Solution
CLASS 11 MATHEMATICS TEST
Class 11 - Mathematics
1. (a) N
Explanation: We have,A′ ∪ (A ∪ B) ∩ B′
= A′ ∪ [(B’ ∩ A) ∪ (B’ ∩ B)] {∵ Distributive property of set: (A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C)}
= A′ ∪ [(A ∩ B’) ∪ Φ] {∵ (B’ ∩ B) = ϕ}
= A′ ∪ (A ∩ B’) = (A’ ∪ A) ∩ (A’ ∪ B’) {∵ Distributive property of set: (A ∪ B) ∩ (A ∪ C) = A ∪ (B ∩ C)}
= Φ ∩ (A’ ∪ B’) {∵ (A’ ∩ A) = ϕ}
= (A’ ∪ B’) = (A ∩ B)’ {∵ (A’ ∪ B’)
= (A ∩ B)’} A′ ∪ (A ∪ B) ∩ B′ = (A ∩ B)’
A contains all odd numbers and B contains all even numbers
Therefore, A ∩ B = ϕ
⇒ A′ ∪ (A ∪ B) ∩ B′ = {ϕ}’
⇒ A′ ∪ (A ∪ B) ∩ B′ = N
2.
(c) A ∩ Bc
Explanation: A ∩ Bc
A and B are two sets.
A ∩ B is the common region in both the sets.
(A ∩ Bc) is all the region in the universal set except A ∩ B
Now, A ∩ (A ∩ B)c = A ∩ Bc
3.
(d) {x : x ∈ R, 4 ≤ x < 5}
Explanation: Set A represents the elements which are greater or equals to 4 and the elements are real no. A[4, ∞)
Set B represents the elements which are less than 5 and are real no. B(−∞, 5)
So if we represent these two in number line we can see the common region is between 4(included) and 5(excluded).
4.
(d) Domain = R, Range = (−∞, 2]
Explanation: We have, f(x) = 2 − |x − 5|
Clearly, f(x) is defined for all x∈R
∴ Domain of f = R
Now, |x − 5| ≥ 0, ∀x ∈ R
⇒ −|x − 5| ≤ 0
⇒ 2 − |x − 5| ≤ 2
∴ f (x) ≤ 2
∴ Range of f = (−∞, 2]
5. (a) [-1, 1/3]
Explanation: we know, −1 ≤ cos x ≤ 1
−2 ≤ −2 cos x ≤ 2
−1 ≤ (1 − 2 cos x) ≤ 3
1 1
−1 ≤ ( ) ≤
1−2 cos x 3
So, R(f) = [-1, 1/3]
6.
(b) [−1, 2) ∪ [3, ∞)
(x+1)(x−3)
Explanation: Here (x−2)
≥ 0
But x ≠ 2
so, x ∈ [−1, 2) ∪ [3, ∞)
1/5
OR
(c) 2-1
1
Explanation: fof(x) = f(f(x)) = {f (25 − x 4
) 4
}
1 1
= {25 − (25 − x 4
)} 4 = (x 4
) 4 =x
∴ f (f (
1
2
)) = 1
= 2-1
7. (a) {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 2), (2, 3)}
Explanation: R = {(x, y) : |x2 - y2| < 7}
R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 2), (2, 3)}
8.
x1 + x2
(c) f (x 1) + f (x2 ) = f (
1+x1 x2
)
1+x1 1+x2
Explanation: f (x 1) + f (x2 ) = log(
1−x1
) + log(
1−x2
)
= log(1 + x1 ) − log(1 − x1 ) + log(1 + x2 ) − log(1 − x2 )
x1 +x2
1+
x1 + x2
⎛ 1+x x ⎞ 1+ x1 x2 + x1 + x2
1 2
f ( ) = log = log[ ]
1+x1 x2 x1 +x2 1+ x1 x2 − x1 − x2
⎝ 1− ⎠
1+x x
1 2
(1+ x1 )(1+ x2 )
= log[ ]
(1− x1 )(1− x2 )
= log(1 + x1 )(1 + x2 ) − log(1 − x1 )(1 − x2 )
= log((1 + x1 ) + log(1 + x2 ) − log(1 − x1 ) − log(1 − x2 )
x1 + x2
∴ f (x1 ) + f (x2 ) = f ( )
1+x1 x2
OR
(d) 17
Explanation: f(-2) + f(0) + f(2) + f(5) = -2 + 0 + 4 + 15 = 17
9. (a) x > 5
Explanation: x - 5 > 0
⇒ x>5
⇒ x ∈ (5, ∞)
2x−4
Now x+2
<2
2x−4
x+2
-2<0
2x−4−2(x+2)
⇒
x+2
<0
2x−4−2x−4
⇒
x+2
<0
−8
⇒
x+2
<0
a
⇒ x + 2 > 0 [∵ b
< 0, a < 0 ⇒ b > 0]
⇒ x > -2
⇒ x ∈ (−2, ∞)
Hence the solution set is (5, ∞) ∩ (−2, ∞) = (5, ∞)
Which means x > 5.
