SOLUTIONS CLASS-XII MATHS SET-1 CBSE 2014 · 2018-12-18 · CBSE SAMPLE PAPER SOLUTIONS CLASS-XII...
Transcript of SOLUTIONS CLASS-XII MATHS SET-1 CBSE 2014 · 2018-12-18 · CBSE SAMPLE PAPER SOLUTIONS CLASS-XII...
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CBSE 2014
1. Given that A2 = A.
We need to find the value of 7A – (I + A)3, where I is the identity matrix.
Thus,
7A - (I + A)3 = 7A - (I3 + 3I2A + 3I2A + A3)
⟹ 7A - (I + A)3 = 7A – (I3 + 3A + 3A2 + A2 x A) [I3 = I, I2A = A, IA2 = A2]
⟹ 7A – (I + A)3 = 7A – (I + 3A + 3A + A) [∵A2 = A]
⟹ 7A – (I + A)3 = 7A – I – 3A – 3A – A
⟹ 7A – (I + A)3 = 7A – I – 7A
⟹ 7A – (I + A)3 = - I.
2. Given that
50
41
2 wyx
zyx
We need to find the value of x + y.
50
41
2 wyx
zyx
Two matrices A and B are equal to each other, if they have the same dimensions and the same
elements aij = bij, for I = 1, 2, …, n and j = 1, 2, …,m.
x – y = -1…(1)
2x – y = 0…(2)
Equation (2) – (1) is x = 1
Substituting the value of x = 1 in equation (1), we have
1 – y = -1
⟹ y = 2
Therefore, x + y = 1 + 2 = 3
3. Given that tan-1x + tan-1y = 4
and xy < 1.
We need to find the value of x + y + xy.
tan-1x + tan-1y = 4
CBSE SAMPLE PAPER SOLUTIONSCLASS-XII
MATHS SET-1
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CBSE
11
11
4tan
1tantan
141
tan
1
1
xyyxxyyx
xy
yx
xy
yx
xyxy
yx
4. Given that46
78
42
73
x
We need to find the value of x
46
78
42
73
x
⟹ 12x – (– 14) = 32 – 42
⟹ 12x + 14 = – 10
⟹ 12x = – 10 – 14
⟹12x = – 24
⟹ x = – 2
5. Since differentiation operation is the inverse operation of integration, we have xxxf sin)('
Let x
tdttxf0
sin)(
Let us do this by integration by parts.
Therefore assume u = t; du = dt
dvtdtsin
−cost = v
Therefore,
Cxxxxf
dttttxf
xx
sincos
coscos0
0
Differentiating the above function with respect to x,
xxxxxxxf sincoscossin'
6. Since the vectors are parallel, we have
2014
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CBSE
kjpikji
kjpikji
ba
ˆ3ˆ2ˆˆ9ˆ2ˆ3
)ˆ3ˆ2ˆ(ˆ9ˆ2ˆ3
Comparing the respective coefficients, we have
⟹ λ = 3;
−2 λp = 2
⟹ −2 x 3 x p = 2
⟹ p = 3
1
7. The set of natural numbers, N = {1,2,3,4,5,6…}
The relation is given as
R = {(x, y) : x + 2y = 8}
Thus, R = {(6,1),(4,2),(2,3)}
Domain = {6,4,2}
Range = {1,2,3}.
8. Given that the cartesian equation of the line as
4
62
7
4
5
3
zyx
That is,
2
3
7
)4(
5
3
4
)3(2
7
)4(
5
)3(
zyx
zyx
Any point on the line is of the form :
32,47,35
Thus, the vector equation is of the form:
bar
, where a
is the position vector of any point on the line and b
is the vector parallel to
the line.
Therefore, the vector equation is
kjikjir
kjikjir
kjir
ˆ2ˆ7ˆ5ˆ3ˆ4ˆ3
ˆ3ˆ4ˆ3ˆ2ˆ7ˆ5
ˆ32ˆ47ˆ35
9. Given that
a
x
dx
0
2 84
We need to find the value of a.
2014
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CBSE
Let I =
a
x
dx
0
2 84
Thus , I = 82
tan2
1
0
1
ax
212
42tan
82
2tan
82tan
2
1
1
1
1
aa
a
a
a
10. Given that ba
and are two perpendicular vectors.
Thus, 0. ba
Also given that, .5 and 13 aba
We need to find the value of b
.
