GUIDELINES FOR
BOARD EXAM (CLASS-XII)
Guidelines for Board Exam
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 : Sol. :
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. We have, R = {(a, b) : a < b2}, where a, b Î R. 2
5 æ5ö æ5 5ö Reflexivity: We observe that £ ç ÷ is not true. Therefore, ç , ÷ Î/ R . è2 2ø 2 è2ø So, R is not reflexive. Symmetry :We observe that – 1 < 3 2 but 3 £/ (-1)2 i.e. (–1, 3) Î R but (3, –1) Ï R. So, R is not symmetric. Transitivity : We find that
Ex.2 : Sol. :
2.
2 < (–3)2 and –3 < 12 but 2 £/ 12 i.e., (2, –3) Î R and (–3, 1) Î R but (2, 1) Ï R. So, R is not transitive. Show that the relation R on R defined as R = {(a, b) : a < b} is reflexive and transitive but not symmetric. 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
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 3
Guidelines for Board Exam y æyö æ yö Î R for y Î R such that ƒ ç ÷ = 2 ç ÷ = y . 2 è2ø è2ø y Thus, for each y Î R (co-domain) there exists x = Î R (domain) such that ƒ(x) = y. 2 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. Show that the function ƒ : R ® R, defined as ƒ(x) = x2, is neither one-one nor onto. 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
Clearly,
Ex.2 : Sol. :
y (–1,1)
(1,1)
ƒ(–1) x
ƒ(1)
(–1,0)
(1,0)
x
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. :
2p ö 2p 2p p p æ as sin -1 ç sin ÷ ¹ does not lie between - and . 3 ø 3 3 è 2 2 2p ö p öü æ æ -1 ì Now, sin -1 ç sin ÷ = sin ísin ç p - ÷ ý è 3 ø 3 øþ î è Þ
pö 2p ö æ æ sin -1 ç sin ÷ = sin -1 ç sin ÷ 3 ø 3ø è è
é 2p p öù æ êQ sin 3 = isn çè p - 3 ÷ø ú ë û [Q sin(p – q) = sinq]
2p ö p æ sin -1 ç sin ÷ = 3 ø 3 è 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.
Þ
2.
ì st ï1 ï ïï For example : tan -1 x + tan -1 y = í2nd ï ï ï3rd îï
3.
æ x+y ö tan -1 ç ÷, è 1 - xy ø æ x+y ö p + tan -1 ç ÷ è 1 - xy ø
if xy < 1 if x > 0, y > 0 & xy > 1 ,
æ x+y ö -p + tan -1 ç ÷ if x < 0, y < 0 & xy > 1 è 1 - xy ø
2nd and 3rd results go beyond the scope of text book. Focus on existence of a value, which is determined. 4
Guidelines for Board Exam p 2
Ex.2 :
Solve : sin -1 (1 - x ) - 2sin -1 x =
Sol. :
We have, sin -1 (1 - x ) - 2sin -1 x =
Þ
æp ö 1 - x = sin ç + 2sin -1 x ÷ è2 ø –1 1 – x = cos{cos (1 – 2x2)}
Þ
1 – x = (1 – 2x2) Þ
Þ
For, x =
p 2
x =2x 2
p + 2 sin -1 x 2
Þ
sin -1 (1 - x ) =
Þ
1 - x = cos ( 2sin -1 x ) 2sin–1x = cos–1(1 – 2x2)] 1 x(2x – 1) = 0 Þ x = 0, 2 [Q
Þ
1 , we have LHS = sin–1(1 – x) – 2sin–1x 2 p 1 1 1 x = sin -1 - 2sin -1 = - sin -1 = - ¹ R.H.S. 2 2 2 6
1 is not a root of the given equation. 2 Clearly, x = 0 satisfies the equation. Hence, x = 0 is a root of the given equation.
So, x =
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 é 4 6 2ù F= ê B ë 2 2 4úû and the recommended daily allowance of calories and proteins for each member can be represented by 3 × 2 matrix Calories Proteins
M é 2400 55 ù ê F = W ê1900 45 úú C êë1800 33 úû The total requirement of calories and proteins for each of the two families is given by the matrix multiplication : é 2400 55 ù A é 24600 556 ù é 4 6 2ù ê FR = ê 1900 45úú = ê ú ê B ë15800 332 úû ë 2 2 4û ê ú 1800 33 ë û Hence, family A requires 24600 calories and 556 gm proteins and family B requires 15800 calories and 332 gm proteins. 5
Guidelines for Board Exam 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 f ormula A -1 =
adjA and firstly check the existence of A–1 through |A| ¹ 0. |A|
é 3 1ù Third type Question : If A = ê show that A2 – 5A + 7I = 0. Hence find A–1. In this case for ú ë -1 2 û finding A–1 use property [A.A–1 = A–1A = 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. 2 -3
For example : D = 6 1
0 5
5 4 -7 -4 -3
Here firstly on applying R1 ® R1 – R2, we get D = 6 1 -4
Secondly, on applying C2 ® C2 – C1, we get D = 6 1 2. 3.
