Mathematics: CBSE Sample Question Paper (2020-21)- 3 | Mathematics (Maths) Class 12 - JEE PDF Download

Class - XII
Math
TIME: 3 Hrs.
M.M: 80

General Instructions:
1. This question paper contains two parts A and B. Each part is compulsory. Part A carries 24 marks and Part B carries 56 marks.
2. Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.
3. Both Part A and Part B have choices.

Part – A :
1. It consists of two sections- I and II.
2. Section I  comprises of 16 very short answer type questions.
3. Section II contains 2 case studies. Each case study comprises of 5 case-based MCQs. An examinee is to attempt any 4 out of 5 MCQs.

Part – B :
1. It consists of three sections-III, IV and V.
2. Section III comprises of 10 questions of 2 marks each.
3. Section IV comprises of 7 questions of 3 marks each.
4. Section V comprises of 3 questions of 5 marks each.
5. Internal choice is provided in 3 questions of Section –III, 2 questions of Section -IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

Question numbers 1 to 16 are very short answer type questions.
Q.1. What is the principal value branch of cosec^–1x ?
Ans. As we know that the principal value of cosec^–1x is
y = cosec^–1x

Q.2. If the set A contains 5 elements and the set B contains 6 elements, then find the number of one-one and onto mappings from A to B.
Ans. We know that, if A and B are two non-empty finite sets containing m and n elements, respectively, then the number of one-one and onto mapping from A to B is
n! if m = n
0, if m ≠ n
Given that, m = 5 and n = 6
∴ m ≠ n
Number of one-one and onto mapping = 0

Q.3. If f(x) 2x and g(x) = x²/2 + 1 and are continuous function then what can be a discontinuous function?
Ans. Since f(x) = 2x and g(x) = x²/2 + 1 are continuous functions, then by using the algebra of continuous functions , the functions f(x) + g(x), f(x) – g(x), f(x).g(x) are also continuous functions but g(x)/f(x) is discontinuous function at x = 0.

Q.4. For what value of x, y = x(x – 3)² decreases ?
Ans.
Given that,
y = x(x - 3)²
∴ dy/dx = x.2(x - 3).1 + (x - 3)².1
= 2x² - 6x + x² + 9 - 6x
= 3x² - 12x + 9
= 3(x² - 3x - x + 3)
= 3(x - 3) (x - 1)

So, y  = x(x – 3)² decreases for (1, 3).
[Since, y’ < 0 for all x ∈ (1, 3), hence y is decreasing on (1, 3)].

Q.5. If y = Ae^5x + Be^–5x, then find d²y/dx².
Ans. y = Ae^5x + Be^–5x

= 25y

Q.6. Find the area of a rectangle having vertices A, B, C and D with position vectors  respectively ?
Ans. The position vectors of vertices A, B, C and D of rectangle ABCD are given as :

The adjacent sides  and of the given rectangle are given as :

Now its, known that the area of parallelogram whose adjacent sides are
So that, the area of the given rectangle is = 2sq. units.
OR
Find a vector in the direction of that has magnitude 7 units.
Ans. 

Q.7. Find the degree of the differential equation

Ans. The degree of above differential equation is not defined because when we expand sin dy/dx we get an infinite series in the increasing powers of dy/dx. Therefore its degree is not defined.

Q.8. Find the interval on which the function f(x) = 2x³ + 9x² + 12x – 1 is decreasing.
Ans. 

Given that,
f(x) = 2x³ + 9x² + 12x - 1
f'(x) = 6x² + 18 x + 12
= 6(x² + 3x + 2)
= 6(x + 2) (x + 1)
So, f'(x) ≤ 0, for decreasing.

On drawing number lines as below :

We see that f’(x) is decreasing in (−2, −1).

Q.9. If A and B are two events such that P(A) ≠ 0 and P(B|A) = 1, then show that A ⊂ B.
Ans. P(A) ≠ 0 and P(B|A) = 1

P(A) = P(B ∩A)
∴ A ⊂ B

Q.10. If the curve ay + x² = 7 and x³ = y, cut orthogonally at (1, 1), then what is the value of a?
And. Given that, ay + x² = 7  and x³ = y On differentiating with respect to x in both equations, we get

Since, the curve cuts orthogonally at (1, 1).
∴ m₁m₂ = -1

∴ a = 6

Q.11. What is the number of points of discontinuity of f defined by f(x) = |x| – |x + 1| ?
Ans. f(x) = |x| – |x + 1|

Here, at x = 0, –1 f(x) is continuous.

Hence, there is no point of discontinuity.

