cbse cbse class 12th mathematics solved question paper 2011 set ii

8/12/2019 CBSE CBSE Class 12th Mathematics Solved Question Paper 2011 Set II

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Chemistry Class 12th

Mathematics Solved Question Papers 2011

Get SOLVED & UNSOLVED question papers, updated Syllabus, Sample papers and study material

and much more


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and much more

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and much more

Solved Question Paper 2011

Mathematics

Class

XIISet - II

1. Write the intercept cut off by the plane 2x + y- z = 5 on x-axis.

Answer

2x + y z = 5

Divide the equation by 5

2 1 11

5 5 5

1 1 11

5 5 5

2

x y z

x y z

Thus, the intercepts cut off by the plane are =5

,5, 5

2

2. Write the direction cosines of the vector 2 5i j k

Answer

2 22

2 5

2 1 5

4 1 25

30

2 1 5Thus, direction cosines of are , ,

30 30 30

a i j k

a

a

a

a


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and much more

3. For what value of a the vectors 2 3 4 & 6 8i j k ai j k are collinear?

Answer

Let 2 3 4

6 8

2 3 42

,

22

4

P i j k

Q ai j k

aQ i j k

Thus

a

a

4. Write the value of:

216

dx

x

Answer

2

2 2

1

16

4

1tan

4 4

dx

x

dx

x

xC

5. Write A-1 for:

2 5

1 3A

Answer


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and much more

1 1

2 2 1

1 1 2

2 2

1

2 5

1 3

2 5 1 0

1 3 0 1

1

2

5 11 0

2 2

1 3 0 1

5 11 0

2 2

1 10 1

2 2

5

1 0 3 5

1 10 1

2 2

2

1 0 3 5

0 1 1 2

3 5

1 2

A

A IA

A

R R

A

R R R

A

R R R

A

R R

A

A

6. For what value of x, the matrix5 1

2 4

x xA

is singular?

Answer


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and much more

5 1

2 4

Matrix A is singular. Thus,

0

5 10

2 4

4 5 2 1 0

20 4 2 2 0

6 18 0

3

x xA

A

x x

x x

x x

x

x

7. For a 2x 2 matrix,ij

A a whose elements are given by iji

aj

, write the value of

12a

Answer

ijA a

Here, 1, 2

,

ij

ia

j

i j

Thus

12

1

2a

8. State the reason for the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)) notto be transitive.

Answer


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and much more

Let A= {1, 2, 3}.

R = {(1, 2), (2, 1)}.

We know that,

(1, 1), (2, 2), (3, 3) R

Hence, R is not reflexive.

Now,

As (1, 2) R and (2, 1) R, then R is symmetric.

(1, 2) and (2, 1) R

Also,

(1, 1) R

Thus, R is not transitive.

Therefore, R is symmetric but neither reflexive nor transitive.

9. Write the value of

1 3tan tan

4

Answer

1 3tan tan

4

1

1

3tan tan

4

tan tan4


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and much more

10.Write the value of:

2

2

sec

cos

xdx

ec x

Answer

2

2

2

2

2

2

sec

cos

sin

cos

tan

sec 1

tan

xdx

ec x

xdx

x

xdx

x dx

x x C

11.Probabilities of solving a specific problem independently by A and B are & 1/3and respectively. If both try to solve the problem independently, find the probability

that (i) the problem is solved (ii) exactly one of them solves the problem.

Answer

P (A) =1

2

P (B) =1

3

Problem is solved independently by A and B,

Thus,

1

1

tan tan4

tan tan4

4


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and much more

' 1

1 1

1 2 2

' 1

1 2 1

3 3

.

1 1 1 .

