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Q: 3. How can I apply for the KVPY exam?

Ans. To apply for the KVPY exam, follow these steps:1. Visit the official website of KVPY.2. Click on the 'Registration' or 'Apply Online' link.3. Fill in the required details such as personal information, educational qualifications, and contact details.4. Upload the necessary documents, such as a scanned photograph and signature.5. Pay the application fee online through the available payment methods.6. Review the application form and submit it.7. Take a printout of the submitted application form for future reference.

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 Page 1


M a thema tics
M a thema tics
Stream : SA
KVPY
Page 2


M a thema tics
M a thema tics
Stream : SA
KVPY
h
 a pter
Contents
KVPY MATHEMATICS
01. NUMBER SYSTEM 1
02. ALGEBRA 3
03. TRIGONOMETRY 10
04. INEQUALITIES 12
05. GEOMETRY 14
Page 3


M a thema tics
M a thema tics
Stream : SA
KVPY
h
 a pter
Contents
KVPY MATHEMATICS
01. NUMBER SYSTEM 1
02. ALGEBRA 3
03. TRIGONOMETRY 10
04. INEQUALITIES 12
05. GEOMETRY 14
node05\B0B0-BA\KVPY\KVPY Maths Module\01-Maths [Th].P65
Mathematics ????? 1
E
 For integers a, b, c
(i) If a | b then a | b c.
(ii) If. a | b and b | c then a | c.
(iii) If a, b are natural numbers and if a | b and b | a, then a = b
(iv) If a | b and a | c then a | (b p + c q) for all integral values of p and q.
b | a means b is a factor of a (or a is a multiple of b).
 Number of divisors of a natural number
If N =
3 12k
n nnn
123k
p p p .......p then the number of divisors of N, d(N) = (n
1
 +1) (n
2
+1) (n
3
+1)..... (n
k
+1)
 Sum of divisors of a natural number? (N)
?(
1 2k
n 1 n 1 n1
1 2k
1 2k
1 p 1p 1p
N ..........
1 p 1p 1p
* **
,,,
?<
,,,
 Product of divisors of N
Product of divisors of N =
1 2k
1
(n 1)(n 1).......( n 1)
2
N
* **
[d(N) divisors of N can be grouped in to
1
2
d(N) pairs such that product of each pair is equal to n.]
 Perfect number
A natural number N is said to be perfect if sum of all divisors of N, ?(N) = 2N
Eg:– (i) 28 is a perfect number.
1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 × 28
(ii) 496 is a perfect number.
 496 = 2
4
 × 31
? (
52
12 131
496 992 2 496
1 2 1 31
,,
? < = < <=
,,
 Sum of the reciprocals of divisors of a perfect number
Let d
1
, d
2
, d
3
,......, d
n
 be all divisors of the perfect number N[including 1 and N].
12n
123n
d d .......... d 1 1 1 1 2N
........ 2
ddd d NN
***
*** * < < <
[L.C.M. of d
1
, d
2
, d
3
,.........., d
n
 = N and d
1
 + d
2
 + d
3
 +........ + d
n
 = 2N]
 Sum of reciprocals of factors of 28
1 1 1 1 1 1 28 14 7 4 2 1 56
2
1 2 4 7 14 28 28 28
* * * **
** ** * < < <
 Amicable numbers
Two numbers are said to be amicable if the sum of the divisors of one, excluding itself is equal to the other.
220 and 284 are amicable numbers.
220 = 1 + 2 + 4 + 7 1 +142
284 = 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110
Sum of divisors of 220 (excluding 220) is 284 and Sum of divisors of 284(excluding 284) is 220.
CHAPTER # 01 NUMBER SYSTEM
Page 4


