Class 12 Mathematics Sample Paper With Solutions Set A

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Sample Paper Class 12 Mathematics With Solutions Set A

PART – A
Section – I

1. If the matrix A =

Class 12 Mathematics Sample Paper With Solutions Set A

is singular, then find x.
Answer : Given that matrix A is singular ⇒ |A| = 0

⇒ 1(–6 – 2) + 2(–3 – x) + 3(2 – 2x) = 0
⇒ –8 – 6 – 2x + 6 – 6x = 0
⇒ –8x – 8 = 0 ⇒ x = –1

What positive value of x makes the following pair of determinants equal ?

Answer : We have,

⇒ 2x² – 15 = 32 – 15 ⇒ 2x² = 32 ⇒ x² = 16
⇒ x = 4 [∵ x > 0]

2. If A=

then find the matrix X, such that 2A + X = 5B.
Answer : We have, 2A + X = 5B
⇒ X = 5B – 2A

3. Determine the order and degree of differential equation d3x/dt3 + d2x/dt2 + (dx/dt)2 = et
Answer : Given differential equation is of order 3 and degree 1.

What is the degree of the differential equation (d2y/dx2) + 4 – 3dy/dx = 0?
Answer : We have,

4. If R = {(x, y) : x + 2y = 8} is a relation on N, then find the range of R.
Answer : Here, R = {(x, y) : x + 2y = 8}, where x, y ∈ N.
For x = 1, 3, 5, … ; x + 2y = 8 has no solution in N.
For x = 2, we have 2 + 2y = 8 ⇒ y = 3
For x = 4, we have 4 + 2y = 8 ⇒ y = 2
For x = 6 , we have 6 + 2y = 8 ⇒ y = 1
For x = 8, 10, … ; x + 2y = 8 has no solution in N.
∴ Range of R = { 1, 2, 3}

5. If a line makes angles 90°, 60° and 30° with the positive directions of x, y and z-axis respectively, then find its direction cosines.
Answer : Let the direction cosines of the line be l, m, n. Then,
l = cos 90° = 0, m = cos60° = 1/2 and n = cos30° = √3/2
So, direction cosines are < 0 1/2 , √3/2 >.

Find the direction cosines of the line passing through two points (2, 1, 0) and (1, –2, 3).
Answer : Here, P(2, 1, 0) and Q(1, –2, 3)

6. Find the area enclosed between the curve x2 + y2 = 16 and the coordinate axes in the first quadrant.
Answer : Given curve is a circle with centre (0, 0) and radius 4.
∴ Required area

7. Evaluate: ∫xe^x2 dx
Answer :

Evaluate :

Answer :

8. If α, β, γ are the angles made by a line with the co-ordinate axes. Then find the value of sin2α + sin2β + sin2γ.
Answer : a, b and g are the angles made by line with the co-ordinate axes.
∴ cos2 α + cos2 β + cos2 γ = 1
⇒ 1– sin2 α + 1 – sin2 β + 1 – sin2 γ = 1
⇒ sin2 α + sin2 β + sin2 γ = 2

9. Check whether the relation R = {(1, 1), (2, 2) (3, 3), (1, 2), (2, 3), (1, 3)} on set A = {1, 2, 3} is an equivalence relation or not.
Answer : Reflexive : (1, 1), (2, 2), (3, 3) ∈ R, R is reflexive
Symmetric : (1, 2) ∈R but (2, 1) ∉ R, R is not symmetric.
Transitive : (1, 2) ∈ R and (2, 3) ∈ R ⇒ (1, 3) ∈ R, R is transitive.
Since, R is not symmetric. So, R is not an equivalence relation.

Find the domain of the function f(x) = 1/√[{sin x} + {sin Π + x}] where {·} denotes fractional part.
Answer :

10. Find the value of (ā.î,)²+(ā.ĵ)²+(ā.k̂)².
Answer :

11. A die, whose faces are marked 1, 2, 3 in red and 4, 5, 6 in green, is tossed. Let A be the event “number obtained is even” and B be the event “number obtained is red”. Find P(A ∩ B).
Answer : We have, S = {1, 2, 3, 4, 5, 6}
Let A be the event that number is even = {2, 4, 6}
and B be the event that number is red = {1, 2, 3}
Now, A ∩ B = {2}
∴ P(A ∩ B) = 1/6

12. Find the value of p for which p(î + ĵ + k̂) is a unit vector.
Answer : Let ā =( î + ĵ + k̂)

13. Let f : N → N be defined by f(n) =

for all n ε N. . Find whether the function f is bijective or not.
Answer :

⇒ f(2k – 1) = f(2k), where k ∈ N
But, 2k – 1 ≠ 2k ⇒ f is not one-one.
Hence, f is not bijective.

