3

2

2

(2)

3

4

2

(3)

3

4

2

(4)

3

2

2

[JEE (Main) 2014]

Q.8 The area (in sq. units) of the region described by {(x, y) : y2 2x and y 4x – 1} is

(1)64

15(2)

32

9(3)

32

7(4)

64

5

[JEE (Main) 2015]

Q.9 The area (in sq. units) of the region {(x, y): y2 2x and x2 + y2 4x, x 0, y 0} is

(1)3

22

2

(2) –

3

4(3) –

3

8(4) –

3

24

[JEE (Main) 2016]

Q.10 The area (in sq. units) of the region {(x, y) : x 0, x + y 3, x2 4y and y 1 + x } is

(1)12

59(2)

2

3(3)

3

7(4)

2

5[JEE (Main) 2017]

Q.11 Let g(x) = cos x2 , f(x) = x and , ( < ) be the roots of the quadratic equation18x2 – 9x + 2 = 0. Then the area (in sq. units) bounded by the curve y = (gof) (x) and the lines x =,x = and y = 0, is

(1) 232

1 (2) 12

2

1

(3) 132

1 (4) 13

2

1 [JEE (Main) 2018]



EXERCISE-4

SECTION-A(CBSE Previous Year's Questions)


Q.1 Solve the following differential equation)xy2(x

)xy2(x

dx

dy

if y= 1 when x = 1. [CBSE Delhi 2008]

Q.2 Form thedifferential equation representing theparabolahavingvertex at the origin and axisalongpositivedirection of x-axis. [CBSE (AI) 2008]

Q.3 Solve the following differential equation xsinxcosydx

dy . [CBSE Delhi 2009]

Q.4 Find theparticular solution, satisfying the given condition, for the followingdifferential equation

0x

yeccos

x

y

dx

dy

, y = 0 when x = 1. [CBSE Delhi 2009]

Q.5 Form the differential equation of the familyof circles touching the y-axis at origin.[CBSE (AI) 2009]

Q.6 Form the differential equation representing the familyof curves given by (x – a)2 + 2y2 = a2, where a isan arbitrary constant. [CBSE (AI) 2009]

Q.7 Solve :

x

ytanxy

dx

dyx . [CBSE (AI) 2009, 2002]

Q.8 Solve the following differential equations x log xdx

dy+ y= 2 log x.

[CBSE Delhi 2009, (AI) 2008]

Q.9 What is the degree of the following differential equation xlogy6dx

yd

dx

dyx5

2

22

.

[CBSE Delhi 2010]

Q.10 Solve the following differential equationdx

dyxyyxyx1 2222 = 0. [CBSE (AI) 2010]

Q.11 Show that the differential equation (x – y)dx

dy= x + 2y, is homogeneous and solve it.

[CBSE (AI) 2010]



Q.12 Solve the following differential equation (x3 + x2 + x + 1)dx

dy= 2x2 + x. [CBSE (AI) 2010]

Q.13 Show that the following differential equation is homogeneous and then solve it.

y dx + x log

x

ydy – 2x dy = 0 [CBSE (AI) 2010, CBSE (F) 2010]

Q.14 Solve the following differential equation (x2 + 1)dx

dy+ 2xy = 4x2 . [CBSE (AI) 2010]

Q.15 Solve the following differential equation x dy – (y + 2x2) dx = 0. [CBSE (AI) 2011]

Q.16 Solve the differential equation (x2 – 1)dx

dy+ 2xy =

1x

12

.

x (– , – 1) (1, ) [CBSE (F) 2011; (AI) 2010; (F) 2009 ]

Q.17 Solve the following differential equation cos2 xdx

dy+ y= tan x.

[CBSE Delhi 2011, (AI) 2009, 2008]

Q.18 Find theparticular solution of thedifferential equation satisfying thegiven condition

(3x2 + y)dy

dx= x, x > 0, when x = 1, y = 1. [CBSE (AI) 2011, (F) 2010]

Q.19 Solve thefollowingdifferential equationdx

dy+ 2y tan x = sin x, given that y = 0, when x =

3

.

