Sample Paper (CBSE) - smartlearning.com€¦ · Sample Paper (CBSE) Series SC/SP Code No. SP-16 CBSE-12- Mathematics ©Educomp Solutions Ltd. 2015-16 Mathematics Time Allowed: 3 hours

Sample Paper (CBSE) Series SC/SP

Code No. SP-16

CBSE-12- Mathematics ©Educomp Solutions Ltd. 2015-16

Mathematics

Time Allowed: 3 hours Maximum : 100

General Instructions:

(i) There are 26 questions in all.

(ii) All questions are compulsory.

(iii) Marks for each question are indicated against it.

(iv) Question numbers 01 to 06 are very short answer questions carrying 01 mark

each.

(v) Question numbers 07 to 19 are short answer questions carrying 04 marks each.

Out of which one question is a value based question.

(vi) Question numbers 20 to 26 are long answer questions.

1 Evaluate: sin(2푐표푠 (- )).

1

2 State the reason for the following Binary Operation *, defined on the set Z of integers, to be not commutative. 푎*푏 =푎푏

1

3 Give an example of a skew symmetric matrix of order 3.

1

4 Using derivative, find the approximate percentage increase in the area of a circle if its radius is increased by 2% .

1

Section A (1 mark each)

CBSE-12-Mathematics ©Educomp Solutions Ltd. 2015-16

5 Find the derivative of 푓 (푒 ) w.r. to 푥 at 푥 = 0. It is given that 푓′(1) = 5.

1

6 If the line = = and = = are perpendicular to

each other, then find the value of 푝.

1

7 Let 푓:푊 → 푊 be defined as 푓(푛) = , ,

. Then show that f is

invertible. Also, find the inverse of 푓. OR

Show that the relation 푅 in the set defined by 푁 × 푁 (푎,푏) 푅 (푐, 푑) iff 푎 + 푑 = 푏 + 푐 ∀ 푎,푏, 푐, 푑 ∈ 푁 is an equivalence relation.

4

8 Prove that: 푡푎푛 √ √√ √

= − , where 휋 < 푥 < .

4

9 Let 퐴 = −2 13 4 . Then verify the following: 퐴(푎푑푗퐴)(푎푑푗퐴)퐴 = |퐴| 퐼,

where 퐼 is the identity matrix of order 2.

4

10 Using properties of determinants, prove that

1 1 + 푝 1 + 푝 + 푞3 4 + 3푝 2 + 4푝 + 3푞4 7 + 4푝 2 + 7푝 + 4푞

=

1 . OR

Without expanding the determinant at any stage, prove that 0 2 −3−2 0 43 −4 0

= 0 .

4

11 Let 퐴 = 2 31 2 , 퐵 = 4 −6

−2 4 . Then compute 퐴퐵. Hence, solve the 4

Section B (3 marks each)


following system of equations: 2푥 + 푦 = 4, 3푥 + 2푦 = 1 .

12 If the following function is differentiable at 푥 = 2, then find the values of 푎 and 푏.

푓(푥) = , ,

.

4

13 Let 푦 = log푥 + 푥 . Then find .

OR

If 푥 = 푎 푠푖푛 푝푡, 푦 = 푏 푐표푠 푝푡. Then find at 푡 = 0

4

14 Evaluate the following indefinite integral: ∫ 푑푥.

OR

Evaluate the following indefinite integral: ∫ √

푑 .

4

15 Evaluate the following definite integral: ∫( )

푑푥 . 4

16 Solve the following differential equation: = + , 푥 < 0 .

4

17 Solve the following differential equation: (1 + 푦 )푑푥 = (푡푎푛 푦 − 푥)푑푦 .

4

18 Find the shortest distance between the following pair of skew lines:

= = , = =

4

19 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 and 2/3. What is the probability that (i) the problem will be solved?

4


(ii) at most one of them will solve the problem?

20 Find the intervals in which the following function is strictly increasing or strictly decreasing. Also, find the points of local maximum and local minimum, if any.