2 3 6

P A P A

P B P B

P AB P A P B

a) Probability (problem is solved) = P (A B)

P (A B) = P (A) + P (B) P (AB)

1 1 1

2 3 6

2

3

b) Probability (exactly one of them solves the problem) =

. ' . 'P A P B P B P A

1 2 1 1

2 3 2 3

1 1

3 6

1

2

12. Find the angle between the following pair of lines :

2 1 3 2 2 8 5&

2 7 3 1 4 4

x y z x y z

Answer


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and much more

2 1 3 2 2 8 5&

2 7 3 1 4 4

x y z x y z

1

2

1 2

1 2

1 2

2 1 3 2 4 5and

2 7 3 1 2 4

2 7 3

2 4

.Angle between the given pair of lines,cos

. 2 7 3 4 4

2 1 7 4 3 4

2 28 12

x y z x y z

a i j k

a i j k

a a

a a

a a i j k i j k

28 14

14

22 2

1

2 2 2

2

1

2 7 3 4 49 9 62

1 2 4 1 4 16 21

14cos

62 21

14

93

14cos

93

a

a

13.Solve the following differential equation:

2cos tandy

x y xdx


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and much more

Answer

2

2

2 2

2 2

sec tan

2 tan tan 2

tan tan 2 tan

cos tan

sec sec .tan

Where sec , sec tan

.

Now, multiplying the equation by I.F

sec sec .tan

sec s

Pdx xdx x

x x

x x x

dyx y x

dx

dyxy x x

dx

dyPy Q

dx

P x Q x x

I F e e e

dyxy e e x x

dx

dye e xy e

dx

2

tan tan 2

2

tan

ec . tan

. sec .tan

Let tan

sec

. . ...................................... 1

Let .

.

.

.

x x

x t

t

t t

t t

t t

x x

y e e x x dx

x t

xdx dt

y e t e dt

I t e dt

dI t e dt t e dt dt

dt

I t e e dt

I t e e C

tan

tan tan tan

tan

Now, equation 1 becomes as

. .

. tan .

tan 1

x t t

x x x

x

y e t e e C

y e x e e C

y x Ce


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and much more

14.Evaluate:

2

5 3

4 10

xdx

x x

OR

Evaluate:

2 22

1 3

xdx

x x

Answer

2

2

5 3

4 10

Let 3 4 10

5 3 2 45 3 2 4

x

x x

dx A x x B

dx

x A x B

x Ax A B

On equating coefficients

5 2

5

2

3 4

53 4

2

7

x Ax

A

A B

B

B

2

2

,

5 3

4 10

5

2 4 72

4 10

Now

xdx

x x

x

dx

x x


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and much more

2 2

2

2

2 45 17

2 4 10 4 10

....................... 1

2 45

2 4 10

Let 4 10

2 4

xdx dx

x x x x

A B Equation

xA dx

x x

x x t

x dx dt

2

2

2

,

2 45

2 4 10

5 1

2

52 4 10

2

5 4 10

Now

xdx

x x

dtt

x x

x x

2

2

22

17

4 10

17

4 4 6

17

2 6

B dx

x x

dx

x x

dx

x

2

2 2

7log 2 4 10

Put value of A & B in equation 1

5 4 10 7log 2 4 10

x x x

A B

x x x x x C


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and much more

OR

2 22

1 3

x

dxx x

Let x2

= t 2x dx= dt

1

1 3

1Let

1 3 1 3

1 3 1

dtt t

A B

t t t t

A t B t

Put t= 3 and t= 1, we obtain

2 2

2

2

1 1, and

2 2

,

1

1 3

1 1

2 1 2 3

1 1log 1 log 3

2 2

1 1log 1 log 3

2 2

1 1log

2 3

A B

Now

dtt t

dxt t

t t C

x x C

xC

x

14 Form the differential equation of the family of parabolas having vertex at the origin

and axis along positive y-axis.

Answer

Vertex = (0, 0)


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and much more

The equation of the parabola

2

2

2

4 ....................................... 1

On differentiation

2 4 '

2 '

2 '

Put value of a in equation 1

42 '

' 2

' 2

' 2 0

x ay

x ay

x ay

xa

y

xx y

y

y x xy

y x y

xy y

This is the required differential equation.

