Solution

CLASS XII THIRD SEMESTER MOCK TEST

Class 12 - Mathematics

1.
(b) None of these
Explanation:
A relation R on a non-empty set A is said to be reflexive iff xRx for all x ∈ R .
A relation R on a non-empty set A is said to be symmetric iff xRy⇔ yRx, for all x , y ∈R .
A relation R on a non-empty set A is said to be transitive iff xRy and yRz ⇒ xRz, for all x ∈ R.
An equivalence relation satisfies all these three properties. .
None of the given relations satisfies all three properties of equivalence relation.

2. (a) one-one and into

Explanation:
f(x) = x2 + x + 1
One-one function
Let p, q be two arbitrary elements in N
Then, f(p) = f(q)
⇒ p2 + p + 1 = q2 + q + 1
⇒ p2 - q2 + p - q = 0
⇒ (p - q) (p + q + 1) = 0
⇒ p = q, p + q + 1 ≠ 0 (∵ p, q ∈ N)
When f(p) = f(q), p = q
thus, f(x) is one-one function.
Onto function
For x = 1, f(x) assumes value 3.
As, f(x) cannot assume value less than 3, for x ∈ N
Thus, f(x) is not onto function. It is into function.
3.
(b) an equivalence relation
Explanation:
R is reflexive since every triangle is congruent to itself.
Further, (T , T ) ∈ R
1 2

⇒ T1 is congruent to T2

⇒ T2 is congruent to T 1 ⇒ (T2 , T1 ) ∈ R .
Hence, R is symmetric.
Moreover, (T1, T2), (T2, T3) ∈ R
⇒ T1 is congruent to T2 and T2 is congruent to T3
⇒ T1 is congruent to T3
⇒ (T1 , T3 ) ∈ R .
Therefore, R is an equivalence relation.

4.
(d) A is false but R is true.
Explanation:
Assertion is false because distinct elements in N has equal images.
(1+1)
for example f(1) = 2
=1

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 f(2) = 2

2
=1
Reason is true because for injective function if elements are not equal then their images should be unequal.

5.
(c) π

Explanation:
−1 1 −1
si n ( ) + co t (3)
√5

⇒ tan
−1
(
1

2
) + tan
−1
(
1

3
) because 1

2
.
1

3
< 1

1 1
( )+( )
−1 2 3
⇒ tan ( )
1 1
1−( )( )
2 3

−1 π
⇒ tan (1) =
4

6.
(b) R - (-1, 1)
Explanation:
We have to find: The range of sec-1(x)
Here, the inverse function is given by y = f-1(x)
The graph of the function y = sec-1(x) can be obtained from the graph of
Y = sec x by interchanging x and y axes.i.e, if a, b is a point on Y = sec x then b, a is the point on the function y = sec-1(x)
Below is the Graph of the range of sec-1(x)

From the graph, it is clear that the domain of sec-1(x) is a set of all real numbers excluding -1 and 1 i.e, R - (-1, 1).
Which is the required solution.

7. (a) [0, π] − { π

2
}

Explanation:
π
[0, π] − { }
2

8.
4 −6
(d) [ ]
6 4

Explanation:
2 −1
Given, A = [ ]
1 2

2 −1 2 −1 4 − 1 −2 − 2
Now, A 2
= [ ][ ]= [ ]
1 2 1 2 2 + 2 −1 + 4

3 −4
2
⇒ A = [ ]
4 3

Now, A2 + 2A - 3I
3 −4 2 −1 1 0
= [ ]+ 2[ ]− 3[ ]
4 3 1 2 0 1

3 + 4 − 3 −4 − 2 − 0 4 −6
= [ ]= [ ]
4 + 2 − 0 3 + 4 − 3 6 4

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 9.
(b) Nilpotent
Explanation:
The given matrix A is nilpotent, because |A| = 0, as determinant of a nilpotent matrix is zero.