Consider : 2
ba
12
144
25169
25169
02513
. 2
2
2
2
222
222
b
b
b
b
b
bbaaba
SECTION – B 11. Given differential equation is:
2
tan
2
tan2
11
1
1
1
x
e
x
y
dx
dy
eydx
dyx
x
x
This a linear equation of the form
2014
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CBSE
x
e
xP
QPydx
dy
x
1Q and
1
1 Where
1tan
2
Therefore,
I.F = xPdx
ee1tan
Thus the solution is,
y(I.F) = dxFIQ .
; Substitute
11
1̀
1
1
tan
tan
2
tan
tan
te
dxex
eey
x
x
x
x
Thus,
C
eey
Ct
ey
tdtey
xx
x
x
2
22
tantan
2tan
tan
1
1
1
1
12. Given position vectors of four A, B, C and D are:
kjiOD
kjiOC
kjOB
kjiOA
ˆˆˆ4
ˆ4ˆ9ˆ3
ˆˆ
ˆˆ5ˆ4
These points are coplanar, if the vectors, ADACAB and , are coplanar.
kjikjikji
OAODAD
kjikjikji
OAOCAC
kjikjikj
OAOBAB
ˆ3ˆˆ8ˆˆ5ˆ4ˆˆˆ4
ˆ3ˆ4ˆˆˆ5ˆ4ˆ4ˆ9ˆ3
ˆ2ˆ6ˆ4ˆˆ5ˆ4ˆˆ
These vectors are coplanar if and only if, they can be expressed as a linear combination of other
two.
2014
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CBSE
kyxjyxiyxkji
ADACxAB
ˆ33ˆ4ˆ8ˆ2ˆ6ˆ4
Comparing the coefficeints, we have,
−x – 8y = – 4;4x – y =– 6; 3x + 3y = –2
Thus, solving the first two equations, we get
3
2 and
3
4
yx
These values of x and y satisfy the equation 3x + 3y = –2.
Hence the vectors are coplanar.
OR
Given that
kjic
kjib
ˆ3ˆ2ˆ
ˆ5ˆ4ˆ2
Now consider the sum of the vectors :cb
kjicb
kjikjicb
ˆ2ˆ6ˆ2
ˆ3ˆ2ˆˆ5ˆ4ˆ2
Let n̂ be the unit vector along the sum of vectors cb
:
222262
ˆ2ˆ6ˆ2ˆ
kjin
The scalar product of na
and is 1. Thus,
1
88
3612444
3612444
6402
6262
262262
262
2.16.1211
262
ˆ2ˆ6ˆ2.ˆˆˆˆ.
22
22
222
222
222
222
kjikjina
2014
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CBSE
kjin
kjin
kjin
kjin
kjin
n
ˆ7
2ˆ7
6ˆ7
3ˆ
7
ˆ2ˆ6ˆ3ˆ
49
ˆ2ˆ6ˆ3ˆ
263
ˆ2ˆ6ˆ3ˆ
2612
ˆ2ˆ6ˆ12ˆ
:is ˆ Thus,
222
222
13. We need to evaluate the integral
0
2)1...(
cos1
sin4dx
x
xxI
Using the property dxxfdxxaf , we have
0
2
0
2
0
2
)2...(cos1
sin4
cos1
sin4
cos1
sin4
dxx
xdx
x
xI
dxxx
xxI
Adding equations (1) and (2), we have,
0
2
0
2
0
2
0
2
0
2
cos1
sin42
cos1
sin42
cos1
sin4
cos1
sin4
cos1
sin42
dxx
xI
dxx
xI
dxx
xdx
x
xdx
x
xxI
Subsititute t = cosx; dt = - sinxdx
When x = , t = 1
When x = π, t = -1
2014
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CBSE
2
1
1
0
1
1
0
2
1
1
2
1
1
2
44
1tan4
tan22
122
12
1
142
I
I
tI
t
dtI
t
dtI
t
dtI
OR
We need to evaluate the integral
dxxx
x
dxxx
x
65
2ILet
65
2
2
2
Consider the integrand as follows:
BAxAxBxAx
xx
Bxxdx
dA
xx
x
522522
65
65
65
2
2
2
2
Comparing the coefficients, we have
2A = 1; 5A + B = 2
Solving the above equations, we have
A = 2
1and B =
2
1
Thus,
2014
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CBSE
dxxx
I
I
xx
t
t
dtI
dtdxxtxx
dxxx
xI
I
dxxx
I
dxxx
xI
III
dxxx
dxxx
x
dxxx
x
dxxx
xI
65
1
:consider Now
652
2
52;65
Subsitute
65
52
:consider Now
65
1 and
65
52 Where
,2
1,
2
1
65
1
2
1
65
52
2
1
65
2
1
2
52
65
2
22
2
2
1
2
21
1
22
21
21
22
2
2
2014
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CBSE
CxxxxxI
III
CxxxI
dx
x
x
dx
x
dx
xx
652
5log
2
165
2
1
2
1 Thus,
652
5log
2
1
2
5
2
4
256
2
5
1
2
56
2
55
1
22
21
2
2
22
2
22
2
14. Given function is
f(x) = [x(x – 2) ]2
⟹f’(x) = x2 x 2(x – 2) + (x – 2)2 x 2x
⟹f’ (x) = 2x (x – 2)[x+(x – 2)]
⟹f’(x) = 2x(x – 2)[2x – 2]
⟹f’(x) = 2x (x – 2)[2(x – 1)]
⟹f’(x) = 4x(x – 1) (x – 2)
Since f’(x) is an increasing function, f’(x) > 0.