15 Find the equation of the plane through the line of intersection of the planes x + y + z =

1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x y + z = 0

Answer

Equations of planes are:

x + y + z = 1 and 2x + 3y + 4z = 5

The equation of the plane passing through the intersection of the given planes is:


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and much more

1 2 3 4 5 0

2 1 3 1 4 1 5 1 0

x y z x y z

x y z

The direction ratios of this plane are:

a1 = (2 + 1)

b1= (3 + 1)

c1= (4 + 1)

The given plane xy + z = 0 is perpendicular to the equation of desired plane

Direction ratios of plane, x y + z = 0 are:

a2= 1

b2= 1

c2= 1

Both planes are perpendicular. Thus,

1 2 1 2 1 2 0

2 1 3 1 4 1 0

3 1 0

1

3

a a b b c c

Equation of desired plane becomes as:

1 1 1 1

2 1 3 1 4 1 5 1 03 3 3 3

1 1 20

3 3 3

2 0

x y z

x z

x z

This is the required equation of the plane.

16 Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone onthe ground in such a way that the height of the cone is always one-sixth of the radius


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and much more

of the base. How fast is the height of the sand cone increasing when the height is

4cm?

OR

Find the points on the curve x2 + y2- 2x-3=0 at which the tangents are parallel to x-

axis.

Answer

The volume of a cone,

21

3V r h

It is given that,

6

6

rh

r h

2

2

3

1

3

16

3

12

V r h

V h h

V h

The rate of change of volume with respect to time is:

3

2

12

36 .

V h

dV dhh

dt dt

We have,

312 /

dVcm s

dt

When h= 4 cm,

2

2

36 .

12 36 4 .

1

48

dV dhh

dt dt

dh

dt

dh

dt


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and much more

Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of

1 /

48

cm s

.

OR

2 22 3 0

On differentiation

2 2 . 2 0

1

x y x

dyx y

dx

dy x

dx y

Now, the tangents are parallel to the x-axis if the slope of the tangent = 0

10

1 0

1

x

y

x

x

Also, x2

+ y2

2x 3 = 0

y2

= 4

2y

Hence, (1, 2) and (1, 2) are the points at which the tangents are parallel to the x-axis.

17 Differentiate:

2

cos

2

1w.r.t

1

x x x

x xx

Answer


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and much more

2cos

2

1Let

1

x x xy xx

y u vdy du dv

dx dx dx

cos

,

Taking log both sides

log cos log

On differentiation

1. .cos . log .log . cos cos .log .

1 1. .cos . .log . sin cos .log .1

1. cos sin log cos log

x x

Now

u x

u x x x

du d d d x x x x x x x x x

u dx dx dx dx

dux x x x x x x

u dx x

dux x x x x x

u dx

du

dx

cos

. cos sin log cos log

. cos sin log cos log

x x

u x x x x x x

du

x x x x x x xdx

Now,

2

2

2

2

2 2

2 2

2 2

2 2

2 2

1

1

Taking log both sides

1log log

1

log log 1 log 1

On differentiation

1 1 1. . 1 . 1

1 1

1 1 1. .2 .2

1 1

1 2 2.

1

xv

x

xv

x

v x x

dv d d x x

v dx x dx x dx

dvx x

v dx x x

dv x x

v dx x x

1


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and much more

2 2

2

2 2 2

2 22

2 2 2

2 3 3

2 2 2

2

2 2 2

2 2.

1 1

1 2 2.

1 1 1

2 1 2 11

1 1 1

1 2 2 2 2

1 1 1

1 4

1 1 1

dv x xv

dx x x

dv x x x

dx x x x

x x x xdv x

dx x x x

dv x x x x x

dx x x x

dv x x

dx x x x

d

22

cos

22

4

1

,

4. cos sin log cos log

1

x x

v x

dx x

Thus

dy du dv

dx dx dx

dy xx x x x x x x

dx x

18. (or) If 2

2sin , 1 cos , find

d yx a y a

dx

Answer

sin

1 cos

x a

dx ad

1 cos

sin

sin

y a

dya

d

dya

d


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and much more

.

sin

1 cos

dy dy d

dx d dx

dy a

dx a

2

2

2

2

2

2

2

2

2

22

2

4

2

sin

1 cos

2sin .cos2 2

2sin2

cos2

sin2

cot2

1cos .