10. (a) no solution
Explanation: We have given: 4x + 3 ≥ 2x + 17
⇒ 4x − 2x ≥ 17 − 3 ⇒ 2x ≥ 1
⇒ x ≥
14
2
[Dividing by 2 on both sides]
⇒ x ≥ 7 ..... (i)
Also we have 3x - 5 < -2
⇒ 3x < −2 + 5 ⇒ 3x < 3
2/5
⇒ x < 1
On combining (i) and (ii), we see that solution is not possible because nothing is common between these two solutions.(i.e., x <
1, x ≥ 7)
11.
3
(c)
√19
Explanation: Using L'Hospital,
2x
2
√x2 +10
lim
1
x→3
2x
2√x2 +10
Substituting x = 3 in 1
We get 3
√19
12. (a) 1
2
tan 2x
x[ −1]
tan 2x−x x
Explanation: Given, lim 3x−sin x
= lim
sin x
x→0 x→0 x[3− ]
x
tan 2x
×2−1
2x 1.2−1 2−1 1
lim = = =
sinx 3−1 2 2
x →0
3−
x
13.
(d) n
n n
(1+x ) −1
Explanation: lim = lim n(1 + x)
n−1
= n
x→0 (1+x)−(1) x→0
14. (a) 100
2 3 n
x+ x + x …x −n
Explanation: lim x−1
= 5050
x→1
2 3
(x−1) (x −1) (x −1) n
x −1
⇒ lim
x−1
+
x−1
+
x−1
…
x−1
= 5050
x→1
n n
= nan-1]
x −a
⇒ 1 + 2 + 3 ... n = 5050 [∵ x−a
n(n+1)
⇒
2
= 5050
⇒ n (n + 1) = 10100
⇒ n(n + 1) = 100 × 101
On comparing:
n = 100
15. (a) 1
Explanation: lim tan x
log(1+x)
⋅
x
x
x→0
tan x 1
⇒ lim ⋅
x log(1+x )
x→0
x
⇒ 1⋅ 1 = 1
16.
(b) -2
sin x+cos x
Explanation: Given, y = sin x−cos x
dy (sinx−cosx)(cosx−sinx)−(sin x+cos x)(cos x+sin x)
=
dx 2
(sin x−cos x)
2 2
−(sin x−cos x ) −(sin x+cos x )
=
2
(sin x−cos x)
2 2 2 2
−[sin x+ cos x−2 sin x cos x+ sin x+ cos x+2 sin x cos x]
=
2
(sin x−cos x)
−2
=
2
(sin x−cos x)
dy −2 −2
∴ ( ) = = = −2
dx 2 2
at x=0 (sin 0−cos 0) (−1)
17. (a) 6[cos(3π/4)+isin(3π/4)]
Explanation:
3/5
.
.
18.
(d) -2
Explanation: -2
502 590 588 586 584
i +i +i +i +i
− 1
582 580 578 576 574
i +i +i +i +i
[∵ i4 = 1 and i2 =-1]
4×148 4×147+2 4×147 4×146+2 4×146
i +i +i +i +i
= − 1
4×145+2 4×145 4×144+2 4×144 4×143+2
i +i +i +i +i
2 2
1+ i +1+ i +1
= − 1
2 2 2
i +1+ i +1+ i
1
= − 1
−1
= -2
19.
3
(d) 2
- 2i
Explanation: Let z = x + iy
Now
|z| = z + 1 + 2i
−−−−−−
2
⇒ √x + y
2
⇒ x + 1 + i(y + 2)
−−−−−−
2
⇒ √x + y
2
= x + 1 and y + 2 = 0
⇒ y = -2 and x2 + 4 = (x + 1)2
⇒ y = -2 and x =
3
Hence z = 3
2
- 2i
20.
