mixture: August 2020

Monday, August 31, 2020

Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.

ncert cbse chapter 10 straight lines miscellaneous exercise

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.

Consider the equation not containing p

3x + y = 2

2x - y =3

solving

adding the

5x = 5

x=1

re substitute in 3x+ y = 2

3(1) + y =2

y = 2-3 = (-1)

point of concurrency is (1,-1)

this should satisfy px + 2 y – 3 = 0

so p -2 -3 = 0

p=5

17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and
(– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle

There are many such right angled triangles which can be found by taking points on the circumference of the circle with (1, 3) and (– 4, 1) as ends of a diameter.

Of these, one of the easiest pair to identify

is to plot the points (1, 3) and (– 4, 1)

and take a line through (1,3) parallel to the y axis which gives x=1

and another perpendicular line through (-4,1) parallel to x axis which

gives y=1.

or vice versa which gives x = (-4) and y=3

ncert cbse chapter 10 straight lines miscellaneous exercise

24. A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

solution

21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

solution

18.Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

solution

17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0

and x – k = 0

solution

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.

solution

ncert cbse chapter 9 sequences and series miscellaneous exercise

32.

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day. 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was finished

solution

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Friday, August 28, 2020

If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

slope of y = mx + c is m.

slope of y=3x+1 is m1 = 3

2y = x + 3 can be rewritten as y =(x/2)+(3/2) so slope is m2 = (1/2)

slope of y =mx +4 is m3 = m

angle u between two lines with slopes m1 and m2 is given by

tan(u) = |[m1-m2] / [ 1+m1*m2 ]|

angle u1 between y = mx + c and y=3x+1 is given by

tan(u1) = |[m-3] / [ 1+3m ]|

angle u2 between y = mx + c and 2y = x + 3 is given by

tan(u1) = |[m-(1/2)] / [ 1+(m/2) ]|

given that y = mx + c is equally inclined to the other two lines

|[m-3] / [ 1+3m ]| = |[m-(1/2)] / [ 1+(m/2) ]|

removing the absolute sign gives two equations

[m-3] / [ 1+3m ] = +{[m-(1/2)] / [ 1+(m/2) ] }

[m-3] / [ 1+3m ] = (-){[m-(1/2)] / [ 1+(m/2) ] }

first equation changes to

[m-3] / [ 1+3m ] = {[2m-1] / [ 2+m] }

[m-3][ 2+m] =[2m-1][ 1+3m ]

(m^2 ) - m -6 = -1-m +6(m^2)

5(m^2) = (- 5)

(m^2) = (- 1) which is not possible.

second equation changes to

[m-3] / [ 1+3m ] = (-){[2m-1] / [ 2+m] }

[m-3][ 2+m] =(-)[2m-1][ 1+3m ]

(m^2 ) - m -6 = +1+m - 6(m^2)

7 (m^2 ) - 2m -7 =0

a = 7 b =(-2) c=(-7)

using quadratic formula

m = { (-b ) + sqrt[(b^2) - 4ac] } / [2a]

m = { (-b ) - sqrt[(b^2) - 4ac} / [2a]

m =[2 + 2sqrt(50)] / [14] or m =[2 - 2sqrt(50)] / [14]

cancelling 2 and factoring

m = [1 + 5sqrt(2)] / [7] or m = [1 - 5sqrt(2)] / [7]

ncert cbse chapter 10 straight lines miscellaneous exercise

solution

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

solution

21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

solution

Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the
line to be a plane mirror.

solution

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0

and x – k = 0

solution

ncert cbse chapter 9 sequences and series miscellaneous exercise

32.

solution

Thursday, August 27, 2020

Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

ncert cbse chapter 10 straight lines miscellaneous exercise

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

Solve the two equations to get the point of intersection

4x + 7y = 3

[2x - 3y = (-1) ] x2

4x + 7y = 3

4x -6y =(-2) subtracting

------------------

13y =5

y =[5/13]

resubstitute

2x -3(5/13) = (-1)

x = (1/13)

point of intersection is ( [1/13] , [5/13] )

using intercept form

let the required line be [x/a] +[y/b] =1

given that the intercepts are equal

put b =a

equation changes to

[x/a] +[y/a] =1

x + y =a

This passes through ( [1/13] , [5/13] )

so that

[1/13] + [5/13] =a

a = [6/13]

equation [x + y =a] changes to

x+y = [6/13]

13x +13y = 6

13x+13y -6 =0

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0

and x – k = 0

solving the equations two by two

first and second

y – x = 0

x + y = 0

adding 2y = 0 ; y =0

substitute back x =0

point of intersection is (0,0)

first and third

y – x = 0

x -k =0

using the second equation x =k

using the first equation y - k =0 or y =k

point of intersection of (k,k)

second and third

x + y = 0

x – k = 0

as before

x = k

k+y = 0 or y = (-k)

point of intersection is (k,-k)

Vertices of the triangle are (0,0) , (k, -k) and (k,k)

[ taking the points in anticlockwise direction assuming k is positive ]

using formula for area of a triangle

{1/2}[x1(y2-y3) + x2(y3-y1) +x3(y1-y2)]

area = [1/2] [0 ( (-k) - (k) ) + k ( k -0 ) +k (0- (-k) ) ]

=[0 + (k^2) + (k^2 ) ] / 2

=[ 2 (k^2)]/2

= (k^2)

ncert cbse chapter 10 straight lines miscellaneous exercise

solution

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

solution

21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the
line to be a plane mirror.

solution

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0

and x – k = 0

solution

ncert cbse chapter 9 sequences and series miscellaneous exercise

32.

solution

Wednesday, August 26, 2020

Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

ncert cbse chapter 10 straight lines miscellaneous exercise

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

let m be the slope of the required line

using formula slope = [(- coefficient of x) / (coefficient of y)]

Slope of the given line [ x – 2y = 3 ]

is m2 = [(-1)/(-2)] = [1/2]

using formula for angle between two lines

tanu = | [m1 - m2] / [ 1 + m1*m2]|

given angle is 45 degrees

tan(45 degrees) = | [m - (1/2)] / [ 1 + m*(1/2)]|

1 = | [2m-1] / [2+m] |

because of the absolute value sign

(+1) = [2m-1] / [2+m] OR (-1) = [2m-1] / [2+m]

solving

2+m = 2m -1 or -2-m = 2m-1

m=3 or m=(-1/3)

