Please refer to Class 12 Mathematics Sample Paper Set G 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
All questions are compulsory. In case of internal choices attempt any one.
Q1. Check whether the function f : N → N : f(x) = x2 + x + 1 is one-one or not.
Answer : f(x1) = f(x2)
⇒ x12 + x1 + 1 = x22 + x2 + 1
⇒ (x12 – x22) + (x1 – x2) = 0
⇒ (x1 – x2) (x1 + x2 + 1) = 0
⇒ x1 – x2 = 0
⇒ x1 = x2
∴ f is one-one.
OR
A relation R in a set A is called _____________, if (a1, a2) ∈ R implies (a2, a1) ∈ R, for all a1, a2 ∈ A.
Answer : Symmetric.
Q2. Let R = {(a, a3) : a is a prime number less than 5}. Find the range of R.
Answer : R = {(2, 8) (3, 27)} \ Range = {8, 27}
Q3. Write the principal value of [tan–1 (– 3 ) + tan–1 (1)]
Answer : tan–1 (– 3 ) + tan–1 (1) = –tan–1 √3 + π/4 = π/3 + π/4 = –π/12
OR
Find the principal value of tan–1 (tan 3π/5)
Answer : − 2π/5
Q4. If A is a matrix of order 3 × 2, then find the order of the matrix A′.
Answer : 2 × 3
Q5. If A is a square matrix of order 3 and |A| = 5, then find the value of |2A′|.
Answer : 40
OR
If A is a square matrix such that A2 = A, then find (I – A)3 + A.
Answer : I
Q6. A square matrix A is said to be skew-symmetric, if ___________.
Answer : A = –A′ (or, A′ = –A)
Q7. Find ∫x4 log x dx .
Answer :
OR
Find
Answer :
Q8. Evaluate
Answer :
Q9. How many arbitrary constants in the particular solution of a differential equation of second order?
Answer : 0
OR
Determine the order of the differential equation y′ + 5y = 0.
Answer : The differential equation y′ + 5y = 0 can be written as dy/dx + 5y = 0.
As the order of the highest order derivative is 1, so the order of the given differential equation is 1.
Q10. If the projection of a̅ = 𝑖̂ − 2𝑗̂ + 3𝑘̂ on b = 2𝑖̂ + λ𝑘̂ , is zero, then find the value of l. 1
Answer : −2/3
Q11. Find the magnitude of the vector a̅ = 2𝑖̂ − 6𝑗̂ − 3𝑘̂ . 1
Answer : We have a̅ = 2 − 6ˆj − 3𝑘̂
Q12. If | a̅ | = √3, | b̅| = 2
a̅.b̅ = 3 find the angle between a̅ and b̅.
Answer : Let q be the angle between a̅ and b̅ , then
Hence, angle between a̅ and b̅ is 30° or p/6.
Q13. Find the d.c’s of a line whose direction ratios are 2, 3, – 6
Answer : a = 2, b = 3, c = –6
Q14. Find the vector equation of the line passing through the point (–1, 5, 4) and perpendicular to the plane z = 0.
Answer : r̅ = î + 5ĵ + (4 + λ) k̂.
Q15. If A and B are two independent events with P(A) = 1/3 and P(B) = 1/4 , then find P(B′ | A).
Answer : 3/4.
Q16. Let E and F be events with P(E) = 3/5 , P(F) = 3/10 and P(E ∩ F) = 1/5 . Are E and F independent?
Answer : P(E) . P(F) = 3/5 3/10 9/50 and P(E ∩ F) = 1/5 , since P(E ∩ F) π P(E) . P(F)
∴ Events E and F are not independent.
Section-II
Both the Case study-based questions are compulsory. Attempt any 4 sub-parts from each question (17–18). Each sub-part carries 1 mark.
Q17. A window is in the form of semi-circle with a rectangle on its diameter. The total perimeter of the window is 10 m.
Based on the above information answer the following questions:
(i) From the figure, the perimeter (P) of the window is 1
(a) 2x + 4y + px
(b) x + 4y + px
(c) 2x + y + px
(d) 2x + 4y + x
Answer : A
(ii) Area(A) of whole window is given by
Answer : C
(iii) Area(A) of window in terms of x is given by
Answer : D
(iv) For maximum / minimum value of area, values of x and y respectively should be
Answer : A
(v) Dimensions (length and breadth) of window is
Answer : B
Q18. In a hostel, 60% of the student read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers.
Based on the above information answer the following:
(i) A student is selected at random, find the probability that a student reads both Hindi and English newspaper.
(a) 1/5
(b) 2/5
(c) 3/5
(d) 4/5
Answer : A
(ii) A student is selected at random, find the probability that a student reads Hindi or English newspaper.
(a) 1/5
(b) 4/5
(c) 3/5
(d) 2/5
Answer : B
(iii) A student is selected at random. Find the probability that student neither reads Hindi or English newspaper.
