Solution

SAMPLE PAPER -2

Class 11 - Mathematics
Section A
2

1. (a) 2 n

Explanation: Number of relations that can be defined on A = 2 n

2 2
(b − a )

2. (a)
2 2
(b + a )

Explanation: On dividing num. and denom. by cos θ, we get

b 2 2
(b× )−a (b − a )
b tan θ−a a
the given exp. = =
b
=
b tan θ+a 2 2
(b× )+a (b + a )
a

3.
(c) no solution
x+7
Explanation: x−8
> 2

x+7
⇒ − 2 > 0
x−8

x+7−2(x−8)
⇒
x−8
>0
x+7−2x+16
⇒
x−8
>0
(23−x)
⇒
x−8
> 0 [∵ a

b
> 0 ⇒ (a > 0 and b > 0) or (a < 0 and b < 0)]
⇒ (23 - x > 0 and x - 8 > 0) or (23 - x < 0 and x - 8 < 0)
⇒ (x < 23 and x > 8) or (x > 23 and x < 8)

⇒ 8 < x < 23 [Since x > 23 and x < 8 is not possible]

⇒ xϵ (8, 23)
2x+1
Now 7x−1
>5
2x+1
⇒
7x−1
-5>0
2x+1−5(7x−1)
⇒
7x−1
>0
2x+1−35x+5
⇒
7x−1
>0
(6−33x)
⇒
7x−1
> 0 [∵ a

b
> 0 ⇒ (a > 0 and b > 0) or (a < 0 and b < 0)]
⇒ (6 - 33x > 0 and 7x - 1 > 0) or (6 - 33x < 0 and 7x - 1 < 0)
6
⇒ (x < and x > ) or (x > and x < ) 1 2 1

33 7 11 7
1 2
⇒ < x <
7 11

⇒ x ∈ (
1

7
,
11
2
) [Since x > 2

11
and x < 1

7
is not possible]
Hence, the solution of the system x+7

x−8
> 2,
2x+1

7x−1
> 5 will be (8, 23) ∩ ( 1

7
,
2

11
) = ϕ

4.
(b) 120
Explanation: We have, 6 objects {C}, {H}, {E}, {E}, {S}, {E} and there are 3 E's.
So, the number of words that can be formed out of the letters of the word ‘CHEESE’ is
6!
= 3!

= 120
5.
(d) 5n
n
Explanation: ∑ r=0
r
4 .
n
Cr =4 0
⋅
n
C0 + 4
1 n
⋅ C1 + 4
2 n
⋅ C2 + ... + 4
n n
⋅ Cn

= 1 + 4. C + 4 n
1
2
.
n
C2 + .... + 4 n n
⋅ Cn

= (1 + 4)n = 5n
6. (a) 2x + y = 4
Explanation: Suppose the line make intercept ‘a’ on x-axis. Then, it makes intercept ‘2a’ on y-axis.
y
Thuse, the equation of the line is given by + = 1
x

a 2a

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 It passes through (1, 2), so, we get
1 2
+ = 1 or a = 2
a 2a
y
Thus, the required equation of the line is given by x

2
+
4
= 1 or 2x + y = 4
7.
(d) 3

Explanation: Here c = 3, b = 4, from relation a2 = b2 + c2 we get a = 5.

e= = c

a
3

8. (a) 486
Explanation: Given T4 = 54 and T9 = 13122.

