Mathematics Unsolved Sample Paper Solutions CBSE Class 12

Section-A

(Multiple Choice Questions)

Each question carries 1 mark

1. (c) AB and BA both are defined

Explanation: Let  A = [aij]2 × 3 and B = [bij]3 × 2

So, both AB and BA are defined.

2. (b) (n – 1)

Explanation: Minor of an element is the determinant obtained by leaving the elements of the column and row containing that element. Therefore, minor of an element of a determinant of order n is a determinant of order n – 1.

3. (b) 5 sq. units

Explanation:

diagram 87

= 5 sq. units.

4. (b) (cos x − 2/3)

Explanation: Given,   2x + 3y = sin x

On differentiating both sides w.r.t. x, we get

(d/dx)(2x+3y)=(d/dx)(sinx)

⇒ 2 + 3(dy/dx)=cosx

⇒ 3(dy /dx)=cosx-2 

(dy/dx)=(cos x-2/3).

5. (d) k =(-1/7)

Explanation: Put    5x7 = t
⇒   35x6 dx = dt
⇒  x6 dx =(dt/35)
∴ ∫ x6 sin (5x7 ) dx = ∫ sin t ·(dt/35)

= -(cos t/35)+c

= (-cos 5x7/35)+C

∴ k =(-1/7).

6. (c) tan y – cot x = C

diagram 93
⇒ sec2 y dy = – cosec2 x dx

On integrating both sides, we get

tan y = cot x + C

⇒ tan y – cot x = C.

7. (b) a function to be optimised

Explanation: The objective of an LPP is a function to be optimised.

8. (a) tan θ

Explanation:

diagram94
9. (c) (8 /3)a2 sq. units
diagram95

10. (d) 4 elements

Explanation: Since, 3 × 4 signifies that there are 3 rows and 4 columns. Then, each row has 4 elements.

11. (c) non-negative restrictions

Explanation: The conditions x ≥ 0, y ≥ 0 are called non-negative restrictions.

12. (a) a2 + b2 – c2 – d2
diagram96

13. (d) None of these

14. (c) (1/3)

Explanation: A = {7, 8, 9}

B = {8}

P(A) =(3/9), P(B) =(1/9), A ∩ B={8}, P(A ∩ B) =(1/9)

P(A) =(3/9), P(B) =(1/9), A ∩ B={8}, P(A ∩ B) =(1/9)

diagram98

15. (a) x2 + 1

16. (c) x = 2

diagram100

17.

18. (a) (a, 1, c)

Explanation: Given, x = ay + b, z = cy + d

diagram102

Hence, direction ratios are (a, 1, c).

19. (b) Both A and R are true and R is not the correct explanation of A.

Explanation: ·.· f(x) is odd.

⇒ f(– x) = – f(x)

and g(x) is even ⇒ g(– x) = g(x)

Let F(x) = f(x) + g(x)

= f(– x) + g(– x)

= – f(x) + g(x)

≠ ± F(x)

∴ F(x) is neither even nor odd.

Reason is also true, because this a factual statement.

20. (c) A is true and R is false.

diagram103

Section-B

This section comprises of very short answer type-questions (VSA) of 2 marks each

21. Given function f : N → N such that

diagram104

For one-one: From the given function we observe that

Case I: When x is odd.

Let f(x1) = f(x2)
⇒ x1 + 1 = x2 + 1
⇒ x1 = x2
·.· f(x1) = f(x2)
⇒ x1 = x2 ∀ x1, x2 ∈ N

So, f(x) is one-one.

Case II: When x is even.

Let f(x1) = f(x2)
⇒ x1 – 1 = x2 – 1
⇒ x1 = x2
·.· f(x1) = f(x2)
⇒ x1 = x2 ∀ x1, x2 ∈ N

So f(x) is one-one.

OR

Given, f : R → R is defined as f(x) = 10x + 7.

For one-one: Let f(x) = f(y) where x, y ∈ R

⇒ 10x + 7 = 10y + 7

⇒ x = y

Therefore, f is a one-one function.

For onto: For y ∈ R

Let y = 10x + 7

x =(y − 7/10)

Therefore for any y ∈ R, there exists.

diagram105

Therefore, f is onto.

Hence, f is one-one and onto.

22.

diagram106

23. If l, m, n are the direction cosines of the line then

l : m : n = 2 : – 3 : 6

diagram107

24. Given differential equation is

diagram108

On integrating both sides, we get

Put y = 1 and x = 0 in equation (i)

tan– 1 1 = 0 + 0 + C 

∴ Equation (i) becomes

diagram111

25.

diagram112

We know that, if three points with position vectors

diagram113

are collinear then

diagram114

= 0, which satisfies the condition of collinearity.

Hence, the given points are collinear.

Section-C

(This section comprises of short answer type questions (SA) of 3 marks each)

26. Put

diagram115

⇒  3x + 4 = A(x + 2) (x – 3) + B(x – 1) (x – 3) + C(x + 2) (x – 1) …(ii)

Put x = 1 in (ii)

​⇒ 7 = A(3) (– 2) + B(0) + C(0)

⇒ 7 = – 6A

Put x = 3 in (ii)

⇒ 13 = 0 + 0 + C(5) (2)

diagram117

Put x = – 2 in (ii)

⇒ – 2 = B(– 3) (– 5)

diagram118

Put the value in (i)

diagram119
27. Let E1 be the probability that he is a cyclist, E2 be the probability that he is a scooter driver and E3 be the probability that he is a car driver.

