General
Easy
Maths-

The condition for the lines c subscript 1 over r equals a subscript 1 c o s space theta plus b subscript 1 s i n space theta and c subscript 2 over r equals a subscript 2 c o s space theta plus b subscript 2 s i n space theta to be perpendicular is

Maths-General

  1. a subscript 1 end subscript a subscript 2 end subscript plus b subscript 1 end subscript text end text b subscript 2 end subscript equals 0    
  2. a subscript 1 end subscript text end text b subscript 1 end subscript plus a subscript 2 end subscript text end text b subscript 2 end subscript equals 0    
  3. r left parenthesis a c o s space theta plus b s i n space theta right parenthesis equals a b    
  4. r left parenthesis a c o s space theta minus b sin space theta right parenthesis equals a b    

    Answer:The correct answer is: a subscript 1 end subscript a subscript 2 end subscript plus b subscript 1 end subscript text end text b subscript 2 end subscript equals 0

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    Related Questions to study

    General
    maths-

    If f : R →R; f(x) = sin x + x, then the value of not stretchy integral subscript 0 end subscript superscript pi end superscript blank (f-1 (x)) dx, is equal to

    If f : R →R; f(x) = sin x + x, then the value of not stretchy integral subscript 0 end subscript superscript pi end superscript blank (f-1 (x)) dx, is equal to

    maths-General
    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
    General
    maths-

    The polar equation of the straight line with intercepts 'a' and 'b' on the rays theta equals 0 and theta equals fraction numerator pi over denominator 2 end fraction respectively is

    The polar equation of the straight line with intercepts 'a' and 'b' on the rays theta equals 0 and theta equals fraction numerator pi over denominator 2 end fraction respectively is

    maths-General
    General
    maths-

    The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is

    The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is

    maths-General
    General
    maths-

    The polar equation of x to the power of 3 end exponent plus y to the power of 3 end exponent equals 3 axy is

    The polar equation of x to the power of 3 end exponent plus y to the power of 3 end exponent equals 3 axy is

    maths-General
    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 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-

    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-

    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-

    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-

    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
    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-

    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-

    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-

    If x, y, z are integers and x greater or equal than 0, y greater or equal than 1, z greater or equal than 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -

    Let y = p + 1 and z = q + 2.
    Then x greater or equal than 0, p greater or equal than 0, q greater or equal than 0 and x + y + z = 15
    rightwards double arrow x + p + q = 12
    therefore The reqd. number of values of (x, y, z) and hence of (x, p, q)
    = No. of non-negative integral solutions of x + p + q= 12
    = Coeff. of x12 in (x0 + x1 + x2 + ……)3
    = Coeff. of x12 in (1 – x)–3
    = Coeff. of x12 in [2C0 + 3C1 x + 4C2 x2 + ….]
    = 14C12 = fraction numerator 14 factorial over denominator 2 factorial 12 factorial end fraction = fraction numerator 14 cross times 13 over denominator 2 end fraction = 91.

    If x, y, z are integers and x greater or equal than 0, y greater or equal than 1, z greater or equal than 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -

    maths-General
    Let y = p + 1 and z = q + 2.
    Then x greater or equal than 0, p greater or equal than 0, q greater or equal than 0 and x + y + z = 15
    rightwards double arrow x + p + q = 12
    therefore The reqd. number of values of (x, y, z) and hence of (x, p, q)
    = No. of non-negative integral solutions of x + p + q= 12
    = Coeff. of x12 in (x0 + x1 + x2 + ……)3
    = Coeff. of x12 in (1 – x)–3
    = Coeff. of x12 in [2C0 + 3C1 x + 4C2 x2 + ….]
    = 14C12 = fraction numerator 14 factorial over denominator 2 factorial 12 factorial end fraction = fraction numerator 14 cross times 13 over denominator 2 end fraction = 91.