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INTEGRATION
CONCEPT MAPPING:
CONCEPT MAPPING – INTEGRATION
1.Integration- as an inverse process of differentiation
𝒅
𝒅𝒙
(sinx) = cosx
𝒅 π’™πŸ‘
( ) = π’™πŸ
𝒅𝒙 πŸ‘
𝒅
𝒅𝒙
(𝒆𝒙 ) =𝒆𝒙
We observe that the function cosx is the derivative function of sinx.we say
that sinx is an antiderivative ( or an integral)of cosx.similarly
π’™πŸ‘
πŸ‘
and 𝒆𝒙 are
the antiderivatives (or integrals) of π’™πŸ and 𝒆𝒙 .we note that for any real
number c,treated as constant function,its derivative is0and hence we can
write
𝒅
𝒅𝒙
(sinx +c) = cosx,
𝒅 π’™πŸ‘
𝒅
𝒅𝒙 πŸ‘
𝒅𝒙
( + 𝒄) = π’™πŸ , (𝒆𝒙 + 𝒄) =𝒆𝒙
Thus antiderivative or integrals of the above functions are not
unique.Actually there exists infinitely many antiderivatives of each of these
functions which can be obtained by choosing C arbitrarily from the set of
real numbers( also called constant of integration).for this reason C is
customarily referred to as arbitrary constant.
More generally
𝒅
𝒅𝒙
(F(x) +c)= f(x),x belongs to I
2.Symbol
∫ 𝒇(𝒙)𝒅𝒙 -- Meaning
3.Comparison between differentiation and integration
a. Both are operations on function
b. All functions are not differentiable
c.All functions are not integrable
d .Derivative of a function when it exists it is unique function but the
integral of of a function is not so. However they are unique upto an
additive constant
ie.any two integrals of a function differ by a constant.
e. When a polynomial function P(x) is differentiated the result is a
polynomial whose degree is one less than the degree of p(x)
eg.
𝒅
𝒅𝒙
(π’™πŸ ) = 2x
When a polynomial function is integrated the result is a polynomial
whose Degree is one more than that of p(x)
π’™πŸ’
Eg.∫ π’™πŸ‘ dx = + c
πŸ’
f. The derivative of a function has a geometrical meaning- the slope of the
tangent to the corresponding curve at a point g.Integration- family of