1
0 5 1
4 -7 1
-6 4 4 -7
Focus on writting (±) sign, when area is given. Do not use CRAMER'S RULE for solving a solution of the system of linear equations. Use matrix method (X = A–1B)
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 = lim ƒ(a + h) = LHL = lim ƒ(a - h) = ƒ(a) h ®0
for diff. (at a point x = a) 2.
h ®0
RHD = lim ƒ(a + h) - ƒ(a) = LHD = lim h ®0 h ®0 h
ƒ(a - h) - ƒ(a) -h
Focus on second order derivative of parametric function. For
example : If x = acos3q and y = asin3q, then find the value of
d2y p . 2 at q = dx 6
Sol. x = acos3q ; y = asin3q dx = 3cos 2 q( - sin q) dq dy dy / dq = from dx dx / dq
........(i) ;
dy = 3a sin 2 q cos q dq
........(ii)
6
Guidelines for Board Exam dy 3a sin 2 q cos q = = - tan q dx -3a cos 2 q sin q
æ d2y ö 1 ç 2 ÷= è dx ø 3a.cos 4 q.sin q
3.
Þ
1 d2 y dq = - sec 2 q ´ = - sec 2 q´ 2 -3a cos2 q sin q dx dx
Þ
æ d2y ö ç 2÷ è dx øq=
1
=
p 6
=
4
32 27a
æ 3ö 1 3a ´ ç ÷ ´ è 2 ø 2 Proving questions based on second order derivative, apply oppropriate method which takes minimum time.
Ex.1 : Sol. :
d2 y dy If y = sin–1x, show that (1 - x 2 ) 2 - x =0 dx dx dy 1 = or We have y = sin–1x. Then dx (1 - x 2 )
d æ dy ö 2 ç (1 - x ). ÷ = 0 dx è dx ø
So
(1 - x 2 )
(1 - x 2 )
or
dy =1 dx
d 2 y dy d + . (1 - x 2 ) = 0 2 dx dx dx
d 2 y dy 2x (1 - x ) 2 - . =0 dx dx 2 1 - x 2 2
or
d2 y dy Hence (1 - x 2 ) 2 - x =0 dx dx 1
Alternatively, Given that y = sin–1x, we have y1 =
1- x
2
, i.e., (1 - x 2 ) y12 = 1
(1 – x2).2y1y2 + y (0 - 2x) = 0 (1 – x2)y2 – xy1 = 0 2 1
So Hence
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 : 1 1 x -a dx = log + C (x > a) 2 x -a 2a x+a
(i)
ò
(iii)
òx
(iv)
ò
(vi)
ò
(ii)
2
1 1 a+x dx = log + C (x < a) 2 a -x 2a a-x
ò
2
1 1 1 æxö æxö dx = tan -1 ç ÷ + C = - cot -1 ç ÷ + C 2 a +a a èaø èaø 1 1 x 1 æxö æxö dx = sec -1 + C (v) ò dx = sin-1 ç ÷ + C = - cos-1 ç ÷ + C 2 2 2 2 a a x x -a èaø èaø a -x 2
1 x +a 2
2
dx = log | x + x 2 + a 2 | + C
(vii)
ò
1 x -a 2
2
dx = log | x + x 2 - a 2 | + C 7
Guidelines for Board Exam 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. a
a
0
0
for example : ò ƒ(x)dx = ò ƒ(a - x)dx 4.
Focus on limit, while applying substitution method
5.
Focus on taking square root in definite integration. p
For example : Evaluate :
ò 0
2
xö æ ç 1 - 2sin ÷ dx . è 2ø
xö xö æ æp ù æ é pö In the given interval, we take + ç 1 - 2sin ÷ for x Î ê 0, ÷ and - ç 1 - 2 sin ÷ for x Î ç , p ú è 2ø 2ø è è3 û ë 3ø
\
p/3
I=
ò 0
6.
p
xö xö æ æ ç 1 - 2 sin ÷ dx + ò - ç1 - 2 sin ÷ dx è 2ø 2ø p/3 è
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. 2.