∴ ar3 = 54 and ar = 13122 8

8
13122
⇒
ar

3
=
54
= 243 ⇒ r
5 5
= 3 ⇒ r=3
ar

∴ a × 33 = 54 ⇒ a = 54

27
=2
Therefore, a = 2 and r = 3.
∴ T6 = ar5 = (2 × 35) = (2 × 243) = 486.
Therefore, the required 6th term is 486.
9.
(d) (–2, 0, 7)
Explanation: Let 'L' is the point of foot of perpendicular of (-2, 8, 7) on the xz-plane on the xz plane, y = 0
So, co-ordinate of L is: (–2, 0, 7)
10.
(d) 2
Explanation: lim sin 2x

x
x→0

sin 2x
= lim 2 ( )
2x
x→0

= 2 × 1

=2
11.
(b) 8
Explanation: Arrange the given data in ascending order, we get
3, 6, 6, 7, 7, 7, 8, 9, 9, 10, 10, 10, 12
Total terms, n = 13 (odd)
n + 1
∴ Median = ( 2
) th term
13 + 1
=( 2
)th term = 7th term = 8

12.
(d) 10
1

Explanation: Clearly, the sample space is given by

S = {1, 2, 3, 4, 5,… 97, 98, 99, 100}
∴ n(S) = 100

Let E = event of getting a square.

Then E = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
∴ n(E) = 10
n(E) 10
Hence, the required probability = n(S)
=
100
=
1

10

13. (a) sin 4β

Explanation: It is given that tan α = 1

7
and tan β = 1

3
2 tan β
Now, tan 2β = 2
1− tan β

1
2×

=
3

1
1−
9

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 2

= 3

= 3

4
tan α+tan 2β
∴ tan(α + 2β) =
1−tan α tan 2β
1 3 25
+

= 7

1
4

3
= 28

25
1− ×
7 4 28

=1
tan(α + 2β) = 1 = tan π

4
π
⇒ α + 2β =
4
π
⇒ α = − 2β
4
π
⇒ 2α = − 4β
2
π
⇒ cos 2α = cos( 2
− 4β) = sin 4β
∴ cos 2α = sin 4β ​
14.
(b) -23 < x ≤ 2
3(x−2)
Explanation: −15 < 5
≤ 0

5 3(x−2) 5 5
⇒ −15 ⋅ < ⋅ ≤ 0 ⋅
3 5 3 3

⇒ -25 < (x - 2) ≤ 0 + 2
⇒ -25 + 2 < x - 2 + 2 ≤ 2

⇒ -23 < x ≤ 2

15.
(b) 350
Explanation: Total no of Men = 7
Total no. of women = 5
Required number of ways = C × 7
3
5
C2
7! 5! 7×6×5 5×4
= × = ×
3!4! 2!3! 3×2 2

= 7 × 5 × 10 = 35 × 10 = 350

16.
(c) x2 = ab
Explanation: Here, a, x, b are in GP ⇒ x

a
=
b

x
2
⇒ x = ab .
17.
(c) 3 units
Explanation: Given point is P(3, 4, 5)
∴ Distance of P from yz-plane
−−−−−−−−−−−−−−−−−−−−−− −
2 2
= √(0 − 3) + (4 − 4) + (5 − 5)
2
[using distance formula]
–
= √9 = 3 units
18.
(d) 10

3
(1−cos 2x) sin 5x
Explanation: lim 2
x→0 x sin 3x

2
2 sin x sin 5x
= lim
2
x→0 x sin 3x

2
2 sin x sin 5x
( )×5
x2 5x
= limx→0
sin 3x
( )3
3x
2
10× 1 ×1
=
1×3
10
=
3

19. (a) Both A and R are true and R is the correct explanation of A.
Explanation: 8x + y - 4 = 0, 4x + 7y - 4 = 0
−8 −4
m1 = 1
= -8, m2 = 7

m1 − m2
tanθ = 1+m1 m2

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 4 −56+4
−8+

tanθ = 7

4
= 7

7+32
1+(−8)(− )
7 7

−52
tanθ = 39
−4
tanθ = 3

−1 −4
θ = tan ( )
3

20. Given : P(A) = 0.3, P(B) = 0.2 , P(A ∩ B) = 0.1

To find : P (A
¯
∩ B)

We know : P (A ¯
∩ B) = P(B) - P(A ∩ B)

