HSC Arts (English Medium) 11th Standard - Maharashtra State Board Question Bank Solutions

Answer the following:

Find the middle term (s) in the expansion of (x²+ 2y²)⁷ 

Answer the following:

Find the middle term (s) in the expansion of `((3x^2)/2 - 1/(3x))^9`

Answer the following:

If the middle term in the expansion of `(x + "b"/x)^6` is 160, find b

Evaluate the following limit :

`lim_(theta -> 0) [(sin("m"theta))/(tan("n"theta))]`

Evaluate the following limit :

`lim_(theta -> 0) [(1 - cos2theta)/theta^2]`

Evaluate the following limit :

`lim_(x -> 0) [(x*tanx)/(1 - cosx)]`

Evaluate the following limit :

`lim_(x ->0)((secx - 1)/x^2)`

Evaluate the following limit :

`lim_(x -> 0)[(1 - cos("n"x))/(1 - cos("m"x))]`

Evaluate the following limit :

`lim_(x -> pi/6) [(2 - "cosec"x)/(cot^2x - 3)]`

Evaluate the following limit :

`lim_(x -> pi/4) [(cosx - sinx)/(cos2x)]`

Evaluate the following limit :

`lim_(x -> 0) [(cos("a"x) - cos("b"x))/(cos("c"x) - 1)]`

Evaluate the following limit :

`lim_(x -> pi) [(sqrt(1 - cosx) - sqrt(2))/(sin^2 x)]`

Evaluate the following limit :

`lim_(x -> pi/4) [(tan^2x - cot^2x)/(secx - "cosec"x)]`

Evaluate the following limit :

`lim_(x -> pi/6) [(2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)]`

Select the correct answer from the given alternatives.

`lim_(x → π/3) ((tan^2x - 3)/(sec^3x - 8))` =

Select the correct answer from the given alternatives.

`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =

Select the correct answer from the given alternatives.

`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =

Evaluate the following :

`lim_(x -> 0)[(secx^2 - 1)/x^4]`

Evaluate the following :

`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`

Evaluate the following :

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