⟹ f’(x) = 4x (x – 1) (x – 2) > 0
⟹x(x – 1) (x – 2) > 0
⟹0 < x < 1 or x > 2
⟹ x ∈(0, 1) ∪ (2, ∞)
OR
Let 12
2
2
2
b
y
a
xbe the equation of the curve.
Rewriting the above equation as,
22
2
22
2
2
2
2
1
bxa
by
a
x
b
y
Differentiating the above function w.r.t.x, we get,
2014
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CBSE
a
b
b
a
a
b
dx
dy
y
x
a
b
dx
dy
xa
b
dx
dyy
ba
22
22
2
2
,2
2
2
2
2
Slope of the tangent is m = a
b2
Equation of the tangent is
(y – y1) = m (x – x1)
02
022
22
22
abaybx
ababaybx
axbbya
axa
bby
Slope of the normal is
a
b2
1
Equation of the normal is
022
0222
22
22
22
22
11
babyax
abbyax
axabyb
axb
aby
xxmyy
15. Given that f(x) = x2 + 2 and g(x) =1x
x
Let us find f °g:
f °g = f(g(x))
⟹ f °g = (g(x))2 + 2
⟹ f °g = 21
2
x
x
⟹ f °g =2
22
)1(
)1(2
x
xx
2014
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CBSE
⟹ f °g = 12
)12(22
22
xx
xxx
⟹ f °g = 12
2432
2
xx
xx
Therefore, (f°g)(2) = 1222
224232
2
⟹ (f °g)(2) = 6144
2812
Now let us find g °f:
g °f = g (f(x))
⟹ g °f = 1)(
)(
xf
xf
⟹ g °f = 12
22
2
x
x
⟹ g °f = 1
22
2
x
x
Therefore, (g°f)(-3) = 10
11
19
29
1)3(
2)3(2
2
16. We need to prove that
2014
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CBSE
R.H.Scos2
1
4
4
4tantan
tan4
tan1
tan4
tan
tan
tan1
tan1tan
sin2cos2
sin2cos2tan
2cos12cos1
2cos12cos1tan L.H.S
;2cosConsider
12
1,cos
2
1
411
11tan
1
1
1
1
1
1
11
x
t
t
t
t
t
t
tt
tt
tt
tt
tx
xxxx
xx
OR
Given that 44
2tan
4
2tan 11
x
x
x
x
We need to find the value of x.
2
2
42
12162
112
162
1416
8282
1)2)(2()4)(4(
)4)(2()4)(2(
4tan
4
2
4
21
4
2
4
2
4
4
2
4
21
4
2
4
2
tan
44
2tan
4
2tan
2
2
2
2
22
22
1
11
x
x
x
x
x
xx
xxxx
xxxx
xxxx
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
2014
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CBSE
17. An experiment succeeds thrice as often as it fails. Therefore, there are 3 successes and 1 failure.
Thus the probability of success =4
3
And the probability of failure =4
1
We need to find the probability of atleast 3 successes in the next five trials.