[CBSE (F) 2011]

Q.20 Solve : x dy – y dx = 22 yx dx [CBSE (AI) 2011, 2005]

Q.21 Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given thaty = 1 when x = 0. [CBSE (F) 2011, 2005,2004,1996 ]

Q.22 Form the differential equation of the familyof circles in the second quadrant and touching the coordinateaxes. [CBSE 2012]

Q.23 Find the particular solution of the differential equation x (x2 – 1)dx

dy= 1 ; y = 0 when x = 2.

[CBSE 2012]



Q.24 Solve the following differential equation : (1 + x2) dy + 2xy dx = cot x dx ; x 0. [CBSE 2012]

Q.25 Write the differential equation representing the familyof curves y=mx, where m is an arbitraryconstant.[CBSE 2013]

Q.26 Find the particular solution of the differential equationdy

dx+ x cot y = 2y + y2 cot y, (y 0),

given that x = 0 when y =2

. [CBSE 2013]

Q.27 Find the particular solution of the differential equationdy

1 x y xydx

, given that y= 0 when x = 1.

[CBSE 2014]

Q.28 Solve the differential equation12 tan xdy(1 x ) y e

dx

[CBSE 2014]

Q.29 Find the sum of the order and the degree of the following differential equation:

y =

3 2

2

dy d yx

dx dx

[CBSE 2015]

Q.30 Find thesolution of the followingdifferential equation

0dy)x1(ydx)y1(x 22 [CBSE 2015]

Q.31 Show that the differential equation (x – y)dy

dx= x + 2y is homogeneous and solve it also.

[CBSE 2015]

Q.32 Find the differential equation of the family of curves (x – h)2 + (y – k)2 = r2, where h and k are arbitraryconstants. [CBSE 2015]

Q.33 Find the particular solution of differential equation :xsin1

xcosyx

dx

dy

given that y = 1 when x = 0.[CBSE 2016]

Q.34 Find the particular solution of the differential equation 2yex/y dx + (y – 2x ex/y) dy= 0Given that x = 0 when y = 1.

[CBSE 2016]

Q.35 Find the general solution of the differential equation.

dx

dy– y = sin x [CBSE 2017]



Q.36 Find the particular solution of the differential equation (x – y)dx

dy= (x + 2y), given that y = 0 when

x = 1.[CBSE 2017]

Q.37 Find the differential equation representing the family of curves y = aebx + 5, where a and b are arbitraryconstants.

[CBSE 2018]

Q.38 Find the particular solution of the differential equation ex tan y dx + (2 – ex) sec2y dy = 0, given that

y =4

when x = 0. [CBSE 2018]

Q.39 Find the particular solution of the differential equation ,xsinxtany2dx

dy given that y= 0 when

x =3

. [CBSE 2018]


Q.1 Find the area of the region enclosed between the two circles x2 + y2 = 1 and (x – 1)2 + y2 = 1.[CBSE (AI) 2008, 1999; 1997 C, 1996 C, 1991]

Q.2 Find the area lying above x-axis and included between the circle x2 + y2 = 8x and the parabolay2 = 4x. [CBSE Delhi 2008]

Q.3 Using integration find the area of the region {(x, y) : 25x2 + 9y2 225 and 5x + 3y 15 }.[CBSE (F) 2009]

Q.4 Using the method of integration, find the area of the region bounded by the lines 2x + y = 4,3x – 2y = 6 and x – 3y + 5 = 0. [CBSE Delhi 2009]

Q.5 Prove that the curves y2 = 4x and x2 = 4y divide the area of the square bounded by x = 0, x = 4,y = 4 and y = 0 into three equal parts. [CBSE Delhi 2009]

Q.6 Find the area of the region enclosed between the two circles x2 + y2 = 9 and (x – 3)2 + y2 = 9.[CBSE Delhi 2009]

