Please refer to Class 12 Mathematics Sample Paper Set A with solutions below. The following CBSE Sample Paper for Class 12 Mathematics has been prepared as per the latest pattern and examination guidelines issued by CBSE. By practicing the Mathematics Sample Paper for Class 12 students will be able to improve their understanding of the subject and get more marks.
PART – A
Section – I
1. Evaluate :
Answer :
OR
Evaluate :
Answer :
2. If
and A2 – kA – 5I = O, then find the value of k.
Answer :
3. A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Answer : Let E : ‘a total of 8’ and F : ‘red die resulted in a number less than 4’
i.e., E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
and F = {(x, y) : x ∈ {1, 2, 3, 4, 5, 6}, y ∈ {1, 2, 3}}
i.e., F = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1),
(3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3),
(6, 1), (6, 2), (6, 3)}
Hence, E ∩ F = {(5, 3), (6, 2)}
P(E) = 5/36,
P(F) = 18/36, P(E ∩ F) = 2/36
∴ Required probability = P (E | F)
OR
If P(A) = 0.4, P(B) = 0.8 and P(B | A) = 0.6, then find P(A ∪ B).
Answer : Given, P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6
Clearly, P(A ∩ B) = P(B|A)P(A) = 0.6 × 0.4 = 0.24
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.4 + 0.8 – 0.24 = 0.96
4. Differentiate the function
Answer :
5. Find the cofactors of the element of third row and second column of the following determinant
Answer :
OR
If A is a matrix of order 3 × 3 and |A| = 5, then find the value of |adj A|.
Answer : |adj A| = |A|n–1
= 5 (3–1) = 52 = 25
6. Set A has three elements and set B has four elements. Find the number of injections that can be defined from A to B.
Answer : Since 3 < 4, injective functions from A to B are defined and the total number of such functions is 4P3
= 4!/(4-3)! = 4 × 3 × 2 × 1 = 24.
7. Find the solution of the differential equation dy/dx = x3e-2y.
Answer :
OR
Find the solution of y′ = y cot 2x.
Answer :
8. Find the principal value of cot-1 (-√3).
Answer :
9. Find the direction cosines of a line, for which a = b and g = 45°.
Answer : Since, cos2 α + cos2 β + cos2 γ = 1
⇒ 2cos2 a + cos2 45° = 1
OR
If P = (–2, 3, 6), then find the d.c.’s of OP.
Answer : Here, O ≡ (0, 0, 0) and P ≡ (–2, 3, 6)
Direction ratios of OP are –2 –0, 3 – 0, 6 – 0 i.e., –2, 3, 6
∴ Direction cosines of OP are
10. How many equivalence relations on the set {1, 2, 3} containing (1, 2) and (2, 1) are there in all ?
Answer : Possible equivalence relations are {(1, 2), (2, 1),
(1, 1), (2, 2), (3, 3)} and {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}
Hence, there are two possible equivalence relations.
11. If the plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(a) with x-axis, then find the value of a .
Answer : Direction ratios of x-axis is (1, 0, 0) and direction
ratios of the normal to the plane 2x – 3y + 6z = 11 is (2, –3, 6).
12. If A and B are two independent events such that P(A ∪ B) = 0.6 and P(A) = 0.2, then find P(B).
Answer : If A and B are two independent events, then
P(A ∩ B) = P(A) × P(B)
It is given that P(A ∪ B)= 0.6, P(A) = 0.2
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P(A ∪ B) = P(A) + P(B) – P(A) × P(B)
⇒ 0.6 = 0.2 + P(B)(1 – 0.2)
⇒ 0.4 = P(B) (0.8)
⇒ P(B) = 0.4/0.8 ⇒ P (B) = 1/2 = 0.5
13. If
Answer :
By equality of two matrices, we have
2x + y = 6 and 3y = 6 ⇒ y = 2.
Putting the value of y, we get
2x + 2 = 6 ⇒ 2x = 4 ⇒ x = 2.
14. If A and B are events such that P(A) > 0 and P(B) ≠ 1, then prove that P(A′ | B′) = 1–P(A∪B)/P(B’).
Answer :
15. Find the value of k in the following probability distribution.
| X = x | 0.5 | 1 | 1.5 | 2 |
| P(X = x) | k | k2 | 2k2 | k |
Answer : Since P(X) is a probability distribution of X,
⇒ P(X = 0.5) + P(X = 1) + P(X = 1.5) + P(X = 2) = 1
⇒ k + k2 + 2k2 + k = 1 ⇒ 3k2 + 2k – 1 = 0
⇒ (3k – 1) (k + 1) = 0
⇒ k = 1/3 or -1
But P(X = 0.5) = k = –1, which is not possible
∴ k=1/.3
16. If the angle between î+k̂and î + ĵ +ak̂ is π/3 then find the value of a.
Answer :
Section – II
Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark.
17. A poster is to be formed for a company advertisement. The top and bottom margins of poster should be 4 cm and the side margins should be 6 cm. Also, the area for printing the advertisement should be 384 cm2.
Based on the above answer the following :
(i) If a be the width and b be the height of poster, then the area of poster, expressed in terms of a and b, is given by
(a) 288 + 8a + 12b
(b) 8a + 12b
(c) 384 + 8a + 12b
(d) none of these
Answer : A
(ii) The relation between a and b is given by
(a) a=288+12b/b-8
(b) a=12b/b-8
(c) a=12b/b+8
(d) none of these
Answer : A
(iii) Area of poster in terms of b is
(a) 12b2/b-8
(b) 288b+12b2/b-8
(c) 288b+12b2/b+8
(d) none of these
Answer : B
(iv) The value of b, so that area of the poster is minimized, is
(a) 24
(b) 36
(c) 18
(d) 22
Answer : A
(v) The value of a, so that area of the poster is minimized, is
(a) 24
(b) 36
(c) 18
(d) 22
Answer : B
18. Consider the earth as a plane having points A(3, –1, 2), B(5, 2, 4) and C(–1, –1, 6) on it. A mobile tower is tied with 3 cables from the point A, B and C such that it stand vertically on the ground. The peak of the tower is at the point (6, 5, 9), as shown in the figure.
Based on the above answer the following :
(i) The equation of plane passing through the points A, B and C is
(a) 3x – 4y + 3z = 0
(b) 3x – 4y + 3z = 19
(c) 4x – 3y + 3z = 0
(d) 4x – 3y + 3z = 19
Answer : B
(ii) The height of the tower from the ground is
(a) 6 units
(b) 5 units
(c) 6/√34 units
(d) 5/√34 units
Answer : C
(iii) The equation of line of perpendicular drawn from its peak to the ground is
Answer : B
(iv) The coordinates of foot of perpendicular are
Answer : A
(v) The area of DABC is
(a) √34 sq. units
(b) 2√34 sq. units
(c) √17 sq. units
(d) 2√7 sq. units
Answer : B