Required Probability=P(x = 3) + P(x = 4) + P( x = 5)
= 5C3p3q2 + 5C4p
4q1 + 5C5p5q0
051323
05
5
5
14
4
5
23
3
5
4
1
4
3
4
1
4
35
4
1
4
310
4
1
4
3
4
1
4
3
4
1
4
3
CCC
512
459
1024
918
18. Given that
y = Peax + Qebx
Differentiating the above function w.r.t.x,
bxax QbePaedx
dy
Differentiating once again, we have,
bxax eQbePadx
yd 22
2
2
Let us now find (a + b) dx
dy:
(a + b)dx
dy= (a + b) bxax QbePae
bxbxax
bxbxbxax
eQbabeQPePadx
dyba
eQbPabeQabeePadx
dyba
22
22
)()(
)(
Also we have,
aby = ab(Peax + Qebx)
Thus, abydx
dyba
dx
yd )(
2
2
= Pa2eax + Qb2ebx – Pa2eax – (P + Q)abebx – Qb2ebx + abPeax + abQebx
2014
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CBSE
= 0
19. Consider the determinant
c
b
a
111
111
111
Taking abc common outside, we have
1111
11
11
111
1
cba
cba
cba
abc
Apply the transformation , C1 → C1 + C2 + C3,
111
1
11
11
111
1111
111111
1
11
11111
111111
cb
cb
cb
cbaabc
cbcba
cbcba
cbcba
abc
Apply the transformations, R2 → R2 – R1 and R3 → R3 – R1
100
010
111
1111
cb
cbaabc
Expanding along C1, we have
cabcababccba
abc
cbaabc
1111
10
011
1111
20. x = cost(3 – 2cos2t)
and
y = sint (3 – 2sin2t)
2014
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CBSE
We need to find dx
dy:
dt
dxdt
dy
dx
dy
Let us find dt
dx:
X = cost (3 – 2cos2t)
dt
dx=cost(4costsint) + (3 – 2cos2t) (- sint)
tttttdt
dycossin2cossin4cos3 22
14
cot
cot
sin211cos2sin
cos
)cos21(sin3
)sin21cos(3
sincos6sin3
cossin6cos3
sincos2sincos4sin3
cossin2cossin4cos3Thus,
4
22
2
2
2
2
22
22
tdx
dy
tdx
dy
ttt
t
dx
dy
tt
t
dx
dy
ttt
tt
dx
dy
ttttt
ttttt
dt
dy
21. Consider the differential equation,
yxdx
dy43log
Taking exponent on both sides, we have
dxee
dy
eedx
dy
edx
dy
ee
x
y
yx
yx
yxdx
dy
3
4
43
43
43log
.
Integration in both the sides, we have
2014
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CBSE
Cee
dxee
dy
xy
x
y
34
34
3
4
We need to find the particular solution.
We have, y = 0, when x = 0
12
7
12
43
3
1
4
1
3
1
4
1
C
C
C
Thus, the solution is 12
7
43
43
yx ee
22. The equation of line L1:
)1....(2
3
7
2
3
1
2
3147
3
1
z
p
yx
z
p
yx
The equation of line L2:
)2...(5
6
1
5
7
3
1
5
6
1
5
3
77
zy
p
x
zy
p
x
Since line L1 and L2 are perpendicular to each other, we have
7
7010
1077
9
052177
33
p
p
pp
pp
Thus equations of lines L1 and L2 are:
5
6
1
5
3
1
2
3
1
2
3
1
zyx
zyx
2014
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CBSE
Thus the equation of the line passing through the point (3, 2, -4) and parallel to the line L1 is:
2
4
1
2
3
3
zyx
SECTION – C
23. Equation of the plane passing through the intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5
is:
(x + y + z – 1) + λ (2x + 3y + 4z – 5)
( 1 +2 λ) x + (1 + 3 λ) y + ( 1 + 4 λ) z – (1+5 λ) = 0
This plane has to be perpendicular to the plane x – y + z = 0
Thus,
( 1 +2 λ) x + (1 + 3 λ)(-1) + ( 1 + 4 λ)1 = 0
1 +2 λ – 1 – 3 λ + 1 + 4 λ = 0
1 + 3 λ = 0
λ = 3
1
Thus, the equation of the plane is:
22
2
101
)2(
222
OR
Any point in the line is
2 + λ, -4 + 4 λ, 2 + 2 λ
The vector equation of the plane is given as
0ˆˆ2ˆ. kjir
Thus the cartesian equation of the plane is x – 2y + z = 0
Since the point lies in the plane
(2 + 3 λ)1 + (-4 + 4 λ)(-2)+(2 + 2 λ) = 0
⟹ 2 + 8 +2 + 3 λ – 8 λ + 2 λ = 0
⟹12 - 3 λ
⟹ λ = 4
Thus, the point of intersection of the line and the plane is: 2 + 3 x 4, -4 + 4 x 4, 2 + 2 x 4
⟹ 14, 12, 10
Distance between (2, 12,5) and (14, 12, 10)is:
units 13
169
25144
5101212214222
d
d
d
d
2014
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CBSE
24. Consider the vertices, A(-1, 2), B(1, 5) and C(3, 4)
Let us find the equation of the sides of ∆ABC.
