Mathematics | Class 12th

CBSE Board Paper 2018

2

CBSE Board Paper 2018 Set - 1

General Instructions:

1. All questions are compulsory.

2. The question paper consists of 29 questions divided into four

sections A, B, C and D. Section A comprises of 4 questions of one

mark each, Section B comprises of 8 questions of two marks

each, Section C comprises of 11 questions of four marks each

and Section D comprises of 6 questions of six marks each.

3. All questions in Section A are to be answered in one word, one

sentence or as per the exact requirement of the question.

4. There is no overall choice. However, internal choice has been

provided in 3 questions of four marks each and 3 questions of

six marks each. You have to attempt only one of the alternatives

in all such questions.

5. Use of calculators is not permitted. You may ask for logarithmic

tables, if required.

Time allowed: 3 Hours Max Marks: 100

3

1. Write the value of tan–1(√3 ) – cot–1(–√3 ).

Section A (Very Short Answer Based Questions)

1

2. If the matrix A =

is skew symmetric, find the

values of ‘a’ and ‘b’.

1

3. Find the magnitude of each of the two

vectors and having the same magnitude such that the angle between them is 60° and their scalar product is 9/2.

1

4. If a * b denotes the larger of ‘a’ and ‘b’ and if a∘b = (a*b) + 3, then write the value of (5)∘(10), where * and ∘ are binary operations.

1

5. Prove that : 3 sin–1x = sin–1 (3x – 4x3), x ∈

.

2

6. Given A =

compute A–1 and show that 2 A–1 = 9I – A.

2

Section B (Short Answer Based Questions)

4

2

8. The total cost C(x) associated with the production of x units of an item is given by C(x) = 0.005x3 – 0.02x2 + 30x + 5000. Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output.

2

9. Evaluate

.

2

10. Find the differential equation representing the family of curves y = aebx+5 , where a and b are arbitrary constants .

11. If θ is the angle between two vectors and .

Find sin θ.

12. 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.

2

7. Differentiate tan–1

w.r.t x.

2

2

5

Section C (Long Answer Questions)

13. Using properties of determinants, prove

that = 9(3xyz + xy + yz + zx).

4

14. If , find . 4

15. If y = sin(sinx), prove that

4

OR

16. Find the equations of the tangent and the normal, to the curve 16x2 + 9y2 = 145 at the point (x1, y1) where x1 = 2 and y1 > 0.

4

OR

If and , find

when

.

Find the intervals in which the function f(x)

= is

(a) Strictly increasing,

(b) Strictly decreasing.

6

17. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when depth of the tank is half of its width. If the cost is to be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question?

4

18. Find

4

19. Find the particular solution of the differential

equation

, given that y = π/4.

4

OR

Find the particular solution of the differential

equation

, given that y = 0 when x = π/3.

20. Let

, and

Find a

vector which is perpendicular to

both and and

4

21. Find the shortest distance between the lines

and .

4

7

22. Suppose a girl throws a die. If she gets 1 or 2 she tosses a coin three times and notes the number of tails. If she gets 3,4,5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw 3,4,5 or 6 with the ride ?

4

23. Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X.

4

24. Let A = such that is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2].

6

Show that the function f:R R defined

by

is neither one – one nor onto. Also

if g: R → R is defined as g(x) = 2x – 1 find f∘g(x)

OR

Section D (Source Based Questions)

8

6

26. Using integration, find the area of the region in the first quadrant enclosed by the x – axis, the line y = x and the circle x2 + y2 = 32.

27. Evaluate

6

25. If matrix A =

, find A–1. Use it to solve the

system of equations.

2x – 3y + 5z = 11

3x + 2y – 4z = 0

x + y – 2z = – 3. OR

Using elementary row transformations, find the inverse

of the matrix A =

.

OR

Evaluate

, as the limit of the sum.

6

9

28. Find the distance of the point ( – 1, – 5, – 10) from the

point of intersection of the line

and the plane

6

29. A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand – operated. It takes 4 minutes on the automatic and 6 minutes on the hand operated machines to manufacture a packet of screws ‘B’. Each machine is availble for at most 4 hours on any day. The manufacturer can sell a packet of screws ‘A’ at a profit of 70 paise and screws ‘B’ at a profit of Rs. 1. Assuming that he can sell all the screws he manufactrures, how many packets of each type should the factory owner produce in a day in order to maximize his profit ? Formulate the above LPP and solve it graphically and find the maximum profit.

6

10

1. To Find: tan–1(√3 ) – cot–1(–√3 )

Formula Used: Range of principal value for tan-1 θ is , and range of principal value for cot-1 θ is (0, π).

