Maths Guidelines for Board Exams.pdf

GUIDELINES

FOR

BOARD EXAM

(CLASS-XII)

CHAPTER-1 : RELATIONS AND FUNCTIONS

[In 2015, 1 Que. of 6 marks]

1.

Relations :

In case of neither nor condition we take examples, otherwise we give a general proof.

Ex.1 : Show that the relation R on the set R of all real numbers, defined as R = {(a, b) : a < b2_{} is}

neither reflexive nor symmetric nor transitive.

Sol. : We have, R = {(a, b) : a < b2_{}, where a, b Î R.}

Reflexivity: We observe that

2

5 5

2 2

æ ö

£ ç ÷_{è ø} is not true. Therefore, 5 5, R 2 2

æ _öÎ/

ç ÷

è ø .

So, R is not reflexive.

Symmetry :We observe that – 1 < 32 _{but 3 ( 1)£ -/} 2 _{i.e. (–1, 3) Î R but (3, –1) Ï R.}

So, R is not symmetric. Transitivity : We find that

2 < (–3)2 _{and –3 < 1}2 _{but 2 £/ 1}2

i.e., (2, –3) Î R and (–3, 1) Î R but (2, 1) Ï R. So, R is not transitive.

Ex.2 : Show that the relation R on R defined as R = {(a, b) : a < b} is reflexive and transitive but not symmetric.

Sol. : We have, R = {(a, b) : a < b}, where a, b Î R. Reflexivity: For any a Î R, we have a < a.

Þ (a, a) Î R for all a Î R. Þ R is reflexive.

Symmetry :We observe that 2 < 3 but 3 £/ 2 i.e. (2, 3) Î R but (3, 2) Ï R. So, R is not symmetric.

Transitivity : Let (a, b) Î R and (b, c) Î R. Then, (a, b) Î R and (b, c) Î R

Þ a < b and b < c Þ a < c

Þ (a, c) Î R So, R is transitive.

2.

Functions :

Graphical representation of functions must be avoided. If it can not be avoided, then the full explanation of the graph must be done.

Ex.1 : Prove that ƒ : R ® R, given by ƒ(x) = 2x, is one-one and onto.

Sol. : We observe the following properties of ƒ :

Injectivity : Let x₁, x₂ Î R such that ƒ(x₁) = ƒ(x₂). Then, ƒ(x₁) = ƒ(x₂)

Þ 2x₁ = 2x₂ Þ x₁ = x₂

So, ƒ : R ® R is one-one.

Surjectivity : Let y be any real number in R(co-domain). Then, ƒ(x) = y Þ 2x = y Þ x y

2 =

Clearly, y R

2Î for y Î R such that

y y ƒ 2 y 2 2 æ ö₌ æ ö₌ ç ÷ ç ÷ è ø è ø .

Thus, for each y Î R (co-domain) there exists x y R 2

= Î (domain) such that ƒ(x) = y. This mean that each element in co-domain has its pre-image in domain.

So, ƒ : R ® R is onto.

Hence, ƒ : R ® R is bijection.

Ex.2 : Show that the function ƒ : R ® R, defined as ƒ(x) = x2_{, is neither one-one nor onto.} Sol. : We observe that ƒ(–1) = ƒ(1)

So, ƒ is not one-one.

Since ƒ(x) assumes only non-negative values. So, no negative real number in R (co-domain) has its pre-image in domain (R). Consequently ƒ is not onto.

These facts are evident from the graph of ƒ(x) as shown in Figure

(–1,1) (1,1) ƒ(–1) ƒ(1) (1,0) (–1,0) x x y y

CHAPTER-2 : I.T.F.