5

21 If 푎, 푏, 푐 are unit vectors such that 푎.푏 = 푎. 푐 = 0 and the angle between

푏 and 푐 is

, then prove that

(i) 푎 = ±2 푏 × 푐

(ii) 푎 + 푏, 푏 + 푐, 푐 + 푎 = ±1

5

22 Using integration, find the area bounded by the tangent to the curve 4푦 = 푥 at the point (2, 1) and the lines whose equations are 푥 = 2푦 and 푥 = 3푦 − 3.

5

23 Find the distance of the point 3 횤 – 2 횥 + 푘 from the plane 3 푥 + 푦 – 푧 +

2 = 0 measured parallel to the line = = . Also, find the

foot of the perpendicular from the given point upon the given plane. OR

Find the equation of the plane passing through the line of intersection of the planes

푟. 2 횤 + 3 횥 − 푘 = −1 and 푟. 횤 + 횥 − 2푘 = 0 and passing through

the point (3,−2,−1). Also, find the angle between the two given planes.

5

24 A Bag I contains 5 red and 4 white balls and a Bag II contains 3 red and 3 white balls. Two balls are transferred from the Bag I to the Bag II and then one ball is drawn from the Bag II. If the ball drawn from

5

Section C (5 marks each)


the Bag II is red, then find the probability that one red ball and one white ball are transferred from the Bag I to the Bag II.

OR Find the mean, the variance and the standard deviation of the number of doublets in three throws of a pair of dice.

25 A farmer wants to construct a circular garden and a square garden in his field. He wants to keep the sum of their perimeters 600 m. Prove that the sum their areas is the least, when the side of the square garden is double the radius of the circular garden. Do you think that a good planning can save energy, time and money?

5

26 A manufacturing company makes two models A and B of a product. Each piece of model A requires 9 hours of labour for fabricating and 1 hour for finishing. Each piece of model B requires 12 hours of labour for fabricating and 3 hours for finishing. The maximum number of labour hours, available for fabricating and for finishing, are 180 and 30 respectively. The company makes a profit of Rs 8000 and Rs 12000 on each piece of model A and model B respectively. How many pieces of each model should be manufactured to get maximum profit? Also, find the maximum profit.

8


1 −2425

2 We have 1, 2 ∈ 푍 such that 1*2 = 8 and 2*1 = 2. This implies that 1*2 ≠ 2*1.

Hence, * is not commutative.

3 0 1 −3−1 0 −23 2 0

4 4% 5 5 6 −14

7 Here, 푓표푓 ∶ 푊 → 푊 is such that, if 푛 is odd, 푓표푓(푛) = 푓 푓(푛) = 푓(푛 − 1) = 푛 −

1 + 1 = 푛

and if 푛 is even, 푓표푓(푛) = 푓 푓(푛) = 푓(푛 + 1) = 푛 + 1 − 1 = 푛

Hence, 푓표푓 = 퐼. This implies that f is invertible and 푓 = 푓 OR

Let (푎,푏) ∈ 푁 × 푁 Then ∵ 푎 + 푏 = 푏 + 푎 ∴ (푎, 푏) 푅 (푎,푏) Hence,푅 is re lexive. Let(푎,푏), (푐, 푑) ∈ 푁 × 푁 be such that (푎, 푏)푅(푐, 푑) 푎 + 푑 = 푏 + 푐 푐 + 푏 = 푑 + 푎 (푐, 푑)푅(푎, 푏) Hence, 푅 is symmetric.

ANSWERS

Section A (1 mark each)

Section B (4 marks each)


Let(푎,푏), (푐, 푑), (푒,푓) ∈ 푁 × 푁 be such that (푎,푏)푅(푐,푑), (푐,푑)푅(푒, 푓) 푎 + 푑 = 푏 + 푐 and 푐 + 푓 = 푑 + 푒

푎 + 푑 + 푐 + 푓 = 푏 + 푐 + 푑 + 푒 푎 + 푓 = 푏 + 푒 (푎,푏)푅(푒, 푓) Hence, 푅 is transitive. Since 푅 is reflexive, symmetric and transitive. Therefore 푅 is an equivalence relation.