curves placed parallel to each other having parallel tangents at the point of
intersection of the curves of the family ,with the lines perpendicular to the
axis representing the variable of integration.
4.Methods of integration
Methods of integration 1. Integration by direct method
↓
↓
Integration by
Integration by Substitution
parts
↓
Integration using
Partial fraction
↓
↓
Application of
Function
↓
Partial fraction
∫ 𝒖𝒅𝒗=uv-∫ 𝒗𝒅𝒖
𝒑𝒙+𝒒
Trigonometric function(𝒙−𝒂)(𝒙−𝒃)
(𝒑𝒙+𝒒)
In integrals
(𝒙−𝒂)𝟐
𝑨
𝑨
(𝒙−𝒂)
Integration of some
+
𝑩
(𝒙−𝒃)
+
π’‘π’™πŸ +𝒒𝒙+𝒓
-
+𝒃𝒙+𝒄
-
𝑨
(𝒙−𝒂)
𝑨
(𝒙−𝒂)
π’™πŸ ±π’‚πŸ
𝒅𝒙
∫ π’‚πŸ−π’™πŸ,∫ π’‚π’™πŸ+𝒃𝒙+𝒄
∫√
𝒅𝒙
π’‚π’™πŸ +𝒃𝒙+𝒄
,∫
(𝒙−𝒃)
(𝒙−𝒄)
𝒅𝒙
𝒅𝒙
𝑩
π‘ͺ
π’‘π’™πŸ +𝒒𝒙+𝒓
Particular function(𝒙−𝒂)(π’™πŸ
𝒅𝒙
(𝒙−𝒂)
+
𝑩
+
(𝒙−𝒂)𝟐 (𝒙−𝒃)
∫ π’™πŸ±π’‚πŸ,∫ √
𝑨
(𝒙−𝒂) (𝒙−𝒂)𝟐
π’‘π’™πŸ +𝒒𝒙+𝒓
↓ (𝒙−𝒂)(𝒙−𝒃)(𝒙−𝒄) -
-
(𝒑𝒙+𝒒)𝒅𝒙
π’‚π’™πŸ +𝒃𝒙+𝒄
,∫
SOME SPECIAL INTEGRALS
(𝒑𝒙+𝒒)𝒅𝒙
√π’‚π’™πŸ +𝒃𝒙+𝒄
+
𝑩
(𝒙−𝒂)𝟐
+
+
π‘ͺ
(𝒙−𝒃)
𝑩𝒙+π‘ͺ
π’™πŸ +𝒃𝒙+𝒄
1.∫ √π’‚πŸ ± π’™πŸ dx
2.∫ √π’™πŸ − π’‚πŸ dx
3.∫ √π’‚π’™πŸ + 𝒃𝒙 + 𝒄dx
4.∫(𝒑𝒙 + 𝒒)√π’‚π’™πŸ + 𝒃𝒙 + 𝒄dx
Definite Integrals
𝒃
∫𝒂 𝒇(𝒙)𝒅𝒙 - Area of the region bounded by the curve y=f(x) and the
ordinates x=a,x=b and x-axis
𝒃
Calculating ∫𝒂 𝒇(𝒙)𝒅𝒙
Find the indefinite integral ∫ 𝒇(𝒙)𝒅𝒙 .Let this be F(x)
𝒃
∫𝒂 𝒇(𝒙)𝒅𝒙 = F(b) –F(a)
Definite integrals
↓↓↓↓
Definite integrals based on
Definite integrals
Properties of
integration types of indefinite integrals as a limit of sum definite
integral of modulus fn .
Some important results.
1.∫ 𝒙𝒏 𝒅𝒙 =
𝒙𝒏+𝟏
𝒏+𝟏
+c
2.∫ πŸπ’…π’™ = x+c
3.∫ π’”π’Šπ’π’™ 𝒅𝒙 = -cosx +c
4.∫ 𝒄𝒐𝒔𝒙𝒅𝒙 = sinx +c
5.∫ π’”π’†π’„πŸ x dx = tanx + c
6.∫ π’„π’π’”π’†π’„πŸ x dx = -cotx +c
7.∫ 𝒔𝒆𝒄𝒙 𝒕𝒂𝒏𝒙 𝒅𝒙 = secx + c
8.∫ 𝒄𝒐𝒔𝒆𝒄𝒙 𝒄𝒐𝒕𝒙 𝒅𝒙 = - cosecx + c
9.∫
√
𝟏
𝟏− π’™πŸ
dx = -𝐜𝐨𝐬 −𝟏 𝒙 +c OR 𝐬𝐒𝐧−𝟏 𝒙 +c
𝟏
10.∫
dx =𝐭𝐚𝐧−𝟏 𝒙 +c OR -𝐜𝐨𝐭 −𝟏 𝒙 +c
𝟏+π’™πŸ
11.∫
𝒅𝒙
𝒙√π’™πŸ −𝟏
= 𝐬𝐞𝐜 −𝟏 𝒙 +c OR -𝐜𝐬𝐜 −𝟏 𝒙 +c