Required line passes through (3, 2)

using point slope form

equation is

[y-2] =[3][x-3] or [y-2] =[(-1/3)][x-3]

y-2 =3x-9 or 3y-6 =(- x)+3

3x-y-7=0 or x+3y-9=0

ncert cbse chapter 10 straight lines miscellaneous exercise

solution

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

solution

21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the
line to be a plane mirror.

solution

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

ncert cbse chapter 9 sequences and series miscellaneous exercise

32.

solution

Tuesday, August 25, 2020

Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

image of a point in a straight line

ncert cbse chapter 10 straight lines miscellaneous exercise

Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the
line to be a plane mirror.

or find the foot of the perpendicular of (3, 8) in the line x +3y = 7

let P=(3,8)

let M be its foot of the perpendicular on the given line x +3y = 7

PM perpendicular to x +3y = 7 means

equation of PM is of the form 3x - y +k =0 ----------(1)

[ because bx-ay+k = 0 is perpendicular to ax+by+c=0 ]

{3x - y +k =0 } passes through P(3,8)

means 3(3)-8+k = 0 so that k =(-1) so that

equation of PM is 3x-y-1=0 using equation(1).

now solving the two equations

[ x + 3y = 7 ] x3

3x - y = 1

solving

3x + 9y = 21

3x - y = 1

subtracting

10y = 20

y = 2

x +3(2)=7

x =1

so that M=(1,2) which is the foot of the perpendicular.

let Q(h,k) be the image of P(3,8) in the given line

Mid point of PQ is [{(h+3)/2} , {(k+8)/2}]

but the midpoint is M(1,2)

therfore

{(h+3)/2} = 1 and {(k+8)/2} = 2

h+3 =2 and k+8 =4

h=(-1) and k =(-4)

image is ( -1 , -4 )

ncert cbse chapter 10 straight lines miscellaneous exercise

solution

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

solution

21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the
line to be a plane mirror.

solution

ncert cbse chapter 9 sequences and series miscellaneous exercise

32.

solution

Monday, August 24, 2020

A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

ncert cbse chapter 10 straight lines miscellaneous exercise

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

Since A lies on the x axis, the y coordinate of A is 0

Let A =(k,0)

let P=(1,2)

Q =(5,3)

both PA and AQ make equal angles with the normal line (perpendicular to the x axis) at A due to the law of reflection.

If AQ makes an angle u with the positive direction of the x axis, then PA makes an angle of (180-u) with the positive direction of the x axis.

using the definition of slope

for PA, P=(1,2), A =(k,0)

tan(u) =[0-2] / [k-1] --------------------(1)

for QA , Q =(5,3) , A =(k,0)

tan[180-u] = [0-3] / [k-5]

but tan[180-u] = {-tan (u)}

we get

{-tan (u)} = [0-3] / [k-5] ------------(2)

using eqn (1) in (2)

-{[0-2] / [k-1] } = [0-3] / [k-5]

2(k-5) = (-3)(k-1)

2k-10 = -3k+3

5k =13

k =[13/5]

therefore

A=( (13/5) , 0 )

-----------------------------------------------------------------------------------------------------------------

21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

first line is

[ 9x + 6y – 7 = 0 ] dividing by 3 to make the coefficients in the two given equations the same

3x +2y -(7/3) =0 -------------(1)

second line is

3x + 2y + 6 =0 --------------(2)

A line parallet to these lines will be of the form

3x+2y +k =0 ----------------(3)

distance between (1) and (3)

d1 = | k +(7/3) | / sqrt[(3^2) +(2^2)]

distance between (2) and (3)

d2 = | k -6 | / sqrt[(3^2) +(2^2)]

for equidistant line d1 = d2

equating d1 and d2 and cancelling off the denominator

| k +(7/3) | = | k -6 |

k +(7/3) = k -6 OR k +(7/3) = -(k -6 )

(7/3) =(-6) which is not possible

k +(7/3) = -(k -6 )

2k =6 -(7/3)

2k = (11/3)

k = (11/6)

substitute in (3)

3x+2y +(11/6) =0

18 x + 12y +11 =0

ncert cbse chapter 10 straight lines miscellaneous exercise

solution

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

solution

21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

ncert cbse chapter 9 sequences and series miscellaneous exercise

32.

solution

Friday, August 21, 2020

A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

ncert cbse chapter 10 straight lines miscellaneous exercise

We have to find the point of intersection of the lines

2x – 3y + 4 = 0

3x + 4y – 5 = 0

[ 2x - 3y = (-4) ] x 3

[ 3x + 4y = 5 ] x 2

6x -9y = (-12)

6x +8y =10

subtracting

-17y =-22

y=(22/17)

substitute in

3x + 4y = 5

3x = 5 -{88/17}=[-3/17]

x =(-1/17)

The person is standing at [ (-1/17) , (22/17) ]

The shortest path is perpendicular to 6x – 7y + 8 = 0

Perpendicular line is 7x+6y+k=0

This passes through [ (-1/17) , (22/17) ] if

7 (-1/17) +6(22/17) +k =0

k = (-125/17)

Perpendicular line is 7x+6y+k=0

changes to 7x+6y+(-125/17)=0

or 119x+102y-125=0

119x+102y=125 is the required path.

ncert cbse chapter 10 straight lines miscellaneous exercise

solution

ncert cbse chapter 9 sequences and series miscellaneous exercise

32.

solution

Wednesday, August 19, 2020

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day. 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was finished

ncert cbse chapter 9 sequences and series miscellaneous exercise

32.

Let k = original number of days to finish the work with 150 workers

With workers dropping out , the number of days changed to (k+8) days.