8 푡푎푛 √ √

√ √= 푡푎푛

푡푎푛√ √

√ √ 휋 < 푥 < ⟹ < < ⟹ 푐표푠 < 0, 푠푖푛 < 0

푡푎푛1 − 푡푎푛

1 + 푡푎푛= 푡푎푛 푡푎푛

휋4 −

푥2

− − > − > −

9 Adj 퐴 = 4 −3

−1 −2′= 4 −1−3 −2

(퐴푑푗 퐴)퐴 = −11 00 −11 = −11 1 0

0 1

퐴(퐴푑푗 퐴) = −11 00 −11 = −11 1 0

0 1

|퐴| = −2 13 4 = −11

Hence, 퐴(푎푑푗퐴) = (푎푑푗퐴)퐴 = |퐴| 퐼 verified.

10 퐿퐻푆 =

1 1 + 푝 1 + 푝 + 푞3 4 + 3푝 2 + 4푝 + 3푞4 7 + 4푝 2 + 7푝 + 4푞

=1 1 + 푝 1 + 푝 + 푞0 1 −1 + 푝4 3 −2 + 3푝

(푅 → 푅 − 3푅 ,푅 → 푅 − 4푅 )

=1 1 + 푝 1 + 푝 + 푞0 1 −1 + 푝0 0 1

(푅 → 푅 − 3푅 )


= 1=푅퐻푆. Hence, proved. OR

Let A = 0 2 −3−2 0 43 −4 0

= 0 2 −3−2 0 43 −4 0

(퐼푛푡푒푟푐ℎ푎푛푔푖푛푔 푟표푤 푎푛푑 푐표푙푢푚푛푠)

= (−1)(−1)(−1)0 2 −3−2 0 43 −4 0

= −1 ⟹ 2 = 0 ⟹ 0 = 0

11 퐴퐵 = 2 0

0 2 = 21

⟹ 퐴12퐵 = 퐼 ⟹ 퐴 =

12퐵 = 2 −3

−1 2

The given system of equations is equivalent to 퐴′푋 = 퐶, where 푋 = 푥푦 , 퐶 = 41

퐶 = 퐴′ 퐶 = 퐴′ 퐶

= 2 −1−3 2

41 = 7

−10 ⟹ 푥 = 7, 푦 = −10

12 Since, 푓 is differentiable at x = 2, therefore, 푓 is continuous at 푥 = 2

⟹ lim 푓(푥)= lim 푓(푥) = 푓(2) ⟹ lim 푥 = lim (푎푥 + 푏) = 4 ⟹ 4 = 2푎 + 푏 Since, 푓 is differentiable at 푥 = 2,

∴ L푓′(2) = R푓′(2) ⟹ lim →( ) ( ) = lim →

( ) ( ) (ℎ > 0)

⟹ lim→

(2 − ℎ) − 4−ℎ = lim

→

푎(2 + ℎ) + 푏 − 4−ℎ

⟹ lim→

(−ℎ + 4) = lim → ⟹ 4 = 푎

푏 = −4

13 Let 푢 = (푙표푔푥) . then 푙표푔 푢 = 푥 푙표푔(log푥) ⟹ = + 푙표푔(log푥)

⟹푑푢푑푥 = (푙표푔푥)

1푙표푔푥 + 푙표푔(log푥)

Let 푣 = 푥 . Then


푙표푔 푣 = 푥 푐표푠 푥 푙표푔 푥 ⟹ = + cos푥 (푙표푔 푥)푥 sin 푥 푙표푔 푥

⟹푑푣푑푥 = 푥 [푐표푠 푥 + 푐표푠 푥(푙표푔 푥)− 푥 sin 푥 푙표푔 푥]

푦 = 푢 + 푣 ⟹ = +

= (푙표푔푥) + 푙표푔(log푥) + 푥 [푐표푠 푥 + 푐표푠 푥(푙표푔 푥)− 푥 sin 푥 푙표푔 푥]