12.∫ 𝒆𝒙 dx = 𝒆𝒙 + c
𝟏
13.∫ dx =log x +c
𝒙
𝒂𝒙
14.∫ 𝒂𝒙 dx =
+c
π’π’π’ˆπ’‚
15.∫ 𝒆𝒙 ( f(x) +f ’(x) ) dx = 𝒆𝒙 f(x) +c
Some Properties of Definite Integrals
𝒃
𝒃
π‘·πŸŽ :∫𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒕)𝒅𝒕
𝒃
𝒂
π‘·πŸ :∫𝒂 𝒇(𝒙)𝒅𝒙 = − ∫𝒃 𝒇(𝒙)𝒅𝒙
𝒃
𝒄
𝒃
π‘·πŸ :∫𝒂 𝒇(𝒙)𝒅𝒙=∫𝒂 𝒇(𝒙)𝒅𝒙 + ∫𝒄 𝒇(𝒙)𝒅𝒙
𝒃
𝒃
π‘·πŸ‘ :∫𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒂 + 𝒃 − 𝒙)𝒅𝒙
𝒂
𝒂
π‘·πŸ’ ∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒂 − 𝒙)𝒅𝒙
πŸπ’‚
𝒂
𝒂
π‘·πŸ“ :∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒙)𝒅𝒙+∫𝟎 𝒇(πŸπ’‚ − 𝒙)𝒅𝒙
πŸπ’‚
𝒂
π‘·πŸ” :∫𝟎 𝒇(𝒙)𝒅𝒙 = 𝟐 ∫𝟎 𝒇(𝒙)𝒅𝒙,If f(2a-x)=f(x) =0 if f(2a-x)=-f(x)
π‘·πŸ•:(π’Š) ∫𝒂
−𝒂
𝒂
𝒇(𝒙)𝒅𝒙=𝟐 ∫𝟎 𝒇(𝒙)𝒅𝒙,
If f is an even function
𝒂
(ii)∫−𝒂 𝒇(𝒙)𝒅𝒙 = 𝟎 𝑰𝒇 π’Šπ’” 𝒂𝒏 𝒐𝒅𝒅 π’‡π’–π’π’„π’•π’Šπ’π’
Limit as a sum
𝒃
3. ∫𝒂 𝒇(𝒙)dx = π₯𝐒𝐦 𝒉 ( f(a) + f(a+h)+f (a+2h)+…………+f(a+(n-1)h)
𝒉→𝟎
∑ 𝒏 = 𝟏 + 𝟐 + πŸ‘ + πŸ’ + β‹― = n (n+1)/2
∑ π’πŸ = 𝟏𝟐 + 𝟐𝟐 + πŸ‘πŸ + πŸ’πŸ + β‹― + π’πŸ = n (n+1) (2n+1)/6
∑ π’πŸ‘ = πŸπŸ‘ + πŸπŸ‘ + πŸ‘πŸ‘ + πŸ’πŸ‘ + β‹― + π’πŸ‘ = [ 𝐧(𝐧 + 𝟏)/𝟐]𝟐
𝒂 + 𝒂𝒓 + π’‚π’“πŸ + π’‚π’“πŸ‘ + β‹― + 𝒂𝒓𝒏 = a(𝒓𝒏 − 𝟏)/(𝒓 − 𝟏)
r≠ 1
LEVEL 1.
𝟏
√𝟐
1. Evaluate:∫𝟎
𝟏
𝒅𝒙.
√𝟏−π’™πŸ
𝝅
ANS. πŸ’
2. Evaluate:∫
𝒄𝒐𝒔√𝒙
√𝒙
3. Evaluate:∫
4.
𝐝𝐱.
π’”π’Šπ’√𝒙
√𝒙
Ans.
𝒅𝒙.
Write a value of ∫
Ans.
𝟏
√πŸ’−π’™πŸ
𝒅𝒙. Ans. 𝐬𝐒𝐧−𝟏 πŸπ’™ + 𝒄
5. Find f(x) satisfying the following :∫ 𝒆𝒙 (π’”π’†π’„πŸ 𝒙 + 𝐭𝐚𝐧 𝒙)𝒅𝒙 = 𝒆𝒙 𝒇(𝒙) + 𝒄.
Ans. F(x) = tan x
πŸ“π’™
6. Evaluate : ∫ πŸ“π’™ 𝒅𝒙. Ans. π’π’π’ˆπŸ“ + 𝒄
𝝅/𝟐
7. Write the value of ∫−𝝅/𝟐 π’”π’Šπ’πŸ“ 𝒙𝒅𝒙 Ans. zero
8.
(𝒂𝒙+𝒃)
∫(𝒂𝒙 + 𝒃)πŸ‘ 𝒅𝒙 Ans. πŸ’π’‚
9.
Evaluate : ∫
√
10.
Evaluate : ∫
𝟐
𝒅𝒙
𝟏−π’™πŸ
(π₯𝐨𝐠 𝒙)𝟐
𝒙
+𝒄
Ans. 𝐬𝐒𝐧−𝟏 𝒙 +
𝒅𝒙
11. Evaluate:∫𝟎 √πŸ’ − π’™πŸ 𝒅𝒙
𝒙
πŸ’
𝒄
Ans. (π’π’π’ˆπ’™)πŸ‘ + 𝒄
Ans
𝝅
x
12. Given ∫ 𝒆 (tan x + 1) sec x dx = e f(x) + c. then f(x) =? Ans. secx
LEVEL II:
πœ‹ π‘₯ 𝑠𝑖𝑛π‘₯
𝑑π‘₯
1+π‘π‘œπ‘ 2 π‘₯
1. Evaluate.∫0
πœ‹
π‘₯ tan π‘₯
𝑑π‘₯
2.
