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

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

 

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

or 

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

or

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

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

 

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.

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

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

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

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

 

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.

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

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

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

OR 

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

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

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

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

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

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.

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

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

 

 

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

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

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  

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

 

 

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 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

 

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

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

 

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

2

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

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

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

ncert cbse chapter 7 permutations miscellaneous exercise

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

First select the king. Since there are four kings, this can be done in

C(4,1) ways.

Now after removing the 4 kings there are 52-4 = 48 cards which will be used to select the remaining 4 non king cards which can be done in C(48,4)

Answer = C(4,1) * C(48,4)

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.

The required arrangement is like

M W M W M W M W M

The men can be arranged in ( 5! ) =120 ways

Now arrange the ladies in their even positions in (4!) = 24 ways

Answer = 120*24 =2880 ways

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.

First possibility

All 3 of the friends join the party

Now all we have to do is to select the remaining 7 participants using the remaining (25-3)=22 students in C(22,7) ways

OR

Second possibility

None of the three friends join the party

Now we have to select the entire 10 participants using the remaining (25-3)=22 students in C(22,10) ways

Answer  = C(22,7) + C(22,10)

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

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

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

ncert cbse chapter 7 permutations miscellaneous exercise

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

Because there are 5 vowels,

the 2 vowels can be selected in C(5,2) = [(5*4)/(1*2)] = 10 ways

Because there are 21 consonants

the 2 consonants can be selected in C(21,2) = [(21*20)/(1*2)] = 210 ways

After selecting the 4 required letters now we can arrange them in

(4!) = 4*3*2*1 = 24 ways

Required number of words = 10*210*24 =50400

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?

let the selections be denoted by (x,y) where x denotes the questions from part I

and y the questions from part II

atleast 3 means 3 or more

(3,5) or (4,4) or (5,3) are the different possibilities with their respective number of

possible selections being

C(5,3) * C(7,5) = [(5*4)/(1*2)]*[(7*6)/(1*2)]=210 ways

C(5,4) * C(7,4) = [5]*[(7*6*5)/(1*2*3)]=175 ways

C(5,5) * C(7,3) = [1]*[(7*6*5)/(1*2*3)]=35 ways

Total number of ways = 210+175+35 =420 ways

Here note that C[n,r] =C[n,(n-r)]

gives

C[5,3]=C[5,2]

C[7,5]=C[7,2]

C[5.4]=C[5,1]=5

C[7,4]=C[7,3]

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

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

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

ncert cbse chapter7permutations miscellaneous exercise

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

first arrange the 11 letters in alphabetical order

AA   2 A's

E

II  2 I's

M

NN 2 N's

O

T

X

Therefore before the first word starting with E, words starting with the letter

A will be arranged in alphabetical order in the dictionary.

fix one A in the first slot

remaining 10 letters contain 2 I's and 2 N's [ one of the 2 A's is used for the first slot ]

So number of words starting with the letter A

= [(10!)/ {(2!)*(2!)}] =907200

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?

The one's place should contain 0 for the number to be divisible by 10

Fix the digit 0 in the one's place Therefore the one's place can be filled in only one way.

The other 5 slots can be filled in 5! = 120 ways

Answer = 1*120 = 120

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

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

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

ncert cbse 7th chapter permutations miscellaneous exercise

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

The vowels are A,U,E

We have to select 2 out of the 3 vowels in C(3,2)= [3*2]/[1*2]= 3 ways

The consonants are D,G,H,T,R

We have to select 3 out of the 5 vowels in C(5,3)= [5*4] / [1*2] =10ways

because C(n,r) = C[n,(n-r)]  gives C(5,3)=C(5,2)

now the 2+3= 5 selected letters can be arranged among themselves in

5! =5*4*3*2*1=120 ways

Required number of words = 3*10*120 =3600

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

 EQUATION  so that the vowels and consonants occur together?