OR

푑푥푑푡 = 푎푝 cos푝푡

푑푦푑푡 = −푏푝 sin푝푡

푑푦푑푥 = = −

푏푎 tan 푝푡

= − 푝 푠푒푐 푝푡 ×

푑 푦푑푥 = −

푏푎 푐표푠 푝푡

⟹푑 푦푑푥 = −

푏푎

14 Given integral =

1sin 푥 − sin 2푥 푑푥 =

1sin 푥 (1− 2cos푥)푑푥 =

푠푖푛 푥(1 + cos 푥)(1− cos푥)(1− 2cos푥)푑푥

= −∫ ( )( )( )푑푥 (cos푥 = 푡 ⟹ −푠푖푛푥 푑푥 = 푑푡)

( )( )( ) = ( ) + ( ) + ( )

∴ 1 = 퐴(1− 푡)(1− 2푡) + 퐵(1 + 푡)(1 − 2푡) + 퐶(1 − 푡 ) (퐴푛 퐼푑푒푛푡푖푡푦)

Putting, 푡 = 1, 12 ,−1 we get 퐴 = 1

6 , 퐵 = −16 ,퐶 = 4

3

Therefore, the given integral

= −16 푙표푔

|1 + 푡| −12 푙표푔

|1 − 푡| +46 푙표푔

|1− 2푡| + 푐

= −16 푙표푔

|1 + cos푥|−12 푙표푔

|1− cos푥| +32 푙표푔

|1− 2 cos푥| + 푐

OR sin

√sin − 2cos + 3푑 =

sin √1− cos − 2cos + 3

푑

= ∫ √

푑 = ∫ √ 푑푡 (푐표푠 = 푡 ⟹ −푠푖푛 푑 = 푑푡)


= 1

√−5 −(푡 − 1)푑푡

= 푠푖푛푡 − 1√5

+ 푐 = 푠푖푛푐표푠 − 1√5

+ 푐

15 Let

I =∫( )

푑푥 = ∫

푑푥 + ∫ 푑푥

= 0 + 2 ∫ 푑푥 푎푠

푖푠 표푑푑 푎푛푑

푖푠 푒푣푒푛

= 4 푥 sin 푥

1 + 푐표푠 푥 푑푥

Let

퐼 =푥 sin 푥

1 + 푐표푠 푥 푑푥 =(휋 − 푥) sin 푥(휋 − 푥)

1 + 푐표푠 (휋 − 푥) 푑푥

퐼 =(휋 − 푥) sin 푥

1 + 푐표푠 푑푥

Adding,

2퐼 = 휋(휋 − 푥) sin 푥

1 + 푐표푠 푑푥

= −휋 ∫ 푑푥 (cos푥 = 푡 ⟹− 푠푖푛 푥 푑푥 = 푑푡)

= 휋 [푡푎푛 푡]

퐼 = . Hence, 퐼 = 휋

16 Given differential equation is

= + , 푥 < 0 or = + 1 + = 푓 , hence, homogeneous.

Put 푦 = 푣 ⟹ = 푣 + 푥 . The differential equation becomes 푣 + 푥 = 푣 +

√1 + 푣

Or, √

=

Integrating, we get 푙표푔 푣 + √1 + 푣 = 푙표푔|푥| + 푙표푔푘


⟹ 푙표푔 푣 + 1 + 푣 = 푙표푔|푥|푘 ⟹ 푣 + 1 + 푣 = |푥|푘

⟹ 푣 + 1 + 푣 = ±푘푥 ⟹ 푦푥 + 1 +

푦푥 = 푐푥

⟹ 푦 + 푥 + 푦 = 푐푥 ,

Which gives the general solution.

17 We have the following differential equation: = or, + =

( ) ,

which is linear in 푥

퐼.퐹. = 푒∫ = 푒

⟹ 푥푒 = 푒 푡푑푡 푡푎푛 푦 = 푡 ⟹1

1 + 푦 푑푦 = 푑푡

⟹ 푥푒 = 푡푒 − 푒 + 푐 ⟹ 푥푒 = 푡푎푛 푦푒 − 푒 + 푐 Which gives the general solution of the differential equation.