(a) 2/5
(b) 3/5
(c) 1/5
(d) 4/5
Answer : C
(iv) If student reads Hindi newspaper, find the probability that she reads English newspaper.
(a) 2/5
(b) 3/5
(c) 4/5
(d) 1/3
Answer : D
(v) If student reads English newspaper, find the probability that she reads Hindi newspaper.
(a) 1/2
(b) 2/5
(c) 1/3
(d) 3/5
Answer : A
Part–B
Section–III
All questions are compulsory. In case of internal choices attempt any one.
Q19. Prove that sin–1
Answer : Put x = cos q ⇔ q = cos–1 x
L.H.S. = sin–1 (2x 1− x2 )
= sin–1 (2 cos q sin q) = sin–1 (sin 2q) = 2q = 2 cos–1 x = R.H.S.
Q20. What positive value of x makes following pair of determinant equal?
Answer :
On expanding, we get
2x2 – 15 = 32 – 15
⇒ 2x2 – 15 = 17
⇒ 2x2 = 32
⇒ x2 = 32 ÷ 2 = 16
∴ x = ± 16
⇒ x = ±4
The positive value of x = 4
Hence, for x = 4, the given pair of determinant is equal.
OR
Find the value of
Answer :
Q21. Examine the continuity of the function f (x) = 2x2 – 1 at x = 3.
Answer :
Q22. Find the slope of the tangent to the curve y = 3x4 – 4x at x = 4. 2
Answer : Equation of the curve is y = 3x4 – 4x
Differentiating y w.r.t. x, we get
Q23. Evaluate:
Answer :
OR
Evaluate:
Answer :
Q24. Find the area of the region between the curves y2 = 4x and x = 3.
Answer : The equation of parabola is y2 = 4x. It contains only even powers of y.
∴ The parabola is symmetrical about x-axis.
In the first quadrant, y > 0
y = 2√x
Required area = Area OBB′ = 2 × Area OAB (shown shaded)
= 2 × Area under y
= 2√x between x = 0 and x = 3
Q25. Solve dy/dx = 2x3 – x , given y = 1, when x = 0
Answer : We have dy/dx = 2x3 – x
Integrating both sides, we get
Q26. Find a unit vector perpendicular to each of the vectors a̅ and b̅
where a̅ = 5iˆ + 6 ˆj − 2kˆ and b̅ = 7iˆ + 6 ˆj + 2kˆ
Answer :
Q27. Find the volume of the parallelopiped whose adjacent edges are represented by
Answer : Volume of the parallelopiped
Q28. Find the probability distribution of the number of heads in two tosses of a coin.
Answer : Let X be a random variable which is the number of heads obtained in two independent tosses of a coin S = {HH, HT, TH, TT}
Then, X(HH) = 2, X(HT) = 1, X(TH) = 1, X(TT) = 0
∴ X can take values 0, 1, 2.
Now, P(X = 0) = P(No head) = 1/4
P(X = 1) = P(Exactly 1 head) =2/4 = 1/2
P(X = 2) = P(Exactly 2 heads) = 1/4
Hence, the probability distribution of the R.V is given by
OR
The probability of finding a green signal on a busy crossing X is 30%. What is the probability of finding a green signal on X on two consecutive days out of three?
Answer :
Section-IV
All questions are compulsory. In case of internal choices attempt any one.
Q29. Let N be the set of natural numbers and R be the relation on N × N defined by (a, b) R (c, d) iff ad = bc for all a, b, c, d ∈ N. Show that R is an equivalence relation. 3
Answer : (i) Reflexive: For any (a, b) ∈ N × N
a ⋅ b = b ⋅ a \ (a, b) R (a, b) thus R is reflexive
(ii) Symmetric: For (a, b), (c, d) ∈ N × N
(a, b) R (c, d) ⇒ a ⋅ d = b ⋅ c ⇒ c ⋅ b = d ⋅ a
(c, d) R (a, b) \ R is symmetric
(iii) Transitive: For any (a, b), (c, d), (e, f ) ∈ N × N
(a, b) R (c, d) and (c, d) R (e, f )
⇒ a ⋅ d = b ⋅ c and c ⋅ f = d ⋅ e
⇒ a ⋅ d ⋅ c ⋅ f = b ⋅ c ⋅ d ⋅ e ⇒ a ⋅ f = b ⋅ e
∴ (a, b) R (e, f ), \ R is transitive.
∴ R is an equivalance relation.
Q30. If y = ex2 cos x + (cos x)x, then find dy/dx .
Answer : Let u = (cos x)x ⇒ y = ex2 cos x + u
∴ dy/dx = ex2 cos x (2x ⋅ cos x – x2 ⋅ sin x) + du/dx
log u = log (cos x)x ⇒ log u = x log(cos x)
Differentiating w.r.t. ‘x’
Q31. Prove that f (x) = x|x| is differentiable for all real values of x.