Assuming total work is 1 unit,

150 workers one day's work = [ 1/k ]

1 workers one day's work = [1 / (150k)]

on the first day, 150* [1 / (150k)] = [150 / (150k)] was finished

on the second day only 150-4 = 146 workers remained

so on the second day, 146 [1 / (150k)] = [146 / (150k)] was finished

on the second day only 150-8 = 142 workers remained

so on the second day, 142 [1 / (150k)] = [142 / (150k)] was finished

This went on for (k+8) days by which the entire 1unit of work was finished

therefore

[150 / (150k)] + [146 / (150k)] + [142 / (150k)] +...(k+8)terms = 1 [full work]

[1/(150k)]{150 + 146 +142 + ... (k+8)terms} = 1

[150 + 146 +142 + ... (k+8)terms ] = 150k

LHS is sum of (k+8)terms of an AP with

a =150 ,

d = t2 -t1 = 146-150 = (-4)

n = (k+8)

using formula for sum of n terms of an AP,

Sn = [n/2][ 2a +(n-1)d ] on the LHS

[(k+8)/2]*[ 2(150) +(k+8-1)(-4) ] = 150k

[(k+8)/2]* [300 + (k+7)(-4) ] = 150k

[(k+8)/2]* [272-4k] = 150k

[(k+8)/2] *[4(68-k)]=150k

cancelling 2

[(k+8)] *[2(68-k)]=150k

dividing 2

(k+8) (68-k) = 75k

68k- (k^2) +544 -8k =75k

(k^2) + 15k -544 = 0

(k+32)(k-17)=0

k =(-32) reject

or k = 17

Required number of days = (k+8) = (17+8) = 25

ncert cbse chapter 9 sequences and series miscellaneous exercise

32.

solution

chapter 8 binomial theorem miscellaneous exercise

1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

2. Find a if the coefficients of (x^2) & (x^3) in the expansion of {(3+ax)^9} are equal

solution

3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution

6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

solution

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution

8.find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of [(fourth root of 2) + {1/(fourth root of 3)}]^n is (sqrt6):1

solution

exercise 8.2

5. find the 4th term in the expansion of (x-2y)^12

solution

7. Find the middle terms in the expansion of [3 - ((x^3) / 6)]^7

solution

Q8) Find the middle terms in the expansion of [(x/3)+9y)]^10

solution

10.The coefficients of the (r-1)th, rth, (r+1)th terms in the expansion of [(x+1)^n] is in the ratio 1:3:5. Find n and r.

solution

exercise 8.1

8. evaluate (101)^4

solution

13.show that 9^(n+1) - 8n -9 is divisible by 64 whenever n is a positive integer

solution

Monday, August 17, 2020

show that 9^(n+1) - 8n -9 is divisible by 64 whenever n is a positive integer

ncert cbse chapter 8 binomial theorem exercise 8.1

13.show that 9^(n+1) - 8n -9 is divisible by 64 whenever n is a positive integer

(x+1)^(n+1) = [x^(n+1)] +C[(n+1),1][x^(n)]+...+C[(n+1),(n-1)][x^2] +C[(n+1),n][x] + 1

C[(n+1),n] =C[(n+1),1] = (n+1)

C[(n+1),(n-1)]= C[(n+1),2] = (n+1)(n)/2

(x+1)^(n+1) = [x^(n+1)] +(n+1)[x^(n)]+...+[(n+1)(n)/2][x^2]+(n+1)[x] + 1

put x=8

(8+1)^(n+1) = [8^(n+1)] +(n+1)[8^(n)]+...+[(n+1)(n)/2][8^2]+(n+1)[8] + 1

concentrating on the last 2 terms only

9^(n+1) = { [8^(n+1)] +(n+1)[8^(n)]+...+[(n+1)(n)/2][8^2]} + [8n+8]+1

9^(n+1) = { [8^(n+1)] +(n+1)[8^(n)]+...+[(n+1)(n)/2][8^2]} + 8n+9

re-arranging

9^(n+1) - 8n - 9 ={ [8^(n+1)] +(n+1)[8^(n)]+...+[(n+1)(n)/2][8^2]}

note that C[(n+1),(n-1)]= C[(n+1),2] = (n+1)(n)/2 is an integer

so after taking factor of (8^2) = 64 out,

9^(n+1) - 8n - 9 = 64*[8^(n-1)+ ...+[(n+1)(n)/2] ] =divisible by 64

8. evaluate (101)^4

using expansion with

C(4,1) =4=C(4,3)

C(4,2)=6

(x+y)^4 = (x^4)+4(x^3)(y)+6(x^2)(y^2)+4(x)(y^3)+(y^4)

x=100

y=1

(100+1)^4 = (100^4)+4(100^3)(1)+6(100^2)(1^2)+4(100^1)(1^3)+(1^4)

(101)^4 =100000000+4000000+60000+400+1

(101)^4 =104060401

chapter 8 binomial theorem miscellaneous exercise

1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

2. Find a if the coefficients of (x^2) & (x^3) in the expansion of {(3+ax)^9} are equal

solution

3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution

6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

solution

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution

8.find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of [(fourth root of 2) + {1/(fourth root of 3)}]^n is (sqrt6):1

solution

exercise 8.2

5. find the 4th term in the expansion of (x-2y)^12

solution

7. Find the middle terms in the expansion of [3 - ((x^3) / 6)]^7

solution

Q8) Find the middle terms in the expansion of [(x/3)+9y)]^10

solution

10.The coefficients of the (r-1)th, rth, (r+1)th terms in the expansion of [(x+1)^n] is in the ratio 1:3:5. Find n and r.

solution

exercise 8.1

8. evaluate (101)^4

solution

13.show that 9^(n+1) - 8n -9 is divisible by 64 whenever n is a positive integer

solution

Friday, August 14, 2020

updates about cbse, university exams, results and other news

Cbse annonces tentative dates for class xii practicals exams 2020-2021

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The IIM CAT 2020 admit cards can be dowloaded 28th october 2020 evening onwards and the exam may be held on 29th november 2020

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The online exam application forms for private students appearing in cbse board 10th and 12th exams in 2021 is available at cbse website

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Anna university has given 75 days to engineering colleges to finish the theory portion of this semester ( using online ? mode.) by October 26

*CBSE has released Teacher energised manuals for science and maths for 6th to 10th it is available on http://cbseacademic.nic.in/

* There seems to be a one time exemption this year for students who have cleared only basic math paper in class 10 cbse board exam and who want to take up math in 11th standard in 2020 -2021 session.