Vowels are E,U,A,I,O

Treat the 5 vowels as a single unit. Inside this the 5 vowels can be arranged

among themselves in 5!=5*4*3*2*1=120 ways

Consonants are Q,T,N

Treat the 3 consonants as a single unit. Inside this the 3 consonants can be arranged among themselves in 3!=3*2*1=6 ways

Now the two units can be arranged among themselves in 2!=2*1 = 2 ways

Required number of words =120*6*2=1440 words

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

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

find the derivative of [(x^2)+1][cosx]

ncert cbse 11th limits and derivatives miscellaneous exercise

23.find the derivative of [(x^2)+1][cosx]

y=[(x^2)+1][cosx]

using product rule of differentiation

y = uv

dy/dx = uv' +vu'

u =[(x^2)+1]

v=cosx

u ' =2x+0 = 2x

v ' = (-sinx)

dy/dx = uv' +vu'

dy/dx = [(x^2)+1](-sinx)) +[cosx][2x]

dy/dx = -(x^2)sinx -sinx +2x cosx

24. find the derivative  of (a(x^2)+sinx )[p+qcosx]

y=[a(x^2)+sinx ] [p+qcosx]

using product rule of differentiation

y = uv

dy/dx = uv' +vu'

u =[a(x^2)+sinx ]

v =[p+qcosx]

u ' = a(2x)+cosx = 2ax+cosx

v ' =0 - qsinx

dy/dx = uv' +vu'

dy/dx = [a(x^2)+sinx ][ - qsinx] + [p+qcosx][ 2ax+cosx]

13. limits and derivatives
miscellaneous exercise

1.

(i)  find the derivative of (-x) using first principles
solution

(ii)   find the derivative of [ (-x)^(-1) ] using first principles

solution

(iii)find the derivative of sin(x+1) using first principles

solution

(iv) find the derivative of cos[x-(pi/8)] using first principles
solution

23.find the derivative of [(x^2)+1][cosx]

solution

24. find the derivative  of (a(x^2)+sinx )[p+qcosx]

solution

28.

find the derivative of x / (1+tanx)

solution

29.

find the derivative of (x+secx)(x-tanx)

solution

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

find the derivative of (x+secx)(x-tanx)

ncert cbse 11th limits and derivatives miscellaneous exercise

29.

find the derivative of (x+secx)(x-tanx)

y = (x+secx)(x-tanx)

using product formula for differentiation

y = uv

dy/dx = uv' +vu'

u =(x+secx)

v=(x-tanx)

taking derivative

u' = 1 + secxtanx

v' = 1 - [(secx)^2]

dy/dx = uv' +vu'

dy/dx =(x+secx) {1 - [(secx)^2]} +(x-tanx){1 + secxtanx ]

28.

find the derivative of x / (1+tanx)

using quotient formula for differentiation

y = u/v

dy/dx = { vu' - uv' } / {v^2}

u=x

v=(1+tanx)

taking derivative

u' = 1

v' = 0 + [(secx)^2] =[(secx)^2]

dy/dx = { (1+tanx)(1) - x[(secx)^2] } / {(1+tanx)^2}

dy/dx = { (1+tanx) - x[(secx)^2] } / {(1+tanx)^2}

dy/dx = { vu' - uv' } / {v^2}

13. limits and derivatives
miscellaneous exercise

1.

(i)  find the derivative of (-x) using first principles
solution

(ii)   find the derivative of [ (-x)^(-1) ] using first principles

solution

(iii)find the derivative of sin(x+1) using first principles

solution

(iv) find the derivative of cos[x-(pi/8)] using first principles
solution

28.

find the derivative of x / (1+tanx)

solution

29.

find the derivative of (x+secx)(x-tanx)

solution

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies.

find the derivative of sin(x+1) using first principles

ncert cbse 11th limits and derivatives miscellaneous exercise

1
(iii)find the derivative of sin(x+1) using first principles

f(x) = sin(x+1)

f(x+h) = sin[(x+h) +1]

f(x+h) - f(x) = sin[(x+h) +1] - sin[x+1]

using trigonometry formula trigonometry identities

sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]

f(x+h) - f(x) = sin[(x+h) +1] - sin[x+1]

