Class 12 Mathematics Sample Paper With Solutions Set C

Please refer to Class 12 Mathematics Sample Paper With Solutions Set C 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 C

PART – A
Section – I

1. If the function

Answer : Since, f (x) is continuous at x = 2.

OR

If y = log7 (log x), then find dy/dx.
Answer : 

2. If tan^–1(cotq) = 2q, then find the value of q.
Answer : We have, tan^–1(cotθ) = 2θ ⇒ cot θ = tan 2θ

3. Find the value of (î + ĵ)x(ĵ +k̂))x(k̂+î).
Answer : (î + ĵ)x(ĵ +k̂))x(k̂+î) = (î + ĵ + î x k̂+ ĵ x k̂) . (k̂+ î)
= ( k̂–ĵ – î )(k̂+ î) = k̂.k̂+ î . î                (∵ î . ĵ = ĵ.k̂= k̂.î)
= |k̂| + |î| = 1+1 = 2

OR 

If lines x  

Answer : 

4. If a line makes angles 90°, 135°, 45° with the X, Y, Z axes respectively, then find its direction cosines.
Answer : Here α = 90°, β = 135°, γ = 45°
Direction cosines are l = cos α = cos 90° = 0,

5. Evaluate : 

Answer : 

OR

Evaluate : 

Answer : 

6. For matrix 

Answer : 

7. Find the direction cosines of the side AC of a △ABC whose vertices are given by A (3, 5, 4), B (–2, –2, –2) and C (3, –5, 4).
Answer : The direction cosines of the line AC are

OR

Show that three points A(–2, 3, 5), B(1, 2, 3) and C(7, 0, –1) are collinear.
Answer : Direction ratios of the line AB = 3, –1, –2,
Direction ratios of the line BC = 6, –2, –4
Now, 3/6 = −1/−2 = −2/−4
Since the direction cosines of the line AB and BC are proportional and B is the common point. Hence, the points are collinear.

8. If A = {1, 5, 6}, B = {7, 9} and R = {(a, b) ∈A × B : |a – b| is even}. Then write the relation R.
Answer : We have, A × B = {(1, 7), (1, 9), (5, 7), (5, 9), (6, 7), (6, 9)}
∴ R = {(1, 7), (1, 9), (5, 7), (5, 9)}

9. Find the degree and order of the differential equation : 

Answer : Here, highest order derivative is d²y/dx² , so its order is 2 and power of d²y/dx² is one, so its degree is 1.

OR

Solve the differential equation (1+x²)dy/dx = e^x.
Answer : 

10. If A and B are the points (– 3, 4, – 8) and (5, – 6, 4) respectively, then find the ratio in which yz-plane divides the line joining the points A and B.
Answer : Let λ be the ratio in which yz-plane divides the line joining the points (–3, 4, –8) and (5, –6, 4). The co-ordinates of any point on the line joining the two

11. If A is a square matrix such that A² = A, then find (I + A)³ – 7A.
Answer : We have, A² = A …(i)
Now, (I + A)³ – 7A = I³ + A³ + 3A²I + 3AI² – 7A
= I + A²A + 3A²I + 3AI – 7A
= I + AA + 3A + 3A – 7A [Using (i)]
= I + A² – A = I + A – A [Using (i)]
= I

12. A line makes an angle of p/4 with each of X-axis and Y-axis. What angle does it make with Z-axis?
Answer : Let g be the required angle. Then
cos²α + cos²β + cos2γ = 1
cos²γ = 1–1/2–1/2 = 0 ⇒ cosγ = 0
⇒ γ = π/2

13. If  

Answer : 

14. Write the projection of b̅ +c̅ ona, where a̅ = 2î – ĵ+k̂.b̅ = î – 2ĵ +2k̂ and c̅ = 2î – ĵ +4k̂.
Answer : 

15. Let n(A) = 4 and n(B) = 6, then find the number of one-one functions from A to B.
Answer : Number of one-one functions from A to B
= ⁶P₄ = 6 · 5 · 4 · 3 = 360

16. A line makes 45° with OX, and equal angles with OY and OZ. Find the sum of these three angles.
Answer : Here α = 45° and β= γ

Section – II

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

17. A card is lost from a pack of 52 cards. From the remaining cards of pack two cards are drawn and are found to be both spades.