10. (a) skew-symmetric matrix

Explanation:
1 5 1 5
A - A' = [ ]− [ ]
6 7 6 7

1 5 1 6
= [ ]− [ ]
6 7 5 7

0 −1 0 1 0 −1
⇒ A - A' = [ ] and (A - A')' = [ ]= −[ ]
1 0 −1 0 1 0

= -(A - A') [∵ A' = - A, so A is a skew-symmetric matrix]

Hence, (A - A') is skew-symmetric matrix
11.
−4 6
(d) [ ]
−3 −3

Explanation:
1 −2
(A − 2B) = ( )
3 0

Multiplying equation by 2
2 −4
2A − 4B = ( ) ...(i)
6 0

−2 2
2A − 3B = ( ) ...(ii)
3 −3

(ii) - (i)
−2 2 2 −4
B= ( ) − ( )
3 −3 6 0

−4 6
= ( )
−3 −3

12.
(b) skew-symmetric matrix
Explanation:
0 −5 8
⎡ ⎤

We have A = ⎢ 5 0 12 ⎥
⎣ ⎦
−8 −12 0

0 5 −8
⎡ ⎤

∴ A' = ⎢ −5 0 −12 ⎥ =-A

⎣ ⎦
8 12 0

So, matrix A is skew-symmetric

13.
(d) (-1)i + j Mij
Explanation:
Cofactor of an element aij denoted by Aij is defined by Aij = (-1)i + j Mij where Mij is minor of aij.

14.
(d) 1, -7, 1
Explanation:

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 The minors of the diagonal elements of the given determinant are
∣ −1 3 ∣
M11 = ∣ ∣ = 1 - 0 = 1,
∣ 0 −1 ∣

∣3 ∣2
M22 = ∣ ∣ = -3 - 4 = -7
∣2 −1 ∣

∣3 −1 ∣
M33 = ∣ ∣ = -3 + 4 = 1
∣4 −1 ∣

15.
(c) I3
Explanation:
Because, the inverse of an identity matrix is an identity matrix itself.

16.
33
(d) 2

Explanation:
The given system of equations has a non-trivial solution if :
∣1 k 3 ∣
∣ ∣ 33
3 k −2 = 0 ⇒ 1(−4k + 6) − k(−12 + 4) + 3(9 − 2k) = 0 ⇒ k =
∣ ∣ 2

∣2 3 −4 ∣

17.
(c) (a) - (iii), (b) - (i), (c) - (ii)
Explanation:
(a) - (iii), (b) - (i), (c) - (ii)

18.
(c) 0
Explanation:
Since, f is continuous at x = π

2
sin(cos x−cos x)
∴ f( ) =
π

2
lim
π 2
x→ (π−2x)
2

sin(cos x−cos x)
i.e. k = lim
π 2
x→ (π−2x)
2

Let x = π

2
- h,
π π
sin(cos( −h)−cos( −h)

⇒ k = lim 2

2
2

h→0 π
(π−2( −h))
2

sin(sin h)−sin h
= lim 2
h→0 4h
3 5

Using sin x = x - x

3!
+ x

5!
+ ......
3 5
sin h sin h
(sin h− + ....)−sin h

⇒ k = lim 3!

2
5!

h→0 4h
3 5
− sin h sin h
= lim (
2
+
2
. . . .)
h→0 3!×4h 5!×4h

=0
∴ lim
π
f(x) = 0 = k
x→
2

⇒ k=0

19.
(b) − sin x

1+cos y
, y ≠ (2n + 1)π

Explanation:

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 We have, y + sin y = cos x On differentiating both sides w.r.t. x, we get
dy

dx
+
d

dx
(sin y) = dx
d
(cos x)

dy dy dy

dx
+ cos y ⋅
dx
= -sin x ⇒ dx
=− sin x

1+cos y

where, y ≠ (2n + 1)π

2
sin (a+y)
20. (a) sin a

Explanation:
sin y
x sin(a + y) = sin y ⇒ x =
sin(a+y)

dx sin(a+y) cos y−sin y cos(a+y)