Now,

and,

As the value for cot does not lie in range, we will use the following concept,

Therefore, principal value = π - θ

Therefore,

Thus,

Hence,

tan–1(√3 ) – cot–1(–√3 ) = -π/2

Solutions (Set-1)

Section A (Solutions)

11

2. If the transpose of a matrix is equal to the negative of itself, the matrix is said to be skew symmetric.

∵ A is a skew symmetric matrix

a12 = - a21

a = – 2

Also,

a31 = - a13

b = 3.

Hence, a = -2 and b = 3.

3. Given: and θ = 60° also

Formula used:

Putting the values, we get,

9

2= |�⃗�|2 ×

1

2

[Since, we know that, |�⃗�| = |�⃗⃗�| and cos 60° = ½ ]

|�⃗�|2 = 9

|�⃗⃗⃗�| = 𝟑, |�⃗⃗⃗�| = 𝟑

4. Given: a*b denotes the larger of ‘a’ and ‘b’ and a∘b = (a * b) + 3

To Find: (5)∘(10),

So,

5∘10 = (5 * 10) + 3

(5*10 = 10, as 10 is larger than 5)

12

5∘10 = 10 + 3

5∘10 = 13.

Hence,

(5)∘(10) = 13.

5. To Prove: 3 sin–1x = sin–1 (3x – 4x3)

Proof:

we know that,

sin3θ = 3 sinθ - 4 sin3θ.....(1)

Let sinθ = x, then

θ = sin-1 x

Put the value in equation 1, we get,

sin3θ = 3x - 4x3

3θ = sin-1(3x - 4x3)

3 sin–1x = sin–1 (3x – 4x3)

Hence, Proved.

Section B (Solutions)

13

6. A =

|A| = 2 × 7 - (-3) × (-4) = 14 – 12 = 2.

We will calculate the co-factors of the matrix A,

∴ A11 = 7, A12 = 4, A21 = 3, A22 = 2

adj(A) = =

Now,

To Prove: 2 A–1 = 9I – A

Proof:

and,

L.H.S = R.H.S

Hence, Proved.

7.

Using formula:

1 + cosx = 2 cos2(x/2) and

Now, we have,

14

[As we know that tanθ = cot (π/2 - θ)]

Now differentiating both sides we have,

8. C(x) = 0.005x3 – 0.02x2 + 30x + 5000

Marginal cost is given by,

Cm = 𝒅

𝒅𝒙(𝑪(𝒙))

Cm = 𝑑

𝑑𝑥(0.005𝑥3 − 0.02𝑥2 + 30𝑥 + 5000)

Cm = (0.005 × 3 × x2) – (0.02 × 2 × x) + 30

Now it is given that number of units = 3

Therefore, x = 3,

Cm = (0.005 × 3 × 9) – (0.02 × 2 × 3) + 30

Cm = 30.135 – 0.12

Cm = 30.015

9. Let I =

Applying the formula: cos 2x = 1 - 2sin2x, we get,

I =

I = ⇒ I =

we know that,

= tan x + c

Therefore,

I = tan x + C.

15

10. Given: Equation of curve: y = aebx+5

y = aebx . e5

y = αebx

where, ae5 = α

Differentiate with respect to ‘x’

𝑑𝑦

𝑑𝑥= 𝛼𝑏𝑒𝑏𝑥

∴𝑑𝑦

𝑑𝑥= 𝑏𝑦

1

𝑦

𝑑𝑦

𝑑𝑥= 𝑏

Again differentiating with respect to x,

11. Let �⃗� = 𝑖̂ − 2𝑗̂ + 3�̂� and �⃗⃗� = 3𝑖̂ − 2𝑗̂ + �̂�

Formula Used:

Therefore,

= 3 + 4 + 3

= 10

Now we know that,

Therefore,

16

Putting in formula of dot product, we get,

We also know the formula,

Therefore,

12. S = {(1,1), (1,2), .....(6,6)}

∴ n(S) = 36

Let the first die be black die and second die be red die,

A = Red die resulted in a number less than 4.

= { (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)}

Therefore, we can see that there are 18 possible outcomes when red die resulted in less than 4.

n(A) = 18

B = sum of number is 8

B = {(4,4), (6,2), (2,6), (5,3), (3,5)}

There are 5 possible outcomes when sum of number = 8

n(B) = 5

From A and B we can say that,

A ∩ B = {(5,3), (6,2)}

17

n(A ∩ B) = 2

∴ 𝑃 (𝐵

𝐴) = Probability of sum of numbers 8 when Red die

resulted in a number less than 4

𝑃(𝐴∩𝐵)

𝑃(𝐴)=

𝑛(𝐴∩𝐵)

𝑛(𝐴)=

2

18=

1

9.