M a thema tics
M a thema tics
Stream : SA
KVPY
h
 a pter
Contents
KVPY MATHEMATICS
01. NUMBER SYSTEM 1
02. ALGEBRA 3
03. TRIGONOMETRY 10
04. INEQUALITIES 12
05. GEOMETRY 14
node05\B0B0-BA\KVPY\KVPY Maths Module\01-Maths [Th].P65
Mathematics ????? 1
E
 For integers a, b, c
(i) If a | b then a | b c.
(ii) If. a | b and b | c then a | c.
(iii) If a, b are natural numbers and if a | b and b | a, then a = b
(iv) If a | b and a | c then a | (b p + c q) for all integral values of p and q.
b | a means b is a factor of a (or a is a multiple of b).
 Number of divisors of a natural number
If N =
3 12k
n nnn
123k
p p p .......p then the number of divisors of N, d(N) = (n
1
 +1) (n
2
+1) (n
3
+1)..... (n
k
+1)
 Sum of divisors of a natural number? (N)
?(
1 2k
n 1 n 1 n1
1 2k
1 2k
1 p 1p 1p
N ..........
1 p 1p 1p
* **
,,,
?<
,,,
 Product of divisors of N
Product of divisors of N =
1 2k
1
(n 1)(n 1).......( n 1)
2
N
* **
[d(N) divisors of N can be grouped in to
1
2
d(N) pairs such that product of each pair is equal to n.]
 Perfect number
A natural number N is said to be perfect if sum of all divisors of N, ?(N) = 2N
Eg:– (i) 28 is a perfect number.
1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 × 28
(ii) 496 is a perfect number.
 496 = 2
4
 × 31
? (
52
12 131
496 992 2 496
1 2 1 31
,,
? < = < <=
,,
 Sum of the reciprocals of divisors of a perfect number
Let d
1
, d
2
, d
3
,......, d
n
 be all divisors of the perfect number N[including 1 and N].
12n
123n
d d .......... d 1 1 1 1 2N
........ 2
ddd d NN
***
*** * < < <
[L.C.M. of d
1
, d
2
, d
3
,.........., d
n
 = N and d
1
 + d
2
 + d
3
 +........ + d
n
 = 2N]
 Sum of reciprocals of factors of 28
1 1 1 1 1 1 28 14 7 4 2 1 56
2
1 2 4 7 14 28 28 28
* * * **
** ** * < < <
 Amicable numbers
Two numbers are said to be amicable if the sum of the divisors of one, excluding itself is equal to the other.
220 and 284 are amicable numbers.
220 = 1 + 2 + 4 + 7 1 +142
284 = 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110
Sum of divisors of 220 (excluding 220) is 284 and Sum of divisors of 284(excluding 284) is 220.
CHAPTER # 01 NUMBER SYSTEM
node05\B0B0-BA\KVPY\KVPY Maths Module\01-Maths [Th].P65
2
KVPY ?????
E
 Highest power of a prime number p contained in N!
The highest power of a prime number p contained in N ! = 23
N NN
.........
p pp
? ?? ?? ?
* **
? ?? ?? ?
? ?? ?? ?
 Cube Root:
3
c b a *
 =
3
c b,
 +
c
 = x + y , where x
3
 + 3xy = a
Note: 1) Above formula can be applied only if x
3
 + 3xy = a
2)
3
y ) y x 3 ( x ) y 3 x ( * * * = y x *
 Important Results:
(1)
3 3
b a
1
*
 =
2/3 1/3 2/3
a (ab)b
ab
,*
*
(2)
3 3
b a
1
,
 =
2/3 1/3 2/3
a (ab)b
ab
**
,
(3) If
k x
2
) b a (
,
*
 +
k x
2
) b a (
,
,
 = 2a and a
2
 – b = 1, then x
2
 – k = ± 1
(4) If
k x
2
) b a (
,
*
 +
k x
2
) b a (
,
,
  = 2(a
2
 + b), then 1
2
k x
2
° <
,
(5)
/ * * * ...... a a a
 =
2
1 a 4 1 * *
 (a > 0)
(6)
/ , , , ...... a a a
 =
2
1 1 a 4 , *
 (a > 0)
(7)
/ ....... a a a
 = a
(8) times n ....... a a a =
n
n
2
1 2
a
,
 Greatest integer function :
(a) [x] represents the greatest integer less than or equal to x.
f(x) = [x] is called the greatest integer function.
(b) {x} is fractional part of x and is defined as {x} = x – [x].
 Properties of Greatest integer function :
(a) [x]' x < [x] + 1 and x –1 < [x]' x, 0' x (b) If x? 0, [x] = 1p ix ''
?
(c) [x + m] = [x] + m, if m is an integer. (d) [x] + [y] ' [x + y]' [x] + [y] + 1
(e) [x] + [–x] = 0, if x is an integer and –1 other wise.
(f) The number of positive integers less than or equal to n and divisible by m is given by
n
m
?)
??
??
(g) If p is prime number and e is the largest exponent of p such that p
e
 /n! then i
i1
n
e
p
/
<
??
<
??
??
?
(h) The number of zeroes at the end of n! is given by the least of the Highest powers of 2 and 5.
Page 5