Based on the above information, answer the following questions :
(i) The probability of drawing two spades, given that a card of spade is missing, is
(a) 21/425
(b) 22/425
(c) 23/425
(d) 1/425
Answer : B

(ii) The probability of drawing two spades, given that a card of club is missing, is
(a) 26/425
(b) 22/425
(c) 19/425
(d) 23/425
Answer : A

(iii) Let A be the event of drawing two spades from remaining 51 cards and E1, E2, E3 and E4 be the events that lost card is of spade, club, diamond and heart respectively, then the value of 

is
(a) 0.17
(b) 0.24
(c) 0.25
(d) 0.18
Answer : B

(iv) All of a sudden, missing card is found and, then two cards are drawn simultaneously without replacement.
Probability that both drawn cards are aces is
(a) 1/52
(b) 1/221
(c) 1/121
(d) 2/221
Answer : B

(v) If two card are drawn from a well shuffled pack of 52 cards, with replacement, then probability of getting not a king in 1st and 2nd draw is
(a) 144/169
(b) 12/169
(c) 64/169
(d) none of these
Answer : A

18. Arun got a rectangular parallelopiped shaped box and spherical ball inside it as his birthday present. Sides of the box are x, 2x, and x/3, while radius of the ball is r cm.
Based on the above information, answer the following questions :

(i) If S represents the sum of volume of parallelopiped and sphere, then S can be written as 

Answer : C

(ii) If sum of the surface areas of box and ball are given to be constant, then x is equal to 

Answer : A

(iii) The radius of the ball, when S is minimum, is 

Answer : B

(iv) Relation between length of the box and radius of the ball can be represented as 
(a) x = 2r
(b) x = r/2
(c) x = r/2
(d) x = 3r
Answer : D

(v) Minimum volume of the ball and box together is  

Answer : C

PART – B
Section – III

19. Find the intervals on which the function f(x) = 2x³ + 9x² + 12x + 20 is increasing.
Answer : Given, f(x) = 2x³ + 9x² + 12x + 20
⇒ f ′(x) = 6x² + 18x + 12
= 6(x² + 3x + 2) = 6(x + 1)(x + 2)
For f (x) to be increasing, f ′(x) > 0
⇒ 6(x + 1)(x + 2) > 0
⇒ (x + 1) (x + 2) > 0
⇒ x + 1 > 0, x + 2 > 0 or x + 1 < 0, x + 2 < 0
⇒ x > –1 or x < –2
⇒ x ∈( –1, ∞) or x ∈ (– ∞, – 2)
∴ f is increasing in (–∞, –2) ∪ (–1, ∞).

20. A vector r̅ is inclined at equal angles to OX, OY and OZ. If the magnitude of r̅ is 6 units, then find r̅ .
Answer : Suppose r̅ makes an angle a with each of the axes OX, OY and OZ. Then, its direction cosines are
l = cos a, m = cos a, n = cos a ⇒ l = m = n

OR

Find the value of λ such that the vectors  a̅ = 2î +λĵ+k̂ and b̅ = î – 2ĵ +3k̂ are perpendicular to each other.
Answer : If the vectors a̅ and b̅ are perpendicular to each other, then a̅.b̅ = 0 .
(2î + λĵ + k̂) . (î − 2ĵ + 3k̂) = 0 
⇒ (2) (1) + λ(–2) + (1) (3) = 0
⇒ –2l + 5 = 0 ⇒ λ = 5/2

21. If A and B are two independent events, such that P(A) = 1/2 and P(B) = 1/5,  then find the value of P(A|A ∪ B).
Answer : 

22. If x ∈ [0, 1], then find the value of 1/2 cos^-1(1–x/1+x).
Answer : 

23. Evaluate 

Answer : 

OR

Evaluate :

Answer : 

24. Solve the differential equation : 

Answer : 

25. A and B are two events such that P(A) ≠ 0. Find P(B/A) if
(i) A is a subset of B (ii) A ∩ B = φ
Answer : (i) Since, A is a subset of B.   ∴ A ⊂ B
⇒ A ∩ B = A

26. Find the derivative of [√1-x² sin-1 x–] w.r.t.x
Answer : 

27. Find the area bounded by the curve x² + y² = 1 in the first quadrant.
Answer : We have, x² + y² = 1, a circle with centre (0, 0) and radius = 1.

28. Compute the adjoint of the matrix 

Answer : 

OR

If the matrix 

Answer : 

Section – IV

29. Let A = R – {2} and B = R – {1}. If f : A → B is a mapping defined by f(x) = x-1/x-2 then show that f is bijective.
Answer : 

30. Consider

If f (x) is continuous at x = 0, then find the value of k.

Answer : f(0) = k (Given)                  …(i)
Since, f(x) is continuous at x = 0.