Answer :
OR
If x = a (q – sin q), y = a (1 + cos q), find d2y/dx2 at θ = π/2
Answer :
Q32. Find the intervals on which the function f (x) = sin4 x + cos4 x, 0 < x < 2p is strictly increasing or decreasing. 3
Answer : f (x) = sin4 x + cos4 x, 0 < x < 2p
∴ f ′(x) = 4 sin3 x ⋅ cos x + 4 cos3 x(– sin x)
= –4 sin x cos x (cos2 x – sin2 x)
⇒ f ′(x) = –2 sin 2x cos 2x = –sin 4x
f ′(x) = 0 ⇒ – sin 4x = 0
Q33. Find the general solution of the differential equation yeydx = (y3 + 2xe y)dy. 3
Answer :
Q34. Find the area included between the two curves y2 = 9x and x2 = 9y.
Answer : As shown in the figure, we have to find the area OAPBO.
Solving the given two equations simultaneously, we have
x4 = 81y2 = 81(9x) = 729 x
⇒ x4 – 729x = 0 ⇒ x(x3 – 729) = 0
⇒ x = 0, x3 = 729 = (9)3 ⇒ x = 9
∴ x = 0 at O and x = 9 at P
Now, Area OAPBO = Area OAPMO – Area OBPMO
OR
Using integration, find the area of the region bounded by the triangle whose vertices are (2, –2), (4, 5) and (6, 2).
Answer : Let A(2, –2) ; B(4, 5) ; C(6, 2)
Q35. Find the particular solution of the differential equation:
Answer : The differential equation can be written as:
Section-V
All questions are compulsory. In case of internal choices attempt any one.
Q36. 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 :
OR
Using properties of determinants, prove that
Answer :
Expand along C1
= (a2 + b2 + c2)(b – a)(c – a)(–b2 – ab + c2 + ac)
= (a – b)(b – c)(c – a)(a + b + c)(a2 + b2 + c2)
Q37. Find the equation of the plane that contains the point A(2, 1, –1) and is perpendicular to the line of intersection of the planes 2x + y – z = 3 and x + 2y + z = 2. Also find the angle between the plane thus obtained and the y-axis.
Answer : Let equation of the required plane be:
a(x – 2) + b(y – 1) + c(z + 1) = 0
Also : 2a + b – c = 0
a + 2b + c = 0
OR
Find the distance of the point P(–2, –4, 7) from the point of intersection Q of the line
r̅ = (3iˆ − 2 ˆj + 6kˆ) + λ(2iˆ − ˆj + 2kˆ) and the plane r̅. (iˆ − ˆj + kˆ) = 6 . Also write the vector equation of the line PQ.
Answer :
Q38. Solve the following linear programming problem (L.P.P.) graphically.
Maximize Z = 5x + 3y such that
3x + 5y ≤ 15
5x + 2y ≤ 10
x, y ≥ 0
Answer : Equations corresponding to the given constraints are
3x + 5y = 15
5x + 2y = 10
x = 0, y = 0
Drawing the bounding lines corresponding to the given inequalities and considering their common solution space, we find the feasible region is given by the shaded area OABC.
Now, solving the equations 3x + 5y = 15 and 5x + 2y = 10 simultaneously, we get the co‑ordinates of B as (20/19 , 15/19)
The vertices of the convex polygon OABC are O(0, 0), A(2, 0), B(20/19 , 15/19) and C(0, 3).
OR
The corner points of the feasible region determined by the system of linear constraints are as shown below:
Answer each of the following:
(i) Let Z = 5x + 7y be the objective function. Find the maximum and minimum value of Z and also the corresponding points at which the maximum and minimum value occurs.
(ii) Let Z = px + qy, where p, q > 0 be the objective function. Find the condition on p and q so that the maximum value of Z occurs at A(7, 0) and B(3, 4). Also mention the number of optimal solutions in this case.
Answer : (i) Feasible region is bounded.
Maximum of Z must occur at the corner point of the feasible region.
We have Z = 5x + 7y
Corner points of the feasible region are
(0, 0), (0, 2), (3, 4) and (7, 0)
At (0, 0) Z = 5(0) + 7(0) = 0
At (0, 2) Z = 5(0) + 7(2) = 14
At (3, 4) Z = 5(3) + 7(4) = 43
At (7, 0) Z = 5(7) + 7(0) = 35
We see that maximum value of Z is 43 and minimum value of Z is 0.
(ii) Since maximum value of Z occurs at A(7, 0) and B(3, 4)
∴ 7p + 0 = 3p + 4q
⇒ 7p = 3p + 4q
⇒ 7p – 3p = 4q
⇒ 4p = 4q
⇒ p = q
Number of optimal solutions are infinite.
