Download - Kendriya Vidyalaya No.1 Sagar

First Pre-Board Examination 2014 -2015
Class – XII (Mathematics)
Time- 3 Hrs.
Max. Marks. 100
General Instructions:1. All questions are compulsory.
2. All questions of one mark each are to be answered in one word, one sentence or as per the exact requirement of
the question.
3. All questions of Section A carry 1 mark each, Questions of Section B carry 4 marks each and Questions of
Section C carry 6 marks each.
SECTION - A
1. If f(x) = x +7 and g(x) = x – 7, then find fog(x).
2. If A = [1 2 3], then find AAT, where AT is transpose of A.
̂ ) × (𝐢̂ + 𝐣̂ + 𝐤
̂ ) = ⃗𝟎.
3. Find the values of  and  for which (𝟐𝐢̂ + 𝟔𝐣̂ + 𝟐𝟕𝐤
4. If A is a square matrix of order 3 such that │A │ = 225, then find │AT│.
5. Write the principal value of 𝐜𝐨𝐬 −𝟏 (𝐜𝐨𝐬
𝟕𝛑
𝟔
).
6. Write the direction cosines of a line equally inclined with coordinate axes.
SECTION – B
7. Solve the equation 𝐭𝐚𝐧−𝟏 𝟐𝐱 + 𝐭𝐚𝐧−𝟏 𝟑𝐱 =
𝛑
𝟒
.
OR
Prove that 𝐭𝐚𝐧−𝟏 {
8.
√𝟏 + 𝐱 − √𝟏 − 𝐱
}
√𝟏 + 𝐱 + √𝟏 − 𝐱
𝝅/𝟒
Evaluate the integral ∫𝟎
𝟏
𝟒
𝟐
𝒍𝒐𝒈(𝟏 + 𝒕𝒂𝒏𝒙) dx
𝟏−𝐜𝐨𝐬𝟒𝐱
𝐱𝟐
9. If the function 𝐟(𝐱) =
𝛑
= + 𝐜𝐨𝐬 −𝟏 𝐱
𝐚,
√𝐱
{√𝟏𝟔+ √𝐱− 𝟒
, 𝐱<0
𝐱 = 𝟎 is continuous at x = 0, then find the value of a.
, 𝐱>0
10. Sand is pouring from a pipe at the rate of 12 cm3/sec. The falling sand forms a cone on the ground in
such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the
height of the sand cone increasing when the height is 4 cm ?
11. If y = 𝐞𝐚𝐜𝐨𝐬
−𝟏 𝐱
𝐝𝟐 𝐲
𝐝𝐲
, then show that (1−x2)𝐝𝐱𝟐 − x 𝐝𝐱 − a2y = 0.
𝒚
𝒅𝒚
𝒙
𝒅𝒙
12. If log(x + y ) = 2 𝐭𝐚𝐧−𝟏 ( ) , then show that
2
2
=
𝒙+𝒚
𝒙−𝒚
13. In a backward state, there are 729 families having 6 children each. If the probability of survival of a
girl is 1/3 and that of a boy is 2/3, find the number of families having 2 girls and 4 boys. Do you
believe that girls are neglected in backward areas ? What steps will you take to restore respect of
girls ?
OR
Two numbers are selected at random from the integers 1 through 9. If the sum is even, find the
probability that both the numbers are odd.
14. Evaluate the integral ∫ 𝒆𝒙 (
𝟏−𝒔𝒊𝒏𝒙
𝟏−𝒄𝒐𝒔𝒙
)dx
15. Find the shortest distance between the lines
OR
𝐱+𝟏
𝟕
=
𝟒
Evaluate ∫𝟏 (𝒙𝟐 − 𝒙) dx as limit of sums.
𝐲+𝟏
−𝟔
=
𝐳+𝟏
𝟔
and
𝐱−𝟑
𝟏
=
𝐲−𝟓
−𝟐
=
𝐳−𝟕
𝟏
.
