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General
Easy

Question

The conic with length of latus rectum 6 and eccentricity is

  1. 3 over r equals square root of 2 plus C o s space theta    
  2. 3 over r equals 1 plus square root of 2 C o s space theta    
  3. 6 over r equals square root of 2 plus C o s space theta    
  4. 6 over r equals 1 plus square root of 2 C o s space theta    

The correct answer is: 3 over r equals 1 plus square root of 2 C o s space theta

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Related Questions to study

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For the circle r equals 6 C o s space theta centre and radius are

For the circle r equals 6 C o s space theta centre and radius are

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The polar equation of the circle of radius 5 and touching the initial line at the pole is

The polar equation of the circle of radius 5 and touching the initial line at the pole is

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The circle with centre at and radius 2 is

The circle with centre at and radius 2 is

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General
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Statement-I : If fraction numerator x squared plus 3 x plus 1 over denominator x squared plus 2 x plus 1 end fraction equals A plus fraction numerator B over denominator x plus 1 end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis squared end fraction text  then  end text bold italic A plus bold italic B plus bold italic C equals bold 0
Statement-II :If fraction numerator x squared plus 2 x plus 3 over denominator x cubed end fraction equals A over x plus B over x squared plus C over x cubed text  then  end text A plus B minus C equals 0

Which of the above statements is true

Statement-I : If fraction numerator x squared plus 3 x plus 1 over denominator x squared plus 2 x plus 1 end fraction equals A plus fraction numerator B over denominator x plus 1 end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis squared end fraction text  then  end text bold italic A plus bold italic B plus bold italic C equals bold 0
Statement-II :If fraction numerator x squared plus 2 x plus 3 over denominator x cubed end fraction equals A over x plus B over x squared plus C over x cubed text  then  end text A plus B minus C equals 0

Which of the above statements is true

Maths-General
General
Maths-

If x, y are rational number such that x plus y plus left parenthesis x minus 2 y right parenthesis square root of 2 equals 2 x minus y plus left parenthesis x minus y minus 1 right parenthesis square root of 6 Then

If x, y are rational number such that x plus y plus left parenthesis x minus 2 y right parenthesis square root of 2 equals 2 x minus y plus left parenthesis x minus y minus 1 right parenthesis square root of 6 Then

Maths-General
General
maths-

A class contains 4 boys and g girls. Every Sunday five students, including at least three boys go for a picnic to Appu Ghar, a different group being sent every week. During, the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was 85, then value of g is -

Number of groups having 4 boys and 1 girl
= (4C4) (gC1) = g
and number of groups having 3 boys and 2 girls
= (4C3) (gC2) = 2g(g – 1)
Thus, the number of dolls distributed
= g(1) + (2)[2g (g – 1)]
= 4g2 – 3g
We are given 4g2 – 3g = 85 rightwards double arrow g = 5.

A class contains 4 boys and g girls. Every Sunday five students, including at least three boys go for a picnic to Appu Ghar, a different group being sent every week. During, the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was 85, then value of g is -

maths-General
Number of groups having 4 boys and 1 girl
= (4C4) (gC1) = g
and number of groups having 3 boys and 2 girls
= (4C3) (gC2) = 2g(g – 1)
Thus, the number of dolls distributed
= g(1) + (2)[2g (g – 1)]
= 4g2 – 3g
We are given 4g2 – 3g = 85 rightwards double arrow g = 5.
General
maths-

There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls, is -

Let the number of yellow balls be x, that of black be 2x and that of green be y. Then
x + 2x + y = 20 or 3x + y = 20
rightwards double arrowy = 20 – 3x.
As 0 less or equal than y 20, we get 0 less or equal than 20 – 3x less or equal than 20
rightwards double arrow 0 less or equal than 3x less or equal than 20 or 0 less or equal thanless or equal than 6
thereforeThe number of ways of selecting the balls is 7.

There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls, is -

maths-General
Let the number of yellow balls be x, that of black be 2x and that of green be y. Then
x + 2x + y = 20 or 3x + y = 20
rightwards double arrowy = 20 – 3x.
As 0 less or equal than y 20, we get 0 less or equal than 20 – 3x less or equal than 20
rightwards double arrow 0 less or equal than 3x less or equal than 20 or 0 less or equal thanless or equal than 6
thereforeThe number of ways of selecting the balls is 7.
General
maths-

If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a3 + b3 + c3 must be divisible by -

We have (a3 – a) + (b3 – b) + (c3 – c)
= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)
Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,
6 | {(a3 – a) + (b3 – b) + (c3 – c)}
rightwards double arrow6 | (a3 + b3 + c3) as 6|(a + b + c).