Evaluate: ∫0
3.
Evaluate: ∫02 √π‘‘π‘Žπ‘›π‘₯ + √π‘π‘œπ‘‘π‘₯
sec π‘₯ π‘π‘œπ‘ π‘’π‘ π‘₯
πœ‹
πœ‹
4. Evaluate:∫02 π‘™π‘œπ‘” sin π‘₯ 𝑑π‘₯ .
𝑒π‘₯
5.Evaluate:∫ √5−4𝑒 π‘₯ 2π‘₯ dx
−𝑒
(π‘₯−4)
6. Evaluate: ∫
𝑒 π‘₯ 𝑑π‘₯
(π‘₯−2)3
πœ‹
7. Evaluate :∫0
8. Evaluate: ∫
𝑒 π‘π‘œπ‘ π‘₯
𝑑π‘₯
𝑒 π‘π‘œπ‘ π‘₯ +𝑒 −π‘π‘œπ‘ π‘₯
𝑑π‘₯
√5−4π‘₯−π‘₯ 2
9. Evaluate:∫ π‘₯ sin−1 π‘₯ 𝑑π‘₯
LEVEL III:
π‘Ž
π‘Ž−π‘₯
1. Evaluate: ∫−π‘Ž √
𝑑π‘₯
π‘Ž+π‘₯
πœ‹ π‘₯𝑠𝑖𝑛π‘₯
2. Evaluate: ∫0
1+π‘π‘œπ‘  2 π‘₯
𝑑π‘₯
3.
1
Evaluate: ∫0 cot −1 (1 − π‘₯ + π‘₯ 2 ) 𝑑π‘₯
πœ‹
4. Evaluate: ∫0
π‘₯ π‘‘π‘Žπ‘›π‘₯
𝑠𝑒𝑐π‘₯+π‘‘π‘Žπ‘›π‘₯
𝑑π‘₯
Evaluate: ∫
π‘₯ 𝑑π‘₯
π‘Ž2 π‘π‘œπ‘ 2 π‘₯+𝑏2 𝑠𝑖𝑛2 π‘₯
πœ‹
𝑑π‘₯
3
πœ‹
1+√π‘‘π‘Žπ‘›π‘₯
6
5. Evaluate: ∫
π’™πŸ’
6. Evaluate:∫ (𝒙−𝟏)(π’™πŸ 𝒅𝒙
+𝟏)
7.
1
1
Evaluate:∫0 log ( − 1) 𝑑π‘₯
π‘₯
9. Evaluate: ∫ π‘₯ 2 tan−1 π‘₯ 𝑑π‘₯
3π‘₯+5
10. Evaluate: ∫ 2
𝑑π‘₯
√π‘₯ −8π‘₯+7
3
11. Evaluate ∫1 (3π‘₯ 2 + 2π‘₯) 𝑑π‘₯ using limit as sum.
11.
LEVEL 2: PRACTICE QUESTIONS
πœ‹
2
1. Evaluate:∫0
π‘₯
sin π‘₯+cos π‘₯
𝑑π‘₯
2. Evaluate : ∫
𝑑π‘₯.
√5−4π‘₯−2π‘₯ 2
−1
3. Evaluate: ∫ π‘₯ 𝑠𝑖𝑛 π‘₯ 𝑑π‘₯
4. Evaluate:∫
(sin π‘₯+ cos π‘₯)
πœ‹
5.Evaluate :∫0
6. Evaluate :∫
dx
√sin 2π‘₯
π‘₯ tan π‘₯
sec π‘₯ π‘π‘œπ‘ π‘’π‘ π‘₯
𝑑π‘₯ .