Substituting in the above formula we get,

P (A ∩ B) = 0.2 - 0.1
¯

P (A ∩ B) = 0.1
¯

Section B
21. We have to prove cos 8A cos 5A−cos 12A cos 9A
= tan 4A
sin 8A cos 5A+cos 12A sin 9A

Take LHS
2 cos 8A cos 5A−2 cos 12A cos 9A
= 2 sin 8A cos 5A+2 cos 12A sin 9A

{cos(8A+5A)+cos(8A−5A)}−{cos(12A+9A)+cos(12A−9A)}
⇒ LHS = {sin(8A+5A)+sin(8A−5A)}+{sin(9A+12A)+sin(9A−12A)}

{cos 13A+cos 3A}−(cos 21A+cos 3A}

⇒ LHS =
(sin 13A+sin 3A)+(sin 21A+sin(−3A)

(cos 13A+cos 3A)−(cos 21A+cos 3A) cos 13A−cos 21A

⇒ LHS = =
(sin 13A+sin 3A)+(sin 21A−sin 3A) sin 13A+sin 21A

13A+21A 21A−13A
2 sin( ) sin( )

LHS =
2 2
⇒
3A+21A 21A−13A
2 sin( ) cos( )
2 2

sin 17A sin 4A

= sin 17A cos 4A

= tan 4A = RHS
Hence proved.
22. Let AB be the given line segment making intercepts a and b on the x-axis and y-axis respectively.

x y
Then, the equation of line AB is a
+
b
= 1

Moreover, these points are A(a, 0) and B( 0, b).

Let M( 2, 3) be the midpoint of AB. Then,
a+0 0+b

2
= 2 and 2
= 3

⇒ a = 4 and b = 6
Hence, the required equation of the given line is
x y
+ = 1
4 6

⇔ 3x + 2y - 12 = 0
Hence, the required equation is 3x + 2y - 12 = 0
√3x2 −1+ √2x2 −1
23. We have to find lim 4x+3
x→∞

Dividing each term in the numerator and denominator by x, we get

3−1 2−1
√ +√
√3x2 −1+ √2x2 −1 x2 x2 √3+ √2
lim
4x+3
= lim
4+3/x
= 4
x→∞ x→∞

OR
We have,
(LHL of f (x) at x = 1)
= lim f (x) = lim f (1 - h) = lim 5(1 - h) - 4 = lim 1 - 5h = 1
− h→0 h→0 h→0
x→1

( RHL of f(x) at x = 1)

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 = lim f(x)
+
x→1

= lim f(1 + h) = lim 4 (1 + h)3 - 3 (1 + h) = 4(1)3 - 3(1) = 1

h→0 h→0

Clearly, lim f(x) = lim f(x).

− +
x→1 x→1

Hence, lim f(x) exists and is equal to 1.

x→1

24. Total number of words that can be made with the letters of the word STRANGE = 7! = 5040
Number of words in which vowels always come together = 1440
Therefore the number of words in which vowels do not come together = 5040 - 1440 = 3600
25. We have to find the probability of getting all the four cards of the same suit.
Since 4 cards can be drawn at a time from a pack of 52 cards in 52C4 ways.

Therefore, the total number of elementary events = 52C4

Consider the following events:
A = Getting all spade cards; B = Getting all club cards;
C = Getting all diamond cards, and D = Getting all heart cards.
Clearly, A, B, C and D are mutually exclusive events such that
13C4 13C4 13C4 13C4
P (A) =
52
, P (B) =
52
, P (C ) =
52
and P (D) = 52
C4 C4 C4 C4

Required probability = P ( A ∪ B ∪ C ∪ D )
= P(A) + P(B) + P(C) + P (D) [By addition Theorem]
13C4 44
= 4( ) =
52 4165
C4

OR
Suppose E be the event that a doublet of prime number appear.
∴ E = {(2,2), (3,3), (5,5)}

n(E) = 3
3 1
∴ P (E) = =
36 12

Section C
26. We have, f(x) = x

(1+ x2 )

Clearly, f(x) is defined for all x. So, dom (f) = R.