̂ , then express b
⃗ = 𝟐𝐢̂ + 𝐣̂ − 𝟑𝐤
⃗ in the form b
⃗ = ⃗⃗⃗⃗
⃗ = 3𝐢̂ − 𝐣̂ and b
⃗
16. For 𝛂
b𝟏 + ⃗⃗⃗⃗
b𝟐, where ⃗⃗⃗⃗
b𝟏 is parallel to 𝛂
⃗.
and ⃗⃗⃗⃗
b𝟐 is perpendicular to 𝛂
OR
⃗⃗⃗ , ⃗⃗⃗
⃗⃗⃗ + ⃗⃗⃗
If three vectors 𝐚
𝐛 and 𝐜⃗⃗ prove that [𝐚
𝐛
⃗⃗⃗
𝐛 + 𝐜⃗⃗
⃗⃗⃗ ] = 2 [𝐚
⃗⃗⃗
𝐜⃗⃗ + 𝐚
⃗⃗⃗
𝐛
17. Let A = R – {3} and B = R – {1}. Consider the function f : A  B defined by f(x) =
𝐜⃗⃗ ].
𝐱−𝟐
. Show
𝐱−𝟑
that f is bijective and hence find 𝐟 −𝟏 .
18. Evaluate
∫√
𝐱+𝟐
𝐱 𝟐 + 𝟐𝐱+𝟑
dx
19. Using the properties of determinant, prove that
𝐱+𝐲
| 𝟓𝐱 + 𝟒𝐲
𝟏𝟎𝐱 + 𝟖𝐲
𝐱
𝟒𝐱
𝟖𝐱
𝐱
𝟐𝐱|= x3
𝟑𝐱
SECTION – C
20. Find the area of the region enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12.
21. From the point P(1, 2, 4), a perpendicular is drawn on the plane 2x + y – 2z + 3 = 0. Find the equation and
length of perpendicular and coordinates of foot of perpendicular.
OR
̂ ) =4 and
Find the equation of the plane which contains the line of intersection of the planes 𝐫. (𝐢̂ + 𝟐𝐣̂ + 𝟑𝐤
̂ ) + 𝟓 = 𝟎 and which is perpendicular to the plane𝐫. (𝟓𝐢̂ + 𝟑𝐣̂ − 𝟔𝐤
̂ ) + 𝟖 = 𝟎.
⃗⃗⃗𝐫. (𝟐𝐢̂ + 𝐣̂ − 𝐤
22. An Apache helicopter of enemy is flying along the curve y = x 2 + 7. A soldier, placed at (3, 7), wants to
shoot down the helicopter when it is nearest to him. Find the nearest distance of the helicopter from the
soldier.
OR
Prove that the height of right circular cylinder of maximum volume, inscribed in a sphere of radius R
is
𝟐𝐑
√𝟑
. Also find the maximum volume of the cylinder.
23. Find the particular solution of the differential equation (x – siny)dy + tany dx = 0, given that y = 0, x = 0.
24. From a pack of 52 cards, a card is lost. From the remaining 51 cards, two cards are drawn at random
(without replacement) and are found to be both diamonds. What is the probability that the lost card was a
card of diamond ?
25. A dealer in rural area wishes to purchase a number of sewing machines. He has only Rs. 5760 to invest
and has space for at most 20 items. An electronic sewing machine costs him Rs. 360 and a manually
operated sewing machine Rs. 240. He can sell an electronic sewing machine at a profit of Rs 22 and a
manually operated sewing machine at a profit of Rs. 18. Assuming that he can sell all the items that he
can buy. How should he invest his money in order to maximize his profit ? Make it as a linear
programming problem and solve it graphically. Keeping the rural background in mind, justify the
‘values’ to be promoted for the selection of the manually operated machine.
26. Three classmates A, B and C visited a Super Market for purchasing fresh fruits. A purchased 1 kg
apples, 3 kg grapes and 4 kg oranges and paid Rs. 800, B purchased 2 kg apples, 1 kg grapes and 2 kg
oranges and paid Rs. 500, while C purchased 5 kg apples, 1 kg grapes and 1 kg oranges and paid Rs.
700. Find the cost of each fruit per kg by matrix method. Why fruits are good for health ?
Marking Scheme
1. fog(x) = x.
2. AAT= [14]
3.  = 3 and  = 27/2
4. │AT│= 225.
5. cos−1 (cos
7π
6
7π
) = 2π –
6. Direction cosines = ±
=
6
𝟏
√𝟑
7. tan−1 2x + tan−1 3x =
5π
6
, ±
π
.