13. LHS =

=

Taking 3 common from C1 and C3, we get,

=

Solving the determinant we get,

= 9 [ – 1(yz – 0) + x(y + 3zy + z)] = 9(yz + xy + 3xyz + xz)

9(3xyz + xy + yz + xz) = RHS

∵ LHS = RHS. Hence, Proved.

Section C (Solutions)

18

14. Given:

Expanding by the formula: (a + b)2 = a2 + b2 + 2ab

Differentiating with respect to x,

,

∴

,

Now

at π/3,

∴ .

OR

19

15. Given: y = sin(sinx) …… (i)

Differentiating both sides, we get,

…… (ii)

Differentiating again with respect to x on both sides, we get

…… (iii)

Putting (i) and (ii) in (iii), we get

Hence proved.

16. Given: P (x1, y1) = (2, y1) lies on

9y1

2 = (145 – 64)

Since its been given that y1 > 0

So, y1 = 3.

∴ P = (2, 3)

16x2 + 9y2 = 145 …… (i)

Differentiating both sides, we get,

20

Slope of tangent =

Slope of normal =

Equation of tangent is,

(y – 3) = (x – 2)

27y – 81 = – 32x + 64

32x + 27y - 145 = 0

Equation of normal is,

(y – 3) = (x – 2)

32y – 96 = 27x – 54

27x – 32y – 54 + 96 = 0

27x – 32y – 54 + 96 = 0

27x – 32y + 42 = 0

Given: f(x) =

Differentiating w.r.t x, we get,

f(x) is strictly increasing if f'(x)>0,

∴

f(x) is strictly decreasing if f'(x) <0

∴ .

OR

21

17. Let the length, width and height of the open tank be x, x and y units respectively. Then, its volume is , x2y and the total suface area is x2 + 4xy.

It is given that the tank can hold a given quantity of water. This means that its volume is constant. Let it be V.

Then,

V = x2y

The cost of the material will be least if the total surface area is least. Let S denote the total surface area.

Then,

S = x2 + 4xy

We have to minimize S object to the condition that the volume V is constant.

Now,

Differentiating w.r.t x, we get,

Differentiating again w.r.t x, we get,

The critical numbers of S are given by

Now,

22

x = 2y

Clearly

> 0, for all x

Hence, S is minimum when x = 2y, i.e. the depth (height) of the tank is half of its width.

Value – In the above mentioned question, the values that are being shown are of care, concern towards the weaker section of society.

18. Let I =

Let sin x = t

Differentiating both sides, we get,

cos x dx = dt

∴

We will solve this by integration by parts.

Therefore,

2 = (A – B)t2 + (B – C)t + (A + C)

Comparing the coefficients we get,

∴ A – B = 0, B – C = 0, A + C = 2.

23

A = 1, B = 1, C = 1.

+

19. Given:

Separating the variables, we get,

ln |ex - 2| = ln |tan y| + ln C

ln |ex - 2| = ln (C tan y )

ex - 2 = C tan y

Given x = 0, y =

e0 - 2 = C tan

1 – 2 = C × 1

C = – 1

ex - 2 = – tan y {since C = – 1}

ex - 2 + tan y = 0.

Let P = 2tanx and Q = sinx

So, this is a linear equation in one-variable of the form,

OR

24

We know that the solution of this equation is given by,

y(I.F) =

I.F = = = =

y.sec2x = ∫sinx × sec2x.dx

y.sec2x = ∫tanx.secx.dx

y.sec2x = secx + C

Given y = 0 when x =

∴ y.sec2x = secx – 2 {Since C = -2}

y.sec2x – secx + 2 = 0

20. Since is perpendicular to both and , therefore, its parallel

to

∴

=

= =

It’s been given that,

∴

.

25

21. Let,

and

∴ =

∴ = =

Also,

We know by the formula that,

Shortest distance =

= units.

22. Let A be the event that girl will get 1 or 2

P(A) = =

Let B be the event that girl will get 3, 4, 5 or 6

P(B) = =

P(T/A) = Probability of exactly one till given she will get 1 or 2

=

P(T/B) = Probability of exactly one till given she will get 3, 4, 5 or 6

= ½

26

Hence the probability that she threw 3,4,5 or 6 with the ride is

23. X can take values as 2,3,4,5 such that P(X = 2) = probability that the larger of two number is 2

∴ P(X = 2) =

Similarly,

∴ P(X = 3) =

∴ P(X = 4) =

∴ P(X = 5) =

E(X) = 2

E(X2) = 4

Therefore, Variance, V(X) = E(X) – E(X2)

= 17 – 16

= 1.

27

24. We have,

R = {(a, b): a – b is a multiple of 4}, where a,b ∈ A = {x ϵ Z: 0 ≤ x ≤ 12} = {0,1,2,...,12}.

We observe the following properties of relation R.

Reflexivity : For any a ϵ A, we have

|a – a| = 0, which is a multiple of 4.

