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HSC Science (General) 11th Standard - Maharashtra State Board Question Bank Solutions

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Use binomial theorem to evaluate the following upto four places of decimal

`root(4)(16.08)`

[2.4] Methods of Induction and Binomial Theorem
Chapter: [2.4] Methods of Induction and Binomial Theorem
Concept: undefined >> undefined

Use binomial theorem to evaluate the following upto four places of decimal

(1.02)–5 

[2.4] Methods of Induction and Binomial Theorem
Chapter: [2.4] Methods of Induction and Binomial Theorem
Concept: undefined >> undefined

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Use binomial theorem to evaluate the following upto four places of decimal

(0.98)–3 

[2.4] Methods of Induction and Binomial Theorem
Chapter: [2.4] Methods of Induction and Binomial Theorem
Concept: undefined >> undefined

Show That C1 + C2 + C3 + .... Cn = 2n − 1

[2.4] Methods of Induction and Binomial Theorem
Chapter: [2.4] Methods of Induction and Binomial Theorem
Concept: undefined >> undefined

Show That C0 + 2C1 + 3C2 + 4C3 + ... + (n + 1)Cn = (n + 2)2n−1

[2.4] Methods of Induction and Binomial Theorem
Chapter: [2.4] Methods of Induction and Binomial Theorem
Concept: undefined >> undefined

Answer the following:

Expand `((2x)/3 - 3/(2x))^4`

[2.4] Methods of Induction and Binomial Theorem
Chapter: [2.4] Methods of Induction and Binomial Theorem
Concept: undefined >> undefined

Answer the following:

Using binomial theorem, find the value of `root(3)(995)` upto four places of decimals

[2.4] Methods of Induction and Binomial Theorem
Chapter: [2.4] Methods of Induction and Binomial Theorem
Concept: undefined >> undefined

Answer the following:

Find approximate value of `1/4.08` upto four places of decimals

[2.4] Methods of Induction and Binomial Theorem
Chapter: [2.4] Methods of Induction and Binomial Theorem
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> pi/2) [("cosec"x - 1)/(pi/2 - x)^2]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> "a") [(sinx - sin"a")/(root(5)(x) - root(5)("a"))]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> pi) [(sqrt(5 + cosx) - 2)/(pi - x)^2]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> pi/6) [(cos x - sqrt(3) sinx)/(pi - 6x)]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> 1) [(1 - x^2)/(sinpix)]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following:

`lim_(x→π/6) [(2sinx − 1)/(π − 6x)]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> pi/4) [(sqrt(2) - cosx - sinx)/(4x - pi)^2]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> pi/6) [(2 - sqrt(3)cosx - sinx)/(6x - pi)^2]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> "a") [(sin(sqrt(x)) - sin(sqrt("a")))/(x - "a")]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following:

`lim_(x -> pi/2) [(cos3x + 3cosx)/(2x - pi)^3]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> 0)[((1 - x)^5 - 1)/((1 - x)^3 - 1)]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> 1) [(2^(2x - 2) - 2^x + 1)/(sin^2 (x - 1))]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined
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