If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a3 + b3 + c3 must be divisible by -

maths-General
We have (a3 – a) + (b3 – b) + (c3 – c)
= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)
Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,
6 | {(a3 – a) + (b3 – b) + (c3 – c)}
rightwards double arrow6 | (a3 + b3 + c3) as 6|(a + b + c).
General
maths-

For x element of R, let [x] denote the greatest integer less or equal than x, then value ofopen square brackets negative fraction numerator 1 over denominator 3 end fraction close square brackets+open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 1 over denominator 100 end fraction close square brackets+ open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 2 over denominator 100 end fraction close square brackets+…+open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 99 over denominator 100 end fraction close square bracketsis -

For 0 less or equal thanless or equal than 66, 0 less or equal than fraction numerator r over denominator 100 end fraction< fraction numerator 2 over denominator 3 end fraction
rightwards double arrowfraction numerator 2 over denominator 3 end fraction < – fraction numerator r over denominator 100 end fraction less or equal than 0
rightwards double arrowfraction numerator 1 over denominator 3 end fractionfraction numerator 2 over denominator 3 end fraction< –fraction numerator 1 over denominator 3 end fractionfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 1 over denominator 3 end fraction
thereforeopen square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= –1 for 0 less or equal thanless or equal than 66
Also, for 67 less or equal thanless or equal than 100, fraction numerator 67 over denominator 100 end fractionless or equal than fraction numerator r over denominator 100 end fraction less or equal than 1
rightwards double arrow–1 less or equal thanfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 67 over denominator 100 end fraction
rightwards double arrowfraction numerator 1 over denominator 3 end fraction – 1 less or equal thanfraction numerator 1 over denominator 3 end fractionfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 1 over denominator 3 end fractionfraction numerator 67 over denominator 100 end fraction
thereforeopen square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= –2 for 67 less or equal thanless or equal than 100
Hence, not stretchy sum subscript r equals 0 end subscript superscript 100 end superscript open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= 67(–1) + 2(–34) = –135.

For x element of R, let [x] denote the greatest integer less or equal than x, then value ofopen square brackets negative fraction numerator 1 over denominator 3 end fraction close square brackets+open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 1 over denominator 100 end fraction close square brackets+ open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 2 over denominator 100 end fraction close square brackets+…+open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 99 over denominator 100 end fraction close square bracketsis -

maths-General
For 0 less or equal thanless or equal than 66, 0 less or equal than fraction numerator r over denominator 100 end fraction< fraction numerator 2 over denominator 3 end fraction
rightwards double arrowfraction numerator 2 over denominator 3 end fraction < – fraction numerator r over denominator 100 end fraction less or equal than 0
rightwards double arrowfraction numerator 1 over denominator 3 end fractionfraction numerator 2 over denominator 3 end fraction< –fraction numerator 1 over denominator 3 end fractionfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 1 over denominator 3 end fraction
thereforeopen square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= –1 for 0 less or equal thanless or equal than 66
Also, for 67 less or equal thanless or equal than 100, fraction numerator 67 over denominator 100 end fractionless or equal than fraction numerator r over denominator 100 end fraction less or equal than 1
rightwards double arrow–1 less or equal thanfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 67 over denominator 100 end fraction
rightwards double arrowfraction numerator 1 over denominator 3 end fraction – 1 less or equal thanfraction numerator 1 over denominator 3 end fractionfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 1 over denominator 3 end fractionfraction numerator 67 over denominator 100 end fraction
thereforeopen square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= –2 for 67 less or equal thanless or equal than 100
Hence, not stretchy sum subscript r equals 0 end subscript superscript 100 end superscript open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= 67(–1) + 2(–34) = –135.
General
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The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, z less or equal than –1

Let t = z + 1
Equation reduces to x + y + t = 25
less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, t greater or equal than 0
Required number of ways
= Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
= Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
= Coefficient of x12 in fraction numerator left parenthesis 1 minus x to the power of 6 end exponent right parenthesis over denominator left parenthesis 1 minus x right parenthesis end fraction (1 – x)–1
= Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
= Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
= 12+3–1C3–17+3–1C3–16+3–1C3–1 + 1+3–1C3–1
= 14C29C28C2 + 3C2
= 91 – 36 – 28 + 3 = 30.