⇒ =
dy 2
sin (a+y)

sin(a+y−y) sina
= =
2 2
sin (a+y) sin (a+y)
2
dy sin (a+y)
⇒ =
dx sin a

21.
(d) 1
Explanation:
u = sin-1 ( 2x

2
) = 2 tan −1
x ⇒
du

dx
= 2

2
1+x 1+x

v = tan-1 ( 2x

2
) = 2 tan −1
x ⇒
du

dx
= 2

2
1+x 1+x
2
du du/dx 2/1+x
∴
dv
= dv/dx
= 2
=1
2/1+x

22.
y(1−x)
(c) x(y−1)

Explanation:
Given that xy = ex + y
Taking log both sides, we get
logexy = x + y (Since logabc = cloga b)
Since logabc = loga b + loga c, we get
loge x + logey = x + y
Differentiating with respect to x, we obtain
1 1 dy dy
+ = 1 +
x y dx dx

Or
dy y−1 1−x
( ) =
dx y x

dy y(1−x)
Therefore, dx
=
x(y−1)

23.
(c) sec x
Explanation:
sec x

24.
√ex −1
(c) 2

Explanation:
Given, x = log (1 + t2) and y = t - tan-1 t
On differentiating both sides w.r.t.x, we get
dy 2
dx

dt
= 1

2
(2t) and dt
=1- 1

2
= t

2
1+t 1+t 1+t

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 2
t

(1+t2 )
dy
∴
dx
= 2t
= t

2
...(i)
2
(1+t )

Also, x = log (1 + t2)

⇒ t2 = ex-1 ...(ii)
From Eqs. (i) and (ii), we get
dy √ex −1
=
dx 2

25.
(c) (1 + sin 2x) y1
Explanation:
y = etanx
y1 = sec2 x etanx

⇒ cos2 xy1 = etanx

Again differentiating w.r.t. x we get
cos2 (x) .y2 - 2 cos x sin xy1 = sec2 xetanx

⇒ cos2 (x).y2 = y1 sin 2x + y1.

26.
(d) a function of y only
Explanation:
y = ax2 + bx + c
dy
= 2ax + b
dx
2
d y
= 2a
2
dx
2
d y
y
3
= 2ay
3
= A function of y only
dx2

27.
(c) f is differentiable at x = 0 but not at x = 1
Explanation:
f (x)−f (1)
We observe that, lim x−1
x→1
1
(x−1) sin( )−0

= lim
x −1

x−1
x→1

= lim sin( x−1

1
)
x→1

= An oscillating number between -1 and 1

f (x)−f (1)
∴ lim
x−1
does not exist.
x→1

f (x)−f (0)
∴ f(x) is not differentiable at x = 1. and lim x−0
x→0

1
(x−1) sin( )−sin 1

= lim
x −1

x
x→0
1 1
x sin( ) sin( )+sin 1

= lim - lim
x −1 x −1

x x
x→0 x→0

x 2−x
2 sin cos{ }
2(x −1) 2(x −1) C+D C−D
= -sin 1 - lim [∵ sin C + sin D = 2 sin
2
cos
2
]
x→0 x
{ }2(x−1)
2(x −1)

2−x
2 cos{ }
2(x −1)

= -sin 1 - lim 2(x−1)

x→0

2−0
cos( )
2(0−1)

= -sin 1 - (0−1)

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 = -sin 1 + cos 1 [∵ cos(−θ) = cos θ]
∴ f(x) is differentiable at x = 0.

28.
(c) -(1 + e) sin e
Explanation:
To find [cos(log x + e
d

dx
x
)] at x = 1 , use the chain rule:
d x 1 x
= − sin(log x + e ) ⋅ ( + e ).
dx x

At x = 1, log 1 = 0 and e 1
= e , so the expression simplifies to − sin e ⋅ (1 + e) . Thus, the answer is −(1 + e) sin e .