14. If P(A) = 7/13 , P(B) = 9/13 and P(A∪B) = 12/13 then evaluate P(A|B).
Answer : Given,

15. If ā,Ђ,č are unit vectors such that ā +Ђ + č = ô then write the value of ā.Ђ +Ђ. č +č.ā.
Answer : We have ā,Ђ,č are unit vectors.
Therefore, |ā| =1 ,|Ђ| =1 and ,|č|=1
Also,ā+Ђ+č = ô (given)
|ā+Ђ+č|²=0

16. Write a 3 × 2 matrix whose elements in the i^th row and j^th column are given by a_ij = (2i-j)/2.
Answer :

Section – II

Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark.

17. Arun and Richa decided to play with dice to keep themselves busy at home as their schools are closed due to coronavirus pandemic. Arun throw a dice repeatedly untiln a six is obtained. He denote the number of throws required by X. Based on this information, answer the following questions.

(i) The probability that X = 3 equals
(a) 1/6
(b) 5²/6³
(c) 5/3⁶
(d) 1/6³
Answer : B

(ii) The probability that X = 5 equals
(a) 1/6⁴
(b) 1/6⁶
(c) 5⁴/6⁵
(d) 5/6⁴
Answer : C

(iii) The probability that X ≥ 3 equals
(a) 25/216
(b) 1/36
(c) 5/36
(d) 25/36
Answer : D

(iv) The value of P(X > 3) + P(X ≥ 6) is
(a) 5³/6⁵
(b) 1− 5³/6²
(c) 5/6
(d) 1/6
Answer : C

18. Peter’s father wants to construct a rectangular garden using a rock wall on one side of the garden and wire fencing for the other three sides as shown in figure. He has 100 ft of wire fencing. Based on the above information, answer the following questions.

(i) To construct a garden using 100 ft of fencing, we need to maximise its
(a) volume
(b) area
(c) perimeter
(d) length of the side
Answer : B

(ii) If x denote the length of side of garden perpendicular to rock wall and y denote the length of side parallel to rock wall, then find the relation representing total amount of fencing wall.
(a) x + 2y = 100
(b) x + 2y = 50
(c) y + 2x = 100
(d) y + 2x = 50
Answer : A

(iii) Area of the garden as a function of x i.e., A(x) can be represented as
(a) 100 + 2x²
(b) x – 2x²
(c) 100x – 2x²
(d) 100 – x²
Answer : C

(iv) Maximum value of A(x) occurs at x equals
(a) 25 ft
(b) 30 ft
(c) 26 ft
(d) 31 ft
Answer : A

(v) Maximum area of garden will be
(a) 1200 sq. ft
(b) 1000 sq. ft
(c) 1250 sq. ft
(d) 1500 sq. ft
Answer : C

PART – B
Section – III

19. Evaluate :

Answer : Using integration by parts, we get

20. The equation of a line is 5x – 3 = 15y + 7 = 3 –10z. Write the direction cosines of the line.
Answer : The given line is 5x – 3 = 15y + 7 = 3 –10z

i.e., its direction ratios are proportional to 6, 2, –3.

Show that the lines x+1/3 = y+3/5 = z+5/7 and x-2/1 = y-4/3 = z-6/5 intersect. Also find their point of intersection.
Answer : Any point on the line

is (k + 2, 3k + 4, 5k + 6)
For lines (i) and (ii) to intersect, we must have
3r – 1 = k + 2, 5r – 3 = 3k + 4, 7r – 5 = 5k + 6
On solving these, we get r = 1/2,k = -3/2
∴ Lines (i) and (ii) intersect and their point of intersection is (1/2 ,-1/2 ,-3/2)

21. Find the area of the region bounded by the curve y = x² and the line y = 4.
Answer : We have, y = x² and y = 4

22. Prove that : 3sin–1x = sin–1(3x – 4x³), x ε [-1/2 , 1/2]
Answer : Put sin–1x = θ. Then x = sinθ
Now, sin3θ = (3sinθ – 4sin3θ) = (3x – 4x³)
⇒ 3θ = sin^–1(3x – 4x³)
⇒ 3sin^–1x = sin^–1(3x – 4×3) [ θ = sin^–1x]
Hence, 3sin^–1x = sin–1(3x – 4x³)

23. An unbiased dice is thrown twice. Let the event A be ‘odd number on the first throw’ and B be the event ‘odd number on the second throw’. Check the independence of the events A and B.
Answer : If all the 36 elementary events of the experiment are considered to be equally likely, then we have
P(A) = 18/36 = 1/2 and P(B) = 18/36 =1/2
Also, P(A ∩ B) = P(odd number on both throws)
9/36 = 1/4
Now, P(A).P(B) = 1/2×1/2=1/4
Clearly, P(A ∩ B) = P(A) × P(B)
Thus, A and B are independent events.