2 2

1 1cos .

2 2 1 cos

1 1cos .

2 22 sin

2

1cos

4 2

dy

dx

dy

dx

dy

dx

dy

dx

d y dec

dx dx

d yec

dx a

d yec

dxa

d yec

dx a

19. If the function f(x) given by:

3 , if 1

11, if 1

5 2 , if 1

ax b x

f x x

ax b x

Answer

3 , if 1

11, if 15 2 , if 1

ax b x

f x xax b x


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and much more

Function fis defined at all points of the real line.

Iffis a continuous function, thenfis continuous at all real numbers.

Sincefis continuous atx= 1,

1 1

1 1

lim lim 1

lim 5 2 lim 3 11

5 2 3 11

5 2 11

3 11

x x

x x

f x f x f

ax b ax b

a b a b

a b

a b

On solving these two equations

5 11

2

3 11

5 113 11

2

6 5 11 22

11 33

3

5 11

2

5 3 112

2

ab

a b

axa

a ax

a

a

ab

b

Thus,

a= 3

b= 2

20. Using properties of determinants, prove the following:

2 2 23 3 3

x y z

x y z xyz x y y z z x

x y z


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and much more

Answer

2 2 2

3 3 3

2 2 2

2 2 1

3 3 1

1 1 1

LHS

x y z

x y z

x y z

xyz x y z

x y z

C C C

C C C

2 2 2 2 2

2

1 0 0

1 0 0

1 1

xyz x y x z x

x y x z x

xyz y x z x x

x y x z x

xyz y x z x z x y x

xyz y x z x z y

xyz x y y z z x

RHS

21. Prove the following:

1 1 sin 1 sincot , 0,

2 41 sin 1 sin

x x xx

x x

OR

21. Find the value of

1 1

tan tanx x y

y x y


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and much more

Answer

1

2 2

2

2 2

2

1

1 sin 1 sincot

1 sin 1 sin

We know that,

1 sin cos sin 2sin cos2 2 2 2

cos sin2 2

1 sin cos sin 2sin cos2 2 2 2

cos sin2 2

,

cos si2cot

LHS

x x

x x

x x x xx

x x

x x x xx

x x

Now

x

1

1

n cos sin2 2 2

cos sin cos sin2 2 2 2

2cos2cot

2sin2

cot cot2

2

Hence Proved

x x x

x x x x

x

x

x

x

RHS

OR


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and much more

1 1

1 1

1 1

1 1 1

1 1 1

1

1

tan tan

1

tan tan

1

1

tan tan

1

tan tan tan 1

tan tan tan 1

tan 1

tan

x x y

y x y

x

x y

xy

y

x

x y

xy

y

x x

y y

x x

y y

tan4

4

22. Consider the binary operation * on the set (1, 2, 3, 4, 5}defined by a * b = min. {a, b}

write the operation table of the operation *.

Answer

a* b= min {a,b}

a,b {1, 2, 3, 4, 5}.

Hence, the operation table for the given operation * is:

* 1 2 3 4 5

1 1 1 1 1 1

2 1 2 2 2 2

3 1 2 3 3 3


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and much more

4 1 2 3 4 4

5 1 2 3 4 5

23. Bag I contains 3 red and 4 black balls and Bag II contains 5 red and 6 black balls.

One ball is drawn at random from one of the bags and is found to be red. Find theprobability that it was drawn from Bag II.

Answer

Let

E1: Event of selecting first bag

E2: Event of selecting second bag

A: Event of getting a red ball

1

2

1

2

1

2

12

Red ball from first bag

3

7

Red ball from second bag

5

11

P E

P E

P A E P

P A E P

P (E2|A) = probability of drawing a red ball from the second bag

By Bayes theorem,


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and much more

2 2

2

1 1 2 2

.

. .

1 5.

2 11

1 3 1 5. .