Let y = f(x) . Then, we have
y= x

2
⇒ x2y - x + y = 0
(1+ x )

2
1± √1−4y

⇒ x =
2y
................(i)
It is clear from equation (i) that x will take real values, when
(1 - 4y2) ≥ 0 and y ≠ 0
⇒ (4y2- 1 ) ≤ 0 and y ≠ 0
⇒ (2y + 1)(2y - 1) ≤ 0 and y ≠ 0

) ≤ and y ≠ 0
1 1
⇒ (y + ) (y −
2 2

⇒ −
1

2
≤ y ≤
1

2
and y ≠ 0
−1 1
⇒ y ∈ [ , ] − {0}
2 2

Also, x = 0 ⇒ y = 0
−1 1
∴ range (f) = [ 2
,
2
]

−1
Hence, dom (f) = R and range (f) = [ 2
,
1

2
]

27. Let x and x + 2 be two consecutive odd positive integers

Then x + 2 < 10 and x + x + 2 > 11.
⇒ x < 8 and 2x + 2 > 11

⇒ x < 8 and 2x > 9

⇒ x <8 and 2x > 9

⇒ x < 8 and x >

9

2
9
⇒ < x < 8
2

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 ⇒ x = 5 and 7
Thus required pairs of odd positive integers are 5, and 7.
OR
Here 3x - 7 > 5x - 1
⇒ 3x − 5x > −1 + 7

⇒ −2x > 6

Dividing both sides by -2, we have

−2x 6
< ⇒ x < −3
−2 −2

Thus the solution set is (−∞, −3).

W e have

5 5 5 5 5 3 2 5 1 4
(x + y) + (x − y) = 2 [ C0 x + C2 x y + C4 x y ]

5 3 2 4
= 2 (x + 10x y + 5xy
–
P utting x = √2 and y = 1, we get

– 5 – 5 – 5 – 3 –
(√2 + 1) + (√2 − 1) = 2 [ (√2) + 10(√2) + 5√2]
28.
– – –
= 2 [4 √2 + 20 √2 + 5√2]

–
= 58√2

OR
To prove: (23n - 7n -1) is divisible by 49, where n ∈ N
(23n - 7n -1) = (23)n – 7n - 1
= 8n - 7n - 1
= (1 + 7)n - 7n - 1
Now using binomial theorem..
nC 1n + nC 1n-17 + nC21n-272 + …… +nCn-17n-1 + nCn7n - 7n-1
⇒ 0 1

= nC0 + nC17 + nC272 + …… +nCn-17n-1 + nCn7n - 7n-1

= 1 + 7n + 72[nC2 + nC37 + … + nCn-1 7n-3 + nCn 7n-2] - 7n-1

= 72[nC2 + nC37 + … + nCn-1 7n-3 + nCn 7n-2]

= 49[nC2 + nC37 + … + nCn-1 7n-3 + nCn 7n-2]

= 49K, where K = (nC2 + nC37 + … + nCn-1 7n-3 + nCn7n-2)

Now, (23n - 7n -1) = 49K

Therefore (23n - 7n -1) is divisible by 49.
29. Given: 4x2+ 9y2 = 1
2 2
y
x

1
+
1
= 1 ...(i)
4 9

Now, above equation is of the form,

2 2
x y

2
+
2
= 1 ...(ii)
a b

Comparing eq. (i) and (ii), we get

−
− −
−
a
2
=
1

4
and b2 = 1

9
⇒ a= √
1

4
and b = √
1

9
⇒ a=
1

2
and b = 1

i. Length of major axes

∴ Length of major axes = 2a = 2 × = 1 units
1

ii. Length of minor axes

∴ Length of minor axes = 2b = 2 × = units
1 2

3 3

iii. Coordinates of the Vertices

∴ Coordinate of vertices = (a, 0) and (-a, 0) = ( , 0) and (- , 0) 1 1

2 2

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 iv. Coordinates of the foci
As we know that
Coordinates of foci = (± c, 0)
Now c2 = a2 - b2 =
1 1 9−4 5 √5