A bag contains 4 balls. Two balls are drawn at random (without replacement) and are found to be white. What is the probability that all balls in the bag are white ?
Answer : Consider the following events.
E : Two balls drawn are white
A : There are 2 white balls in the bag
B : There are 3 white balls in the bag
C : There are 4 white balls in the bag

24. Determine the value of ‘k’ for which the following function is continuous at x = 3.

Answer : Given, f(x) is continuous at x = 3.

25. Solve the differential equation :
dy/dx = 1+x²+y²+x²y² , given that y = 1 when x = 0.
Answer : We have,

26. Find (AB)^–1, if A =

Answer :

27. Show that the function f(x) = x³ – 3x² + 6x – 100 is increasing on R.
Answer : We have, f(x) = x³ – 3x² + 6x – 100 …(i)
Differentiating (i) w.r.t. x, we get
f ′(x) = 3x² – 6x + 6 = 3(x² – 2x + 1) + 3
= 3(x – 1)² + 3 > 0
For all values of x, (x – 1)² is always positive
∴ f ′ (x) > 0
So, f(x) is an increasing function on R.

28. Prove that the points A, B and C with position vectors ā ,Ђ ,č respectively are collinear if and only if āxЂ+č+čxā = ô.
Answer : The points A, B and C are collinear

OR î, ĵ, k̂

If ā = 7î + ĵ – 4k̂and Ђ = 2î + 6ĵ + 3k̂ then find the projection of Ђ on ā.
Answer :

Section – IV

29. Let A = R−{2} and B =R−{1}If f :A → B is a function defined by f (x) = x−1/x−2 ,then show that f is one-one and onto.
Answer : Here, f : A → B is given by f(x) = x-1/x-2

∴ f(x) = y when x = 2y-1/y-1 ε A (as y ≠ 1)
Hence, f is onto.
Thus, f is one–one and onto.

30. Using integration, find the smaller area enclosed by the circle x2+y2 = 4 and the line x + y = 2.
Answer : The given curves are
x² + y² = 4 and x + y = 2

31. Find the equation of tangent to the curve x = sin 3t, y = cos 2t at t = Π/4.
Answer : The given curve is x = sin 3t; y = cos 2t

Find the intervals in which the function f(x) = 3x⁴−4x³−12x²+5 s
(a) strictly increasing (b) strictly decreasing
Answer : We have, f(x) = 3x⁴ – 4x³ – 12x² + 5
f ′(x) = 12x³ – 12x² – 24x = 12x(x² – x – 2)
⇒ f ′(x) = 12x(x + 1)(x – 2)
Now, f ′(x) = 0
⇒ 12x(x + 1)(x – 2) = 0
⇒ x = –1, x = 0 or x = 2
Hence these points divide the whole real line into four disjoint open intervals namely (–∞, –1),
(–1, 0), (0, 2) and (2, ∞)

Interval	Sign of f ′(x)	Nature of function
(– ∞, –1)	(–) (–) (–) < 0	Strictly decreasing
(–1, 0)	(–) (+) (–) > 0	Strictly decreasing
(0, 2)	(+) (+) (–) < 0	Strictly decreasing
(2, ∞)	(+) (+) (+) > 0	Strictly decreasing

(a) f(x) is strictly increasing in (–1, 0) ∪ (2, ∞)
(b) f(x) is strictly decreasing in (–∞, –1) ∪ (0, 2)

32. Evaluate :

Answer :

33. For what value of a is the function f defined by

is continuous at x = 0 ?
Answer : For f(x) to be continuous at x = 0, we have

34. Solve the following differential equation :

Answer : We have,

Solve the differential equation x(dy/dx) + y = x cos x + sin x, given that y = 1 when x = Π/2.
Answer :

35. Show that the function f(x) = |x – 1| + |x + 1|, for all x ∈ R, is not differentiable at the points x = –1 and x = 1.
Answer : The given function is f(x) = |x – 1| + |x + 1|

Section – V

36. If

, then find A–1 and use it to solve the following system of the equations :
x + 2y – 3z = 6, 3x + 2y – 2z = 3
2x – y + z = 2
Answer :

The given system of equations is
x + 2y – 3z = 6
3x + 2y – 2z = 3
2x – y + z = 2
The system of equations can be written as AX = B

∵ A^–1 exists, so system of equations has a unique solution given by X = A^–1B

Determine the product

and use it to solve the system of equations x – y + z = 4, x – 2y – 2z = 9, 2x + y + 3z = 1.
Answer :

The given system of equations is
x – y + z = 4, x –2y – 2z = 9, 2x + y + 3z = 1 and it can
be written as AX = B
where,

Here, |A| = 1(–6 + 2) + 1(3 + 4) + 1(1 + 4)
= –4 + 7 + 5 = 8 ≠ 0
So, the given system of equations has a unique solution
given by X = A^–1B.