[In 2015, 1 Que. of 4 marks]

1. The principal value of I.T.F. always exist in the principal value branch (range)

Ex.1 : cot–1_{(cotq) = q, for all q Î (0, p).} Sol. : _sin 1 _sin2 2

3 3 - æ pö ¹ p ç ÷ è ø as 2 3 p

does not lie between

2 p - and 2 p . Now, _sin 1 _sin2 _sin 1 _sin

3 3 - æ pö₌ - ì æ_{p -}p üö ç ÷ í ç ÷ý è ø _î è ø_þ 2 sin isn 3 3 é p₌ æ_{p -}p ùö ç ÷ ê _{è ø}ú ëQ û

Þ _sin 1 _sin2 _sin 1 _sin

3 3 - æ pö₌ - æ pö ç ÷ ç ÷ è ø è ø [Q sin(p – q) = sinq] Þ 1 2 sin sin 3 3 - æ pö = p ç ÷ è ø

2. Page no. 42 of NCERT : We will not go in to the details of these values of x in the domain as this discussion goes beyond the scope of this text book.

For example : st 1 1 1 nd 1 rd 1 x y 1 tan , if xy 1 1 xy x y

tan x tan y 2 tan if x 0, y 0 & xy 1

1 xy x y 3 tan if x 0, y 0 & xy 1 1 xy -- - -ì _{æ + ö} < ï ç _- ÷ è ø ï ï ₊ ï æ ö + =_í p + _{ç ÷} > > > -è ø ï ï _{æ + ö} ï -p + _{ç ÷} < < > -ï è ø î ,

2nd _{and 3}rd _{results go beyond the scope of text book.} 3. Focus on existence of a value, which is determined.

Ex.2 : Solve : _sin 1(_{1 x}) _{2sin x}1

2

- _{- -} - ₌p

Sol. : We have, _sin 1(_{1 x}) _{2sin x}1

2 - _{- -} - ₌p _{Þ sin} 1_{(1 x) 2sin x}1 2 - _{- = +}p -Þ 1 x sin 2sin x1 2 -p æ ö - = ç + ÷ è ø Þ 1 x cos 2sin x- =

(

-1

)

Þ 1 – x = cos{cos–1_{(1 – 2x}2_{)} [}Q 2sin–1_{x = cos}–1_{(1 – 2x}2_)]

Þ 1 – x = (1 – 2x2_{) Þ x 2x=} 2 _{Þ x(2x – 1) = 0 Þ x 0,} 1

2 =

For, x 1 2

= , we have LHS = sin–1_{(1 – x) – 2sin}–1_x

x sin 11 2sin 11 sin 11 R.H.S.

2 2 2 6

- - - p

= - = - = - ¹

So, x 1 2

= is not a root of the given equation. Clearly, x = 0 satisfies the equation.

Hence, x = 0 is a root of the given equation.

CHAPTER-3 : MATRICES

[In 2015, 3 Que. of (1 + 4 + 4) = 9 marks]

1. Focus on presentation of word problems.

Ex.1 : There are two families A and B. There are 4 men, 6 women and 2 children in family A and 2 men, 2 women and 4 children in family B. The recommended daily allowance for calories is : Man : 2400, Woman : 1900, Child : 1800 and for proteins in Man : 55gm, Woman : 45 gm and Child : 33 gm.

Represent the above information by matrices. Using matrix multiplication, calculate the total requirement of calories and proteins for each of the two families.

Sol. : The members of the two families can be represented by the 2 × 3 matrix

M W C A 4 6 2 F B 2 2 4 é ù = ê ú ë û

and the recommended daily allowance of calories and proteins for each member can be represented by 3 × 2 matrix Calories Proteins M 2400 55 F W 1900 45 C 1800 33 é ù ê ú = _{ê ú} ê ú ë û

The total requirement of calories and proteins for each of the two families is given by the matrix multiplication : 2400 55 4 6 2 A 24600 556 FR 1900 45 2 2 4 B 15800 332 1800 33 é ù é ù_{ê ú} é ù =_{ê úê ú}= _{ê ú} ë û_{ê ú} ë û ë û

Hence, family A requires 24600 calories and 556 gm proteins and family B requires 15800 calories and 332 gm proteins.

2. Read the question carefully based on finding the inverse matrix (There are three types of questions)

(a) First type question : find the inverse matrix using elementary transformation.

In row ele. trans. In column ele. trans

We write A = IA We write A = AI.

Do not mix. row and column ele. trans.