1
cos(π‘₯−π‘Ž)cos(π‘₯−𝑏)
2
𝑑π‘₯
7. Evaluate : ∫ π‘₯(π‘™π‘œπ‘” π‘₯) 𝑑π‘₯.
8. Evaluate : ∫
π‘₯
π‘₯ 3 −1
𝑑π‘₯
πœ‹
9. Using properties of definite integral , evaluate:∫0
10. Evaluate:∫
𝒙+𝟐
√(𝒙−𝟐)(𝒙−πŸ‘)
πŸ“π’™πŸ
𝟐
11. Evaluate :∫𝟏
π’™πŸ +πŸ’π’™+πŸ‘
𝒅𝒙
𝒅𝒙
𝟏+𝐬𝐒𝐧 πŸπ’™
12. Evaluate: ∫ π’†πŸπ’™ (𝟏+𝐜𝐨𝐬 πŸπ’™)𝒅𝒙.
𝟏
13. Evaluate: ∫ 𝐬𝐒𝐧 𝒙−𝐬𝐒𝐧
𝒅𝒙.
πŸπ’™
πœ‹/2 π‘₯+sin π‘₯
14. Evaluate :∫0
πœ‹/4
15. Evaluate :∫0
1+cos π‘₯
𝑑π‘₯ .
log(1 + tan π‘₯) 𝑑π‘₯
π‘₯
4−π‘π‘œπ‘  2 π‘₯
𝑑π‘₯.
LEVEL 3: PRACTICE QUESTIONS
1. Evaluate:∫
π’™πŸ +𝒙+𝟏
(𝒙+𝟐)(π’™πŸ +𝟏)
𝐝𝐱
𝟏
2. Evaluate: ∫
𝐬𝐒𝐧 𝒙(πŸ“−πŸ’ 𝐜𝐨𝐬 𝒙)
πŸ“π’™+πŸ‘
3. Evaluate ∫
𝒅𝒙.
𝒅𝒙 .
√π’™πŸ +πŸ’π’™+𝟏𝟎
𝟏−√𝒙
4. Evaluate: ∫ √𝟏+ 𝒙 𝒅𝒙.
√
π‘Ίπ’Šπ’ 𝒙+𝐜𝐨𝐬 𝒙
5. Evaluate: ∫
√𝐬𝐒𝐧 𝒙 .𝐜𝐨𝐬 𝒙
𝒅𝒙.
𝟐
𝒙 +𝟏
6. Evaluate: ∫ 𝒆𝒙 ((𝒙+𝟏)
𝟐 )𝒅𝒙.
𝟐
7. Evaluate:∫𝟏 (π’™πŸ + 𝒙 + 𝟐)dx as a limit of sums.
𝟏
8. Evaluate :∫𝟎 𝐬𝐒𝐧−𝟏(𝒙√𝟏 − 𝒙 − √𝒙√𝟏 − π’™πŸ ) 𝒅𝒙 , 𝟎 ≤ 𝒙 ≤ 𝟏.
πŸ‘
9. Evaluate :∫𝟏 (πŸπ’™πŸ + πŸ‘)𝒅𝒙 as limit of sums.
𝝅
𝒙 𝐬𝐒𝐧 𝒙 𝐜𝐨𝐬 𝒙
10. Evaluate :∫𝟎𝟐 π’”π’Šπ’πŸ’π’™+π’„π’π’”πŸ’π’™ 𝒅𝒙.
𝝅/πŸ‘
11. Evaluate :∫𝝅/πŸ”
12. Evaluate:∫
𝒅𝒙
𝟏+√𝒕𝒂𝒏𝒙
πŸ”π’™+πŸ•
√(𝒙−πŸ“)(𝒙−πŸ’)
𝟏
𝒅𝒙.
𝟏
13. Evaluate:∫𝟎 π₯𝐨𝐠 ( − 𝟏) 𝒅𝒙.
𝒙
14. Evaluate: ∫
𝒙 𝐬𝐒𝐧−𝟏 𝒙
𝒅𝒙
√𝟏−π’™πŸ
(π’™πŸ +𝟏)(π’™πŸ +πŸ’)
15. Evaluate :∫ (π’™πŸ+πŸ‘)(π’™πŸ−πŸ“) 𝒅𝒙.
𝟐
𝟏
16. Evaluate:∫𝟎 𝒙(𝒕𝒂𝒏 −𝟏 𝒙) 𝒅𝒙.
𝒙
17. Evaluate:∫πŸŽπ’‚ 𝐬𝐒𝐧−𝟏 √𝒂+𝒙
𝒅𝒙.
