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

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

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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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

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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

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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

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integral using substitution and then integration by parts

integral using substitution and then integration by parts

integral of [1 / (x^4)][sqrt( 1 + (x^2))][ log( 1 + (x^2)) - 2log( x)]

first simplify using property of logarithms

take x^2 common from the sqrt term obtain [1+1/(x^2)] and try to get the same term inside the log expression

cancel off the x to get 1 /[x^3]   then use substitution

then use integration by parts with log(t) as the first function

formulae on integration
 
PAGE 1 BASIC INTEGRATION

PAGE 2 INTEGRATION BY SUBSTITUTION

 PAGE 3 INTEGRATION BY COMPLETION OF SQUARES

PAGE 4 INTEGRATION BY PARTS

PAGE 5 INTEGRATION BY MANIPULATION OF NUMERATOR IN TERMS OF DENOMINATOR


PAGE 6 INTEGRATION USING PARTIAL FRACTIONS

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integral of sqrt{[1-sqrt(x)] / [1+sqrt(x)]}

integral of sqrt{[1-sqrt(x)] / [1+sqrt(x)]}

use the substitution sqrt(x) = cost

use trigonometric formulae to simplify [1-cos t]  and [ 1+ cos t], sint using the half angle formulae

cancel off the common factors

simplify then again use trigonometric formulae to change the square terms to first degree expressions before integrating.

again use trigonometric formulae to change the variable back to x

trigonometric identities 

formulae on integration
 
PAGE 1 BASIC INTEGRATION

PAGE 2 INTEGRATION BY SUBSTITUTION

 PAGE 3 INTEGRATION BY COMPLETION OF SQUARES

PAGE 4 INTEGRATION BY PARTS

PAGE 5 INTEGRATION BY MANIPULATION OF NUMERATOR IN TERMS OF DENOMINATOR


PAGE 6 INTEGRATION USING PARTIAL FRACTIONS

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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