M a thema tics
M a thema tics
Stream : SA
KVPY
h
 a pter
Contents
KVPY MATHEMATICS
01. NUMBER SYSTEM 1
02. ALGEBRA 3
03. TRIGONOMETRY 10
04. INEQUALITIES 12
05. GEOMETRY 14
node05\B0B0-BA\KVPY\KVPY Maths Module\01-Maths [Th].P65
Mathematics ????? 1
E
 For integers a, b, c
(i) If a | b then a | b c.
(ii) If. a | b and b | c then a | c.
(iii) If a, b are natural numbers and if a | b and b | a, then a = b
(iv) If a | b and a | c then a | (b p + c q) for all integral values of p and q.
b | a means b is a factor of a (or a is a multiple of b).
 Number of divisors of a natural number
If N =
3 12k
n nnn
123k
p p p .......p then the number of divisors of N, d(N) = (n
1
 +1) (n
2
+1) (n
3
+1)..... (n
k
+1)
 Sum of divisors of a natural number? (N)
?(
1 2k
n 1 n 1 n1
1 2k
1 2k
1 p 1p 1p
N ..........
1 p 1p 1p
* **
,,,
?<
,,,
 Product of divisors of N
Product of divisors of N =
1 2k
1
(n 1)(n 1).......( n 1)
2
N
* **
[d(N) divisors of N can be grouped in to
1
2
d(N) pairs such that product of each pair is equal to n.]
 Perfect number
A natural number N is said to be perfect if sum of all divisors of N, ?(N) = 2N
Eg:– (i) 28 is a perfect number.
1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 × 28
(ii) 496 is a perfect number.
 496 = 2
4
 × 31
? (
52
12 131
496 992 2 496
1 2 1 31
,,
? < = < <=
,,
 Sum of the reciprocals of divisors of a perfect number
Let d
1
, d
2
, d
3
,......, d
n
 be all divisors of the perfect number N[including 1 and N].
12n
123n
d d .......... d 1 1 1 1 2N
........ 2
ddd d NN
***
*** * < < <
[L.C.M. of d
1
, d
2
, d
3
,.........., d
n
 = N and d
1
 + d
2
 + d
3
 +........ + d
n
 = 2N]
 Sum of reciprocals of factors of 28
1 1 1 1 1 1 28 14 7 4 2 1 56
2
1 2 4 7 14 28 28 28
* * * **
** ** * < < <
 Amicable numbers
Two numbers are said to be amicable if the sum of the divisors of one, excluding itself is equal to the other.
220 and 284 are amicable numbers.
220 = 1 + 2 + 4 + 7 1 +142
284 = 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110
Sum of divisors of 220 (excluding 220) is 284 and Sum of divisors of 284(excluding 284) is 220.
CHAPTER # 01 NUMBER SYSTEM
node05\B0B0-BA\KVPY\KVPY Maths Module\01-Maths [Th].P65
2
KVPY ?????
E
 Highest power of a prime number p contained in N!
The highest power of a prime number p contained in N ! = 23
N NN
.........
p pp
? ?? ?? ?
* **
? ?? ?? ?
? ?? ?? ?
 Cube Root:
3
c b a *
 =
3
c b,
 +
c
 = x + y , where x
3
 + 3xy = a
Note: 1) Above formula can be applied only if x
3
 + 3xy = a
2)
3
y ) y x 3 ( x ) y 3 x ( * * * = y x *
 Important Results:
(1)
3 3
b a
1
*
 =
2/3 1/3 2/3
a (ab)b
ab
,*
*
(2)
3 3
b a
1
,
 =
2/3 1/3 2/3
a (ab)b
ab
**
,
(3) If
k x
2
) b a (
,
*
 +
k x
2
) b a (
,
,
 = 2a and a
2
 – b = 1, then x
2
 – k = ± 1
(4) If
k x
2
) b a (
,
*
 +
k x
2
) b a (
,
,
  = 2(a
2
 + b), then 1
2
k x
2
° <
,
(5)
/ * * * ...... a a a
 =
2
1 a 4 1 * *
 (a > 0)
(6)
/ , , , ...... a a a
 =
2
1 1 a 4 , *
 (a > 0)
(7)
/ ....... a a a
 = a
(8) times n ....... a a a =
n
n
2
1 2
a
,
 Greatest integer function :
(a) [x] represents the greatest integer less than or equal to x.
f(x) = [x] is called the greatest integer function.
(b) {x} is fractional part of x and is defined as {x} = x – [x].
 Properties of Greatest integer function :
(a) [x]' x < [x] + 1 and x –1 < [x]' x, 0' x (b) If x? 0, [x] = 1p ix ''
?
(c) [x + m] = [x] + m, if m is an integer. (d) [x] + [y] ' [x + y]' [x] + [y] + 1
(e) [x] + [–x] = 0, if x is an integer and –1 other wise.
(f) The number of positive integers less than or equal to n and divisible by m is given by
n
m
?)
??
??
(g) If p is prime number and e is the largest exponent of p such that p
e
 /n! then i
i1
n
e
p
/
<
??
<
??
??
?
(h) The number of zeroes at the end of n! is given by the least of the Highest powers of 2 and 5.
node05\B0B0-BA\KVPY\KVPY Maths Module\01-Maths [Th].P65
Mathematics ????? 3
E
 Number of functions from A to B:
Let n(A) = m and n(B) = n. Under a function f, an element of A can be associated to any of the
n elements of the set B.
i.e., each element of A can be associated to an element of B in n ways.
And A contains m elements.
Therefore total number of functions from A to B =
m
m times
n n n ...... n n === = <
033 31333 2
NOTE : Number of functions from A to A = [n(A)]
n(A)
 Finding remainders when divisor is of second or third degree.
Remainder is always one degree less than divisor.
If the divisor is linear expression of the form (ax + b), the remainder is a numerical number i.e. a constant.
But when the divisor is of second degree like (x – a)(x – b) or third degree like (x – a)(x – b) (x – c)
the remainders will be (px + q) or (px
2
 + qx + r) respectively.
In the above cases, the division identity will be
f(x) = (x – a)(x – b)e (x) + (px + q) (or)
f(x) = (x – a)(x – b)(x – c)e (x) + (px
2
 + qx + r)
We can find the value of p, q and r by substituting x = a, x = b and x = c in the above identities. We
shall get equations involving p, q and r.
The values of p, q, r will enable us to write the required remainder.
Using factor theorem the following concepts can be proved.
1. (a
k
 + b
k
) is divisible by (a + b) if k is odd.
2. (a
k
 – b
k
) is divisible by (a – b) if k ? N
Many problems can be solved using these two concepts, but they need tactful algebraical manipulation
as can be seen from the following examples.
As per binomial theorem
(1 + x)
n
= 1 +
n
C
1
x
1
 +
n
C
2
x
2
 + .....+
n
C
n
x
n
= 1 + nx +
n
C
2
 x
2
 + ....+
n
C
n
x
n
 (
n
C
1
 = n)
This concept is used occasionally in coordination with the concepts given in the above model.
CHAPTER # 02 ALGEBRA