(b) Second type question : Find the inverse matrix.

in this case use formula _A 1 adjA

| A |

- _{= and firstly check the existence of A}–1 _{through |A| ¹ 0.} Third type Question : If A = 3 1

1 2

é ù

ê_- ú

ë û show that A

2 _{– 5A + 7I = 0. Hence find A}–1_{. In this case for}

finding A–1 _{use property [A.A}–1 _{= A}–1_{A = I].}

CHAPTER-4 : DETERMINANT

[In 2015, 1 Que. of 4 marks]

1. Focus on applying only one either row or column operation at a time. For example : 2 3 5 6 0 4 1 5 7 -D =

-Here firstly on applying R₁ ® R₁ – R₂, we get

4 3 1 6 0 4 1 5 7 - -D =

-Secondly, on applying C₂ ® C₂ – C₁, we get

4 1 1 6 6 4 1 4 7 -D = -2. Focus on writting (±) sign, when area is given.

3. Do not use CRAMER'S RULE for solving a solution of the system of linear equations. Use matrix method (X = A–1_B)

CHAPTER-5 : CONTINUITY & DIFFERENTIABILITY

[In 2015, 3 Que. of 4+4+4 = 12 marks]

1. Use limit for any type of discussion of continuity and differentiability (avoid the graph) for continuity (at a point x = a) RHL =

h 0

lim ƒ(a h)

® + = LHL = lim ƒ(a h)h®0 - = ƒ(a)

for diff. (at a point x = a) RHD =

h 0 ƒ(a h) ƒ(a) lim h ® + - _{= LHD =} h 0 ƒ(a h) ƒ(a) lim h ® - -2. Focus on second order derivative of parametric function.

For example : If x = acos3_{q and y = asin}3_{q, then find the value of}

2 2 d y dx at 6 p q = .

Sol. x = acos3_{q ; y = asin}3_q

2 dx 3cos ( sin ) dq= q - q ...(i) ; 2 dy 3a sin cos dq = q q ...(ii) from dy dy / d dx dx / d q = q

2 2 dy 3a sin cos tan dx 3a cos sin q q = = - q - q q Þ 2 2 2 d y d sec dx dx q = - q´ = - q´ - q q 2 2 1 sec 3a cos sin 2 2 4 d y 1 dx 3a.cos .sin æ ö = ç ÷ _{q q} è ø Þ 2 4 2 6 d y 1 32 dx _{3 1} 27a 3a 2 2 p q= æ ö = = ç ÷ è ø _{æ ö} ´ç_è ÷_ø ´

3. Proving questions based on second order derivative, apply oppropriate method which takes minimum time.

Ex.1 : If y = sin–1_{x, show that}

(

2

)

2 2

d y dy

1 x x 0

dx dx

- - =

Sol. : We have y = sin–1_{x. Then}

2 dy 1 dx = _{(1 x )-} or 2 dy (1 x ) 1 dx - = So d (1 x ).2 dy 0 dx dx æ _- ö₌ ç ÷ è ø or 2 2 2 2 d y dy d (1 x ) . (1 x ) 0 dx dx dx - + - = or 2 2 2 ₂ d y dy 2x (1 x ) . 0 dx _{dx 2 1 x} - - = -Hence

(

2

)

2 2 d y dy 1 x x 0 dx dx - - =

Alternatively, Given that y = sin–1_{x, we have}

(

2

)

2

1 ₂ 1 1 y , i.e., 1 x y 1 1 x = - = -So (1 – x2_).2y 1y2 +y (0 2x) 012 - = Hence (1 – x2_)y 2 – xy1 = 0

CHAPTER-6 : AOD

[In 2015, 1 Que. of 6 marks]

Maxima and minima word problems : Attempt these question with following steps : Step 1 : Draw the required named figure.

Step 2 : Let function properly.

Step 3 : Establish a relation between two variables. Step 4 : Differentiate the function and find critical point. Step 5 : Use FODT or SODT for maximum and minimum .