* The dates for application for optional exam for class 12th students who want to better their marks has been announced.The last date seems to be August 22nd .

* The date of application for compartment exams for class 12th students has been announced

* Regular students might have to apply through their schools while private students may have to apply online.

* Students can access more details at http://cbse.nic.in/

or cbse website

*The exams may be held in September.

* cbse marking schemes for various exams in 10th and 12th is available at

https://www.jagranjosh.com/articles/cbse-marking-scheme-2020-for-10th-12th-released-download-pdf-1596814891-1

chapter 8 binomial theorem miscellaneous exercise

1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

2. Find a if the coefficients of (x^2) & (x^3) in the expansion of {(3+ax)^9} are equal

solution

3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution

6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

solution

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution

8.find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of [(fourth root of 2) + {1/(fourth root of 3)}]^n is (sqrt6):1

solution

exercise 8.2

5. find the 4th term in the expansion of (x-2y)^12

solution

7. Find the middle terms in the expansion of [3 - ((x^3) / 6)]^7

solution

Q8) Find the middle terms in the expansion of [(x/3)+9y)]^10

solution

10.The coefficients of the (r-1)th, rth, (r+1)th terms in the expansion of [(x+1)^n] is in the ratio 1:3:5. Find n and r.

solution

The coefficients of the (r-1)th, rth, (r+1)th terms in the expansion of [(x+1)^n] is in the ratio 1:3:5. Find n and r.

ncert cbse chapter 8 binomial theorem exercise 8.2

10.The coefficients of the (r-1)th, rth, (r+1)th terms in the expansion of [(x+1)^n] is in the ratio 1:3:5. Find n and r.

T(r+1) =C(n,r) [x^(n-r) ] [1^r] = C(n,r) [x^(n-r) ]

T(r) = C( n,(r-1) ) [x^(n-(r-1) ) ]

T(r-1)= C( n,(r-2) ) [x^(n-(r-2) ) ]

The coefficients of (r-1)th, rth, (r+1)th terms in the expansion of [(x+1)^n]

are C( n,(r-2) ) , C( n,(r-1) ) , C(n,r)

C( n,(r-2) ) : C( n,(r-1) ): C(n,r) = 1 : 3 : 5

C( n,(r-2) ) / C( n,(r-1) ) = 1/3

[(n!) / {(r-2)! *(n-(r-2))!}] / [(n!) / {(r-1)! *(n-(r-1))!}] =1 / 3

[{ (r-1)! * (n-r+1)! } / { (r-2)! * (n-r+2)! } ] = 1 / 3

using

(r-1)! = (r-1) * [(r-2)!]

(n-r+2)! = (n-r+2) *[(n-r+1)!] and cancelling

[(r-1) / (n-r+2) ] = 1 / 3

3r-3 = n-r+2

4r - n = 5 -------------(1)

C( n,(r-1) ) / C( n,(r) ) = 3/5

[(n!) / {(r-1)! *(n-(r-1))!}] / [(n!) / {(r)! *(n-r)!}] = 3/5

{(r)! *(n-r)!} / {(r-1)! * (n-r+1)!} = 3/5

use

r! = (r) * [(r-1)!]

(n-r+1)! =(n-r+1)* [(n-r)!] and cancelling

[r] / [n-r+1] = 3/5

5r =3n-3r+3

8r-3n =3 -----------------(2)

solving (1) and (2) using elimination method

r=3

n=7

7. Find the middle terms in the expansion of [3 - ((x^3) / 6)]^7

There are (7+1) = 8 terms in the expansion

middle terms are 4th and 5th terms

T(r+1) =C(7,r) [3^(7-r)]*[ [(-x^3) / 6)]^r ]

put r=3

T(4) = C(7,3) [3^(7-3)]*[ [(-x^3) / 6)]^3 ]

T(4) =35* [3^4]* [ [(-x^3) / 6)]^3 ]

T(4)= [- 105/8] * (x^9)

put r=4 in T(r+1) =C(7,r) [3^(7-r)]*[ [(-x^3) / 6)]^r ]

T(5) = C(7,4) [3^(7-4)]*[ [(-x^3) / 6)]^4 ]

T(5) = 35* [3^3] * [ [(-x^3) / 6)]^4 ]

T(5)=[35/48] *(x^12)

middle terms are

[- 105/8] * (x^9) and [35/48] *(x^12)

chapter 8 binomial theorem miscellaneous exercise

1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

2. Find a if the coefficients of (x^2) & (x^3) in the expansion of {(3+ax)^9} are equal

solution

3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution

6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

solution

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution

8.find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of [(fourth root of 2) + {1/(fourth root of 3)}]^n is (sqrt6):1

solution

exercise 8.2

5. find the 4th term in the expansion of (x-2y)^12

solution

7. Find the middle terms in the expansion of [3 - ((x^3) / 6)]^7

solution

Q8) Find the middle terms in the expansion of [(x/3)+9y)]^10

solution

10.The coefficients of the (r-1)th, rth, (r+1)th terms in the expansion of [(x+1)^n] is in the ratio 1:3:5. Find n and r.

solution

find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of [{2^(1/4)}+{3^(-1/4)}]^n is (sqrt6):1

ncert cbse chapter 8 binomial theorem miscellaneous exercise

find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of [(fourth root of 2) + {1/(fourth root of 3)}]^n is (sqrt6):1

in [{2^(1/4)}+{3^(-1/4)}]^n

T(r+1) = C(n.r)[ {2^(1/4)}^(n-r) ][{3^(-1/4)}^(r)]

for fifth term from the beginning

r=4

T(5) = C(n,4)[2^((n-4)/4)][3^(-(4)/4)] =C(n,4)[2^((n-4)/4)] / [3]

For finding the fifth term from the end in the given expansion, we interchange the terms and find the fifth term from the beginning of

[{3^(-1/4)} + {2^(1/4)} ]^n

t ' (5) = C(n,4) [3^( -(n-4)/4)][ 2^((4)/4)]=C(n,4) [3^( -(n-4)/4)] [2]