π
29. (a) increasing in [0, 2
]

Explanation:
f(x) = 2 sin3x - 3 sin2x + 12 sin x + 5
f'(x) = 6 sin2x cos x - 6 sin x cos x + 12 cos x
= 6 cos x (sin2x - sin x + 2}
= 6 cos x {sin2x - 2 sin x × 1

2
+
1

4
−
1

4
+ 2}
2
= 6 cos x {(sin x − 1

2
) +
7

4
} ≥ 0∀x ∈ ⌊0,
π

2
⌋

∴ f(x) is increasing in [0, π

2
]

1
30. (a) a = 2, b = − 2

Explanation:
Let f(x) = alogx + bx2 + x
′ 1
f (x) = a. + 2bx + 1
x

For maximum and minimum value of f(x) we have f'(x) = 0

Therefore, at x = -1 and x = 2 we have 2bx2 + x + a = 0
i.e, a + 2b = 1....(i) and a + 8b = 2....(ii)
(ii) - (i ) gives b = − 1

Now , from (i) we get a = 2

⇒ a = 2, b = −
1

31. (a) local minima at x = 2 and a local maxima at x = – 2

Explanation:
Given , f(x) = x + 4

′ 4
⇒ f (x) = 1 −
x2

′
⇒ f (x) = 0

⇒ x = ±2

′′ 8
⇒ f (x) =
3
x

′′ 8
⇒ f (2) = = 1 > 0
8

′′ 8
⇒ f (−2) = = −1 < 0
−8

So, f(x) has a local minima at x = 2 and a local minima at x= -2.

32.
(c) tan-1 3

Explanation:
tan-1 3

33.
25
(d) 42

Explanation:

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 Here, P(A) = 2

5
, P(B) = 3

10
and P (A ∩ B) = 1

5
′ ′
P (A ∩ B ) 1−P (A∪ B)
′ ′
P (A /B ) = =
′ 1−P (B)
P (B )

1−[P (A)+P (B)−P (A∩ B)]

=
1−P (B)

2 3 1
1−( + − )
5 10 5
=
3
1−
10
4+3−2 1
1−( ) 1−
10 2 5
= = =
7 7 7

10 10
′ ′
P (B ∩ A ) 1−P (A∪ B)
And P (B /A ) = ′ ′
′
=
P (A ) 1−P (A)

1
1−
2 1/2 5 1
= = = [∵ P (A ∪ B) = ]
2 3/5 6 2
1−
5

′ ′ ′ ′ 5 5 25
∴ P (A /B ) ⋅ P (B /A ) = ⋅ =
7 6 42

34. (a) P (A). P (B)

Explanation:
If A and B are independent, then P (A ∩ B) = P (A) ⋅ P (B)
35.
1−P(A∪ B)
(c) ′
P( B )

Explanation:
∵ P (A) > 0 and P (B) ≠ 1
′ ′
P (A ∩ B ) 1−P (A∪ B)
′ ′
P (A /B ) = =
′ ′
P (B ) P (B )

36.
(b) 7

Explanation:
P(A) = , P(A⋂B) = 4

5
7

10
(Given)
Now we know,
P(A∩ B)
P(B|A) =
P(A)

= 10

= 10
7
×
5

= 7

37.
(d) 2

Explanation:
2

38.
(d) 3

Explanation:
′
P (B∩ A ) P (B)−P (B∩ A)
′
P (B/A ) = =
′
P (A ) 1−P (A)

3 3 6−3
−
5 10 10 6 3
= = = =
1 1 10 5
1−
2 2

39.
(c) 7

Explanation:

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 7

40. (a) 5

Explanation:
Here, P (A) = 7

13
, P (B) =
9

13
and P (A ∩ B) = 4

13
′
P (A ∩ B) P (B)−P (A∩ B)
′
∵ P (A |B) = =
P (B) P (B)
9 4 5
−
13 13 13 5
= = =
9 9 9

13 13

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