2 7 2 11

5

22

3 5

14 22

5

22

68

154

35

68

P E P A E P E A

P E P A E P E P A E

24. A factory makes tennis rackets and crickct bats. A tennis racket takes 1.5 hours of

machine time and 3 hours of crafiman's time in its making while a cricket bat takes 3hours of machine time and 1 hour of craftmnan's time. In a day, the factory has theavailability of not more than 42 hours of machine time and 24 hours of craftsman's

time. If the profit on a racket and on a bat is Rs. 20 and Rs. 10 respectively, find thenumber of tennis rackets and crickets bats that the factory must manufacture to earn

the maximum profit. Make it as an L.P.P. and solve graphically.

Answer

Let the number of rackets = x

The number of bats =y

As per the question,

1.5x+ 3y= 42

3x+y= 24

On solving these equations, we have

x= 4 andy= 12

Thus, rackets are 4 and bats are 12


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and much more

Tennis Racket Cricket Bat Availability

Machine Time (h) 1.5 3 42

Craftsmans Time(h) 3 1 24

1.5x+ 3y 42

3x+ y 24

x, y 0

The profit on a racket = Rs 20

Profit on a bat = Rs 10

20 10Z x y

Now,

The corner points are:

A = (8, 0)

B = (4, 12)

C = (0, 14)

O = (0, 0)

Now,

Corner point Z = 20x+ 10y

A(8, 0) 160

B(4, 12) 200

C(0, 14) 140

O(0, 0) 0


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and much more

Thus, the maximum profit of the factory when it works to its full capacity is Rs 200.

25. Find the equation of the plane which contains the line of intersection of the planes

. 2 3 4 0, . 2 5 0 and which is perpendicular to the plane

. 5 3 6 8 0.

r i j k r i j k

r i j k

Answer

The equation of planes are:

. 2 3 4 0

. 2 5 0

The equation of the plane passing through the line intersection

of the given planes is:

. 2 3 4 . 2 5 0

2 1 2 3

r i j k

r i j k

r i j k r i j k

r i j k

5 4 0............ (1)

The plane in equation (1) is perpendicular to the plane . 5 3 6 8 0

5 2 1 3 2 6 3

r i j k


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and much more

7

19

Put value of in equation (1)33 45 50 41

019 19 19 19

33 45 50 41 0

This is the vector equation of required plane.

The cartesian equation of this plane is:

33 45 50 41 0

r i j k

r i j k

x y z

26. Evaluate:

2

1

0

2sin cos tan sinx x x dx

OR

Evaluate:

2

4 4

0

sin cos

sin cos

x x xdx

x x

Answer

21

0

21

0

1

1 1

2sin cos tan sin

Let sin

cos .

2 tan .............................. (1)

,

2 tan

tan 2 tan 2

x x x dx

x t

x dx dt

t tdt

Now

t tdt

dt tdt t tdt dt

dt


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and much more

2

1 2

2

22 1

2

2 1

2

2 1 1

tan1

1 1tan1

1tan 1

1

tan tan

tt t dt

t

tt t dt t

t t dt t

t t t t

21

0

2 1 1 2

0

2 1 1 2

0

2 1

2 1 1

1

1

from 1

2 tan

tan tan

sin tan sin sin tan sin

sin 0.tan sin 0 sin 0sin tan sin sin tan sin

2 2 2 2 tan sin 0

tan 1 1 t

t tdt

t t t t

x x x x

1an 1 0

14 4

12

OR

2

4 4

0

2

4 4

0

2

4 40

2

4 4

0

sin cos

sin cos

sin cosLet ........................... (1)

sin cos

sin cos2 2 2

sin cos2 2

cos sin2

.

cos sin

x x xdx

x x

x x xI dx

x x

x x x

I dx

x x

x x x

I dx

x x

............................. (2)