FAQs on MATHEMATICS FOR KVPY - MBBS

1. What is the eligibility criteria for KVPY exam?

Ans. The eligibility criteria for the KVPY exam include: - The student should be an Indian national. - For Stream SA: Students studying in Class 11 with Science stream are eligible. - For Stream SX: Students studying in Class 12 with Science stream are eligible. - For Stream SB: Students enrolled in the first year of undergraduate course in Basic Sciences (B.Sc./B.S./B.Stat./B.Math./Int. M.Sc./M.S.) are eligible. - The student must have scored a minimum of 75% marks in aggregate in Mathematics and Science subjects in Class 10 (65% for SC/ST/PWD category). - For Stream SB, the minimum marks requirement is 60% (50% for SC/ST/PWD category) in Mathematics and Science subjects in Class 12.

2. What is the syllabus for the KVPY exam?

Ans. The syllabus for the KVPY exam includes topics from Physics, Chemistry, Mathematics, and Biology. The syllabus is based on the curriculum of Class 10, 11, and 12. Some of the topics covered in the syllabus include mechanics, thermodynamics, organic chemistry, algebra, geometry, and genetics. It is important for candidates to thoroughly study these topics to perform well in the exam.

3. How can I apply for the KVPY exam?

Ans. To apply for the KVPY exam, follow these steps: 1. Visit the official website of KVPY. 2. Click on the "Registration" or "Apply Online" link. 3. Fill in the required details such as personal information, educational qualifications, and contact details. 4. Upload the necessary documents, such as a scanned photograph and signature. 5. Pay the application fee online through the available payment methods. 6. Review the application form and submit it. 7. Take a printout of the submitted application form for future reference.

4. Is there any negative marking in the KVPY exam?

Ans. Yes, there is negative marking in the KVPY exam. For every incorrect answer, 0.25 marks will be deducted from the total score. However, there is no negative marking for unanswered questions. It is important for candidates to carefully attempt the questions and avoid random guessing to minimize the negative marking.

5. How can I prepare for the KVPY exam effectively?

Ans. To prepare effectively for the KVPY exam, follow these tips: 1. Understand the exam pattern and syllabus thoroughly. 2. Create a study schedule and allocate time for each subject. 3. Practice previous years' question papers and sample papers to get acquainted with the exam pattern and types of questions. 4. Refer to standard textbooks and study materials recommended by experts. 5. Take regular mock tests to assess your preparation level and identify areas of improvement. 6. Seek guidance from teachers, mentors, or coaching institutes, if required. 7. Stay focused, maintain a positive mindset, and revise regularly to strengthen your concepts.

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