𝟏
√𝟑
,±
𝟏
√𝟑
π
2x+3x
 tan−1 (1−6x2 )=
4
5x
 1−6x2 = tan
4
π
4
=1
 6x2 + 5x – 1 = 0  (6x – 1)(x + 1) = 0
1
1
6
6
 x = or x = –1. But x = –1 does not satisfy the given equation, so x = .
OR
To Prove tan−1 {
√1 + x + √1 − x
}
√1 + x − √1 − x
π
1
4
2
= + cos −1 x
1
Put x = cos2 or  = cos −1 x then L. H. S. =
2
cos − sin
tan−1 {
cos + sin
π
π
4
4
π
} = tan−1 {tan ( + )}
4
1
= +  = + 2 cos −1 x
𝜋/4
8. Given integral I = ∫0
𝜋/4
𝑙𝑜𝑔(1 + 𝑡𝑎𝑛𝑥)dx = ∫0
𝜋/4
= ∫0
𝜋/4
= ∫0
𝑙𝑜𝑔 (1 + [
𝟏−𝒕𝒂𝒏𝒙
])
𝟏+𝒕𝒂𝒏𝒙
𝜋/4
𝑙𝑜𝑔 2 dx – ∫0
𝛑
𝑙𝑜𝑔 (1 + 𝑡𝑎𝑛 [ 𝟒 − 𝒙]) dx
𝜋/4
𝑙𝑜𝑔 ([
dx = ∫0
𝟐
])
𝟏+𝒕𝒂𝒏𝒙
dx
𝑙𝑜𝑔(1 + 𝑡𝑎𝑛𝑥)dx
𝛑
= log2. [ 𝟒 − 0] − I
𝛑
I =
𝟖
log2. .
1−cos4x
x2
a,
9. Given function f(x) =
, x<0
x=0
√x
{√16+ √x− 4 , x > 0
f (0) = a
Now
limx → 0− f(x) = limx → 0−
Also, limx → 0+ f(x) =
limx → 0+
1−cos4x
=
x2
limx → 0−
√x
√16+ √x− 4
×
2si𝑛2 2𝑥
=
x2
√16+ √x+ 4
limx → 0−
= limx → 0+
√16+ √x+ 4
8si𝑛2 2𝑥
(2x)
2
=8
√x(√16+ √x+ 4)
16+ √x − 16
Now, since f(x) is continuous at x = 0,  a = 8.
10. Given
𝑑𝑉
𝑑𝑡
1
= 12 cm3/sec and height h = 6 r or
Now, volume of cone V =
this gives
11. Given y = eacos
dh
dt
−1 x
=
1
48𝜋
1
3
r = 6h. Also given h = 4 cm.
πr2h = 12πh3 
𝑑𝑉
𝑑𝑡
dh
= 36πh2
dt
cm/sec.
 logy = acos−1 x
By differentiating with respect to x, we have
Again differentiating and using above, we get
1 dy
y dx
=
a
√1 − x2
d2 y

√1 − x 2
dy
dy
dx
= ay
(1−x2)dx2 − x dx − a2y = 0.
=8
𝑦
12. Given log(x + y ) = 2 tan−1 ( )
2
2
𝑥
1
Differentiating w.r.t.’x’
𝑑𝑦
By Calculating, we get
1
13. Suppose p = P(girl) =
x2 +𝑦2
=
𝑑𝑥
[2𝑥
𝑥+𝑦
𝑥−𝑦
and q = P(boy) =
3
dy
+ 2𝑦
dx
]=
𝑥
2
[
1+(𝑦/𝑥)2
dy
− 𝑦×1
dx
𝑥2
]
.
2
3
. Clearly p + q = 1, so this is the case of binomial distribution.
Now, using P(X = r) = nCr pr.qn – r, the required probability is given by P(2girls,4boys) = 6C2 p2.q4
P(2girls,4boys) =
80
243
Now, the number of families having 2 girls and 4 boys (out of 6 children) = 729 ×
80
243
= 240 families.
Now, steps restore respect of girls : promotion of girls education, participation of girls in development of
social activities etc.
OR
Let A be the event of selecting two odd numbers and B be the event of selecting two numbers whose
sum is even, then n(A) = 5C2 and n(B) = 5C2 + 4C2 and n(A∩B) = 5C2.