Find order and degree of D.E. carefully. 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 = r2, where h and k are arbitrary constants. [CBSE 2015, 6M] 3
2 æ d2 y ö ïì æ dy ö ïü Ans. í1 + ç ÷ ý = r 2 ç 2 ÷ ïî è dx ø ïþ è dx ø
2
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 (linear D.E.) ; dx
Pdx I.F. = e ò ;
Sol. y(I.F.) = ò Q.(I.F.)dx
8
Guidelines for Board Exam
CHAPTER-10 : VECTOR [In 2015, 3 Que. of (1+1+4) = 6 marks]
uuur Do not calculate directly from coordinate geometry use vector sign AB or r r r r Ex.1 : L and M are two points with position vector 2a - b and a + 2b position vector of a point N which divides the line segment externally. uuuur uuur uuur 2 ´ OM - 1´ OL L M ON = Sol. 2 -1 (2a–b) a+2b r r uuur 2 ´ (ar + 2b) - 1´ (2ar - b) r O ON = = 5b 1
1.
Ex.2
r
Find a vector a of magnitude 5 2 , making an angle of
r a. respectively. Write the LM in the ratio 2 : 1
N
[1]
p p with x-axis, with y-axis and an 4 2
acute angle q with z-axis. Sol.
Here l = cos
p p 1 = and m = cos = 0 2 4 2
Therefore, l2 + m2 + n2 = 1 gives
1 1 + 0 + n2 = 1 Þ n = 2 2
r ˆ is given by Hence, the required vector r = 5 2 (lˆi + mjˆ + nk)
r 1 ˆö r æ 1 ˆ r =5 2ç i + 0j+ k ÷ = r = 5iˆ + 5kˆ . 2 ø è 2 Ex.3 Sol.
.........(1)
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). uuur uuur A(1,2,3) Side BA = - ˆi + 3ˆj - kˆ & [1] BC = 2iˆ + 6ˆj - 5kˆ area of DABC =
1 uuur uuur BA ´ BC 2
ˆi ˆj kˆ C(4,5,–1) B(2,–1,4) uuur uuur ˆ -6 - 6) = -9iˆ - 7ˆj - 12kˆ BA ´ BC = -1 3 -1 = ˆi(-15 + 6) - j(5 + 2) + k( 2 6 -5 uuur uuur 274 BA ´ BC = 81 + 49 + 144 Þ Now area of DABC is = 2.
[1]
S.T.P. in syllabus.
1 274 2
[½]
[1½]
[1]
9
Guidelines for Board Exam
CHAPTER-11 : 3-D [In 2015, 3 Que. of (1 + 4 + 6) = 11 marks] 1. 2.
Find the answer in vector form or cartesian form accordingly question. If we use standard results (formula) directly, then we save the time. for example : r r r r r r (a) Shortest distance between two skew lines r = a 1 + lb1 and r = a 2 + µb 2 r r r r (a 2 - a1 ).(b1 ´ b 2 ) r r S.D. = | b1 ´ b 2 | (b) Equation of a plane passing through three non collinear points (x 1, y1, z1) (x2, y2, z2) (x3 y3 z3)
x - x1 is required plane = x 2 - x1
y - y1 y 2 - y1
z - z1 z 2 - z1 = 0
x 3 - x1
y3 - y1
z 3 - z1
(other results shows in NCERT Page No. 499 to 503)
CHAPTER-12 : LPP [In 2015, 1 Que. of 6 marks] 1.
2.
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. 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.