18 The vector equations of the given lines are

푟 = 횤 + 2 횥 − 푘 + 휆 2 횤 + 3 횥 + 4푘 , 푟 = 2 횤+ 3 횥 + 휇 − 횤 + 2 횥 − 3푘

푎 = 횤 + 2 횥 − 푘, 푏 = 2 횤 − 3 횥 + 4푘, 푎 = −2 횤 + 3 횥, 푏 = − 횤 + 2 횥 − 3푘

푎 − 푎 = −3 횤 + 횥 + 푘, 푏 × 푏 = 횤 횥 푘2 −3 4−1 2 3

= −17횤 − 10 횥 + 푘

The required shortest distance = ( ). ×

×

= √

units

19 Let us define the following events: 퐸 = 퐴 solves the problem, 퐹 = 퐵 solves the

problem, 퐺 = 퐶 solves the problem, 퐻 = 퐷 solves the problem

(i) The required probability = 푃(퐸 ∪ 퐹 ∪ 퐺 ∪ 퐻) =1 − 푃(퐸 ∩ 퐹 ∩ 퐺 ∩ 퐻) =1 − 푃(퐸) × 푃(퐹) × 푃(퐺) × 푃(퐻) =1 − × × × =


(ii) The required probability = 푃(퐸) × 푃(퐹) × 푃(퐺) × 푃(퐻) + 푃(퐸) × 푃(퐹) × 푃(퐺) × 푃(퐻)

+ 푃(퐸) × 푃(퐹) × 푃(퐺) × 푃(퐻) + 푃(퐸) × 푃(퐹) × 푃(퐺) × 푃(퐻)+ 푃(퐸) × 푃(퐹) × 푃(퐺) × 푃(퐻)

=

23 ×

34 ×

45 ×

13 +

13 ×

34 ×

45 ×

13 +

23 ×

14 ×

45 ×

13 +

23 ×

34 ×

15 ×

13 +

23 ×

34 ×

45

×23

=

20 푓′(푥) = (푥 − 1) (푥 + 2)(5푥 + 4)

푓′(푥) = 0 ⟹ 푥 = 1,−2,

Hence, 푓 is strictly increasing in (−∞,−2) and − , ∞ . 푓 is strictly decreasing

in −2,−

In the left nhd of −2 푓′(푥) > 0, in the right nhd of −2 푓′(푥) < 0 and 푓′(−2) = 0,

Section C (5 marks each)


therefore, by the first derivative test, −2 is a point of local maximum. In the left nhd of −4/5, 푓’(푥) < 0, in the right nhd of −4/5,푓’(푥) > 0 and 푓’(−4/5) = 0, therefore, by the first derivative test, −4/5 is a point of local minimum.

21

∴ |푎| = ±2 푏 × 푐

푎 . 푏 × 푐 + 푎 × 푏 . 푐 = 푎 . 푏 × 푐 + 푎 . 푏 × 푐 = 2푎 . 푏 × 푐

We have

푎.푏 = 0, 푎. 푐 = 0 ⟹ 푎 ⊥ both 푏 and 푐 ( as 푎, 푏, 푐 are nonzero vectors)

⟹ 푎 × 푏 × 푐

Let 푎 = 휆 푏 × 푐

Then

|푎| = |휆| 푏 × 푐 ⟹ | | ×

= |휆| ⟹ |휆| = = 2 ⟹ 휆 = ±2

Now 푎 + 푏 , 푏 + 푐 , 푐 + 푎 = 푎 + 푏 × 푏 + 푐 . (푐 + 푎) = 푎 × 푏 . 푐 + 푏 + 푐 . 푎

(As the scaler triple product = 0 if two vectors are equal.)

=2푎 .(± 푎) = ±1

22 We have the curve

4푦 = 푥 ⟹ 4 = 2푥 ⟹ = ⟹ = 1

The equation of the tangent is 푦 = 푥 − 1


The required area = the shaded area =

∫ (푥 − 1)− 푑푥 + ∫ ( ) − 푑푥 = ∫ (푥 − 1)푑푥 + ∫ (푥 + 3)푑푥 −

ݔݔ 6 2 12

− 푥 + + 3푥 − [푥 ]

= 1 square units

23

(3)(2 + 휆) + (−2)(3 + 휆) + (−1)(−1− 2휆) = −1 ⟹ 휆 =−23

The equation of the line passing through the point (3,−2, 1) and parallel to the given line is

= =

Any point on this line is (2휆 + 3,−3휆 − 2, 휆 + 1) If it lies on the plane, we have 3(2휆 + 3),−3휆 − 2 − 휆 − 1) + 2 = 0 ⟹ 휆 = −4) Hence, the point common to the plane and the line is (−5, 10,−3).