Thus, the equation of AB is:
0723
11
1
25
2
yx
xy
Similarly, the equation of BC is:
0112
13
1
54
5
yx
xy
Also, the equation of CA is:
052
31
3
42
4
yx
xy
Now the area of ABC=Area of ∆ADB + Area of ∆BDC
1
12
5
2
73ADB of Area dx
xx
Similarly, Area of ∆ADB =
3
12
5
2
11dx
xx
Thus, Area of ∆ADB + Areaof ∆BDC
2014
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CBSE
units square 4
2
5
2
92
2
13
2
992
23
2
31
2
26
2
22
2
5
2
11
2
5
2
73
3
1
21
1
2
3
1
1
1
1
1
3
1
1
1
3
1
xxx
x
dxxdxx
dxx
dxx
dxxx
dxxx
25. Let x be the number of pieces manufactured of type A and y be the number of pieces manufactured
of type B. Let us summarise the data given in the problem as follows:
Product Time for
Fabricating
(in hours)
Time for
Finishing (in
hours)
Maximum labour
hours available
Type A 9 1 180
Type B 12 3 30
Maximum
Profit (in
Rupees)
80 120
Thus,the mathematical form of above L.P.P. is
Maximise Z = 80x + 120y subject to
9x + 12y ≤ 180
x + 3y ≤ 30
Also, we have x≥ 0, y ≥ 0
Let us now find the feasible region, which is the set of all points whose coordinates satisfy all
constraints. Consider the following figure.
2014
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CBSE
Thus, the feasible region consists of the points A, B and C. The values of the objective function at the
corner points are given below in the following table:
Points Value of Z
A(12, 6) Z = 80 x 12+120 x 6 = Rs. 1680
B(0, 10) Z = 80 x 0+12 x 10 = Rs. 1200
C(20, 0) Z = 80 x 20+120 x 0 = Rs. 1600
Clearly,Z is maximum at x = 12 and y = 6 and the maximum profit is Rs. 1680.
26. Let E1 ,E2 ,E3 and A be the events defined as follows:
E1 = Choosing 2 headed coin
E2 = Choosing coin with 75% chance of getting heads
E3 = Choosing coin with 40% chance of getting heads
A= Getting heads
Then P(E1) = P(E2) = P(E3) = 3
1
Also, P(A/E1) = 1, P(A / E2) = 4
3
100
75 , P(A / E3) =
5
2
100
40
Required probability
= P (E1 / A)
2014
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CBSE
43
20
60
433
1
15
2
4
1
3
13
1
5
2
3
1
4
3
3
11
3
1
13
1
///
/)(
332211
11
EAPEPEAPEPEAPEP
EAPEP
OR
If 1 is the smallest number,
the other numbers are:2,3,4,5,6
If 2 is the smallest number,
the other numbers are:3,4,5,6
If 3 is the smallest number,
the other numbers are:4,5,6
If 4 is the smallest number,
the other numbers are:5,6
If 5 is the smallest number,
the other number is:6
Thus, the sample space is S =
56
46,45
36,35,34
26,25,24,23
16,15,14,13,12
Thus, there are 15 set of numbers in the sample space.