CHAPTER-7 : INTEGRAL

[In 2015, 1 Que. of 4+4+4 = 12 marks]

Indefinite integration

1. Integrals of some particular function use directly :

(i) ₂1 ₂ dx 1 log x a C x a 2a x a -= + - +

ò

(x > a) (ii) ₂ 1 ₂ dx 1 log a x C a x 2a a x + = + -

-ò

(x < a) (iii) 1 2 2 1 1 x dx tan C x a a a - æ ö = _{ç ÷}+ + è ø

ò

= 1 1 x cot C a a - æ ö - _{ç ÷}+ è ø (iv) 1 2 2 1 1 x dx sec C a a x x a -= +

-ò

(v) 1 1 2 2 1 x x dx sin C cos C a a a x - æ ö - æ ö = _{ç ÷}+ = - _{ç ÷}+ è ø è ø

-ò

(vi) 2 2 2 2 1 dx log | x x a | C x +a = + + +

ò

(vii) 2 2 2 2 1 dx log | x x a | C x -a = + - +

ò

2. Use partial fraction method for integration of proper rational function (for this use table 7.2 Page No 317 NCERT)

Definite Integration :

3. If we use properties of D.I., then write standard form of properties. for example :

a a

0 0

ƒ(x)dx= ƒ(a x)dx

-ò

ò

4. Focus on limit, while applying substitution method

5. Focus on taking square root in definite integration. For example : Evaluate :

2 0 x 1 2sin dx 2 p æ _- ö ç ÷ è ø

ò

.

In the given interval, we take 1 2sinx for x 0,

2 3

p

æ ö é ö

+ -ç_è ÷_ø Îê_ë ÷_ø and - -_èæç1 2sin ₂x_ø÷öfor xÎ_èæç₃p,púù_û

\ / 3 0 /3 x x I 1 2sin dx 1 2sin dx 2 2 p p p æ ö æ ö = _ç - _÷ + - -_{ç ÷} è ø è ø

ò

ò

6. Some question can be solved through many method but you need to select the method which takes minimum time.

CHAPTER-8 : AOI

[In 2015, 1 Que. of 6 marks]

Attempt this question with following steps : Step 1 : Draw the required named figure.

Step 2 : Calculate intersection point of given curves

Step 3 : Take the limit for bounded region and draw required vertical (Horizontal) strips. Step 4 : Find area by using suitable formula

CHAPTER-9 : D.E.

[In 2015, 3 Que. of (1 + 1 + 6) = 8 marks]

1. Find order and degree of D.E. carefully.

2. Better practice required for formation of a D.E. (for an idea try yourself )

Q. Find the differential equation of the family of curves (x – h)2 _{+ (y – k)}2 _{= r}2_{, where h and k are}

arbitrary constants. [CBSE 2015, 6M]

Ans. 3 ₂ 2 ₂ 2 2 dy d y 1 r dx dx ì ü æ ö ï ₊æ ö ï ₌ í ç_è ÷_ø ý ç ÷ è ø ï ï î þ

3. Learn to show that D.E. is homogeneous.

4. Focus on standard form of linear D.E., its I.F. and solution. e.g. dy py Q

dx+ = (linear D.E.) ; I.F. =

Pdx

CHAPTER-10 : VECTOR

[In 2015, 3 Que. of (1+1+4) = 6 marks]

1. Do not calculate directly from coordinate geometry use vector sign ABuuur or ar .

Ex.1 : L and M are two points with position vector 2a br-r and a 2br+ r respectively. Write the position vector of a point N which divides the line segment LM in the ratio 2 : 1 externally. Sol. ON 2 OM 1 OL 2 1 ´ - ´ = -uuuur uuur uuur _{L M N} a+2b (2a–b) O 2 (a 2b) 1 (2a b) ON 1 ´ + - ´ -= r r r r uuur =5br [1]

Ex.2 Find a vector ar of magnitude 5 2, making an angle of

4 p

with x-axis,

2 p

with y-axis and an acute angle q with z-axis.

Sol. Here cos 1

4 2 p = = l and m cos 0 2 p = = Therefore, l2 _{+ m}2 _{+ n}2 _{= 1 gives} 1 _{0 n}2 ₁ 2+ + = Þ = 1n 2 Hence, the required vector rr 5 2 ( i mj nk)= lˆ+ ˆ+ ˆ is given by

æ ö = _ç + + _÷= = + è ø r 1 _ˆ 1 _ˆ r _{ˆ ˆ} r 5 2 i 0 j k r 5i 5k 2 2 . ...(1) [1]

Ex.3 Using vectors, find the area of the triangle ABC, whose vertices are A(1, 2, 3), B(2, –1, 4) and C(4, 5,–1).