Q.7 Find the area of the region included between the parabola 4y = 3x2 and the line 3x – 2y + 12 = 0.[CBSE (AI) 2009]

Q.8 Using integration, find the area of the region. {(x, y) : 9x2 + y2 36 and 3x + y 6}.[CBSE (F) 2009]



Q.9 Using integration, find the area of the triangleABC with vertices as A(–1, 0), B(1, 3) and C(3, 2).[CBSE (F) 2009]

Q.10 Find the area enclosed by the parabola y2 = x and line y + x = 2.[CBSE (AI) 2009; CBSE 2005]

Q.11 Find the area of the region included between the parabolas y2 = 4ax and x2 = 4ay, where a > 0.[CBSE (AI) 2009; CBSE 2003, 2004]

Q.12 Find the area of the smaller region bounded by the ellipse4

y

9

x 22 = 1 and line

2

y

3

x = 1.

[CBSE Delhi 2010]

Q.13 Using integration, find the area of the followingregion

2x5y|1x|:)y,x( .

[CBSE Delhi 2010]

Q.14 Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y.[CBSE (AI) 2010]

Q.15 Using integration, find the area of triangleABC, co-ordinates of whose vertices areA(4, 1), B(6, 6) andC(8, 4). [CBSE (AI) 2010]

Q.16 Using integration, find the area of the region bounded by the lines 4x – y + 5 = 0, x + y – 5 = 0 andx – 4y + 5 = 0. [CBSE (F) 2010]

Q.17 Using integration, find the area of the following region.

2x20y|2x|:)y,x(

[CBSE (F) 2010]

Q.18 Find the area bounded by the curve x2 = 4y and the straight line x = 4y – 2.[CBSE Delhi 2010; CBSE 2004, 2005]

Q.19 Using integration find the area of the triangular region whose sides have equations y = 2x + 1,y = 3x + 1 and x = 4. [CBSE Delhi 2011]

Q.20 Sketch the graph of y = |x + 3| and evaluate the area under the curve y = |x + 3| above x-axis andbetween x = –6 to x = 0. [CBSE (AI) 2011]

Q.21 Using the method of integration, find the area of the region bounded by the lines :2x + y = 43x – 2y = 6x – 3y + 5 = 0 [CBSE (F) 2011]



Q.22 Find the area of the region {(x, y) : x2 + y2 4, x + y 2}. [CBSE 2012]

Q.23 Find the area of the region {(x, y) : y2 4x, 4x2 + 4y2 9} using method of integration.[CBSE 2013]

Q.24 Using integration, find the area of the region bounded by the triangle whose vertices are (–1, 2), (1, 5)and (3, 4) [CBSE 2014]

Q.25 If the area bounded bythe parabola y2 = 16ax and the line y= 4mx is2a

12sq. units, then using integration,

find the value of m. [CBSE 2015]

Q.26 Using the method of integration, find the area of the triangular region whose vertices are (2, –2) (4, 3)and (1, 2).

[CBSE 2016]

Q.27 Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices areA(4, 1), B(6, 6) and C(8, 4).

[CBSE 2017]

Q.28 Find the area enclosed between the parabola 4y = 3x2 and the straight line 3x – 2y + 12 = 0.[CBSE 2017]

Q.29 Using integration, find the area of the region in the first quadrant enclosed bythe x-axis, the line y= x andthe circle x2 + y2 = 32. [CBSE 2018]

SECTION-B(Potential Problems Based on CBSE)


Q.1 Write the order and degree of each of the following differential equations :

(i) 0ydx

dy

dx

yd2

dx

yd2

2

3

3

(ii) 0y

dx

dyx

dx

yd3

22

2

Q.2 Solve :

(i) sec2 x tan y dx + sec2 y tan x dy = 0 (ii) 0dyx

ydxy1e 2x .

Q.3 Solve ydx

dy sec x = tan x.



Q.4 Write the integrating factor in each of the following linear differential equations :

xe)x1(yx1

1

dx

dy

Q.5 ydx

dy)xlogx( = 2 log x.