31. Find the values of x for which f (x) = (x (x – 2))² is an increasing function. Also, find the points on the curve, where the tangent is parallel to x-axis.
Answer : Given, f (x) = (x(x – 2))² = x²(x – 2)², Df = R.
Differentiating w.r.t. x, we get
f ′(x) = x² · 2(x – 2) + (x – 2)² · 2x
= 2x(x – 2)(x + x – 2) = 2x(x – 2)(2x – 2) = 4x(x – 1)(x – 2)
Now, the given function f is (strictly) increasing iff
f ′ (x) > 0
⇒ x ∈ (0, 1) ∪ (2, ∞)

Further, the tangents will be parallel to x-axis iff
f ′(x) = 0
⇒ x = 0, 1, 2
The given curve is y = x²(x – 2)²
When x = 0, y = 0;
When x = 1, y = 1²(1 – 2)² = 1 × (–1)² = 1 × 1 = 1;
When x = 2, y = 2² (2 – 2)² = 4 × 0 = 0.
∴ The points on the given curve, where the tangents
are parallel to x-axis are (0, 0), (1, 1) and (2, 0).

OR

An open box with a square base is to be made out of a given quantity of cardboard of area c2 square units.
Show that the maximum volume of the box is c³/6√3 cubic units.
Answer : Let h be height and x be the side of the square base of the open box.
Then its area = x × x + 4 h × x = c²

32. Evaluate : 

Answer : 

33. If 

Answer : 

Again differentiating (ii) w.r.t. x, we get

34. Solve the differential equation  

Answer : 

OR

Find the solution of the equation  

Answer : 

Integrating both sides, we get

35. Find the area enclosed between the curve y = loge (x + e) and the coordinates axes.
Answer : The bounded area is as shown in figure.

Section – V

36. Find the image of the point having position vector î + 3ĵ+4k̂ in the plane r̅.(2î–ĵ+k̂)+3 = 0.
Answer : Let Q be the image of the point P(î + 3ĵ+4k̂) in the plane r̅.(2î–ĵ+k̂)+3 = 0
Then, PQ is normal to the plane. Since PQ passes through P and is normal to the given plane, therefore equation of line PQ is

 r̅ = (î + 3ĵ+4k̂) + λ(2î–ĵ+k̂)+3 = 0
Since Q lies on line PQ, so let the position vector of Q be
 = (î + 3ĵ+4k̂) + λ(2î–ĵ+k̂)+3 = 0
(1+2λ)î + 3(3–λ)ĵ +(4+λ)k̂
Since, R is the mid-point of PQ. Therefore, position vector of R is

OR

Find the points on the line x+2/1 = y+1/2 = z–3/2 at a distance of 2 units from the point (–2, –1, 3).
Answer : 

37. Solve the following linear programming problem (LPP) graphically.
Maximize Z = 4x + 6y
Subject to constraints:
x + 2y ≤ 80, 3x + y ≤ 75 ; x, y ≥ 0
Answer : We have maximize Z = 4x + 6y.
Subject to constraints :
x + 2y ≤ 80, 3x + y ≤ 75 and x ≥ 0, y ≥ 0
Now we draw the graphs of the lines
l1 : x + 2y = 80, l2 : 3x + y = 75 and x = 0, y = 0.

We obtain shaded region as the feasible region.
The lines l₁ and l₂ intersect at Q(14, 33).
Thus, the vertices of the feasible region are P(0, 40),
Q(14, 33), R(25, 0) and O(0, 0).

Corner Points	Value of Z = 4x + 6y
P(0, 40)	240
Q(14, 33)	254 (Maximum)
R(25, 0)	100
O(0, 0)	0

Thus, Z has maximum value 254 at Q(14, 33).

OR
Solve the following linear programming problem (LPP) graphically.
Minimize Z = 30x + 20y
Subject to constraints : x + y ≤ 8, x + 4y ≥ 12, 5x + 8y ≥ 20 ; x, y ≥ 0
Answer : We have minimize Z = 30x + 20y.
Subject to constraints :
x + y ≤ 8, x + 4y ≥ 12, 5x + 8y ≥ 20, x, y ≥ 0
Now, we draw the graphs of
l1 : x + y = 8, l2 : x + 4y = 12, l3 : 5x + 8y = 20 and
x = 0, y = 0

38. If  

then calculate AC, BC and (A + B)C. Also verify that (A + B)C = AC + BC.

Answer : 

OR

Find the matrix A satisfying the matrix equation

Answer :