OR
What is the principal value of cos^–1
Ans.


Q.12. The negative of a matrix is obtained by multiplying it by which number ?
Ans. Let A is a given matrix
∴ – A = – 1[A]
So, the negative of a matrix is obtained by multiplying it by – 1.

Q.13. What is the slope of the tangent to the curve y = x3 – x at the point (2, 6) ?
Ans. y = x³ – x
Differentiate w.r.t. x
dy/dx = 3x² - 1
Slope of the tangent to the curve
y = x³ – x at point (2, 6) is
= 3(2)² – 1 = 11
OR
Show that function y = 4x – 9 is increasing for all x ∈ R.
Ans. Given, y = 4x – 9
dy/dx = 4 > 0 for all x ∈ R.
Hence, function is increasing for all x ∈ R.

Q.14. What are the values of a for which the function f(x) = sin x – ax + b increases on R ?
Ans. The value of ‘a’ for which the function f(x) = sin x - ax+ b increases on R are (∞, −1).
∵ f'(x) = cos x - aa
and f'(x) > 0
⇒ cos x > a
Since, cos x ∈ [-1, 1]
⇒ a < -1
⇒ a ∈(∞, −1)

Q.15. If a line has direction ratios 2, – 1, – 2 then find its direction cosines?
Ans. Since,
∴ Required direction cosines are :

OR
Find the vector equation of the line through the points (3, 4, –7) and (1, –1, 6).
Ans. We know that, vector equation of a line that passes through two points a and b is represented by
Here, and


Q.16. For A =  write A^–1.
Ans.

OR
Matrix A = is given to be symmetric, find values of a and b.

Ans. As A is a symmetric matrix,

∴ By equality of matrices, a = −2/3 and b = 3/2.

Both the case study based questions are compulsory. Attempt any 4 sub parts from each question 17 and 18. Each question carries 1 mark.
Q.17. There is a bridge whose length of three sides of a trapezium other than base are equal to 10 cm.

Based on the above information answer the following:

(i) What is the value of DP ?
(a) 
(b) 
(c) 100 – x²
(d) x² – 100
Ans. a

(ii) What is the area of trapezium A(x) ?
(a)
(b)
(c) (x – 10)(100 – x²)
(d) (x + 10)(100 – x²)
Ans. b

(iii) If A'(x) = 0, then what are the values of x ?
(a) 5, –10
(b) –5, 10
(c) –5, –10
(d) 5, 10
Ans. a

(iv) What is the value of A"(5) ?
(a) 
(b)
(c) 
(d) 
Ans. c
(v) What is the value of maximum area ?
(a) 75√2 cm²
(b) 75√3 cm²
(c) 75√5 cm²
(d) 75√7 cm²
Ans. b

Q.18. A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank Rs 70 per square metre for the base and Rs 45 per square metre for the sides ?

Based on the above information answer the following questions:

(i) What is the cost of base ?
(a) 70 xy
(b) 80 xy
(c) 45 xy
(d) 90 xy
Ans. a

(ii) What is the cost of making all sides ?
(a) 190(x + y)
(b) 200(x + y)
(c) 180(x + y)
(d) 170(x + y)
Ans. c

(iii) If 'C' be the total cost of tank, then dC/dx is :
(a) 180(1 – 4x^–1)
(b) 180(1 – 4x^–2)
(c) 180(1 – 4x^–3)
(d) 180(1 – 4x^–4)
Ans. b

(iv) For what value of x, C is minimum ?
(a) 1
(b) 2
(c) 3
(d) 4
Ans. b

(v) What is the least cost of construction ?
(a) Rs 1000
(b) Rs 2000
(c) Rs 3000
(d) Rs 4000
Ans. a

Question numbers 19 to 28 carry 2 marks each.
Q.19. The two vectors represent the two sides AB and AC, respectively of DABC. Find the length of the median through A.
Ans. 


Now ABEC represent a parallelogram with AE as the diagonal.


Q.20. Find the shortest distance between the lines

Ans.



Q.21. Check if the relation R in the set R of real numbers defined as R = {(a, b): a < b} is (i) symmetric, (ii) transitive.
Ans. (i) It is not symmetric because if a < b then b < a is not true.
(ii) Here, if a < b and b < c then a < c is also true for all a, b, c ∈ Real numbers. Therefore R is transitive.

