# Class 12 Mathematics Sample Paper With Solutions Set D

Please refer to Class 12 Mathematics Sample Paper With Solutions Set D below. These Class 12 Mathematics Sample Papers will help you to get more understanding of the type of questions expected in the upcoming exams. All sample guess papers for Mathematics Class 12 have been designed as per the latest examination pattern issued by CBSE. Please practice all CBSE Sample Papers for Mathematics in Standard 12.

## Sample Paper Class 12 Mathematics With Solutions Set D

PART – A

Section – I

1. Write the cofactor of the element a31 in A =

OR

If A is a square matrix of order 3 and |2A| = k|A|, then find the value of k.

2. Evaluate :

3. Find the integrating factor of the differential equation

OR

Find order and degree of the equation

4. If f (x) = x2 – 4x + 1, find f (A), where

5. Find the unit vector in the direction of vector a̅ = 2î + 3ĵ + 4k̂.
The unit vector in the direction of a vector a̅ is given

OR

Find the projection of the vector 7î + ĵ – 4k̂ on 2î + 6ĵ + 3k̂.

6. Check whether the function f(x) = x3 – 3x2 – x is one-one or not?
We have f(x) = x3 – 3x2 – x
Clearly, f(1) = 1 – 3 – 1 = –3
and f(–1) = –1 – 3 + 1 = –3
⇒ Distinct elements have same image, therefore f is not one-one.

7. Evaluate :

OR

Evaluate :

8. Check whether the lines having direction ratios (√3 −1, − √3 −1, 4) and (−√3 +1, √3 +1, − 4) are perpendicular to each other.

9. If the vectors 3î + 2ĵ − k̂  and 6ɵi − 4xɵj + ykɵ are parallel, then the values of x and y.

OR

Find the point which divides the line segment joining the points (–2, 3, 5) and (1, 2, 3) in the ratio 2 : 3 externally.
Let C(x, y, z) divides the line segment joining the points A(–2, 3, 5) and B(1, 2, 3) in the ratio 2 : 3 externally.

10. Find vector equation of the plane which is at a distance of 6/√29 from the origin and its normal vector from the origin is 2î − 3ĵ + 4k̂.

11. Let A = {1, 2, 3, 4}. Show that f = {(1, 2), (2, 3), (3, 4), (4, 1)} is a bijection from A to A?
Here f (1) = 2, f (2) = 3, f (3) = 4, f (4) = 1
Since no two elements have the same image. So f is one-one. Also, every elements has atleast one preimage.
So, f is onto.
Thus f is bijective.

12. If a̅  and b̅ are unit vectors enclosing an angle q and |a̅+b̅| <1, find the value of q.

13. Find the area of the region bounded between the line x = 2 and the parabola y2 = 8x.
We have, y2 = 8x and x = 2

14. Find equation of a line which passes through the point (1, 2, 3) and is parallel to the vector 3î + 2ĵ − 2k̂.
Let a̅ = î + 2ĵ − 2k̂ and b̅ = 3î + 2ĵ − 2k̂
We know that the line which passes through point a̅ and parallel to b̅ is given by r̅ = a̅ + b̅ , where λ is a constant.
r̅ = î + 2ĵ + 3k̂ + λ(3î + 2ĵ − 2k̂) is the required equation of the line.

15. If

such that A + B + C is a zero matrix, then find the matrix C.
We have, A + B + C = O  C = –[A + B]

16. If the line joining (2, 3, –1) and (3, 5, –3) is perpendicular to the line joining (1, 2, 3) and (3, 5, l), then find the value of λ.
D.R.’s of the two lines are 1, 2, –2 and 2, 3, λ – 3.
Since, lines are perpendicular
∴ a1a2 + b1b2 + c1c2 = 0
⇒ 1 × 2 + 2 × 3 – 2 (l – 3) = 0 ⇒ l = 7

Section – II

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

17. A teacher arranged a surprise game for students of a classroom having 5 students, namely Amit, Aruna, Eklavya, Yash and Samina. He took a bag containing tickets numbered 1 to 11 and told each student to draw two tickets without replacement.

(i) Probability that both ticket drawn by Amit shows even number, is
(a) 1/11
(b) 2/11
(c) 3/11
(d) 4/11
B

(ii) Probability that both tickets drawn by Aruna shows odd number, is
(a) 1/11
(b) 2/11
(c) 3/11
(d) 4/11
C

(iii) When tickets are drawn by Eklavya, find the probability that number on one ticket is a multiple of 4 and on other ticket is a multiple 5.
(a) 4/55
(b) 6/55
(c) 7/55
(d) None of these
A

(iv) When tickets are drawn by Yash, find the probability that number on one ticket is a prime number and on other ticket is a multiple of 4 .
(a) 3/11
(b) 5/11
(c) 6/11
(d) 2/11
D

(v) When tickets are drawn by Samina, find the probability that first ticket drawn shows an even number and second ticket drawn shows an odd number.
(a) 2/11
(b) 3/11
(c) 5/11
(d) 8/11
B

18. An open water tank of aluminium sheet of negligible thickness, with a square base and vertical sides, is to be constructed in a farm for irrigation. It should hold 4000 l of water, that comes out from a tube well.