Sol. Side BAuuur= - + -ˆi 3j kˆ ˆ & uuur_{BC 2i 6 j 5k=} ˆ_{+ -}ˆ ˆ A(1,2,3)

B(2,–1,4) C(4,5,–1) [1] area of ABC 1 BA BC 2 D = uuur uuur´ [½] ˆ ˆ ˆ i j k BA BC 1 3 1 2 6 5 ´ = - -uuur -uuur =ˆ_{i( 15 6) j(5 2) k( 6 6)- + - + + - -}ˆ _{= - -9i 7 j 12k}ˆ ˆ_- ˆ [1½] BA BC´ = 81 49 144+ + uuur uuur Þ 274

Now area of DABC is =1 274

2 [1]

CHAPTER-11 : 3-D

[In 2015, 3 Que. of (1 + 4 + 6) = 11 marks] 1. Find the answer in vector form or cartesian form accordingly question.

2. If we use standard results (formula) directly, then we save the time. for example :

(a) Shortest distance between two skew lines r a= + l1 b1

r r r _and 2 2 r=a +µbr r r 2 1 1 2 1 2 (a a ).(b b ) S.D. | b b | - ´ = ´ r r r r r r

(b) Equation of a plane passing through three non collinear points (x₁, y₁, z₁) (x₂, y₂, z₂) (x₃ y₃ z₃) is required plane = 1 1 1 2 1 2 1 2 1 3 1 3 1 3 1 x x y y z z x x y y z z 0 x x y y z z - - -- - - = - -

-(other results shows in NCERT Page No. 499 to 503)

CHAPTER-12 : LPP

[In 2015, 1 Que. of 6 marks]

1. Focus on optimal value of unbounded region :

To decide the optimal value of function we graph the inequality

[ax + by > M (max. value) or ax + by < m (min. value)] dotted line. See example No. 4 and 6.

2. Draw the named graph properly.

CHAPTER-13 : PROBABILITY

[In 2015, 2 Que. of (4 + 6) = 10 marks]

1. Use formula's of Baye's theorem, Bernoulli Trials, mean, variance and standard deviation accordingly NCERT Tex Book.

IMPORTANT QUESTIONS OF NCERT TEXT BOOK

CHAPTER-1 : RELATIONS AND FUNCTIONS

Example No. Q. No. 12, 23, 45

Exercise # 1.1 Q. No. 2, 9, 14

Exercise # 1.2 Q. No. 9, 10

Exercise # 1.3 Q. No. 8, 9, 12

Exercise # 1.4 Q. No. 5, 6, 11

Miscellaneous Exercise # 1 Q. No. 4, 12

CHAPTER-2 : I.T.F

Examples Q. No. 9, 12, 13

Exercise # 2.1 Q. No. 12, 14

Exercise # 2.2 Q. No. 8, 13, 14, 20

Miscellaneous Exercise Q. No. 2, 8, 10, 11, 16, 17

CHAPTER-3 : MATRICES

Example Q. No. 3, 10, 18, 22, 24

Exercise # 3.1 Q. No. 5 (i), 9, 10

Exercise # 3.2 Q. No. 2 (i), 10, 15, 18

Exercise # 3.3 Q. No. 9, 11

Miscellaneous Exercise Q. No. 7, 13

CHAPTER-4 : DETERMINANT

Examples Q. No. 5, 15, 16, 32 Exercise # 4.2 Q. No. 10, 11, 13 Exercise # 4.3 Q. No. 2, 3 Exercise # 4.4 Q. No. 5 Exercise # 4.5 Q. No. 12, 13, 15, 18 Exercise # 4.5 Q. No. 15

Miscellaneous Exercise Q. No. 5, 6, 11, 12, 16, 19

CHAPTER-5 : CONTINUITY & DIFFERENTIABILITY

Examples Q. No. 37, 41, 45 (iii)