T(5) / t'(5) = [ 6^((n-4)/4) ] / [6^1]

(a^m)/(a^n) =a^(m-n)

[(n-4)/4] -[1] = [(n-8)/4]

T(5) / t'(5) = [6^( (n-8)/4 ) ]

using given ratio

[ sqrt(6) / 1] = [6^( (n-8)/4 ) ]

6^(1/2) = [6^( (n-8)/4 ) ]

(1/2) = ( (n-8)/4 )

4/2 = n-8

2=n-8

n=10

exercise 8.2

5. find the 4th term in the expansion of (x-2y)^12

n=12

T(r+1)=C(12,r) [x^(12-r)][ (-2y)^r]

for 4th term

put r=3

T(3+1) = C(12,3) [x^(12-3)][ (-2y)^3]

T(4) = 220*[x^9][-8y^3]

T(4)=(-1760)[x^9][y^3]

chapter 8 binomial theorem miscellaneous exercise

1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

2. Find a if the coefficients of (x^2) & (x^3) in the expansion of {(3+ax)^9} are equal

solution

3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution

6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

solution

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution

8.find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of [(fourth root of 2) + {1/(fourth root of 3)}]^n is (sqrt6):1

solution

exercise 8.2

Q8) Find the middle terms in the expansion of [(x/3)+9y)]^10

solution

5. find the 4th term in the expansion of (x-2y)^12

solution

Thursday, August 13, 2020

cbse deleted portions

cbse deleted portions in mathematics 2020 2021

For Class 10 mathematics

The data given here may not be correct

Students are asked to consult the official cbse website for the correct deleted portions.

A few of the deleted portions for class 10 mathematics are given below

Euclid's division lemma has been deleted in real numbers.

division algorithm of polynomials has been deleted.

Cross multiplication method has been removed in two variable linear equations

Situational problems have been removed in quadratic equations.

Solution of real life problems based on sum of n terms of an AP has been removed.

Area of triangle has been deleted in Coordinate geometry.

Proof of ratio of areas of similar triangles and proog of if the square on one side is equal to sum of the squares on the other two sides, the angle opposite to the first side is a right angle have been deleted

Construction of similar triangles has been deleted.

Trigonometry ratios of complimentary angles has been removed.

Frustum of cone, step deviation method for mean, cumulative frequency graph have been delted.

chapter 8 binomial theorem miscellaneous exercise

1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

2. Find a if the coefficients of (x^2) & (x^3) in the expansion of {(3+ax)^9} are equal

solution

3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution

6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

solution

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution

exercise 8.2

Q8) Find the middle terms in the expansion of [(x/3)+9y)]^10

solution

ncert cbse 11th mathematics chapter 7 permutations exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

solution

ncert cbse 11th mathematics chapter 7 permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

solution

2. How many words of can be formed using all the letters of the word

EQUATION so that the vowels and consonants occur together?

solution

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

solution

6.How many 6 digit numbers can be formed using 0,1,3,5,7,9 which are divisible by 10 and no digit is repeated?

solution

6.The english alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed

solution

7. In an examination a question paper consists of two parts containing 5 and 7 questions respectively. A student is required to answer 8 questions in all selecting atleast 3 from each part. In how many ways can a student select the questions?

solution

8.determine the number of 5 card selections of a deck of 52 cards if each selection of 5 cards should have exactly one king

solution

9. It is required to seat 5 men and 4 ladies in a row so that the ladies occupy the even positions. How many such arrangements are possible.

solution

10. In a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will go. In how many ways can the party be chosen.

solution

Wednesday, August 12, 2020

Progesterone the pregnancy hormone and PMS like cramps at the start of pregnancy

Progesterone is one of the main hormones that prepares the uterus for implantation.

In case pregnancy does not occur Progesterone levels will fall and menstruation will start.

If pregnancy occurs , the levels of progesterone will rise leading to symptoms similar to that in the second half of the menstrual cycle before onset of period bleeding

Tenderness of breasts, lower stomach cramps,headache, light bleeding / spotting, fatigue, tiredness , constipation, nausea may occur in early pregnancy especially in the first week after the expected day of menstruation / onset of periods in the case of women with regular cycles.

In the first couple of weeks after conception, pregnant women will feel that their breasts are tender, sore, fuller and heavier with a slight nagging pain.

There might be slight bleeding or spotting in the vagina called implantation bleeding around one or two weeks after conception usually around the time you would expect your usual periods to start. This might be caused by the embryo attaching to the mothers uterus.

Trying to get pregnant or trying to concieve ( TTC )

Pregnancy testing

first trimester of pregnancy

second trimester of pregnancy

Third trimester

labour and child birth

post partum

life of a woman after child birth

Male Fertility

Bipolar disorder medicine during pregnancy.

Changes in women after childbirth

find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

ncert cbse chapter 8 binomial theorem miscellaneous exercise

6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

using expansion with

C(4,1) =4=C(4,3)

C(4,2)=6

(x+y)^4 = (x^4)+4(x^3)(y)+6(x^2)(y^2)+4(x)(y^3)+(y^4)

(x-y)^4 = (x^4)-4(x^3)(y)+6(x^2)(y^2)-4(x)(y^3)+(y^4)

adding the two expansions

(x+y)^4 +(x-y)^4= 2[(x^4) +6(x^2)(y^2) +(y^4) ]

use x =(a^2) y = sqrt{(a^2)-1}

[(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

= 2[{(a^2)^4}+6{(a^2)^2}{[ sqrt{(a^2)-1}]^2} +{[ sqrt{(a^2)-1}]^4}]

=2[ (a^8) +6(a^4)((a^2)-1) +{(a^2)-1}^2 ]

=2 [ (a^8) +6(a^6) -6(a^4) +{(a^4) -2(a^2)+1} ]

=2[ (a^8) +6(a^6) -5(a^4)-2(a^2)+1 ]