Based on above information, answer the following questions.
(i) If the length, width and height of the open tank be x, x and y m respectively, then surface area of tank is
given by
(a) S = x2 + 2xy
(b) S = 2x2 + 4xy
(c) S = 2x2 + 2xy
(d) S = 2x2 + 8xy
D

(ii) The relation between x and y is
(a) x2y = 4
(b) xy2 = 4
(c) x2y2 = 4
(d) xy = 4
A

(iii) The outer surface area of tank will be minimum when depth of tank is equal to
(a) half of its width
(b) its width
(c) (1/4) th of its width
(d) (1/3) rd of its width
A

(iv) The cost of material will be least when width of tank is equal to
(a) half of its depth
(b) twice of its depth
(c) (1/4)th of its depth
(d) thrice of its depth
B

(v) If cost of aluminium sheet is ₹ 360/m2, then the minimum cost for the construction of tank will be
(a) ₹ 2320
(b) ₹ 3320
(c) ₹ 4320
(d) ₹ 5320
C

PART – B
Section – III

19. Find the equations of the tangent and the normal to the curve y = x3 at the point P(1, 1).
The given curve is y = x3.

20. Express

21. Find the area of region bounded by the curve y2 = x(1 – x)2, shown in following figure.

Answer : Given curve is y2 = x(1 – x)2
If y = 0, then x(1 – x)2 = 0 ⇒ x = 0, x = 1

22. Suppose 5 men out of 100 and 25 women out of 1000 are good orator. If an orator is chosen at random, find the probability that a male person is selected. Assume that there are equal number of men and women.
Let E1, E2 and A denote the events defined as follows :
E1 = person selected is man
E2 = person selected is woman
A = person selected is good orator

23. Evaluate :

OR

Evaluate :

24. If

25. Find the projection of the vector 3î − 3ĵ − 6k̂ on vector joining the points (5, 6, – 3) and (3, 4, – 2).

OR

If  a̅ = 4î + 3ĵ + 2k̂  find b̅ = 3î + 2k̂ find |b̅ X 2a̅|.

26. Suppose that two cards are drawn at random from a deck of 52 cards. Let X be the number of aces obtained. Then, find the probability distribution of X.
Total no. of aces = 4
Also, X can take the values 0, 1, 2

27. If y = sin–1x, then show that

28. Find the solution of the differential equation

OR

Find the particular solution of the differential equation log (/dx) = 3x+4y ; y = 0 , x= 0.

Section – IV

29. Show that the curve for which the normal at every point passes through a fixed point is a circle.
Let P(x, y) be an arbitrary point on the given curve. The equation of the normal to the given curve at

30. Find the point on the parabola y2 = 2x which is closed to the point (1, 4).
Let A (x, y) be the required point which is closest to the point B(1, 4). Then, the distance AB should be minimum and therefore AB2 should be minimum.

31. If

OR

Differentiate

32. Let f : A → B be a function defined as f(x) = 2x+3/x–3 where A = R − {3} and B = R − {2}. Is the function f one-one and onto?

33. Find the area of the region bounded by the curve y = x2 + x, the x-axis and the lines x = 2, x = 5.
Given curve is y = x2 + x.
Clearly, required area = Area of shaded region

34. Evaluate :

OR

Evaluate :

35. If f(x) is continuous at x = 0, where

Answer : Since, f(x) is continuous at x = 0, therefore

Section-V

36. Find the vector and cartesian equation of the line through the point î + ĵ−3k̂ and perpendicular to the lin r̅ = 2î − 3ĵ + λ(2î + ĵ −3k̂) and r̅ = 3î − 5ĵ + μ(î + ĵ + k̂)
Here we need to find, the equation of the line through the point (1, 1, –3) and perpendicular to the lines

OR

The four points A(3, 2, –5), B(–1, 4, –3), C(–3, 8, –5) and D(–3, 2, 1) are coplanar. Find the equation of the plane containing them.
The equation of the plane passing through the point A(3, 2, –5) is given by
a(x – 3) + b(y – 2) + c(z + 5) = 0 …(i)
If it passes through B(–1, 4, –3) and C(–3, 8, –5),
we get
a(–1 –3) + b(4 – 2) + c(–3 + 5) = 0
⇒ – 4a + 2b + 2c = 0
⇒ 2a – b – c = 0 …(ii)
Also a(–3 – 3) + b(8 –2) + c(–5 + 5) = 0
⇒ –6a + 6b + 0c = 0
⇒ a – b – 0c = 0 ..(iii)
Solving (ii) and (iii) by cross multiplication method,
we get

37. Find the minimum value of Z = 3x + 4y + 270 subject to the constraints
x + y ≤ 60
x + y ≥ 30
x ≤ 40, y ≤ 40
x ≥ 0, y ≥ 0
Converting inequations into equations, we get
x + y = 60 …(i)
x = 40 …(ii)
y = 40 …(iii)
x + y = 30 …(iv)
x = 0 …(v)
and y = 0 …(vi)
Let us draw the graph of equations (i) to (vi). The feasible region is shown in figure.

The coordinates of the corner points of the feasible region are A(30, 0), B(40, 0), C(40, 20), D(20, 40), E(0, 40) and F(0, 30).
Let us evaluate Z at these points.

From the table, the minimum value of Z is 360, which is attained at the point A(30, 0).

OR

Find the point for which the maximum value of Z = x + y subject to the constraints 2x + 5y ≤ 100,  x/25 ≤ 1 x ≥0, y≥0 is obtained.
Answer : Converting inequations into equations, we get

Clearly, the feasible region is OABCO, which is shaded in the figure.
Here, B is the point of intersection of lines 2x + 5y = 100

38. If

2x + 3y + 7z = 12
3x – 2y – z = 0
x + y + 2z = 4

= 2(–4 + 1) – 3(6 + 1) + 7(3 + 2) = –6 – 21 + 35
= 8 ≠ 0. So A–1 exist.
The cofactors of elements of A are
C11 = –3, C21 = 1, C31 = 11
C12 = –7, C22 = –3, C32 = 23
C13 = 5, C23 = 1, C33 = –13

OR

If