Exercise # 5.1 Q. No. 24, 26, 30 Exercise # 5.2 Q. No. 10 Exercise # 5.3 Q. No. 7, 10, 15 Exercise # 5.4 Q. No. 7, 10 Exercise # 5.5 Q. No. 9, 12, 14, 15 Exercise # 5.6 Q. No. 7, 11 Exercise # 5.7 Q. No. 13, 14, 16, 17

Miscellaneous Exercise Q. No. 6, 15, 16, 17, 23

CHAPTER-6 : AOD

Example No. Q. No. 12, 13, 16, 19, 20, 38, 39, 48

Exercise # 6.1 Q. No. 5, 8, 10

Exercise # 6.2 Q. No. 7, 9, 16

Exercise # 6.3 Q. No. 10, 18, 24

Exercise # 6.4 Q. No. 1(iv), 2, 5

Exercise # 6.5 Q. No. 3(iii), (iv), 8, 11, 12, 17, to 26 Miscellaneous Exercise Q. No. 14, 15, 17, 18

CHAPTER-7 : INTEGRAL

Examples Q. No. 4, 6(ii), 10(ii), 12, 15, 21, 22(ii), 26, 30, 32, 36, 39, 40, 43

Exercise # 7.2 Q. No. 24, 36, 37, 38 Exercise # 7.3 Q. No. 14, 19, 22 Exercise # 7.4 Q. No. 7, 15, 19 Exercise # 7.5 Q. No. 6, 9, 12 16, 21 Exercise # 7.6 Q. No. 6, 10, 18, 20, 22 Exercise # 7.7 Q. No. 5, 9 Exercise # 7.8 Q. No. 6 Exercise # 7.9 Q. No. 18 Exercise # 7.10 Q. No. 8, 9, 10 Exercise # 7.11 Q. No. 8, 9, 12, 16, 19

Miscellaneous Exercise Q. No. 5, 10, 18, 19, 21, 24, 31, 33, 40, 43, 44

CHAPTER-8 : AOI

Examples Q. No. 7, 9, 10, 13

Exercise # 8.1 Q. No. 4, 6, 9

Exercise # 8.2 Q. No. 5

Miscellaneous Examples Q. No. 10, 11

Miscellaneous Exercise Q. No. 15

CHAPTER-9 : D.E.

Examples Q. No. 5, 7, 16, 20, 28 Exercise # 9.1 Q. No. 1, 12 Exercise # 9.2 Q. No. 6 Exercise # 9.4 Q. No. 4, 10 Exercise # 9.5 Q. No. 6, 8, 11, 13 Exercise # 9.6 Q. No. 5, 10, 15

Miscellaneous Exercise Q. No. 6, 9, 11, 14

CHAPTER-10 : VECTOR

Examples Q. No. 8, 9, 17, 18, 20, 21, 27, 28

Exercise # 10.2 Q. No. 9, 11, 13

Exercise # 10.3 Q. No. 5, 6, 9, 13

Exercise # 10.4 Q. No. 4, 7, 9, 10, 11

Mise. Exercise Q. No. 9, 13, 15

CHAPTER-11 : 3-D

Examples Q. No. 11, 21, 22, 24, 25

Exercise # 11.1 Q. No. 4, 5

Exercise # 11.2 Q. No. 6, 12, 15, 17

Exercise # 11.3 Q. No. 9

Miscellaneous Exercise Q. No. 6, 7, 10, 13, 14, 15, 17, 18, 19, 20

CHAPTER-12 : LPP

Examples Q. No. 6, 9, 10

Exercise # 12.1 Q. No. 4

Exercise # 12.2 Q. No. 3, 8

Miscellaneous Examples Q. No. 5, 8, 10

CHAPTER-13 : PROBABILITY

Examples Q. No. 7,17,24,27,29,36 Exercise # 13.1 Q. No. 5,12,14 Exercise # 13.2 Q. No. 4,9,14 Exercise # 13.3 Q. No. 4,6,12 Exercise # 13.4 Q. No. 6,11 Exercise # 13.5 Q. No. 4,7,8,12