A
Now required Probability P(B) =
14. Given integral I = ∫ 𝑒 𝑥 (
1−𝑠𝑖𝑛𝑥
1−𝑐𝑜𝑠𝑥
n(A∩B)
n(B)
=
5 𝐶2
5𝐶2 +4𝐶2
)dx = ∫ 𝑒 𝑥 (
1−2𝑠𝑖𝑛(𝑥/2)𝑐𝑜𝑠(𝑥/2)
2𝑠𝑖𝑛2 (𝑥/2)
)dx
1
= ∫ 𝑒 𝑥 ( 𝑐𝑜𝑠𝑒𝑐2 (𝑥/2) − 𝑐𝑜𝑡(𝑥/2))dx = −𝑒𝑥 . 𝑐𝑜𝑡(𝑥/2)+ C
2
By using ∫ 𝑒 𝑥 {f(x) + f ′(x)}dx = ex. f(x) + C where f(x) = −𝑐𝑜𝑡(𝑥/2)
and f ′(x) =
1
2
𝑐𝑜𝑠𝑒𝑐2 (𝑥/2)
OR
4
For the integral ∫1 (𝑥 2 − 𝑥) dx, a = 1, b = 4, then nh = b – a = 3.
f(x) = x2 – x , f(a) = f(1) = 0, f(a + h) = h2 – h ,………., f(a + (n – 1)h ) = (n – 1)2h2 – (n – 1) h
4
Now, ∫1 (𝑥 2 − 𝑥) dx = limh →0 h[f(a) + f(a + h) + f(a + 2h) + ⋯ … f(a + (n − 1)h]
=
15.Given lines
𝐱+𝟏
𝟕
=
𝐲+𝟏
−𝟔
=
27.
2
𝐳+𝟏
𝟔
Here x1 = -1, y1 = -1, z1 = -1,
And x2 = 3, y2 = 5, z2 = 7,
Hence shortest distance =
𝐱−𝟑
𝟏
=
𝐲−𝟓
−𝟐
=
𝐳−𝟕
𝟏
.
a1 = 7, b1 = -6, c1 = 6
a2 = 1, b2 = -2, c2 = 1
x2 − x1
| a1
a2
y2 − y1
b1
b2
z2 − z1
c1 |
c2
√(𝑏1 𝑐2 − 𝑏2 𝑐1 )2 +(𝑐1 𝑎2 − 𝑐2 𝑎1 )2 +(𝑎1 𝑏2 − 𝑎2 𝑏1 )2
3+1
| 7
=
and
1
5+1
−6
−2
7+1
6 |
1
4
|7
√(−6+12)2 +(6−7)2 +(−14+6)2
=
1
 ⃗⃗⃗⃗
b1 =  ⃗α =  (3î − ĵ)
⃗ - ⃗⃗⃗⃗
 ⃗⃗⃗⃗
b2 = b
b1 = (2 - 3)î + (1 + )ĵ −3k̂
⃗ = ⃗⃗⃗⃗
(as b
b1 + ⃗⃗⃗⃗
b2 )
Now since ⃗⃗⃗⃗
b2 is perpendicular to ⃗α  ⃗⃗⃗⃗
b2 . ⃗α = 0   =
 ⃗⃗⃗⃗
b1 =  ⃗α =
1
2
(3î − ĵ) and
⃗⃗⃗⃗
b2 =
1
2
î +
3
2
ĵ
1
2
8
6|
1
√36+1+64
16. For given vectors ⃗α = 3î − ĵ and ⃗b = 2î + ĵ − 3k̂,
Since ⃗⃗⃗⃗
b1 is parallel to ⃗α
6
−6
−2
=
46
√101
OR
⃗⃗⃗
[a⃗⃗ + b
⃗⃗⃗b + c⃗⃗
⃗⃗⃗ ) . {(b
⃗⃗⃗ + c⃗⃗ )c⃗⃗ + a⃗⃗ 
c⃗⃗ + a⃗⃗ ] = (⃗a⃗ + b








⃗⃗⃗ a⃗⃗ c⃗⃗ c⃗⃗ c⃗⃗ a⃗⃗ 
r⃗ (⃗⃗⃗
b c⃗⃗ b
⃗⃗⃗ 
wherer⃗ = a
⃗⃗ + b




⃗⃗⃗ c⃗ ⃗r . (b
⃗⃗ a⃗⃗ r⃗ . (c⃗⃗ a⃗⃗ 
⃗r . (b
as c⃗⃗ c⃗⃗ ⃗⃗⃗
0 



⃗⃗⃗ ) . (c⃗⃗ a⃗⃗ 
⃗⃗ c⃗⃗ (a
⃗⃗ a⃗⃗ (a
⃗⃗ + ⃗⃗⃗
⃗⃗⃗ + b
(a⃗⃗⃗ + ⃗⃗⃗
b ) . (b
b ) . (b


⃗⃗⃗ . (c⃗⃗ a⃗⃗ 
⃗⃗ c⃗⃗ ⃗⃗⃗
⃗⃗ c⃗ a
⃗⃗ a⃗⃗ ⃗⃗⃗
⃗⃗ a⃗⃗ a
a
⃗⃗ . (b
b . (b
⃗⃗ . (b
b . (b
⃗⃗ . (c⃗⃗ a⃗⃗ b




⃗⃗ , c⃗ ,a⃗⃗ [a⃗⃗ , ⃗b⃗ ,c⃗⃗ 
[a⃗⃗ , ⃗⃗⃗
b ,c⃗ [b
17. Given f : A  B by f(x) =
x−2
x−3