The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, z less or equal than –1

maths-General
Let t = z + 1
Equation reduces to x + y + t = 25
less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, t greater or equal than 0
Required number of ways
= Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
= Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
= Coefficient of x12 in fraction numerator left parenthesis 1 minus x to the power of 6 end exponent right parenthesis over denominator left parenthesis 1 minus x right parenthesis end fraction (1 – x)–1
= Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
= Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
= 12+3–1C3–17+3–1C3–16+3–1C3–1 + 1+3–1C3–1
= 14C29C28C2 + 3C2
= 91 – 36 – 28 + 3 = 30.
General
maths-

Ravish write letters to his five friends and address the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes -

Required number of ways not stretchy sum subscript r equals 2 end subscript superscript 5 end superscript 5 C subscript 5 minus r end subscript D(r)
equals sum from r equals 2 to 5 of   fraction numerator 5 factorial over denominator r factorial left parenthesis 5 minus r right parenthesis factorial end fraction r factorial times open curly brackets 1 minus fraction numerator 1 over denominator 1 factorial end fraction plus fraction numerator 1 over denominator 2 factorial end fraction minus fraction numerator 1 over denominator 3 factorial end fraction plus horizontal ellipsis plus fraction numerator left parenthesis negative 1 right parenthesis squared over denominator r factorial end fraction close curly brackets
equals sum from r equals 2 to 5 of   fraction numerator 5 factorial over denominator left parenthesis 5 minus r right parenthesis factorial end fraction=open curly brackets fraction numerator 1 over denominator 2 factorial end fraction – fraction numerator 1 over denominator 3 factorial end fraction plus.... plus fraction numerator left parenthesis negative 1 right parenthesis to the power of r end exponent over denominator r factorial end fraction close curly brackets
= 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)
= 10 + 20 + 45 + 44
= 119.

Ravish write letters to his five friends and address the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes -

maths-General
Required number of ways not stretchy sum subscript r equals 2 end subscript superscript 5 end superscript 5 C subscript 5 minus r end subscript D(r)
equals sum from r equals 2 to 5 of   fraction numerator 5 factorial over denominator r factorial left parenthesis 5 minus r right parenthesis factorial end fraction r factorial times open curly brackets 1 minus fraction numerator 1 over denominator 1 factorial end fraction plus fraction numerator 1 over denominator 2 factorial end fraction minus fraction numerator 1 over denominator 3 factorial end fraction plus horizontal ellipsis plus fraction numerator left parenthesis negative 1 right parenthesis squared over denominator r factorial end fraction close curly brackets
equals sum from r equals 2 to 5 of   fraction numerator 5 factorial over denominator left parenthesis 5 minus r right parenthesis factorial end fraction=open curly brackets fraction numerator 1 over denominator 2 factorial end fraction – fraction numerator 1 over denominator 3 factorial end fraction plus.... plus fraction numerator left parenthesis negative 1 right parenthesis to the power of r end exponent over denominator r factorial end fraction close curly brackets
= 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)
= 10 + 20 + 45 + 44
= 119.
General
maths-

In how many ways can we get a sum of at most 17 by throwing six distinct dice -

x1 + x2 + x3 + x4 + x5 + x6  17
When 1 less or equal than xi less or equal than 6, i = 1, 2, 3, …..6
Let x7 be a variable such that
x1 + x2 + x3 + x4 + x5 + x6 + x7 = 17
Clearly x7 greater or equal than 0 Required number of ways
= Coefficient of x17 in (x1 + x2 + ….. + x6)6
(1 + x + x2 + …..)
= Coefficient of x11 in open parentheses fraction numerator 1 minus x to the power of 6 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent open parentheses fraction numerator 1 over denominator 1 minus x end fraction close parentheses
= Coefficient of x11 in (1– 6C1 x6 + 6C2 x12……) (1 – x)–7
= Coefficient of x11 in (1 – x)–76C1 × coefficient of x5 in (1 – x)–7
= 11+7–1C7–16C1 × 7+5–1C7–1
= 17C6 – 6 × 11C6 = 9604.

In how many ways can we get a sum of at most 17 by throwing six distinct dice -

maths-General
x1 + x2 + x3 + x4 + x5 + x6  17
When 1 less or equal than xi less or equal than 6, i = 1, 2, 3, …..6
Let x7 be a variable such that
x1 + x2 + x3 + x4 + x5 + x6 + x7 = 17
Clearly x7 greater or equal than 0 Required number of ways
= Coefficient of x17 in (x1 + x2 + ….. + x6)6
(1 + x + x2 + …..)
= Coefficient of x11 in open parentheses fraction numerator 1 minus x to the power of 6 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent open parentheses fraction numerator 1 over denominator 1 minus x end fraction close parentheses
= Coefficient of x11 in (1– 6C1 x6 + 6C2 x12……) (1 – x)–7
= Coefficient of x11 in (1 – x)–76C1 × coefficient of x5 in (1 – x)–7
= 11+7–1C7–16C1 × 7+5–1C7–1
= 17C6 – 6 × 11C6 = 9604.
General
maths-