4
−
9
⇒ c
2
=
36
⇒ c
2
=
36
⇒ c =
6
....(iii)
√5
∴ Coordinates of foci = (± 6
, 0)

v. Eccentricity
√5

c √5
As we know that, Eccentricity = [from (i)]
6
⇒ e= =
a 1 3

vi. Length of the Latus Rectum

2
1 2
2 2×( )

As we know that, Length of Latus Rectum = 2b 3 9 4

= = =
a 1 1 9
2 2

OR
Let AB be the parabolic arch having O as the vertex and OY as the axis.
The parabola is of the form x2 = 4ay
Now CD = 5 m ⇒ OD = 2.5 m
BD = 10 m

⇒ Coordinates of point B are (2.5, 10)

Since the point B lies on the parabola x2 = 4ay
2 6.25 625 5
∴ (2.5) = 4a × 10 ⇒ a = = =
40 4000 32

∴ Equation of parabola is x 2
= 4 ×
5

32
y

2 5
⇒ x = y
8

Let PQ = d ⇒ NQ = d

∴ Coordinates of Point Q are ( d

2
, 2)

5
Since point Q lies on the parabola x 2
=
8
y

2 2
d 5 d 5 2 –
∴ ( ) = × 2 ⇒ = ⇒ d = 5 ⇒ d = √5
2 8 4 4

–
Thus width of arc = √5 m = 2.24m approx.
30. Let A(0, 7, -10), B(1, 6, -6) and C(4, 9, -6) be three vertices of triangle ABC. Then
−−−−−−−−−−−−−−−−−−−−−−−−−
2 2 2 −−−−−−− − −− –
AB = √(1 − 0) + (6 − 7) + (−6 + 10) = √1 + 1 + 16 = √18 = 3√2
−−−−−−−−−−−−−−−−−−−−−−−−−
2 2 2 −−−−−− − −− –
BC = √(4 − 1) + (9 − 6) + (−6 + 6) = √9 + 9 + 0 = √18 = 3√2
−−−−−−−−−−−−−−−−−−−−−−−−−
2 2 2 −−−−−−−− − −−
AC = √(4 − 0) + (9 − 7) + (−6 + 10) = √16 + 4 + 16 = √36 = 6

Now AB = BC
Thus, ABC is an isosceles triangle.
31. Let the two positive numbers be a and b
a+b −
−
Therefore A = 2
and G = √ab
−−−−−−−−−−−−− −−− −−−−
Now, A ± √(A + G)(A − G) 2
= A ± √A − G
2

−−−−−−−−−−−−− −
2
a+b a+b −−
= 2
± √( ) − (√ab )
2
2

−−−−
2
−−−−−−−
2

= a+b

2
± √
a + b +2ab

4
− ab

−−−−
2
−−−−−−
2

= a+b

2
± √
a + b +2ab−4ab

4
−−−−−
2
a+b (a−b) a+b a−b
= 2
± √
4
=
2
±
2

a+b a−b a+b a−b

= 2
+
2
and 2
−
2

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 a+b+a−b a+b−a+b
= 2
and 2

= 2a

2
= a and 2b

2
= b

Section D
32. We have,
LHS = cos 20° cos 40° cos 60° cos 80°
⇒ LHS = cos 60° (cos 20° cos 40°) cos 80°

⇒ LHS = 1

2
×
1

2
(2 cos 20° cos 40°) cos 80° [∵ cos π

3
=
1

2
]

⇒ LHS = 1

4
[{cos (40° + 20°) + cos (40° - 20°)} cos 80°] [∵ 2 cos A cos B = cos (A + B) + cos (A - B)]
1
⇒ LHS = 4
{(cos 60° + cos 20°) cos 80°}
⇒ LHS = 1