Let x1, x2  A and f(x1) = f(x2)
𝑥1 −2
=
𝑥1 −3
𝑥2 −2
𝑥2 −3

Proved
 x 1 = x2
hence f is one-one.
Now, let y =
x−2
x−3
x=
2−3y
1−y
 Range f = R – {1} = co-domain. Hence, f is onto and hence bijective.
Now, the inverse is x = f – 1(y) =
18. Given integral I = ∫
x+2
√x2 + 2x+3
dx =
2−3y
1−y
1
2
2x+2+2
∫√
x2 + 2x+3
dx =
1
2
2x+2
∫√
x2 + 2x+3
dx +
1
2
∫√
2
x2 + 2x+3
dx
In first integral, putting x 2 + 2x + 3 = t, then (2x + 2)dx = dt
1
I=
2
∫
dt
√t
+∫
1
√x2 + 2x+1+2
dx =
1
2
×2√t + ∫
1
√(x+1)2 +2
dx
= √x 2 + 2x + 3 + log(𝑥 + 1 + √x 2 + 2x + 3) + C
x+y
19. Given determinant = | 5x + 4y
10x + 8y
x
4x
8x
x
2x|
3x
Using R2 → R2 – 2R1 and R3 → R3 – 3R1
x+y
 = |3x + 2y
7x + 5y
x
2x
5x
x
0|
0
Expanding w. r. t. C3 , = x(15x2 + 10xy – 14x2 – 10xy) – 0 + 0 = x3.
20. Given parabola 4y = 3x2 and the line 2y = 3x + 12
Solving these equations, we get x = 4, -2 and y = 12, 3, so the two curves intersect each other at
points (4, 12) and (-2, 3).
4
4
Now, the required area A = ∫−2[yline − yparabola ] dx = ∫−2 [
3𝑥+12
2
−
3𝑥 2
4
] dx = 27 sq. units.
21. Given plane 2x + y – 2z + 3 = 0, now equation of normal to the plane passing through P(1, 2, 4) is
x− 1
2
=
y−2
1
=
z− 4
−2
= , then (2+1,  + 2, - 2+4) is a general point on the plane, so
2(2+1) + (+2) – 2(- 2+4) + 3 = 0   =
1
9
11 19 34
1
9
3
. Foot of perpendicular ( 9 ,
, ). Length =
9
Equation can be obtained.
OR
Cartesian equations of given planes are x + 2y + 3z = 4 …………….(1)
2x + y – z + 5 = 0……………….(2) and 5x + 3y – 6z + 8 = 0 ………………….(3)
Now equation of plane containing the line of intersection of planes (1) and (2) is
(2x + y – z + 5) + (2x + y – z + 5) = 0………………………..(4)
(1 + 2)x + (2 + )y + (3 - )z + (5 - 4) = 0
since this plane is perpendicular to plane (3)  5(1 + 2) + 3(2 + ) – 6(3 - ) = 0   =
Now, putting  =
7
29
in equation (4), the required plane is 33x + 45y + 50z = 41.