10
Guidelines for Board Exam
IMPORTANT QUESTIONS OF NCERT TEXT BOOK CHAPTER-1 : RELATIONS AND FUNCTIONS
Example No. Exercise # 1.1 Exercise # 1.2 Exercise # 1.3 Exercise # 1.4 Miscellaneous Exercise # 1
Q. No. 12, 23, 45 Q. No. 2, 9, 14 Q. No. 9, 10 Q. No. 8, 9, 12 Q. No. 5, 6, 11 Q. No. 4, 12
CHAPTER-2 : I.T.F Examples Exercise # 2.1 Exercise # 2.2 Miscellaneous Exercise
Q. No. 9, 12, 13 Q. No. 12, 14 Q. No. 8, 13, 14, 20 Q. No. 2, 8, 10, 11, 16, 17
CHAPTER-3 : MATRICES Example Exercise # 3.1 Exercise # 3.2 Exercise # 3.3 Miscellaneous Exercise
Q. No. 3, 10, 18, 22, 24 Q. No. 5 (i), 9, 10 Q. No. 2 (i), 10, 15, 18 Q. No. 9, 11 Q. No. 7, 13
CHAPTER-4 : DETERMINANT Examples Exercise # 4.2 Exercise # 4.3 Exercise # 4.4 Exercise # 4.5 Exercise # 4.5 Miscellaneous Exercise
Q. No. 5, 15, 16, 32 Q. No. 10, 11, 13 Q. No. 2, 3 Q. No. 5 Q. No. 12, 13, 15, 18 Q. No. 15 Q. No. 5, 6, 11, 12, 16, 19
CHAPTER-5 : CONTINUITY & DIFFERENTIABILITY Examples Exercise # 5.1 Exercise # 5.2 Exercise # 5.3 Exercise # 5.4 Exercise # 5.5 Exercise # 5.6 Exercise # 5.7 Miscellaneous Exercise
Q. No. 37, 41, 45 (iii) Q. No. 24, 26, 30 Q. No. 10 Q. No. 7, 10, 15 Q. No. 7, 10 Q. No. 9, 12, 14, 15 Q. No. 7, 11 Q. No. 13, 14, 16, 17 Q. No. 6, 15, 16, 17, 23
CHAPTER-6 : AOD Example No. Exercise # 6.1 Exercise # 6.2 Exercise # 6.3 Exercise # 6.4 Exercise # 6.5 Miscellaneous Exercise
Q. No. 12, 13, 16, 19, 20, 38, 39, 48 Q. No. 5, 8, 10 Q. No. 7, 9, 16 Q. No. 10, 18, 24 Q. No. 1(iv), 2, 5 Q. No. 3(iii), (iv), 8, 11, 12, 17, to 26 Q. No. 14, 15, 17, 18 11
Guidelines for Board Exam Examples Exercise # 7.2 Exercise # 7.3 Exercise # 7.4 Exercise # 7.5 Exercise # 7.6 Exercise # 7.7 Exercise # 7.8 Exercise # 7.9 Exercise # 7.10 Exercise # 7.11 Miscellaneous Exercise
CHAPTER-7 : INTEGRAL
Q. No. 4, 6(ii), 10(ii), 12, 15, 21, 22(ii), 26, 30, 32, 36, 39, 40, 43 Q. No. 24, 36, 37, 38 Q. No. 14, 19, 22 Q. No. 7, 15, 19 Q. No. 6, 9, 12 16, 21 Q. No. 6, 10, 18, 20, 22 Q. No. 5, 9 Q. No. 6 Q. No. 18 Q. No. 8, 9, 10 Q. No. 8, 9, 12, 16, 19 Q. No. 5, 10, 18, 19, 21, 24, 31, 33, 40, 43, 44
CHAPTER-8 : AOI Examples Exercise # 8.1 Exercise # 8.2 Miscellaneous Examples Miscellaneous Exercise
Q. No. 7, 9, 10, 13 Q. No. 4, 6, 9 Q. No. 5 Q. No. 10, 11 Q. No. 15
Examples Exercise # 9.1 Exercise # 9.2 Exercise # 9.4 Exercise # 9.5 Exercise # 9.6 Miscellaneous Exercise
Q. No. 5, 7, 16, 20, 28 Q. No. 1, 12 Q. No. 6 Q. No. 4, 10 Q. No. 6, 8, 11, 13 Q. No. 5, 10, 15 Q. No. 6, 9, 11, 14
CHAPTER-9 : D.E.
CHAPTER-10 : VECTOR
Examples Exercise # 10.2 Exercise # 10.3 Exercise # 10.4 Mise. Exercise
Q. No. 8, 9, 17, 18, 20, 21, 27, 28 Q. No. 9, 11, 13 Q. No. 5, 6, 9, 13 Q. No. 4, 7, 9, 10, 11 Q. No. 9, 13, 15
Examples Exercise # 11.1 Exercise # 11.2 Exercise # 11.3 Miscellaneous Exercise
Q. No. 11, 21, 22, 24, 25 Q. No. 4, 5 Q. No. 6, 12, 15, 17 Q. No. 9 Q. No. 6, 7, 10, 13, 14, 15, 17, 18, 19, 20
Examples Exercise # 12.1 Exercise # 12.2 Miscellaneous Examples
Q. No. 6, 9, 10 Q. No. 4 Q. No. 3, 8 Q. No. 5, 8, 10
Examples Exercise # 13.1 Exercise # 13.2 Exercise # 13.3 Exercise # 13.4 Exercise # 13.5 Miscellaneous Examples
CHAPTER-11 : 3-D
CHAPTER-12 : LPP
CHAPTER-13 : PROBABILITY Q. No. 7,17,24,27,29,36 Q. No. 5,12,14 Q. No. 4,9,14 Q. No. 4,6,12 Q. No. 6,11 Q. No. 4,7,8,12 Q. No. 7,9,13,16 12
