# 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 = [a

_{ij}]

_{2 × 3}and B = [b

_{ij}]

_{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:**

= 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**5x

^{7}= t

^{6}dx = dt

^{6}dx =(dt/35)

^{6}sin (5x

^{7}) dx = ∫ sin t ·(dt/35)

= -(cos t/35)+c

^{7}/35)+C

∴ k =(-1/7).

6. (c) tan y – cot x = C

^{2}y dy = – cosec

^{2}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:**

^{2}sq. units

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.

^{2}+ b

^{2}– c

^{2}– d

^{2}

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)

15. (a) x^{2} + 1

16. (c) x = 2

17.

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

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

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.

## Section-B

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

21. Given function f : N → N such that

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

**Case I:** When x is odd.

_{1}) = f(x

_{2})

_{1}+ 1 = x

_{2}+ 1

_{1}= x

_{2}

_{1}) = f(x

_{2})

_{1}= x

_{2}∀ x

_{1}, x

_{2}∈ N

So, f(x) is one-one.

**Case II:** When x is even.

_{1}) = f(x

_{2})

_{1}– 1 = x

_{2}– 1

_{1}= x

_{2}

_{1}) = f(x

_{2})

_{1}= x

_{2}∀ x

_{1}, x

_{2}∈ 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.

Therefore, f is onto.

Hence, f is one-one and onto.

22.

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

l : m : n = 2 : – 3 : 6

24. Given differential equation is

On integrating both sides, we get

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

tan^{– 1} 1 = 0 + 0 + C

∴ Equation (i) becomes

25.

We know that, if three points with position vectors

are collinear then

= 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

⇒ 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)

Put x = – 2 in (ii)

⇒ – 2 = B(– 3) (– 5)

Put the value in (i)

_{1}be the probability that he is a cyclist, E

_{2}be the probability that he is a scooter driver and E

_{3}be the probability that he is a car driver.

Let E be the event that accident occurs

**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

Hence, E and F are independent events.

28.

Adding (i) and (ii), we get

Put

When x = 0, t = 0

**OR**

^{2}+ xy) dy = (x

^{2}+ y

^{2}) dx

which is a homogeneous differential equation.

Put y = vx

**OR**

Given, differential equation is

It is linear differential equation where,

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

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.

= 20 – (– 7) = 27 sq. units.

^{–}gets oxidised at the anode to produce Cl

_{2}gas.

33. Let x = a cos θ

·.· x = a cos θ

**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].

and ul

^{2}+ vm

^{2}+ wn

^{2}= 0 …(ii)

_{1}l

_{2}+ m

_{1}m

_{2}+ n

_{1}n

_{2}= k(b

^{2}w + c

^{2}v + c

^{2}u + a

^{2}w + b

^{2}u + a

^{2}v)

For perpendicularity, we have

_{1}l

_{2}+ m

_{1}m

_{2}+ n

_{1}n

_{2}= 0 i.e., a

^{2}(v + w) + b

^{2}(w + u) + c

^{2}(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 4u

^{2}b

^{2}c

^{2}– 4(b

^{2}u + a

^{2}v) × (c

^{2}u + a

^{2}w) = 0

⇒ u

^{2}b

^{2}c

^{2}= b

^{2}c

^{2}u

^{2}+ b

^{2}c

^{2}wu + a

^{2}vuc

^{2}+ a

^{4}vw

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

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

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

Mid-point of diagonal BD = Mid-point of diagonal AC

(·.· Diagonals of parallelogram bisect each other)

Comparing corresponding coordinates of

x = 3, y = 4, z = 5

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

35. Given

**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)

(iii) The given differential equation can be rewritten as

⇒ 2^{y} dy = dx

On integrating both sides, we get

∫ 2^{y}.dy = ∫ dx + C

^{y}= 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

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

(iii) We have

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

and F = Event of drawing a second non yellow ball

Here,

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

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

Here,

The dot mark ◉ field are mandatory, So please fill them in carefully**To download the Sample Paper (PDF File), Please fill & submit the form below.**