Maths-

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### Question

#### For x R, let [x] denote the greatest integer x, then value of++ +…+is -

- –100
- –123
- –135
- –153

#### The correct answer is: –135

#### For 0 r 66, 0 <

– < – 0

– – < – – –

= –1 for 0 r 66

Also, for 67 r 100, 1

–1 – –

– – 1 – – – –

= –2 for 67 r 100

Hence, = 67(–1) + 2(–34) = –135.

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### Related Questions to study

maths-

#### The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 x 5, 12 y 18, z –1

Let t = z + 1

Equation reduces to x + y + t = 25

1 x 5, 12 y 18, t 0

Required number of ways

= Coefficient of x

= Coefficient of x

= Coefficient of x

= Coefficient of x

= Coefficient of x

=

=

= 91 – 36 – 28 + 3 = 30.

Equation reduces to x + y + t = 25

1 x 5, 12 y 18, t 0

Required number of ways

= Coefficient of x

^{25}in [(x + x^{2}+ x^{3}+ x^{4}+ x^{5}) (x^{12}+ x^{13}+…..+ x^{18}) (1 + x + x^{2}+ …..)]= Coefficient of x

^{12}in (1 + x + x^{2}+ x^{3}+ x^{4}) (1 + x + x^{2}+ x^{3}+ x^{4}+ x^{5}+ x^{6}) (1 + x + x^{2}+ …..)= Coefficient of x

^{12}in (1 – x)^{–1}= Coefficient of x

^{12}in (1 – x^{5}) (1 – x^{6}) (1 – x)^{–3}= Coefficient of x

^{12}in (1 – x^{5}– x^{6}+ x^{11}) (1 – x)^{–3}=

^{12+3–1}C_{3–1}–^{7+3–1}C_{3–1}–^{6+3–1}C_{3–1}+^{1+3–1}C_{3–1}=

^{14}C_{2}–^{9}C_{2}–^{8}C_{2}+^{3}C_{2}= 91 – 36 – 28 + 3 = 30.

#### The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 x 5, 12 y 18, z –1

maths-General

Let t = z + 1

Equation reduces to x + y + t = 25

1 x 5, 12 y 18, t 0

Required number of ways

= Coefficient of x

= Coefficient of x

= Coefficient of x

= Coefficient of x

= Coefficient of x

=

=

= 91 – 36 – 28 + 3 = 30.

Equation reduces to x + y + t = 25

1 x 5, 12 y 18, t 0

Required number of ways

= Coefficient of x

^{25}in [(x + x^{2}+ x^{3}+ x^{4}+ x^{5}) (x^{12}+ x^{13}+…..+ x^{18}) (1 + x + x^{2}+ …..)]= Coefficient of x

^{12}in (1 + x + x^{2}+ x^{3}+ x^{4}) (1 + x + x^{2}+ x^{3}+ x^{4}+ x^{5}+ x^{6}) (1 + x + x^{2}+ …..)= Coefficient of x

^{12}in (1 – x)^{–1}= Coefficient of x

^{12}in (1 – x^{5}) (1 – x^{6}) (1 – x)^{–3}= Coefficient of x

^{12}in (1 – x^{5}– x^{6}+ x^{11}) (1 – x)^{–3}=

^{12+3–1}C_{3–1}–^{7+3–1}C_{3–1}–^{6+3–1}C_{3–1}+^{1+3–1}C_{3–1}=

^{14}C_{2}–^{9}C_{2}–^{8}C_{2}+^{3}C_{2}= 91 – 36 – 28 + 3 = 30.

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 D(r)

=

= 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)

= 10 + 20 + 45 + 44

= 119.

=

= 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 D(r)

=

= 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)

= 10 + 20 + 45 + 44

= 119.

=

= 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)

= 10 + 20 + 45 + 44

= 119.

maths-

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

x

When 1 x

Let x

x

Clearly x

= Coefficient of x

(1 + x + x

= Coefficient of x

= Coefficient of x

= Coefficient of x

=

=

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6} 17When 1 x

_{i}6, i = 1, 2, 3, …..6Let x

_{7}be a variable such thatx

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6}+ x_{7}= 17Clearly x

_{7}0 Required number of ways= Coefficient of x

^{17}in (x^{1}+ x^{2}+ ….. + x^{6})^{6}(1 + x + x

^{2}+ …..)= Coefficient of x

^{11}in= Coefficient of x

^{11}in (1–^{ 6}C_{1}x^{6}+^{6}C_{2}x^{12}……) (1 – x)^{–7}= Coefficient of x

^{11 }in (1 – x)^{–7}–^{6}C_{1}× coefficient of x^{5}in (1 – x)^{–7}=

^{11+7–1}C_{7–1}–^{6}C_{1}×^{7+5–1}C_{7–1}=

^{17}C_{6}– 6 ×^{11}C_{6}= 9604.#### In how many ways can we get a sum of at most 17 by throwing six distinct dice -

maths-General

x

When 1 x

Let x

x

Clearly x

= Coefficient of x

(1 + x + x

= Coefficient of x

= Coefficient of x

= Coefficient of x

=

=

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6} 17When 1 x

_{i}6, i = 1, 2, 3, …..6Let x

_{7}be a variable such thatx

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6}+ x_{7}= 17Clearly x

_{7}0 Required number of ways= Coefficient of x

^{17}in (x^{1}+ x^{2}+ ….. + x^{6})^{6}(1 + x + x

^{2}+ …..)= Coefficient of x

^{11}in= Coefficient of x

^{11}in (1–^{ 6}C_{1}x^{6}+^{6}C_{2}x^{12}……) (1 – x)^{–7}= Coefficient of x

^{11 }in (1 – x)^{–7}–^{6}C_{1}× coefficient of x^{5}in (1 – x)^{–7}=

^{11+7–1}C_{7–1}–^{6}C_{1}×^{7+5–1}C_{7–1}=

^{17}C_{6}– 6 ×^{11}C_{6}= 9604.maths-

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

3x + y + z = 24, x 0, y 0, z 0

Let x = k y + z = 24 – 3k …(1)

24 – 3k 0 k 8

0 k 8

For fixed value of k the number of solutions of (1) is

=

= 25 – 3k

Hence number of solutions

= 25 × 9 – = 225 – 108 = 117.