37. A variable plane which remains at a constant distance 3p from the origin cuts the coordinates axes at A, B, C.
Show that the locus of the centroid of triangle ABC is 1/x² + 1/y² + 1/z² = 1/p².
Answer : Let the equation of the plane be x/a+ y/b + z/c = 1
where, a, b, c are variables.
This meets X, Y and Z axes at A(a, 0, 0), B(0, b, 0) and C(0, 0, c).
Let (α, β, γ) be the coordinates of the centroid of triangle ABC. Then,

The plane (i) is at a distance 3p from the origin.
∴ 3p = Length of perpendicular from (0, 0, 0) to the plane (i)

From (ii), we have
a = 3α, b = 3β and c = 3γ
Substituting the values of a, b, c in (iii), we get

Find the distance between the lines l1 and l2 given by
l1 : r̄ = î + 2ĵ − 4k̂ + λ( 2î + 3ĵ + 6k̂)
l2 : r̄ = 3î + 3ĵ − 5k̂ + μ( 4î + 6ĵ + 12k̂)
Answer : l1 : r̄ = î + 2ĵ − 4k̂ + λ( 2î + 3ĵ + 6k̂)
l2 : r̄ = 3î + 3ĵ − 5k̂ + μ( 4î + 6ĵ + 12k̂)
∴ We have
ā₁ = î + 2ĵ − 4k̂, Ђ₁ = 2î + 3ĵ + 6k̂ and
ā₂ = 3î + 3ĵ − 5k̂, Ђ₂ = 4î + 6ĵ + 12k̂
So, ā₂ − ā1 = 2î + ĵ − k̂
Also Ђ₂ = 4î + 6ĵ + 12k̂= 2Ђ₁ ⇒ Ђ₁||Ђ₂
Hence l1 and l2 are parallel lines.
Shortest distance between two parallel lines is,

38. Find graphically, the maximum value of z = 2x + 5y, subject to constraints :
2x + 4y ≤ 8, 3x + y ≤ 6, x + y ≤ 4; x ≥ 0, y ≥ 0
Answer : Let l₁ : 2x + 4y = 8, l₂ : 3x + y = 6, l₃ : x + y = 4;
x = 0, y = 0
Solving l₁ and l₂, we get B(8/5 ,6/5)

Shaded portion OABC is the feasible region, where coordinates of the corner points are O(0, 0),
A(0, 2), B(8/5 , 6/5),C(2, 0)
The value of objective function at these points are :

Corner points	Value of the objective function z = 2x + 5y
O(0, 0)	2 × 0 + 5 × 0 = 0
A(0, 2)	2 × 0 + 5 × 2 = 10 (Maximum)
B(8/5 , 6/5),C(2, 0)	2×8/5 + 5×6/5 = 9.2
C(2, 0)	2 × 2 + 5 × 0 = 4

∴ The maximum value of z is 10, which is at A(0, 2).

Maximise z = 8x + 9y subject to the constraints :
2x + 3y ≤ 6, 3x – 2y ≤ 6, y ≤ 1; x, y ≥ 0
Answer : Let l₁ : 2x + 3y = 6, l₂ : 3x – 2y = 6, l₃ : y = 1; x = 0, y = 0

Solving l₁ and l₃, we get D (1.5, 1)
Solving l₁ and l₂, we get C
Shaded portion OADCB is the feasible region, where coordinates of the corner points are O(0, 0),
A(0, 1), D(1.5, 1), C (30/13 , 6/13),B(2, 0).
The value of the objective function at these points are :

Corner points	Value of the objective function z = 2x + 5y
O(0, 0)	8 × 0 + 9 × 0 = 0
A(0, 1)	8 × 0 + 9 × 1 = 9
D (1.5, 1)	8 × 1.5 + 9 × 1 = 21
C(30/13 , 6/13)	8 x 30/13 + 9 x 6/13 = 22.6(Maximum)
B (2, 0)	8 × 2 + 9 × 0 = 16