18. Evaluate: ∫
π₯𝐨𝐠(𝐬𝐒𝐧 𝒙)
𝐭𝐚𝐧 𝒙
𝒅𝒙
−𝟏 𝒙
π’†π­πšπ§
19. Evaluate:∫
dx.
(𝟏+π’™πŸ )𝟐
STRATEGIES TO ACHIEVE QUALITATIVE AND QUANTITAIVE RESULTS IN MATHEMATICS
CLASS XII
Strategies for Slow learners:
1. Identify the slow learners at the beginning of the year. Set achievable targets
and motivate them throughout the year so that they will not be depressed and
discouraged.
2. Question papers of last five years (both main and supplementary examinations)
are to be collected and the list out all repeated, important concepts/problems.
The slow learners are to be given sufficient practice in these areas/concepts.
3. The strengths and weaknesses are to be diagnosed in these areas. Thorough
revision in these concepts is to be given by conducting frequent slip tests and
re-teaching.
4. Preparation of Question-wise analysis of each examination including slip tests to
be done to locate the weak areas and thorough revision is to be conducted.
5. Collect the drilling problems of a particular concept, and solve two or three
problems in the class.
Then allow the slow learners to solve the remaining
problems as per their capacity to attain a good command and confidence over
that particular method/type (Drilling Exercises).
6. Three model papers based on the Sample Papers issued by CBSE (SET I, II,
III) along with marking scheme should be prepared by the teacher. Copies of
these papers are to be issued to all the slow learners. This will help the child to
know the type of questions/methods important for board exams. They will get
more confidence to face the board exam.
7. Concept wise, specially designed home assignments are to be given to students
daily.
The assignments are to be corrected by giving proper suggestions in
front of students.
8. After the completion of each concept/topic allow the low achiever to solve the
problem pertaining to that method. If possible every day at least one low
achiever should come on to the board to solve a problem.
9. Whenever possible, teach Mathematics by using PP Presentations in an effective
way.
10. Weekly test pertaining to these formulae has to be conducted regularly.
11. The students have to be asked to read the entire text book thoroughly.
12. The students are to be made aware about the chapter wise distribution of
marks or marking scheme.
13. Sufficient tips should be given for time management.
14. Few easy topics are to be identified from examination point of view and are to
be assigned to the slow learners. The slow learners are to be prepared for
reduced, identified syllabus.
Strategies for bright and Gifted Student:
15. Bright Children are the back bones to improve the overall Performance Index of
the Vidyalaya. So they should be encouraged by providing concepts wise HOTS
questions. They should be encouraged to solve more challenging questions which
have more concepts and challenging tasks. More thought provoking questions
are to be collected and a question bank is to be given to gifted students to
develop their analyzing and reasoning capabilities.
16. Instead of preparing the PP presentation by the teacher, better to handover all
the necessary content to the students and ask the bright students, to prepare
one PPT each. After submission of completed PP Presentation, check the PPT
and the same can be used effectively in the teaching learning process.
17. On completion of syllabus topic wise revision plan is to be framed for both slow
learners and gifted students.
18. The students have to be asked to read the entire text book thoroughly.
19. The students are to be made aware about the chapter wise distribution of
marks or marking scheme.
20. Sufficient tips should be given for time management.
Revision Plan:
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After completion of coverage of syllabus, proper revision plan is to be prepared
Concept-wise
(questions
for
slow
learners/gifted
students),
HOTS
questions/optional exercises (for gifted students) is to be prepared and given to
the students.
Minimum learning programme for slow learners is to be prepared and
identified/reduced syllabus is to assigned to slow learners.
CBSE Board pattern question papers (at least 10 papers should be solved)
CBSE Board papers 2010
(3 sets)
CBSE Board Compartment Paper 2011
(1 set)
CBSE Board papers 2012. 2013, 2014
(3 sets)
CBSE Board Compartment Paper 2013
(1 set)
Common Pre-board Board Examination 2013, 2014 (2 sets)
CBSE sample papers
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