=2(a^8) +12(a^6) -10(a^4)- 4(a^2)+2

chapter 8 binomial theorem miscellaneous exercise

1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

2. Find a if the coefficients of (x^2) & (x^3) in the expansion of {(3+ax)^9} are equal

solution

3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution

6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

solution

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution

exercise 8.2

Q8) Find the middle terms in the expansion of [(x/3)+9y)]^10

solution

find an approximate value of (0.99^5) using the first three terms of its expansion

ncert cbse chapter 8 binomial theorem miscellaneous exercise

7.find an approximate value of (0.99^5) using the first three terms of its expansion

(1-x)^5 = 1 -C(5,1)x + C(5,2)(x^2)+ higher powers of x

C(5,1)=5

C(5,2)=10

(1-x)^5 = 1 -5x +10(x^2)+ higher powers of x

choose x =0.01

(1-0.01)^5 =1 -5(0.01) +10((0.01)^2)+ higher powers of 0.01

0.99^5 = 1 - 0.05 +0.001 = 0.951 approximately.

exercise 8.2

Q8) Find the middle terms in the expansion of [(x/3)+9y)]^10

There are (10 +1) = 11 terms in the expansion

middle term is the 6th term

T(r+1) = C(10,r) [(x/3)^(10-r)][(9y)^(r)]

put r=5

T(6) =C(10,5) [(x/3)^(10-5)][(9y)^(5)]

= [252]*[(x^5)/243][59049 (y^5)]

=61236(x^5)(y^5)

chapter 8 binomial theorem miscellaneous exercise

1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

2. Find a if the coefficients of (x^2) & (x^3) in the expansion of {(3+ax)^9} are equal

solution

3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution

exercise 8.2

Q8) Find the middle terms in the expansion of [(x/3)+9y)]^10

solution

Tuesday, August 11, 2020

evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

ncert cbse chapter 8 binomial theorem miscellaneous exercise

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

(a+b)^6 = {a^6} + C(6,1){a^5}{b} +C(6,2){a^4}{b^2}+C(6,3){a^3}{b^3} +C(6,4){a^2}{b^4}++C(6,5){a^1}{b^5}+C(6,6)(b)^6

C(6,1) = 6 =C(6,5)

C(6,2)=15=C(6,4)

C(6,3)=20

C(6,6)=1

(a+b)^6 = {a^6} + 6{a^5}{b} +15{a^4}{b^2}+20{a^3}{b^3} +15{a^2}{b^4}+6{a^1}{b^5}+(b)^6

(a-b)^6 = {a^6} - 6{a^5}{b} +15{a^4}{b^2} - 20{a^3}{b^3} +15{a^2}{b^4}- 6{a^1}{b^5}+(b)^6

adding the two equations

{ (a+b)^6} - {(a-b)^6} =

2[6{a^5}{b} +20{a^3}{b^3}+ 6{a^1}{b^5} ]

a=sqrt(3)

b=sqrt(2)

{ (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

= 2[6*{(sqrt(3))^5}{(sqrt(2))} +20{(sqrt(3))^3}{(sqrt(2))^3}+6{(sqrt(3))^1}{(sqrt(2))^5} ]

=2[54+120+24 ] {sqrt(6)}

=396*sqrt(6)

chapter 8 binomial theorem miscellaneous exercise

1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

2. Find a if the coefficients of (x^2) & (x^3) in the expansion of {(3+ax)^9} are equal

solution

3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution

If the equations x^2 -ax+b=0 and x^2-ex+f =0 have a root in common and the second equation has equal roots show that ae =2(b+f)

question on quadratic equations

If the equations (x^2) -ax+b=0 and (x^2)-ex+f =0 have a root in common and the second equation has equal roots show that ae =2(b+f)

Let alpha be the common root

Therefore (alpha) ,(alpha) are the equal roots of the second equation

[(x^2)-ex+f =0]

using sum of roots formula

(alpha) + (alpha) ={ -(-e)} / {1}

2(alpha) = e

(alpha) ={e}/{2} -----------------(1)

using product of roots formula

(alpha) * (alpha) = {f}/{1}

{alpha^2} = f -----------------------(2)

Therefore let

(alpha) ,(beta) be the roots of the first equation [(x^2) -ax+b=0]

again using product of roots formula

(alpha) * (beta) ={b}/{1}

(alpha) * (beta) = b

beta = b/{alpha} -------------------(3)

using sum of roots formula

(alpha) + (beta) ={-(-a)}/{1}

(alpha) + (beta) = a

using equation(3)

(alpha) +{ {b}/{alpha}} = a

{{alpha}^2 + b } / {alpha} = a

{{alpha}^2 + b } = a *{alpha}

use eqn(2) in LHS and eqn(1) in RHS

{f +b}=a*{{e}/{2}}

therfore

2*(f+b) = ae

hence

ae =2(b+f)

Monday, August 10, 2020

find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

ncert cbse chapter 8 binomial theorem miscellaneous exercise

3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

using binomial expansion

(1+2x)^6 = 1 +C(6,1)(2x)+C(6,2)(2x)^2+C(6,3)(2x)^3 +C(6,4)(2x)^4

+C(6,5)(2x)^5+C(6,6)(2x)^6

C(6,1) = 6 =C(6,5)

C(6,2)=15=C(6,4)

C(6,3)=20

C(6,6)=1

(1+2x)^6 =

1 +6(2x) +15(4(x^2))+20(8(x^3))+15(16(x^4))+6(32(x^5))+1(32(x^6))

=1+12x+60(x^2)+160(x^3)+240(x^4)+192(x^5)+32(x^6)

(1-x)^7=

1-C(7,1)(x)+C(7,2)(x^2)-C(7,3)(x^3)+C(7,4)(x^4)

-C(7,5)(x^5)+C(7,6)(x^6)-C(7,7)(x^7)

C(7,1)=7=C(7,6)

C(7,2)=21=C(7,5)

C(7,3)=35=C(7,4)

C(7,7)=1

(1-x)^7=1 -7x+21(x^2)-35(x^3)+35(x^4)-21(x^5)+7(x^6)-1(x^7)

now multiplying the two expansions and concentrate only on the terms containing

(x^5)

1* {-21(x^5)}+{+12x}*{+35(x^4}+{+60(x^2)}*{-35(x^3)}+{+160(x^3)}{+21(x^2)}+{+240(x^4)}{-7x}+{+192(x^5)}{1}

coefficient of (x^5) is

1*(-21)+(12)*(35)+(60)(-35)+(160)(21)+(240)(-7) +192*1

=(-21)+420-2100+3360-1680+192

=171

chapter 8 binomial theorem miscellaneous exercise

1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

2. Find a if the coefficients of (x^2) & (x^3) in the expansion of {(3+ax)^9} are equal

solution

3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

Friday, August 7, 2020

blood tests and injections in early pregnancy

As a woman trying to get pregnant, when you miss your period by a few days. then wait for a few anxious days and then most women are likely to use a urine based pregnancy home test kit for checking if they are pregnant. [or as a joke use the diy pregnancy test based on common salt where you mix your first morning urine with a solution of common salt and water and if it turns milky after some time , it might indicate pregnancy.]