Q.6 ysecx)y(tandy

dx 2

Q.7 Write the solution of the differential equation (ex + e–x) dy = (ex – e–x) dx.

Q.8 Write the general solution of the differential equationx

y

dx

dy .

Q.9 Form thedifferential equation representing thegiven familyof curves 1b

y

a

x byeliminatingarbitrary

constants a and b.

Q.10 Form the differential equation representing the given family of curves y = ae3x + be–2x by eliminatingarbitrary constants a and b.

Q.11 Solve the differential equation

dx

dyya

dx

dyxy 2 .

Q.12 Find the particular solution of the differential equation

dx

dylog = 3x + 4y given that y = 0

when x = 0.

Q.13 Solve : 0dyy

x1edxe1 y

x

y

x

.

Q.14dx

dy)y3x( 2 = y(y > 0).

Q.15 Solve the differential equation (tan–1 y – x) dy = (1 + y2) dx.

Q.16 Find the equation of the curve passing through the point

4,0 whose differential equation is

sin x cos y dx + cos x sin y dy = 0.



Q.17 Show that the given differential equation is homogeneous and solve it (x2 + xy) dy = (x2 + y2) dx.

Q.18 Solve : xdx

dy+ y – x + xy cot x = 0 (x 0).


Q.1 Find the area of the region bounded by the parabola y = x2 and y = |x| .

Q.2 Find the area of the region bounded by the curve y2 = 2y – x and the y-axis.

Q.3 Find the area of the region enclosed between the two circles : x2 + y2 = 4 and (x – 2)2 + y2 = 4.

Q.4 Using integration, find the area of region bounded by the triangle whose vertices are (1, 0), (2, 2) and(3, 1).

Q.5 Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x, and the circlex2 + y2 = 32.

Q.6 Find the area of the region bounded by the two parabolas y = x2 and y2 = x.

Q.7 Find the area of the region enclosed by the curve y = x2 and the line y = x.

Q.8 Using integration find the area of the region bounded by the parabola y2 = 4x and the circle4x2 + 4y2 = 9.

Q.9 Find the area of that part of the circle x2 + y2 = 16 which is exterior to the parabola y2 = 6x.

Q.10 Using integration, find the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7.

EXERCISE-5 (Rank Booster)

Q.1 Find all functions f (x) defined on

2,

2with real values and has a primitive F(x) such that

f (x) + cos x · F(x) = 2)xsin1(

x2sin

. Find f (x).

Q.2 Consider the differential equation,dy

dx+ P(x)y = Q(x)

(i) If two particular solutions of given equation u(x) and v(x) are known, find the general solution of thesame equation in terms of u(x) and v(x).

(ii) If and are constants such that the linear combinations · u(x) +·v(x) is a solution of the givenequation, find the relation betweenand.

(iii) If w(x) is the third particular solution different from u(x) and v(x) then find the ratiov x u x

w x u x

( ) ( )

( ) ( )

.



Q.3 Let f (x) be a differentiable function and satisfy f (0) = 2, f ' (0) = 3 and f '' (x) = f (x). Find(a) the range of the function f (x)(b) the value of the function when x = ln 2(c) the area enclosed by y = f (x) in the 2nd quadrant

Q.4 Given two curves y = f(x) passing through the points (0, 1) & y =

x

f (t)dt passing through the points

(0, 1/2). The tangents drawn to both curves at the points with equal abscissas intersect on the xaxis.Find the curve f(x).

Q.5 A& B are two separate reservoirs of water. Capacity of reservoirAis double the capacity of reservoirB. Both the reservoirs are filled completely with water, their inlets are closed and then the water isreleased simultaneously from both the reservoirs. The rate of flow of water out of each reservoir at anyinstant of time is proportional to the quantity of water in the reservoir at that time. One hour after thewater is released, the quantity of water in reservoirAis 1.5 times the quantity of water in reservoir B.After how manyhours do both the reservoirs have the same quantity of water ?

