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