7
29
22. Given curve y = x2 + 7.
Let P(x, y) be any point on the curve and A(3, 7) be the given point, then the
required distance AP = √(x − 3)2 + (𝑦 − 7)2 = √(x − 3)2 + 𝑥 4
Let f(x) = (x − 3)2 + 𝑥 4
Differentiating f ′(x) = 2(x – 3) + 4x3
and f ′′(x) = 2 + 12x2
For minimum distance, f ′(x) = 0  x = 1 and at x = 1, f ′′(x) > 0 i. e. the distance is minimum when x = 1.
Now the minimum distance AP = √5.
OR
Draw Diagram and suppose radius of cylinder = x and height = 2y, then x2 + y2 = R2.
Now, Vol. of cylinder V = π (R2 – y2).2y = 2π (R2y – y3)
Differentiating ,
For max. vol.
dV
dy
dV
dy
= 2π (R2 – 3y2) and
= 2π (R2 – 3y2)= 0  y =
Now height h = 2y =
2R
√3
R
√3
d2 V
dy2
= –12πy
and
d2 V
= –12π
dy2
and maximum volume V =
R
√3
< 0 so vol. is max. when y =
R
√3
4πR3
3 √3
23. Given differential equation (x – siny)dy + tany dx = 0
or
𝑑𝑦
𝑑𝑦
+ x.coty = siny.coty = cosy
this is linear DE where P = coty, Q = cosy
Now Integrating Factor IF = e∫ P dy = e∫ coty dy = elogsiny = siny
1
Now the general solution is x.IF = ∫ Q IF dy + C = ∫ cosy. siny dy + C = ∫ sin2y dy + C
2
 x.siny =
−cos2y
4
+C
Now putting y = 0, x = 0, then C = ¼ and hence the particular solution is x.siny =
−cos2y
4
+
1
4
24. Total cards = 52, out of which diamonds = 13.
Let E1 = Lost card is diamond
E2 = Lost card is not diamond
A = two diamond cards are drawn. Clearly E1, E2 are mutually exclusive and exhaustive events.
Now P(E1) =
13
52
=
1
4
P(E2) =
39
52
=
3
A
12 ×11
1
51 ×50
P(E )=
4
A
13 ×12
2
51 ×50
P(E )=
A
P(E1 )P( )
E1
Using Baye’s theorem, required prob. = P( A )=
E1
A
A
E1
E2
P(E1 )P( )+ P(E2 )P( )
=
11
.
50
25. Suppose the dealer buys ‘x’ no. of electronic and ‘y’ no. of manually operated sewing machines, then
mathematical form of LPP is to
Max. z = 22x + 18 y
subject to x + y ≤ 20, 360x + 240y ≤ 5760, x ≥ 0, y ≥ 0.
Make correct graph and shade the feasible region.
The feasible region is bounded
With vertices O(0, 0), A(16, 0), P(8, 12) and
The values of z at these vertices are
B(0, 20).
zO = 0, zA = 352, zP = 392, zB = 360
Clearly, Max. z = 352 at vertex P(8, 12), hence the dealer should purchase 8 electronic sewing machines
and 12 manually operated sewing machines to obtain maximum profit.
Now, manually operated machines should be promoted, so that employment in rural areas be given, no
electricity is required, pollution can be prevented.
26. Let the cost per kg of apple, grapes and oranges are Rs. x, y and z respectively, then the given information
can be expressed as x + 3y + 4z = 800,
2x + y + 2z = 500 and 5x + y + z = 700.
3 4 𝑥
800
1
1 2] [𝑦] = [500] or AX = B. where A = [2
1 1 𝑧
700
5
−1
1
1
-1
-1
Now, │A│ = 11 ≠ 0, hence A exists and A = [ 8
−19
11
−3
14
1
Matrix form [2
5
−1
𝑥
1
Hence the solution is given by X = A .B [𝑦] =
[8
11
𝑧
−3
-1
1
−19
14
3 4
1 2]
1 1
2
6]
−5
2 800
100
6 ] [500] = [100]
100
−5 700
 x = 100, y = 100, z = 100. So, the cost of each food per kg . is Rs. 100/-.
Now, fruits are good for health because fruits give us vitamins, carbohydrates, minerals, protein etc.