The number of non negative integral solutions of equation 3x + y + z = 24

3x + y + z = 24, x greater or equal than 0, y greater or equal than 0, z greater or equal than 0
Let x = k rightwards double arrow y + z = 24 – 3k …(1)
rightwards double arrow 24 – 3k greater or equal thanrightwards double arrowless or equal than 8
rightwards double arrow 0 less or equal thanless or equal than 8
For fixed value of k the number of solutions of (1) is
24–3k+2–1C2–1
= 25–3kC1
= 25 – 3k
Hence number of solutions
not stretchy sum subscript k equals 0 end subscript superscript 8 end superscript left parenthesis 25 minus 3 k right parenthesis= 25 × 9 – fraction numerator 3 cross times 8 cross times 9 over denominator 2 end fraction= 225 – 108 = 117.

The number of non negative integral solutions of equation 3x + y + z = 24

maths-General
3x + y + z = 24, x greater or equal than 0, y greater or equal than 0, z greater or equal than 0
Let x = k rightwards double arrow y + z = 24 – 3k …(1)
rightwards double arrow 24 – 3k greater or equal thanrightwards double arrowless or equal than 8
rightwards double arrow 0 less or equal thanless or equal than 8
For fixed value of k the number of solutions of (1) is
24–3k+2–1C2–1
= 25–3kC1
= 25 – 3k
Hence number of solutions
not stretchy sum subscript k equals 0 end subscript superscript 8 end superscript left parenthesis 25 minus 3 k right parenthesis= 25 × 9 – fraction numerator 3 cross times 8 cross times 9 over denominator 2 end fraction= 225 – 108 = 117.
General
maths-

Sum of divisors of 25 ·37 ·53 · 72 is –

Any divisor of 25 · 37 · 53 · 72 is of the type of 2l 3m 5n 7p, where 0less or equal than  l less or equal than 5, 0 less or equal thanless or equal than 7, 0  n  3 and 0 less or equal thanless or equal than 2
Hence the sum of the divisors
= (1 + 2 + …… + 25) (1 + 3 + ……. + 37) (1 + 5 + 52 + 53) (1 + 7 + 72)
= open parentheses fraction numerator 2 to the power of 6 end exponent minus 1 over denominator 2 minus 1 end fraction close parentheses open parentheses fraction numerator 3 to the power of 8 end exponent minus 1 over denominator 3 minus 1 end fraction close parentheses open parentheses fraction numerator 5 to the power of 4 end exponent minus 1 over denominator 5 minus 1 end fraction close parentheses open parentheses fraction numerator 7 to the power of 3 end exponent minus 1 over denominator 7 minus 1 end fraction close parentheses
=fraction numerator left parenthesis 2 to the power of 6 end exponent – 1 right parenthesis left parenthesis 3 to the power of 8 end exponent minus 1 right parenthesis left parenthesis 5 to the power of 4 end exponent minus 1 right parenthesis left parenthesis 7 to the power of 3 end exponent minus 1 right parenthesis over denominator 2 times 4 times 6 end fraction.

Sum of divisors of 25 ·37 ·53 · 72 is –

maths-General
Any divisor of 25 · 37 · 53 · 72 is of the type of 2l 3m 5n 7p, where 0less or equal than  l less or equal than 5, 0 less or equal thanless or equal than 7, 0  n  3 and 0 less or equal thanless or equal than 2
Hence the sum of the divisors
= (1 + 2 + …… + 25) (1 + 3 + ……. + 37) (1 + 5 + 52 + 53) (1 + 7 + 72)
= open parentheses fraction numerator 2 to the power of 6 end exponent minus 1 over denominator 2 minus 1 end fraction close parentheses open parentheses fraction numerator 3 to the power of 8 end exponent minus 1 over denominator 3 minus 1 end fraction close parentheses open parentheses fraction numerator 5 to the power of 4 end exponent minus 1 over denominator 5 minus 1 end fraction close parentheses open parentheses fraction numerator 7 to the power of 3 end exponent minus 1 over denominator 7 minus 1 end fraction close parentheses
=fraction numerator left parenthesis 2 to the power of 6 end exponent – 1 right parenthesis left parenthesis 3 to the power of 8 end exponent minus 1 right parenthesis left parenthesis 5 to the power of 4 end exponent minus 1 right parenthesis left parenthesis 7 to the power of 3 end exponent minus 1 right parenthesis over denominator 2 times 4 times 6 end fraction.
General
maths-

The length of the perpendicular from the pole to the straight line fraction numerator 6 square root of 2 over denominator r end fraction equals C o s space theta plus S i n space theta is

The length of the perpendicular from the pole to the straight line fraction numerator 6 square root of 2 over denominator r end fraction equals C o s space theta plus S i n space theta is

maths-General