what are the laws of addition,subtraction of integers?
associative law of addition:
a + (b+c) = (a+b) + c

commutative law of addition:
a + b = b + a

additive inverse:
a + (-a) = 0

zero property:
a + 0 = a

subtraction is basically the same thing as addition.
a - b = a + (-b)
instead of subtracting b, you add its additive inverse.
write a pair of intergers whose sum gives let i = `(1)/(sinx*(3 + 2cosx))*dx` = `int sinx/(sin^2x*(3 + 2cosx))*dx` = `int sinx/((1 - cos^2x)(3 + 2cosx))*dx` = `int sinx/((1 - cosx)(1 + cosx)(3 + 2cosx))*dx` put cos x = t∴ – sinx.dx = dt∴ sinx.dx = – dt ∴ i = `int (1)/((1 - t)(1 + t)(3 + 2t))*(-dt)` = `int (-1)/((1 - t)(1 + t)(3 + 2t))*dt` let `(-1)/((1 - t)(1 + t)(3 + 2t)) = "a"/(1 - t) + "b"/(1 + t) + "c"/(3 + 2t)` ∴ – 1 = a(1 + t)(3 + 2t) + b(1 - t)(3 + 2t) + c(1 - t)(1 + t)put 1 – t = 0, i.e. t = 1, we get– 1 = a(2)(5) + b(0)(5) + c(0)(2) ∴ – 1 = 10a ∴ a = `(-1)/(10)`put 1 + t = 0, i.e. t = – 1, we get– 1 = a(0)(1) + b(2)(1) + c(2)(0) ∴ – 1 = 2b ∴ b = `-(1)/(2)` put 3 + 2t = 0, i.e. t = `-(3)/(2)`, we get – 1 = `"a"(-1/2)(0) + "b"(5/2)(0) + "c"(5/2)(-1/2)` ∴ – 1 = `-(5)/(4)"c"` ∴ c = `(4)/(5)` ∴ `(-1)/((1 - t)(1 + t)(3 + 2t)) = (((-1)/(10)))/(1 - t) + ((-1/2))/(1 + t) + ((4/5))/(3 + 2t)` ∴ i = `int [(((-1)/10))/(1 - t) + ((-1/2))/(1 + t) + ((4/5))/(3 + 2t)]*dt` = `-(1)/(10) int 1/(1 - t)*dt - (1)/(2) int 1/(1 + t)*dt + (4)/(5) int 1/(3 + 2t)*dt` = `-(1)/(10) (log|1 - t|)/(-1) - (1)/(2) log | 1 + t| + 4/5 (log|3 + 2t|)/(2) + c` = `(1)/(10)log|1 - cosx| - (1)/(2)log|1 + cosx| + (2)/(5)log|3 + 2cos| + c`. a - ii, b - iv, c - i, d - iii multiply 2 in both lhs and rhs,integrating on both the sides, 2, 0 ( any two integers with one of them as 0). >> [x,y] = ismember(a,b);
>> [~,z] = sort(y(x));
>> v = find(x);
>> r = reshape(v(z),2,[]).'
r =
 2 9
 4 8
 5 7 use

summation of series of natural numbers -


 

- wherein

sum of first n natural numbers



 

 and

 

summation of series of natural numbers -


 

- wherein

sum of  squares of first n natural numbers



 

 

3+7+14+24+37 15k

let tn=an2+bn+c            because difference is in a.p: [4,7,10,13]

at n=1     (i)

    n=2      (ii)

    n=3        (iii)

    (ii)-(i)        (iv)

    (iii)-(ii)       (v)

    (v)-(iv)      

                    

 

from (i) 

from (ii) 







 



      

 

put n=15   

               

               

               

               

 




option 1)

126

this option is incorrect

option 2)

122

this option is correct

option 3)

81

this option is incorrect

option 4)

119

this option is incorrect (a) for digit



expression : 5n + 1

where n = no of digits

for 5 digits

no of segments required = 5(5) + 1 = 25 + 1 = 26

for 10 digits

no of segments required = 5(10) + 1 = 50 + 1 = 51

for 100 digits

no of segments required = 5(100) + 1 = 500 + 1 = 501

(b) for digit



expression : 3n + 1

where n = no of digits

for 5 digits

no of segments required = 3(5) + 1 = 15 + 1 = 16

for 10 digits

no of segments required = 3(10) + 1 = 30 + 1 = 31

for 100 digits

no of segments required = 3(100) + 1 = 300 + 1 = 301

(c) for digit



expression : 5n + 2

where n = no of digits

for 5 digits

no of segments required = 5(5) + 2 = 25 + 2 = 27

for 10 digits

no of segments required = 5(10) + 2 = 50 + 2 = 52

for 100 digits

no of segments required = 5(100) + 2 = 500 + 2 = 502 a, b and c 4x3 – 12x2 + 11x – 5 is divided by 2x – 1remainder theorem states that if p(x) is any polynomial and a is any real number and if p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).let p(x) = 4x3 – 12x2 + 11x – 5 and we have (2x – 1)the zero of (2x – 1) is now using remainder theorem,p(x) = 4x3 – 12x2 + 11x – 5 is divided by 2x – 1 then, is the remainderremainder = –2