Q.6 Consider a circle x2 + (y – 1)2 = 1 and the parabola y = –4

x2. The common tangents to the two curves

constitute a triangle ABC, the point A and B being on the x-axis and C on the y-axis. CA producedtouches the parabola at P and CB produced touches the parabola at Q.

(a) Find the equation of the common tangent BC.

(b) Find the area of the portion between the upper arc of the circle and the common tangents QC and PC.

(c) Find the area enclosed by the parabola y = –4

x2, the x-axis and the linesAP and BQ.

Q.7 Consider one side AB of a square ABCD, (read in order) on the line y = 2x – 17, and the other twovertices C, D on the parabola y = x2.

(a) Find the minimum intercept of the line CD on y-axis.

(b) Find the maximum possible area of the squareABCD.

(c) Find the area enclosed by the line CD with minimum y-intercept and the parabola y = x2.

Q.8 Let f : R R be a differentiable function such that f(x + 2y) = f(x) + 4f(y) + 2y (2x – 1)for all x, yR and f '(0) = 1.

(a) Find the value of f(3) + f ' (3).

(b) Find the number of points of non-differentiability of the function y= |x|f)x(f .

(c) Find the area bounded by the curve f(x) and x-axis.



Q.9 Let f(x) be a continuous and positive function in [0, 5] and the area bounded by the graph of y= f(x) andx-axis for 0 x 5 is 8. A(c) denotes the area between the graph of y = f(x) and x-axis for 0 x c

and B(c) denotes the area between the graph of y = f(x) and x-axis for c x 5. Also R(c) =)c(B

)c(A,

R(3) = 1 anddc

dRis equal to 7 at c = 3.

(a) Find the value ofA(3).

(b) Find the value of A'(c) + B'(c) at c = 5.

(c) Find the value of f(3).

Q.10 In the adjacent figure, graphs of two functions y = f(x) andy = sinx are given. y = sinx intersects, y = f(x) at A(a, f(a));B(, 0) and C(2, 0).Ai (i = 1, 2, 3,) is the area bounded bythe curves y = f (x) and y= sinx between x=0 and x= a; i = 1,between x = a and x = ; i = 2, between x = and x = 2;i = 3. IfA1 = 1 –sina+ (a– 1)cosa,determine the function f(x).Hence determine ‘a’andA1.Also calculateA2 andA3.

Q.11 For what value of 'a' is the area of the figure bounded by the lines,

y =1

x, y =

1

2 1x, x = 2 & x = a equal to ln

4

5?

Q.12 Let C1 & C2 be two curves passing through the origin as shown in the figure.A curve C is said to "bisect the area" the region between C1 & C2, if for eachpoint Pof C, the two shaded regionsA& B shown in the figure have equalareas. Determine the upper curve C2, given that the bisecting curve C hasthe equation y = x2 & that the lower curve C1 has the equation y = x

2/2.



EXERCISE-1


SPECIAL DPP-1

Q.1 C Q.2 C Q.3 B Q.4 C Q.5 B

Q.6 C Q.7 D Q.8 B Q.9 A Q.10 0004

SPECIAL DPP-2

Q.1 A Q.2 D Q.3 C Q.4 A Q.5 C

Q.6 C Q.7 A Q.8 AD Q.9 ABC Q.10 0062

SPECIAL DPP-3

Q.1 D Q.2 B Q.3 A Q.4 D Q.5 C

Q.6 D Q.7 A Q.8 C Q.9 AB Q.10 ABD


SPECIAL DPP-1

Q.1 C Q.2 D Q.3 C Q.4 A

Q.5 B Q.6 A Q.7 B Q.8 C

Q.99

2sq. units Q.10 A =

3

4sq. units

SPECIAL DPP-2

Q.1 A Q.2 A Q.3 B Q.4 A Q.5 A

Q.6 C Q.7 B Q.8 A Q.9 BCD Q.10 0002

SPECIAL DPP-3

Q.1 C Q.2 B Q.3 B Q.4 D Q.5 B

Q.6 D Q.7 AC Q.8 0001 Q.9 0002 Q.10 0005



EXERCISE-2


Q.1 (i) order 2 & degree 3 (ii) order 2 & degree 2

Q.2 (a) 12y (y)2 = x[8(y)3 27]; (b) xyd y

dx

2

2 + xdy

dx

2

ydy

dx= 0;