Let x = k y + z = 24 – 3k …(1)

24 – 3k 0 k 8

0 k 8

For fixed value of k the number of solutions of (1) is

^{24–3k+2–1}C_{2–1}=

^{25–3k}C_{1}= 25 – 3k

Hence number of solutions

= 25 × 9 – = 225 – 108 = 117.

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

maths-General

3x + y + z = 24, x 0, y 0, z 0

Let x = k y + z = 24 – 3k …(1)

24 – 3k 0 k 8

0 k 8

For fixed value of k the number of solutions of (1) is

=

= 25 – 3k

Hence number of solutions

= 25 × 9 – = 225 – 108 = 117.

Let x = k y + z = 24 – 3k …(1)

24 – 3k 0 k 8

0 k 8

For fixed value of k the number of solutions of (1) is

^{24–3k+2–1}C_{2–1}=

^{25–3k}C_{1}= 25 – 3k

Hence number of solutions

= 25 × 9 – = 225 – 108 = 117.

maths-

#### Sum of divisors of 2^{5 }·3^{7 }·5^{3 }· 7^{2} is –

Any divisor of 2

Hence the sum of the divisors

= (1 + 2 + …… + 2

=

=.

^{5}· 3^{7}· 5^{3}· 7^{2}is of the type of 2*3*^{l}^{m}5^{n}7^{p}, where 0*l*5, 0 m 7, 0 n 3 and 0 p 2Hence the sum of the divisors

= (1 + 2 + …… + 2

^{5}) (1 + 3 + ……. + 3^{7}) (1 + 5 + 5^{2}+ 5^{3}) (1 + 7 + 7^{2})=

=.

#### Sum of divisors of 2^{5 }·3^{7 }·5^{3 }· 7^{2} is –

maths-General

Any divisor of 2

Hence the sum of the divisors

= (1 + 2 + …… + 2

=

=.

^{5}· 3^{7}· 5^{3}· 7^{2}is of the type of 2*3*^{l}^{m}5^{n}7^{p}, where 0*l*5, 0 m 7, 0 n 3 and 0 p 2Hence the sum of the divisors

= (1 + 2 + …… + 2

^{5}) (1 + 3 + ……. + 3^{7}) (1 + 5 + 5^{2}+ 5^{3}) (1 + 7 + 7^{2})=

=.

maths-

#### The length of the perpendicular from the pole to the straight line is

#### The length of the perpendicular from the pole to the straight line is

maths-General

maths-

#### The condition for the lines and to be perpendicular is

#### The condition for the lines and to be perpendicular is

maths-General

Maths-

#### If f : R →R; f(x) = sin x + x, then the value of (f^{-1} (x)) dx, is equal to

#### If f : R →R; f(x) = sin x + x, then the value of (f^{-1} (x)) dx, is equal to

Maths-General

maths-

#### The polar equation of the straight line with intercepts 'a' and 'b' on the rays and respectively is

#### The polar equation of the straight line with intercepts 'a' and 'b' on the rays and respectively is

maths-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

maths-

#### The polar equation of axy is

#### The polar equation of axy is

maths-General

maths-

#### If x, y, z are integers and x 0, y 1, z 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 0, p 0, q 0 and x + y + z = 15

x + p + q = 12

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 x

= Coeff. of x

= Coeff. of x

=

Then x 0, p 0, q 0 and x + y + z = 15

x + p + q = 12

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 x

^{12}in (x^{0}+ x^{1}+ x^{2}+ ……)^{3}= Coeff. of x

^{12}in (1 – x)^{–3}= Coeff. of x

^{12}in [^{2}C_{0}+^{3}C_{1}x +^{4}C_{2}x^{2}+ ….]=

^{14}C_{12}= = = 91.#### If x, y, z are integers and x 0, y 1, z 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 0, p 0, q 0 and x + y + z = 15

x + p + q = 12

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 x

= Coeff. of x

= Coeff. of x

=

Then x 0, p 0, q 0 and x + y + z = 15

x + p + q = 12

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 x

^{12}in (x^{0}+ x^{1}+ x^{2}+ ……)^{3}= Coeff. of x

^{12}in (1 – x)^{–3}= Coeff. of x

^{12}in [^{2}C_{0}+^{3}C_{1}x +^{4}C_{2}x^{2}+ ….]=

^{14}C_{12}= = = 91.Maths-

#### If then the equation whose roots are

#### If then the equation whose roots are

Maths-General

Maths-

#### Let p, q {1, 2, 3, 4}. Then number of equation of the form px^{2} + qx + 1 = 0, having real roots, is

#### Let p, q {1, 2, 3, 4}. Then number of equation of the form px^{2} + qx + 1 = 0, having real roots, is

Maths-General

Maths-

#### ax^{2} + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –

#### ax^{2} + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –

Maths-General