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and much more

2

4 4

0

2

4 4

0

4

22

4

0

2

2

, 1 2

sin cos22

sin cos

sin cos

4 sin cos

Divide the equation by cos

tan .sec4 tan 1

Put tan

2 tan .sec

Now

x x

I dxx x

x xI dx

x x

x

x xI dxx

x t

x x dt

2

2

0

1 2

0

1 2 2

0

1 2 1 2

1 1

2

8 t 1

tan8

tan tan8

tan tan tan tan 08 2 8

tan tan 08 8

08 2 8

16

dtI

I t

I x

I

I

I

I

27. Using integration find the area of the triangular region whose sides have equations

Y=2x+1, y =3x+l and x = 4


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and much more

2 3 10 4 6 5 6 9 204, 1, 2; , , 0x y z

x y z x y z x y z

OR

Using elementary transformations, find the inverse of the matrix

1 3 2

3 0 1

2 1 0

Answer

2 3 104

4 6 51

6 9 202

x y z

x y z

x y z

1

,

1

2 3 10 41

4 6 5 1

6 9 20 21

2 3 10

4 6 5

6 9 20

2 120 45 3 80 30 10 36 36

150 330 720

1200 0

Now

x

y

z

AX B

X A B

A

A


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and much more

11 12 13

21 22 23

31 32 33

,

75 110 72

150 100 0

75 30 24

Now

C C C

C C C

C C C

75 110 72

150 100 0

75 30 24

75 150 75

110 100 30

72 0 24

T

adjA

adjA

1

1

75 150 751

110 100 301200

72 0 24

AdjAA

A

A

1

,

75 150 75 41

110 100 30 11200

72 0 24 2

300 150 1501

440 100 601200

288 0 48

Now

X A B

X

X

6001

4001200

240

X


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and much more

600

1200

400

1200

240

1200

1 600

1200

1 400

1200

24011200

2, 3, and 5

X

x

y

z

x y z

OR

1 3 2

3 0 1

2 1 0

1 3 2 1 0 0

3 0 1 0 1 0

2 1 0 0 0 1

A

A IA

A

R2 R2+ 3R1

R3 R3 2R1,

2 2

1 3 2 1 0 0

0 9 7 3 1 0

0 5 4 2 0 1

1

9

1 3 2 1 0 0

7 1 10 1 0

9 3 9

0 5 4 2 0 1

A

R R

A


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and much more

1 1 2

3 3 2

3 3

3

5

1 11 0 0 0

3 3

7 1 10 1 0

9 3 9

1 1 50 0 1

9 3 9

9

1 11 0 0 03 3

7 1 10 1 0

9 3 9

0 0 1 3 5 9

R R R

R R R

A

R R

A

1 1 3

2 2 3

1

3

7

9

1 0 0 1 2 3

0 1 0 2 4 7

0 0 1 3 5 9

R R R

R R R

A

1

1 2 3

2 4 7

3 5 9

A

29. Show that of all the rectangles inscribed in a given fixed circle, the square has the

largest area.

Answer

Let length of rectangle

breadth of rectangle

Radius of circle

Diagonal 2

l

b

a

a


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and much more

2 2 2

2 2 2

2 2

2 2

2 2

2 2

2 2

22 2

2 2

By pythagoras theorem,

2

4

4

Area of the rectangle,

. 4

. 4

On differentiation

1 24

2 4

4

4

a l b

a l b

b a l

A l b

l a l

A l a l

ldAl a l

dl a l

dA la l

dl a l

dA

dl

2 2

2 2

2 2 2 2

2 22

2 2 2

2 2 2 22

322 2 2

2 2 3

322 2 2

2 22

322 2 2

4 2

4

On further differentiation,

24 4 4 2

2 4 2

4

4 4 4 2

4

12 2

4

2 6

4

a l

a l

ll a l a l

a ld A

dl a l

l a l a l ld A

dla l

d A a l l

dla l

l a ld A

dla l


39/39



2

2

2 2

,

0

4 0

2

Now

d A

dl

a l

l a

2 2

2 2

2 22

32

2 2 2

2

2

4

4 2

2

,2 2 6 2

4 2

4 0

b a l

b a a

b a

Now

a a ad A

dla a

d A

dl

Here, length=breadth

Thus, square has the maximum area.

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cbse cbse class 12th mathematics solved question paper 2011 set ii

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