(c) [1 + (y)²] .y 3y(y)² = 0 (d)d y

dx

d y

dx

dy

dx

3

3

2

26 11 6y = 0

Q.3 ln

2

yxtan1 = x + c Q.4 y sin y = x² ln x + c Q.5 y = etan(x/2)

Q.6 (a) P = 1000 + 1500e–kt where k =

3

5n

10

1l ; (b) T = 10 log5/3(3); (c) P = 1000 as t

Q.7 m=m0ekt where k =

1

0tln 1

100

Q.8 x2 + y2 = k2 Q.9 19

Q.10 y = kx or xy = c Q.114

1

Q.12 c(x y)2/3 (x² + xy + y²)1/6 = exp

3x

y2xtan

3

1 1 where exp x ex

Q.132

222

x

xyyy = 3

222

x

cxyyn

, where same sign has to be taken.

Q.15 xy cosx

y= c Q.16 x + y +

4

3= ce3(x2y) Q.17 e

yx

223

1tan

= c . (y + 2)

Q.18 0 Q.19 cos x – 1 Q.209

727 minutes

Q.21 y = cex ; y = c +x2

2Q.22 y = 1

xn cxtan Q.23 y =

1

3 e(2 ex e + 1)

Q.24 y = ± 4 22 42

2

x nx

x Q.25 y2 = 2x + 1 e2x


Q.1 5/6 sq. units Q.2 64 Q.3

tan 12 2

3;

tan 1

4 2

3

Q.4

2

1

1;

Q.5

3 2

2

Q.6 (i) m = 1, (ii) m = ; Amin= 4/3

Q.7 e Q.8 2 sq. units Q.9 a= 3/4

Q.10 4 Q.112

1(1 – e–1/2 ) Q.12 2/3

Q.13 12 Q.14 1/2



EXERCISE-3

SECTION-A


Q.1 C

Q.2 (i) (A) p, q, s (B) p, t (C) p, q, r, t (D) s ; (ii) (A) p; (B) q, s ; (C) q, r, s, t ; (D) r

Q.3 9 Q.4(a) 6 ;(b) 0 Q.5 AD Q.6 A Q.7 B

Q.8 AC Q.9 BC Q.10 A Q.11 AD Q.12 C

Q.13 BCD Q.14 0.40

AREAUNDER THE CURVE

Q.1 (a) B, (b) (i) B, (ii) A, (iii) D Q.2 BCD

Q.3 (i) C; (ii) A; (iii) B Q.4(a) B ; (b) C

Q.5 B Q.6 3 Q.7 CD Q.8 C

Q.9 AD Q.10 BC

SECTION-B


Q.1 2 Q.2 3 Q.3 4 Q.4 2 Q.5 2

Q.6 1 Q.7 3 Q.8 2 Q.9 2 Q.10 1

Q.11 1 Q.12 1 Q.13 1

AREAUNDER THE CURVE

Q.1 3 Q.2 3 Q.3 1 Q.4 3 Q.5 1

Q.6 4 Q.7 2 Q.8 2 Q.9 3 Q.10 4

Q.11 3

EXERCISE-4

SECTION-A


Q.17

3tan

7

62logxlog2

x7

xy4tan

7

6

x

xxyy2log 11

2

22

Q.2dx

dy=

x2

yQ.3 y = cos x + Ce– x Q.4 cos

x

DIFFERENTIAL EQUATION AND AREA UNDER THE CURVE · 2020. 2. 1. · DIFFERENTIAL EQUATION & AREA...

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Transcript of DIFFERENTIAL EQUATION AND AREA UNDER THE CURVE · 2020. 2. 1. · DIFFERENTIAL EQUATION & AREA...