4
{( 1

2
+ cos 20°) cos 80°}
⇒ LHS = 1

4
{ 1

2
cos 80° + cos 80° cos 20°}
⇒ LHS = 1

8
{cos 80° + 2 cos 80° cos 20°}
⇒ LHS = 1

8
[cos 80° + {cos (80° + 20°) + cos (80° - 20°)}]
⇒ LHS = 1

8
{cos 80° + cos 100° + cos 60°}
⇒ LHS = 1

8
{cos 80° + cos (180° - 80°) + cos 60°}
⇒ LHS = 1

8
{cos 80° - cos 80° + cos 60°} [∵ cos (180° - x) = -cos x]
⇒ LHS = 1

8
×
1

2
=
1

16
= RHS
OR
LHS = cos2x × cos x

2
−cos3x × cos
9x

2
x 9x
= 1

2
[2 cos 2x × cos 2
- 2 cos 2
× cos 3x] [multiplying numerator and denominator by 2]
= 1

2
[cos (2x + x

2
) + cos (2x − x

2
) - cos ( 9x

2
+ 3x) - cos ( 9x

2
− 3x) ] [∵ 2cos x × cos y = cos(x + y) + cos(x − y) ]
= 1

2
[ cos 5x

2
+ cos 3x

2
- cos 15x

2
- cos 3x

2
]
= 1

2
[cos 5x

2
- cos 15x

2
]
5x 15x 5x 15x
+ −
x+y x−y
= 1

2
[- 2 sin ( 2

2
2
) sin(
2

2
2
)] [∵cos x − cos y = −2sin (
2
) ⋅ sin(
2
)]

−5x
= - sin 5x sin ( 2
) = sin5x × sin
5x

2
[∵ sin(−θ) = −sin θ]
= RHS
∴ LHS = RHS

Hence proved.

Class interval cf Mid value (xi) fi ui =

xi −67.5
fiui fiu 2

33. 15
i

0-15 12 7.5 12 -4 -48 192

15-30 30 22.5 18 -3 -54 162

30-45 65 37.5 35 -2 -70 140

45-60 107 52.5 42 -1 -42 42

60-75 157 67.5 50 0 0 0

75-90 202 82.5 45 1 45 45

90-105 222 97.5 20 2 40 80

105-120 230 112.5 8 3 24 72

Total 230 -105 733

Here, a = 67.5, b = 15, N = ∑ f i = 230 , ∑f i ui = −105 and ∑ f iu
2
i
= 733

1 −105
∴ Mean = a + b ( N
∑ fi ui ) = 67.5 + 15 ( 230
) = 67.5 − 6.85 = 60.65

and Variance (σ 2
)= b
2
[
1

N
∑ fi u
2
i
− (
1

N
∑ fi ui ) ]

= 225 [ 733

230
− (−
105

230
) ]

= 225 [3.187 − (0.45) 2

]

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 = 225(3.187 − 0.2025) = 671.51
−−−− −−−− −−−−−
∴ Standard deviation = √V ariance = √671.51 = 25.91

OR
Let the other two observations be x and y
Therefore, our observations are 5, 7, 9, x and y
Mean = Sum of observations

Total number of observations

5+7+9+x+y
6 =
5
⇒ 6 × 5 = 21 + x + y
⇒ 30 − 21 = x + y or x + y = 9 ...........(i)
Now prepare the following table we have,
xi x1 − x̄ = x1 − 6 (x1 − x̄)
2

5 5 - 6 = -1 (-1)2 = 1

7 7-6=1 (1)2 = 1

9 9-6=3 (3)2 =

x x-6 (x - 6)2

y y-6 (y - 6)2

∑ (xi − x̄)
2
= 11 + (x - 6)2 + (y - 6)2
2
∑ ( x1 − x̄)
So, Variance, σ 2
=
n
2 2
11+(x−6) +(y−6)
4 =
5

⇒ 20 = 11 + (x2 + 36 - 12x) + (y2 + 36 - 12y)