OR
Show that the function f in A = R –  defined as f(x) =  is one-one.
Ans. 

or (4x₁ + 3)(6x₂ – 4)= (6x₁ – 4)(4x₂ + 3)
or 24x₁x₂ – 16x₁ + 18x₂ – 12= 24x₁x₂ + 18x₁ – 16x₂ – 12
or – 16x₁ + 18x₂ = 18x₁ – 16x₂
or –16x₁ – 18x₁ = – 18x₂ – 16x₂
or –34x₁ = – 34x₂
or x₁ = x₂
or f is one-one.

Q.22. Find a matrix A such that 2A – 3B + 5C = 0, where B =
Ans.


Q.23. Find Ans.



Q.24. Evaluate sinx.cos² x.dx
Ans. Let sinx.cos² x.dx

Let f(x) = 1 - x²). sin x cos² x
as f(-x) = - f(x) ⇒ f is odd function.
∴ l = 0
Detailed Solution :

Let f(x) = (1 – x²) sin x cos2 x
Then f(–x) = [1 – (–x)²] sin (–x) [cos (–x)]²
= (1 – x²) (–sin x) cos² x
= –(1 – x²) sin x cos² x
= –f(x)
So, f(x) is an odd function,


Q.25. Let l_i, m_i, n_i i = 1, 2, 3 be the direction cosines of three mutually perpendicular vector in space. Show that AA’ = I₃, where A = .
Ans.

because


Q.26. If A and B are two independent events, prove that A’ and B are also independent.
Ans.
P(A’ ∩ B) = P(B) – P(A ∩ B)
= P(B) – P(A) . P(B)
(∵ A and B are independent events)
= (1 - P(A)) P(B)
= P(A’) P(B) 1
Since, P(A’ ∩ B) = P(A’) P(B)
Therefore A’ and B are independent events.

Q.27. Show that the relation R in the set N × N defined by (a, b) R (c, d) if a² + d² = b² + c² ∀ a, b, c, d ∈ N, is an equivalence relation.
Ans. Let (a, b) ∈ N × N
then,
∵ a² + b² = a² + b²
∴ (a, b) R (a, b)
Hence R is  reflexive.
Let (a, b), (c, d) ∈ N × N be such that
(a, b) R (c, d)
⇒ a² + d² = b² + c² 
⇒ c² + b² = d² + a²
⇒ (c, d) R (a, b)
Hence, R is  symmetric.
Let (a, b), (c, d), (e, f) ∈ N × N be such that (a, b) R (c, d), (c, d) R (e, f).

⇒ a² + d² = b² + c² ...(i)
and c² + f² = d² + e² ...(ii)
Adding eqn. (i) and (ii),
⇒ a² + d² + c² + f² = b² + c² + d² + e²
⇒ a² + f² = b² + e²
⇒ (a, b) R (e, f)
Hence, R is transitive.
Since, R is reflexive, symmetric and transitive. Therefore, R is an equivalence relation.

Q.28. Prove that x² – y² = C(x² + y²)2 is the general solution of the differential equation (x² – 3xy²) dx = (y³ – 3x²y) dy, where C is a parameter.
Ans. x² – y² = C(x² + y²)2
or 2x – 2yy' = 2C(x² + y²)(2x+2yy')

or (y² + x²)(x – yy')= (x² – y²)(2x + 2yy')
or  [–2y(x² – y²) – y(y² + x²)] dy/dx
= 2x(x² – y²) – x(y² + x²)
or (y³ – 3x2y) dy/dx  = (x³ – 3xy²)
or (y³ – 3x²y) dy = (x³ – 3xy²) dx
Hence x² –  y² = C (x² + y²)² is the solution of given differential equation.

OR
If y(x) is a solution of the differential equation = – cos x and y (0) = 1, then find the value of y(π/2)
Ans. 

Integrating, we get
log |1 + y| = – log |2 + sin x| + log C
or (1 + y)(2 + sin x) = C,
Putting y (0) = 1, we get C = 4
∴ (1 + y)(2 + sin x) = 4
Or

∴

Question Numbers 29 to 35 carry 3 marks each.
Q.29. Solve the differential equation:

Given that x = 1 when y = π/2
Ans. We have,

Above differential equation is a homogeneous equation
Put y = vx
Then,

From (i) and (ii)


Now, integrating both sides

Also, given that x = 1, when y = π/2

Therefore log |x| = is the required solution. 

OR
Solve the differential equation:

Ans.

Compare equation with

Integrating factor (I.F.)

Solution of equation is:


Substitute in eqn. (i), we get

∴

Q.30. If then find the value of λ so that are perpendicular vectors.
Ans.

OR
The two adjacent sides of a parallelogram are Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.
Ans.






Q.31. Evaluate:
Ans.

OR
Evaluate: 
Ans.