9
(d) ( 1
4
+
4
i)
1 1 (1+2i) 1 3
Explanation: (1−2i)
= (1−2i)
×
(1+2i)
=( 5
+
5
i)
3 3 (1−i) (3−3i) 3 3
(1+i)
= (1+i)
×
(1−i)
= 2
=( 2
−
2
i)
(1− i )
(3+4i) (3+4i) (2+4i) −10+20i −10 20 −1
(2−4i)
= (2−4i)
×
(2+4i)
= 2
=( 20
+
20
i) =( 2
+ i)
(4−16i )
−1 −1
∴ given expression = {( 1
5
+
3
2
) + (
2
5
−
3
2
i)} (
2
+ i) =( 17
10
−
11i
10
)(
2
+ i)
−17 17 5 45 9
(
20
+
11
10
) + (
10
+
11
20
)i =( 20
+
20
i) =( 1
4
+
4
i)
21. (a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)
22.
1
(c)
√2
4/5
Explanation: cos 15° - sin 15° = cos(45° - 30°) - sin (45° - 30°)
= (cos 45° cos 30° + sin 45° sin 30°) - (sin 45° cos 30° - cos 45° sin 30°)
1 √3 1 1 1 √3 1 1
= {( × ) + ( × )} − {( × ) − ( × )}
√2 2 √2 2 √2 2 √2 2
( √3+1) ( √3−1)
2 1
= − = =
2√2 2√2 2√2 √2
23.
(d) −56
65
Explanation: In quadrant IV, cos θ > 0, sin θ < 0 , cos ϕ > 0 and sin ϕ < 0
Now, cos θ = 4
5
⇒ sin
2
θ = (1 − cos
2
θ) = (1 −
16
25
) =
9
25
−−
9 −3
⇒ sin θ = − √ =
25 5
12 2 2 144 25
cos ϕ = ⇒ sin θ = (1 − cos ϕ) = (1 − ) =
13 169 169
−−
−
25 −5
⇒ sin ϕ = − √ =
169 13
∴ sin(θ + ϕ) = sin θ cos ϕ + cos θ sin ϕ
−3 12 −5 4 −36 20 −56
= ( × ) + ( × )= ( − ) =
5 13 13 5 65 65 65
24. (a) 1
Explanation: π = 180 ∘
Using tan (180 - A) = -tan A, we get;
C = π - (A + B)
Now,
tan A+tan B+tan C
tan A tan B tan C
tan A+tan B+tan[π−(A+B)]
=
tan A tan B tan[π−(A+B)]
tan A+tan B−tan(A+B)
=
− tan A tan B tan(A+B)
tan A+tan B
tan A+tan B−
1−tan A tan B
=
tan A+tan B
− tan A tan B×
1−tan A tan B
2 2
tan A+tan B− tan A tan B−tan A tan B−tan A−tan B
=
2 2
− tan A tan B−tan A tan B
2 2
− tan A tan B−tan A tan B
=
2 2
− tan A tan B−tan A tan B
=1
OR
(b) 1
Explanation: Given exp. = 1
2
(2 sin 70° sin 50°) . sin 10° = 1
2
[cos(70° - 50°) - cos (70° + 50°)] sin 10°
=
1
2
(cos 20° - cos 120°) sin 10° = 1
2
(cos 20
∘
+
1
2
) sin 10
∘
1 1
=
4
( 2 cos 20° sin 10°) + 4
sin 10°
=
1
4
[sin (20° + 10 °) - sin (20° - 10°)] 1
4
sin 10°
1 ∘ 1 1 1
= sin 30 = ( × ) =
4 4 2 8
25.
(b) 2
Explanation: We have:
2 π 2 π 2 7π 2 4π
sin + sin + sin + sin
18 9 18 9
2 π 2 2π 2 7π 2 8π
= sin + sin + sin + sin
18 18 18 18
2 π 2 2π 2 7π 2 8π
= sin + sin + sin ( ) + sin ( )
18 18 18 18
2 π 2 2π 2 π 2π 2 π π
= sin + sin + sin ( − ) + sin ( − )
18 18 2 18 2 18
2 π 2 2π 2 2π 2 π
= sin + sin + cos + cos
18 18 18 18
2 π 2 π 2 2π 2 2π
= sin + cos + sin + cos
18 18 18 18
=1+1
=2
5/5