=2cos{ [(x+h) +1 +x+1 ] /2 } sin{ [(x+h) +1 - (x+1) ] /2}

=2cos{(2x+h+2)/2}sin{h/2}

[f(x+h) - f(x)] / h =[2cos{(2x+h+2)/2}sin{h/2} / h

=cos{(2x+h+2)/2} * [ sin(h/2) / (h/2) ] --------------(1)

using lim [ (sinx)/x ] =1 as x -->0

taking lim as h --> 0 in equation(1)

f ' (x) = cos{(2x+0+2)/2} *[1]

f ' (x) = cos{x+1} cancelling off the 2

(iv) find the derivative of cos[x-(pi/8)] using first principles

f(x) =cos[x-(pi/8)]

f(x+h) =cos[x+h-(pi/8)]

f(x+h) - f(x) = cos[x+h-(pi/8)]  - cos[x-(pi/8)]

 using trigonometry formula trigonometry identities

cosx - cosy = -2sin[(x+y)/2] sin[(x-y)/2]

f(x+h) - f(x) = cos[x+h-(pi/8)]  - cos[x-(pi/8)]

=-2sin{ {[x+h-(pi/8)]+[x-(pi/8)]}/2 }sin{ {[x+h-(pi/8)] - [x-(pi/8)]}/2 }

= (-2)sin{ {2[x-(pi/8)]+h}/2 } sin(h/2)

[f(x+h) - f(x)] / h =[ -2sin{ {2[x-(pi/8)]+h}/2 } sin(h/2) ] / h

= -sin{ {2[x-(pi/8)]+h}/2 }* [ sin(h/2) / (h/2) ] --------------(1)

using lim [ (sinx)/x ] =1 as x -->0

taking lim as h --> 0 in equation(1)

f ' (x)  = - sin[x-(pi/8)]

13. limits and derivatives
miscellaneous exercise

1.

(i)  find the derivative of (-x) using first principles
solution

(ii)   find the derivative of [ (-x)^(-1) ] using first principles

solution

(iii)find the derivative of sin(x+1) using first principles

solution

(iv) find the derivative of cos[x-(pi/8)] using first principles
solution




disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies.

miscellaneous trigonometry question 6 cbse ncert 11th mathematics

miscellaneous trigonometry question 6 cbse ncert 11th mathematics

6. Show that [sin7x+sin5x +sin9x+sin3x] / [cos7x+cos5x+cos9x+cos3x] = tan6x

using trigonometry formula trigonometry identities

sinx + siny =2sin[(x+y)/2] cos[(x-y)/2]

cosx + cosy =2cos[(x+y)/2] cos[(x-y)/2]

LHS =

[(sin7x+sin5x )+(sin9x+sin3x)] / [(cos7x+cos5x)+(cos9x+cos3x)]

= [2sin(12x/2)cos(2x/2)+2sin(12x/2)cos(6x/2)] / [2cos(12x/2)cos(2x/2)+2cos(12x/2)cos(6x/2)]

= [2sin6xcosx +2sin6xcos3x] / [2cos6xcosx +2cos6xcos3x]

={2sin6x[cosx +cos3x]}/{2cos6x[cosx+cos3x]}

=sin6x/cos6x ,cancelling off the common factors

=tan6x











1.Prove that 2cos(pi/13)cos(9pi/13)+cos (3pi/13)+cos(5pi/13) = 0
solution 

2. Prove that ( sin3x + sinx ) sinx + (cos3x - cosx) cosx = 0

solution

3. Prove that (cosx +cosy)^2 + ( sinx - siny )^2 = 4 { cos[(x+y)/2] }^2

solution

4. Prove that (cosx - cosy)^2 + ( sinx - siny )^2 = 4 { sin[(x-y)/2] }^2

solution 




5.Show that sinx +sin3x+ sin5x +sin7x = 4cosx cos2x sin 4x

solution

6. Show that [sin7x+sin5x +sin9x+sin3x] / [cos7x+cos5x+cos9x+cos3x] = tan6x


solution


disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work