⇒ 20 - 11 = x2 + 36 - 12x+ y2 + 36 - 12y
⇒ x2 + y2 + 72 - 12(9) - 9 = 0 from (i)]
⇒ x2 + y2 + 63 - 108 = 0
⇒ x2 + y2 = 45 ............... (ii)
From eq.(i)
Now, x + y = 9
(x + y)2 = (9)2
⇒ x2 + y2 + 2xy = 81
⇒ 45 + 2xy = 81 [from (ii)]

⇒ 2xy = 81 - 45 ⇒ 2xy = 36
18
⇒ xy =18 ⇒ x = y
.......... (iii)
After Putting the value of x in eq. (i) we get
2

y2 + 18 = 9y
18 18+y
⇒ + y = 9⇒ = 9 ⇒
y y

⇒ y2 - 9y + 18 = 0 ⇒ y2 - 6y - 3y + 18 = 0
⇒ y(y - 6) - 3(y - 6) = 0 ⇒ (y - 3) (y - 6) = 0

⇒ y - 3 = 0 and y - 6 = 0 ⇒ y = 3 and y = 6
For y = 3
18 18
x = = = 6
y 3

Hence, x = 6, y = 3 are the remaining two observation

For y = 6
18 18
x = = = 6
y 6

Hence, x = 3, y = 6 are the remaining two observation

Therefore, remaining two observations are 3 and 6
34. From the given question, we can write,
i. Let F be the midpoint of side AB.
Then, CF is the median through C.

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 2+(−2) 1+3
Coordinates of F are F ( 2
,
2
), i.e., F(0, 2)
⇔ 4(y - 5) = 3(x - 4) ⇔ 3x - 4y + 8 = 0
Hence, the equation of median CF is 3x - 4y + 8 = 0.
ii. Draw BL ⊥ AC , Then, BL is the altitude through B.
5−1
Slope of AC = 4−2
= 2

Let the slope BL be m.

−1
Since BL ⊥ AC , we have 2m = -1 and therefore, m = 2
−1
Thus, the slope of BL is = 2

So, the equation of BL is given by

y−3 −1
= ⇔ 2(y − 3) = −(x + 2) ⇔ x + 2y − 4 = 0
x+2 2

Hence, the equation of altitude BL is x + 2y - 4 = 0

iii. Let D be the midpoint of BC.
−2+4 3+5
Then, the coordinates of D are D ( 2
,
2
) , i.e., D(1, 4).
Through D, draw DP ⊥ BC
5−3
Slope of BC = = =
4+2
2

6
1

Let the slope of PD be m.

Since PD ⊥ BC, we have m × 1

3
= −1 ⇒ m = −3

So, the slope of PD is -3.

So, the equation fo PD is given by
y−4

x−1
= −3 ⇔ −3(x − 1) = (y - 4) ⇔ 3x + y - 7 = 0
Hence, the equation of the right bisector of BC is 3x + y - 7 = 0.
35. We have, f(x) = sinx + cosx
By using first principle of derivative
f (x+h)−f (x)
′
f (x) = lim
h
h→0

sin(x+h)+cos(x+h)−sin x−cos x
′
∴ f (x) = lim
h
h→0

[ sin x⋅cos h+cos x⋅sin h+cos x⋅cos h− sin x⋅sin h−sin x−cos x]
= lim h
[∵ sin ( x + y ) = sin x cos y + cos x sin y and cos ( x + y ) = cos x cos
h→0

y - sin x sin y]
[(cos x⋅sin h−sin x⋅sin h)+(sin x⋅cos h−sin x)+(cosx⋅cosh−cosx)]
= lim
h
h→0

sin h(cos x−sin x)+sin x(cos h−1)+cos x(cos h−1)

= lim
h
h→0

sin h sin x(cos h−1) cos x(cos h−1)