If the woman finds that she is pregnant, it is best to book a prenatal appointment with a doctor especially if the woman has a history of thyroid problems, diabetes or previous loss of pregnancy.

Managing diabetes with pregnancy is hard.

Once you go to the doctor , it will be the start of blood tests and injections. This might be bad news for women who are afraid of syringes and for women who have small veins which are hard to find. Some persons faint while drawing blood. Holding hands of their loved ones like spouses or lying down while drawing blood might be helpful for such women.If allowed drinking water to avoid dehydration might also be helpful.

Most doctors will order a HCG blood test to confirm the pregnancy. HCG blood test refers to a blood test to measure the level (quantitative blood test) of human chorionic gonadotropin [ HCG blood test in short or beta hcg blood test or quantitative hcg blood test]. After conception the level of hcg in the blood increases and can usually be detected after two weeks in the blood.

The doctor might schedule another test to confirm that the HCG levels are increasing to confirm a healthy pregnancy. If the doctor feels that the urine tests used were sensitive , the doctor might wait for the ultrasound tests at 7 to 12 weeks.

There might be problems with insurance companies wanting doctor confirmation of the pregnancy before starting prenatal coverage.

The doctor will want to test for blood type and to take precautions in case of

Rh -ve type etc .A CBC might be done to check for anaemia and the doctor might prescribe iron supplements or iron rich foods in case of anaemia.Other screening tests might be for HIV, syphillis, hepatitis B ,genetic disorders , etc .

In some extreme cases with Rh -ve type, the fathers blood test might also be required to screen for complications.

Further diagonostic tests might be ordered based on the results of the screening tests.

Trying to get pregnant or trying to concieve ( TTC )

Pregnancy testing

first trimester of pregnancy

second trimester of pregnancy

Third trimester

labour and child birth

post partum

life of a woman after child birth

Male Fertility

Bipolar disorder medicine during pregnancy.

Changes in women after childbirth

Find a if the coefficients of (x^2) and (x^3) in the expansion of {(3+ax)^9} are equal

ncert cbse chapter 8 binomial theorem miscellaneous exercise

Find a if the coefficients of (x^2) & (x^3) in the expansion of {(3+ax)^9} are equal

using binomial theorem

{(3+ax)^9}= 3^9 + C(9,1)(3^8){(ax)}+ C(9,2)(3^7){(ax)^2}

+ C(9,3)(3^6){(ax)^3} + ...+{(ax)^9}

given coefficients of (x^2) & (x^3) are equal

therfore

C(9,2)(3^7){(a)^2} =C(9,3)(3^6){(a)^3}

36* (3^7){(a)^2}=84*(3^6){(a)^3}

if a cannot be zero

a= {36*(3^7) }/{84*(3^6)}

a =9/7

chapter 8 binomial theorem miscellaneous exercise

1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

2. Find a if the coefficients of (x^2) & (x^3) in the expansion of {(3+ax)^9} are equal

solution

Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

ncert cbse chapter 8 binomial theorem miscellaneous exercise

1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

(a+b)^n = (a^n) + C(n,1)[a^(n-1)][b]++ C(n,2)[a^(n-2)][b^2] +...+(b^n)

using the given data and also

C(n,1) = n

C(n,2) = {n(n-1)}/{2}

we get

a^n = 729--------------------(1)

n[a^(n-1)][b] = 7290--------------------(2)

{n(n-1)}/{2}*[a^(n-2)][b^2] = 30375---------------(3)

(2)/(1) gives {nb} / {a} = 10 ----------------------(4)

(3)/(2) gives {(n-1)b}/{2a} = {25} / {6}-----------(5)

(5)/(4) gives {n-1}/{2n} = {5}/{12}

therefore 12(n-1)=5(2n)

12n-12 =10n

12n-10n =12

2n=12

n=6

use in (1)

a^6 = 729

a = 3

use these values of n and a in (4)

{6*b}/{3} = 10

gives b =5

Thursday, August 6, 2020

Changes in women after childbirth

Women are usually very self conscious about their own body. Usually many women fret and worry too much about the changes in the appearance of their body after childbirth and try to compare that with their body before pregnancy and they wonder about the changes in their body that motherhood has brought with it. Some women are unduly worried about the stretch marks on the stomach / abdomen and saggy / flabby skin. Some of the new mothers get fooled by some celebrities who magically seem to get back to their pre- pregnant body shape very fast, forgetting that most of the celebrities have access to lots of money, specialist trainers, cosmetic surgery etc.

The breasts of a new mother usually become enlarged with milk inside the milk ducts of the breast for breastfeeding the new born child. But after the mother stops breast feeding her child, the size of the breast may not go back to their pre-pregnant size and in some women, the breasts may appear saggy / less firmer than it was before pregnancy . But breast feeding itself might not be the cause of the sagging of the breasts. Many new mothers wonder how to get back their firm breasts after their breast feeding stage. Push-ups and dumbbell fly exercises might be useful to get firmer breasts. The fat for lactation or breast milk may be stored around the hips and those areas are likely to become bigger.

The legs / feet of a woman may swell during pregnancy, but even after pregnancy, the feet / arch of the feet may not go back to their old size / structure and might be flatter / longer and hence the new mother might need bigger footwear after childbirth. Hormones bring about a lot of changes to the skin and the hair during and after pregnancy. Some areas of the skin may darken.