Let X be
X : 2 3 4 5 6
P(X) = 15
5
15
4
15
3
15
2
15
1
We know that,
E(X) = XiP(Xi)
2014
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CBSE
66.4
15
70
15
30201262
15
56
15
45
15
34
15
23
15
12
27. From the given data, we write the following equations:
900
1
1
1
2300
3
1
4
1600
1
2
3
zyx
zyx
zyx
From above system, we get:
3x+ 2y + z = 1600
4x + y + 3z = 2300
x + y + z = 900
Thus we get:
900
2300
1600
111
314
123
z
y
x
AX = B, where
900
2300
1600
and ;
111
314
123
B
z
y
x
XA
05326141342313
111
314
123
A
We need to find A-1 :
C11 = -2; C12 = -1; C13 = 3
C21 = -1; C22 = 2; C23 = -1
C31 = 5; C32 = -5; C33 = -5
2014
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CBSE
Therefore, adj A =
513
521
512
555
121
312T
Thus, A-1 =
513
521
512
5
1
A
adjA
Therefore, X = A-1B
400
300
200
2000
1500
1000
5
1
90052300116003
90052300216001
90052300116002
5
1
900
2300
1600
513
521
512
5
1
z
y
x
z
y
x
z
y
x
z
y
x
Awards can be given for discipline.
28. Let ∆ABC be the right angled triangle with base b and hypotenuse h.
Given that b + h = k
Let A be the area of the right triangle.
4
22
4
24
4
4
1
2
1
32222
22
2222
2
222
2
2222
22
kbkbAkbk
bA
bkbbkb
A
bkhbbkb
A
bhbA
bhbA
Differentiating the above function w.r.t.x, we have
2014
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CBSE
A
kbbk
db
dA
kbbk
db
dAA
2
3
)1....(4
622
22
22
For the area to be maximum, we have
3
3
03
0
2
22
kb
bbk
kbbk
db
dA
Again differentiating the function in equation (1), with respect to b, we have
)2....(4
12222
2
2
22kbk
db
AdA
bd
dA
Now substituting db
dA=0 and b =
3
kin equation (2), we have
04
22
12
1262
4
3122
2
2
2
2
2
2
2
22
2
2
2
2
2
A
k
db
Ad
k
db
AdA
kk
db
AdA
kkk
db
AdA
Thus area is maximum at 3
kb .
Now, h = k 3
2
3
kk
Let θ be the angle between the base of the triangle and the hypotenuse of the right triangle.
Thus, cosθ = 2
1
3
23 k
k
h
b
32
1cos 1
2014
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CBSE
29. We need to evaluate xxxx
dx4224 coscossinsin
Let I = xxxx
dx4224 coscossinsin
Multiply the numerator and the denominator by sec 4x, we have
1tantan
secsec
1tantan
sec
24
24
24
4
xx
xdxxI
xx
xdxI
We know that sec2x = 1 + tan2x
Thus,
1tantan
sectan124
22
xx
xdxxI
Now substitute t = tanx; dt = sec2xdx
Therefore,
42
2
1
1
tt
dttI
Let us rewrite the integrand as
11
1
1
122
2
42
2
tttt
t
tt
t
Using partial fractions, we have
11
1
1
11
1
1
11
11
1
1
111
1
22
23
42
2
22
223223
42
2
22
22
42
2
2242
2
tttt
DBDCBAtDCBAtCAt
tt
t
tttt
DDtDtCtCtCtBtBtAtAtAt
tt
t
tttt
ttDCtttBAt
tt
t
tt
DCt
tt
BAt
tt
t
So we have,
A + C = 0; A + B – C + D = 1; A + B+ C – D = 0; B + D = 1
Solving the above equations, we have
2014
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CBSE
A = C = 0 and B = D = 2
1
3
1tan2tan
3
1
3
12tan
3
1
4
3
2
1
tan
4
3
1
2
1
4
3
2
12
1
4
11
4
12
1
12
1
:Consider
12
1 and
12
1 Where,
1212
1/
1
12
1
1
1
1
1
1
2
2
21
1
2221
22
22
22
42
2
x
t
t
t
dt
tt
dt
tt
dtI
I
tt
dtI
tt
dtI
II
tt
dt
tt
dt
tttt
tt
dttI
Similarly,
Consider I2:
2014
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CBSE
4
3
2
12
1
4
11
4
12
1
12
1
2
2
22
t
dt
tt
dt
tt
dtI
Cxx
I
xxI
III
x
t
t
3
1tan2tan
3
1tan2tan
3
1
3
1tan2tan
3
1
3
1tan2tan
3
1
Thus,
3
1tan2tan
3
1
3
12tan
3
1
4
3
2
1
tan
4
3
1
2
1
11
11
21
1
1
1
2014