= lim (cos x − sin x) + lim + lim
h h h
h→0 h→0 h→0

−(1−cos h) −(1−cos h) sin x

= 1 ⋅ (cos x − sin x) + lim sin x [ ] + lim cos x [ ] [∵ lim = 1]
h h x
h→0 h→0 x→0

1−cos h 1−cos h
= (cos x − sin x) − sin x ⋅ lim ( ) − cos x ⋅ lim ( )
h h
h→0 h→0

2 h 2 h
2sin 2sin
2 h 2 h
= (cos x − sin x) − sin x ⋅ lim × − cos x ⋅ lim ×
h 4 h 4
h→0 h× h→0 h×
4 4

2 2
h h
sin sin
1 2 1 2
= (cos x − sin x) − sin x ⋅ 2 ⋅ lim ( ) × h − cos x ⋅ 2 ⋅ lim ( ) h
4 h 4 h
h h
→0 →0
2 2
2 2

1 1 sin x
= (cos x − sin x) − ⋅ sin x ⋅ (1) × 0 − cos x ⋅ ⋅ (1) × 0 [∵ lim = 1]
2 2 x
x→0

= (cos x - sin x) - 0 - 0
= cos x - sin x

10 / 11
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 Section E
36. Total letters in the word PERMUTATIONS = 12.
Here T = 2
(i) Now first letter is P and last letter is S which are fixed.
So the remaining 10 letters are to be arranged between P and S.
∴ Number of Permutations

=
10!
= = 1814400
10×9×8×7×6×5×4×3×2!

2! 2!

(ii) There are 5 vowels in the word PERMUTATIONS. All vowels can be put together.
∴ Number of permutations of all vowels together = P
5
5

5!
= = 5 × 4 × 3 × 2 × 1 = 120
0!

Now consider the 5 vowels together as one letter. So the number of letters in the word when all vowels are together = 8.
∴ Number of Permutations =
8! 8×7×6×5×4×3×2!
= = 20160
2! 2!

Hence the total number of permutations = 120 × 20160 = 2419200

OR
2 consonants out of 21 consonants can be chosen in 21C2 ways. 3 vowels out of 5 vowels can be chosen in 5C3 ways. Length of

the word is = (2 + 3) = 5 And also 5 letters can be written in 5! Ways. Therefore, the number of words can be formed is = (21C2 ×
5C × 5!) = 252000
3
37. Let AOB be the cable of uniformly loaded suspension bridge. Let AL and BM be the longest wires of length 30 m each. Let OC
be the shortest wire of length 6 m and LM be the roadway.
Now AL = BM = 30 m, OC = 6 m and LM = 100 m.
∴ LC = CM = LM= 50 m 1

Let O be the vertex and axis of the parabola be y-axis. So the equation of parabola in standard form is x2 = 4ay

Coordinates of point B are (50, 24)

Since point B lies on the parabola x2 = 4ay
∴ (50)2 = 4a × 24 ⇒ a = 2500

4×24
=
625

24

So equation of parabola is x 2
=
4×625

24
y ⇒ x
2
=
625

6
y

Let length of the supporting wire PW at a distance of 18 m be h.

∴ OR = 18 m and PR = PQ – QP = PQ - OC = h - 6

Coordinates of point P are (18, h - 6)

625
Since the point P lies on parabola x = 2

6
y

(18)2 = (h - 6) ⇒ 324 × 6 = 625h - 3750

625
∴
6
5694
⇒ 625 h = 1944 + 3750 ⇒ h = 625
= 9.11 m approx.
38. E : Student read Hindi newspaper
F : Student read English newspaper
60 3 40 2
P (E) = = ⋅ P (F ) = =
100 5 100 5
20 1
P (E ∩ F ) = =
100 5

i. P (E ∪ F ) ′
= 1 − P (E ∪ F )

= 1 − (P (E) + P (F ) − P (E ∩ F ))
1
=
5
P (E∩ F)
ii. P ( F

E
) =
P (E)

5 1
= =
3 3

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