Hence, the required distance = (3 + 5) + (−2 − 10) + (1 + 3) units = 4√14

The equation of the line passing through (3,−2, 1) and perpendicular to the plane is

= =

Any point on it is (3휇 + 3, 휇 − 2,−휇 + 1 )

If it lies on the plane, we get 3(3휇 + 3) + 휇 − 2 + 휇 − 1 + 2 = 0 ⟹ 휇 =

The required foot of the perpendicular = , ,

OR Any plane through the line of intersection of the given planes is

푟 . 2 횤 + 3 횥 − 푘 + 1 + 휆 푟. 횤 + 횥 − 2푘 = 0

Or, 푟 . (2 + 휆)횤+ (3 + 휆) 횥 + (−1 − 2휆) = −1 If it contains the point (3,−2,−1), we have

The required equation of the plane is


푟 . 2− 횤 + 3− 횥 + −1 + 푘 = −1 or, 푟 . 4 횤 + 7 횥 + 푘 = −3

If 휃 be the angle between the normals to the two given planes, then 휃��is the angle between the planes and cos휃 = .

| |.| |=

√ √=

√

24

= 20 37⁄

Let us define the following events: 퐸 = Two white balls are transferred, 퐸 = Two red balls are transferred, 퐸 = One red and one white balls are transferred, 퐴 = The ball drawn from the Bag 퐼퐼 is red

푃(퐸 )= = ××

푃(퐸 )= = ××

푃(퐸 )= × = × ××

푃(퐴 퐸⁄ )= , 푃(퐴 퐸⁄ ) = , 푃(퐴 퐸⁄ ) = ,

The required probability, 푃(퐸 퐴⁄ ), by Baye’s Theorem

= ( )× ( ⁄ )( )× ( ⁄ ) ( )× ( ⁄ ) ( )× ( ⁄ )

OR Let 푋 represent the random variable. Then 푋 = 0, 1, 2, 3

푃(푋 = 0) = 푃(푟 = 0) 3 =

푃(푋 = 1) = 푃(푟 = 1) 3 =

푃(푋 = 2) = 푃(푟 = 2) 3 =

푃(푋 = 3) = 푃(푟 = 3) 3 =


Mean = ∑푥 푝 = , 푣푎푟(푋) ∑푥 푝 − (∑푥 푝 ) = , 푣푎푟(푋) = − =

Standard deviation = 푣푎푟(푥) = √

25 Let the radius of the circular garden be 푟 푚 and the side of the square garden be

푥 푚. Then

600 =2휋푟 + 4푥 ⟹ 푥 =

The sum of the areas = 퐴 = 휋푟 + 푥 ⟹ 퐴 = 휋푟 +

=2휋푟 + (600− 2휋푟)(−2휋) = (4푟 − 300 + 휋푟), = 0 ⟹ 푟 =

= (4 + 휋), > 0

Therefore, A is minimum when 푟 = For this value of 푟, 푥 = 2푟

To achieve any goal, there is every possibility that energy, time and money are required to be invested. One must plan in such a manner that least energy, time and money are spent. A good planning and execution, therefore, is essentially required.

26

9 + 12 ≤ 180, , 3 + 4 ≤ 60, + 3 ≤ 30, ≥ 0, ≥ 0

Let the number of pieces of model 퐴 to be manufactured be = 푥 and the number of pieces of model 퐵 to be manufactured be = 푦. Then to maximize the profit, 푃 = Rs (8000 +12000y) subject to the constraints


The number of pieces of model = 12, the number of pieces of model = 6 and the maximum profit = Rs 168000.

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