Let E be the event that accident occurs

diagram121

OR

When a coin is tossed three times, the sample space is given by

S = {HHH, HHT, HTH, THH, HTT, TTH, THT, TTT}

Now E = event that the first throw results in a head

∴ E = {HHH, HHT, HTH, HTT}

and F = event that the last throw results in a tail

∴ F = {HHT, THT, HTT, TTT}

So, E ∩ F = {HHT, HTT}

Clearly, n(E) = 4, n(F) = 4, n(E ∩ F) = 2

and n(S) = 8

diagram122

Hence, E and F are independent events.

28.

diagram123

Adding (i) and (ii), we get

Put

diagram125

When x = 0, t = 0

diagram126
diagram127

OR

diagram128
29. Given, (x2 + xy) dy = (x2 + y2) dx

which is a homogeneous differential equation.

Put y = vx

diagram131

OR

Given, differential equation is

It is linear differential equation where,

diagram133

30. We have to, Maximize, Z = 10x + 6y

Subject to constraints 3x + y ≤ 12

2x + 5y ≤ 34

x ≥ 0, y ≥ 0

Now, consider    3x + y = 12       2x + 5y = 34

diagram134

From graph, it is clear that lines intersects at (2, 6).

So, the maximum value of Z is 56 at point (2, 6).

31. Given

Section-D

(This section comprises of long answer type questions (LA) of 5 marks each)

32.

diagram137
On solving (i) and (ii), we get
x = – 2 or 4

= 20 – (– 7) = 27 sq. units.

E° is preferred. But due to the over potential of oxygen, Cl– gets oxidised at the anode to produce Cl2 gas.

33.  Let x = a cos θ

·.· x = a cos θ

diagram140

OR

Given, R = {(a, b) : a, b ∈ A, |a – b| is divisible by 4}

Reflexivity: For any a ∈ A

| a – a| = 0, which is divisible by 4

⇒ (a, a) ∈ R

So, R is reflexive.

Symmetric: Let (a, b) ∈ R

⇒ |a – b| is divisible by 4.

⇒ |b – a| is divisible by 4.

⇒ (b, a) ∈ R

So, R is symmetric.

Transitivity: Let (a, b) ∈ R and (b, c) ∈ R.

⇒ |a – b| is divisible by 4.

|a – b| = 4k

⇒ a – b = ± 4k   …(i)

Also |b – c| is divisible by 4.

|b – c| = 4m

⇒ b – c = ± 4m   …(ii)

Adding equations (i) and (ii), we get

a – b + b – c = ± 4(k + m)

⇒ a – c = ± 4(k + m)

∴ (a, c) ∈ R

So, R is transitive.

⇒ R is reflexive, symmetric and transitive.

∴ R is an equivalence relation.

Let x be the element of R such that (x, 1) ∈ R then |x – 1| is divisible by 4.

x – 1 = 0, 4, 8, 12, …    [·.· x ≤ 12]

∴ Set of elements of A which are related to 1 are {1, 5, 9}.

Equivalence of class 2 i.e.,

[2] = {(a, 2) : ∈ A, |a – 2| is divisible by 4}

⇒   |a – 2| = 4   [k is whole number, k ≤ 3]

a = 2, 6, 10

Therefore, equivalence class [2] is [2, 6, 10].

34.   Given, al + bm + cn = 0   …(i)
and ul2 + vm2 + wn2 = 0 …(ii)
diagram141
diagram142
So, l1l2 + m1m2 + n1n2 = k(b2w + c2v + c2u + a2w + b2u + a2v)

For perpendicularity, we have

l1l2 + m1m2 + n1n2 = 0 i.e.,   a2(v + w) + b2(w + u) + c2(u + v) = 0 For the given lines to be parallel, the d.c.’s must be equal and the roots of (iii) must be equal. Then, we must have 4u2b2c2 – 4(b2u + a2v) × (c2u + a2w) = 0
⇒ u2b2c2 = b2c2u2 + b2c2wu + a2vuc2 + a4vw
or

OR

First, we find vector equation of AB, where A(4, 5, 10) and B(2, 3, 4).

We know that two points vector formula of line is given by

diagram143

∴ Using equation (i) required equation of line AB is

diagram144

Similarly, vector equation line BC where B(2, 3, 4) and C(1, 2, – 1) is

Now, let the coordinates of D be (x, y, z).
Mid-point of diagonal BD = Mid-point of diagonal AC
(·.· Diagonals of parallelogram bisect each other)
diagram146

Comparing corresponding coordinates of

x = 3, y = 4, z = 5

∴ Coordinates of point D(x, y, z) = (3, 4, 5).

35. Given

diagram148

Hence Proved.

Section-E

(This section comprises of 3 case-study/passage-based questions of 4 marks each with two sub-parts. First two case study questions have three sub-parts (i), (ii), (iii) of marks 1, 1, 2 respectively. The third case study question has two sub-parts of 2 marks each.)

36. (i) Here,

(ii) Solution of given differential equation is given by

y.x = ∫ x.3x dx + C         (·.· I.F. = x)

diagram150

(iii) The given differential equation can be rewritten as

⇒ 2y dy = dx

On integrating both sides, we get

∫ 2y.dy = ∫ dx + C

diagram160
⇒ 2y = x log 2 + C′
where C′ = log 2.

37. (i) Let p be the price per ticket and x be the number of tickets sold.

So, revenue function, R(x) = px

diagram161

(ii) Since, more that 36,000 tickets cannot be sold. So, range is [0, 36000].

(iii) We have

diagram 162
diagram163

38. (i) Let E = Event of drawing a first green ball

and F = Event of drawing a second non yellow ball

Here,

diagram164

(ii) Let E = Event of drawing a first non-blue ball

F = Event of drawing a second non-blue-ball

Here, 

diagram165

CBSE 36 Sample Question Papers Commerce Stream

All Subjects Combined for Class 12 Exam 2023

CBSE 36 Sample Question Papers Science Stream (PCM)

All Subjects Combined for Class 12 Exam 2023

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