(a,a) R

Thus, (a,a) ϵ R for all a ϵ A.

So, R is reflexive.

Symmetry: Let (a,b) ϵ R. Then,

(a,b) ϵ R

|a – b| is a multiple of 4.

|a – b| = 4 λ for some λ ϵ N

|b – a| = 4λ for some λ ϵ N

(b – a) ϵ R

So, R is symmetric.

Transitivity: Let (a,b) ϵ R and (b,c) ϵ R

|a – b| is a multiple of 4 and |b – c| is a multiple of 4

|a – b| = 4λ and |b – c| = 4μ for some λ, μ ∈ N

|a – b| = and b – c =

|a – c| =

|a – c| is a multiple of 4

|a – c|is a multiple of 4

(a,c) ϵ R

Thus, (a,b) ϵ R and (b,c) ϵ R

Section D (Solutions)

28

Hence (a,c) ϵ R

So, R is Transitive.

Hence R is an equivalence relation.

Let x be an element of A such that (x,1) ϵ R. Then,

| x – 1|is a multiple of 4

|x – 1| = 0,4,8,12

|x| = 1,5,9 {13 ∉ A }

Hence the set of all elements of A which are related to 1 is {1,5,9} i.e., [1] = [1,5,9].

f:R → R,

Check for one - one:

∴ f(x) is not one – one.

Also, 𝑦 =𝑥

𝑥2+1

Δ ≥ 0, if x is real

∴ B2 – 4AC ≥ 0

OR

29

1 – 4y2 ≥ 0

(1 – 2y)(1 + 2y) ≥ 0

(2y – 1)(2y + 1) ≤ 0

∴

Co-domain ϵ R

But range

∴ Function is not onto.

as f:R → R

g(x) = 2x – 1 as g: R → R

(fog)(x) = f(g(x)) =

= =

=

25. A =

∴ |A| = 2( – 4 + 4) + 3( – 6 + 4) + 5(3 – 2) = 0 – 6 + 5 = – 1 ≠ 0

Now,

…… (i)

30

Now the given system of equations can be written in the form of AX = B, where

A = , X = , B =

The solution of the system of equations is given by X = A-1B,

X =

[Using (i)]

Hence, x = 1, y = 2, z = 3.

A =

|A| = 1( – 25 + 28) – 2( – 10 + 14) + 3( – 8 + 100)

= 3 – 2(4) + 3(2)

= 9 – 8

= 1 ≠ 0

A-1 exists.

A.A-1 = I

OR

31

=

26. First let us find the intersection points,

Put y = x in x2 + y2 = 32

∴

2x2 = 32

x2 = 16

x = ±4

And as y = x, y = ±4

So, the intersection points are (-4, -4), (4, 4)

32

A =

A =

= 4π sq. Units.

27. Let I =

Here, we express the denominator in terms (sin x – cos x), which is the integration of numerator.

Clearly,

Therefore, becomes,

Let, sin x – cos x = t.

Then, d(sin x – cos x) = dt

(cos x + sin x)dx = dt.

33

Now the limits of integration also change,

at x = 0

t = sin0 – cos0 = – 1 and

at x =

∴

(let)

When ; a = 1, b = 3

By, the definition we know that,

OR

34

As ∴

28. Given line:

Here we have the cartesian equation of a line,

And, the cartesian equation of a plane,

x – y + z – 5 = 0.

35

Let Q(α, β, γ) be the point of intersection of line and plane which will satisfy both equation.

(let)

α = 3λ + 2, β = 4λ – 1, γ = 2λ + 2

also, α - β + γ – 5 = 0

3λ + 2 – 4λ + 1 + 2λ + 2 – 5 = 0

λ = 0

So, putting the value of λ, we get,

α = (3 × 0) + 2 ⇒ α = 2 ⇒ β = (4 × 0) – 1 ⇒ β = -1

And,

γ = (2 × 0) + 2 ⇒ γ = 2

Therefore, we have,

Q = (2, – 1, 2)

l(PQ) =

=

= 13 units.

29. Let the factory manufactures x screws of type A and y screws of type B on each day.

∴ x ≥ 0, y ≥ 0

Given that:

The constraints are

4x + 6y ≤ 240, 6x + 3y ≤ 240

Total profit: Z = 0.70x + y

Subject to,

36

2x + 3y ≤ 120, 2x + y ≤ 120

x ≥ 0, y ≥ 0.

Now, plotting the equations 2x + 3y ≤ 120 and 2x + y ≤ 120 we get,

∴ the common feasible region is OCBAO.

The maximum value of ‘Z’ is 41 at (30,20). Thus the factory

showed produce 30 packages at screw A and 20 packages of screw B to get the maximum profit of Rs.41.

37