Some new mothers find a lot of changes in their metabolism after childbirth. Some mothers find that they have new allergies ( breaking out in hives , intolerance to some food or food allergies , celiac disease or intolerance to wheat gluten etc ) after childbirth because being exposed to allergens when the body / immune system is weak might start off the allergies.

Sometimes women with irregular periods may find that they get regular periods after childbirth and sometimes there will be decrease in the menstrual cramps also.

Some women experience urinary incontinence ( or slight urine leakage during stress, orgasm etc ) after childbirth due to excessive stretching / injury of the ligaments and pelvic area during pregnancy. The vagina will also feel more bigger / looser after vaginal childbirth. Some women also experience dryness in the vagina. Kegel exercises or pelvic floor exercises after child birth may be helpful to strengthen the muscles in the pelvic region and may help with the problem of urinary incontinence.

Bipolar disorder medicine during pregnancy.

Bipolar disorder ( manic depression ) refers to experience of extremes of mood swings from high energy levels / manic euphoria to depressive levels usually with cycles over many days, weeks or months . Some famous actors and celebrities like Carrie Fisher (who is famous for her character of Princess Leia in Star wars ) have been very open about their battle with bipolar disorder. Carrie Fisher has publicly disclosed her diagnosis with bipolar disorder and given many interviews about her experience with bipolar disorder. Drinking alcohol is also likely to negatively affect the mood swings of people having bipolar disorder.

People with bipolar disorder are usually prescribed a combination of different drugs called a drug cocktail of mood stabilizers, anti depressants, anti psychotic ,anti convulsant drugs and(examples are lithium, valproic acid, lamotrigine, carbamazepine , ariprazole , clozapine, lamotragine etc ). Many people wonder if bipolar women on medication can give birth to healthy children. There are many examples over the years of bipolar women on medication giving birth to healthy children. Bipolar disorder in pregnant woman usually needs a lot of attention from properly trained doctors. But some of the good psychiatrists may charge expensive fees and may be out of reach of most pregnant women who may have to consult their insurance companies also . Insurance companies sometimes decline them or charge exorbitant amounts for people with bipolar disorder.

One of the main concerns for pregnant women is if the medications for bipolar disorder can be continued during pregnancy. The doctors you consult will usually show you which medications are usually harmful and which are not usually harmul during pregnancy in various medical studies. There may be a lot of trial and error testing with different combinations, before settling on a good combination of drugs to use during pregnancy. So it is better to get the consultation with the doctor and get a stable combination going before trying to conceive ( ttc ). But that is not always possible because of unplanned pregnancies . ( Bipolar disorder may also be associated with irregular menstrual cycles and some other menstrual abnormalities. Some of the medicines for bipolar disorder may cause menstrual irregularities.

Some of the common medications for bipolar disorder may increase the risk of birth disorders and some other birth anomalies like Ebstein’s anomaly ( especially during the first trimester ). Some studies show that medications involving Lithium may be associated with defects of the baby's heart, while some doctors would like the mother to be hydrated so as to control lithium toxicity. Doctors who prescribe lamotrigine (an anti convulsant or anti epileptic drug ) may ask the women to take higher dosage of folic acid supplement than usual.

Some of the women who stop their medication without consulting the doctor during pregnancy may suffer relapse like being unable to sleep, experience anxiety, depression etc . Pregnant women and their partners are usually concerned about what mood stabilizers are safe during pregnancy. Breast feeding after pregnancy is also challenging for women with bipolar disorder on medication. Some of the medications used for bipolar disorders will also be present in traces in the breast milk for the baby . Even after the breast feeding stage, the weaning stage might also present some hormonal changes in the levels of oxytocin and prolactin in the mother which might be a trigger for emotional swings / changes for the mother

Male Fertility

Male Fertility

Usually the tips for ttc ( trying to concieve ) will be written with the woman in mind consisting of tips about FAM ( fertility awareness methods ), bbt charts ( basal body temperature charts ), thickness of the cervical mucus ( CM / EWCM or egg white cervical mucus ) finding the probable ovulation time etc. The idea is that EWCM is likely to be a prelude to ovulation.

Now for a male , the sperm takes nearly 50 to 75 days to mature. In a single ejaculation, millions of sperms are mixed per ml of the semen and ejaculated. When ejaculated into the female vagina, the sperm will move through the cervix and into the uterus and then to the fallopian tubes where hopefully it might fertilize the egg in the female. The sperm may last theoretically from around 3 days to around 5 days in the nutritive supportive cervical fluids ( not in the vagina ) in the woman's body . So the three days around the probable ovulation time is the likely to be the most fertile period for women

When somebody goes for semen analysis / sperm analysis to check the health of his sperm , they usually check for

semen volume (usually over 2 ml per ejaculation ) ,

sperm count / sperm concentration / sperm density ( usually around 20 million to 200 million sperms per ml. ) ,

sperm motility (progressive motility or the ability of the sperm to move in a relatively straight line ),

sperm velocity for propagation,

sperm morphology ( sperm size/shape expressed as a percentage of sperm which appear normal under a microscope ),

Normozoospermic usually refers to normal sperm quality.
If there is no sperm in the semen the condition is known as azoospermia.

Low sperm count/ motility may be caused by exposure to chemicals (like xenoestrogens which imitate estrogens from some components in plastic bottles ) , mobile phone radiation , prolonged exposure to heat in the testicular region like hot baths in tubs, tobacco and alcohol usage.

Sometimes , overweight / obese men may find it hard to impregante their female partners due to problems related to hormone levels . It is good to decrease body weight to normal range in order to boost male sperm count / motility and hence male fertility. High blood pressure and medications for high blood pressure is likely to have detrimental effect on erection in males and also quality of the sperm and male fertility in general. Diabetes is also likely to have bad effects on male fertility due to diabetes causing damage to the nerves ( diabetic neuropathy ) and reduced blood circulation which might lead to problems of erection ( erectile dysfunction ED ), lower ejaculate volume etc .

Regular moderate exercise ( not intense and strenuous exercise ) may boost male fertility. Excess stress levels might reduce sperm quality. reducing stress levels is also good for boosting male fertility. Spme steroids used for muscle development in body building may lead to decrease in sperm quality