Q.32. A problem in mathematics is given to 4 students A, B, C, D. Their chances of solving the problem, respectively, are 1/3,1/4,1/5. What is the probability that
(i)  the problem will be solved?
(ii) a t most one of them solve the problem?
Ans. Let
E be the event =  A solves the problem
F be the event = B solves the problem
G be the event = C solves the problem
H be the event = D solves the problem

(i) the probability = P(E ∪ F ∪ G ∪ H)

= 13/15
(ii) the required probability



Q.33. Show that the relation S in the set R of real numbers defined as S = {(a, b) : a, b ∈ R and a ≤ b³} is neither reflexive nor symmetric nor transitive.
Ans. S = {(a, b) : a, b ∈ R and a ≤ b³}.
Reflexive As  where  1/2 ∈ R, is not true.
∴
Thus, S is not reflexive.
Symmetric As -2 ≤ (3)³, where -2, 3 ∈ R, is true but 3 ≤ (-2)³ is not true.
i.e. (-2, 3) ∈ S but (3, -2) ∉ S.
Therefore, S is not symmetric.
Transitive As  where  is not true.

Therefore, S is not transitive.
Hence, S is neither reflexive nor symmetric nor transitive.

Q.34. Solve the differential equation (tan^–1x – y)dx = (1 + x²) dy.
Ans. Given differential equation can be written as

⇒ ye^tan−1 x = e^tan−1 x . (tan^–1 x – 1) + c
or y = (tan^–1 x–1) + c.e^–tan–1x

Q.35. Prove that y =  is an increasing function of θ on
Ans. Getting
Equating dy/dθ  to 0 and getting critical point as cos
= 0 i.e., θ = π 2
θ, 0 ≤ θ π/2, dy/dθ ≥ 0,

Hence, y is an increasing function of θ on
Detailed Solution:



(4 – cos θ) is always greater than 0.
Since – 1 ≤ cos θ ≤ 1, (2 + cos θ)² > 0.

Question numbers 36 to 38 carry 5 marks each.
Q.36. Find the foot of perpendicular from P(1, 2, – 3) to the line . Also, find the image of P in the given line.
Ans. Any point on the given line is (2l – 1, – 2l + 3, – l)i if this point is Q then
Since  is perpendicular to the line
2(2λ – 2) – 2(– 2λ + 1) – 1(– λ + 3) = 0
or λ = 1

∴ Foot of perpendicular is Q(1, 1, – 1)
(Let P x, y, z) be the image of P in the line, then

or x = 1, y = 0, z = 1
or Image P is (1, 0, 1).
OR
Show that the lines intersect. Also find their point of intersection.
Ans. 
 
General points on the lines are (3u – 1, 5u – 3, 7u – 5) & (v + 2, 3v + 4, 5v + 6)
Lines intersect if
3u – 1 = v + 2,
5u – 3 = 3v + 4,
7u – 5 = 5v + 6 for some u & v
or 3u – v = 3 ...(i)
5u – 3v = 7....(ii)
7u – 5v = 11 ...(iii)
Solving equations (i) and (ii), we get
u= 1/2, v = - 3/2
Putting u & v in equation (iii),

∴ lines intersect. ½
Putting value of u and v in general points, point of intersection of lines is:


Q.37. Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, – 4, – 5) and B(2, – 3, 1) intersects the plane 2x + y + z = 7.
Ans. The equation of the line passing through A(3, –4, –5) and B(2, –3, 1) is given by

Then co-ordinates of any random point on the line AB is Q(– λ + 3, λ – 4, 6λ – 5)
Line AB interests the plane 2λ + y + z = 7
Then 2(– λ + 3) + (λ – 4) + (6λ – 5) = 7
⇒ –2λ + 6 + λ – 4 + 6λ – 5 = 7
⇒ 5λ – 3 = 7
⇒ 5λ = 10
⇒ λ = 2
Therefore, co-ordinates of the point of intersection of the given line and the plane are Q(1, – 2, 7)
Now the distance between P(3, 4, 4) and Q(1, –2, 7)


∴ PQ = 7 units

OR
A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A, B, C. Show that the locus of the centroid of triangle ABC is
Ans. Equation of plane cutting intercepts (say, a, b, c) on the axes is with A(a, 0, 0), B(0, b, 0) and C(0, 0, c) distance of this plane from origin is


Q.38. Evaluate the following:
Ans. The given definite integral
 
Hence, f is odd.
Therefore,

Hence, g is even. Thus

OR
Find:
Ans.

Dividing numerator and denominator by cos⁴x,