HSC Commerce (English Medium) ११ वीं कक्षा - Maharashtra State Board Question Bank Solutions

By using properties of determinants, prove that `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|` = 0.

Without expanding the determinants, show that `|("b" + "c", "bc", "b"^2"c"^2),("c" + "a", "ca", "c"^2"a"^2),("a" +  "b", "ab", "a"^2"b"^2)|` = 0

Without expanding the determinants, show that `|(x"a", y"b", z"c"),("a"^2, "b"^2, "c"^2),(1, 1, 1)| = |(x, y, z),("a", "b", "c"),("bc", "ca", "ab")|`

Without expanding the determinants, show that `|(l, "m", "n"),("e", "d", "f"),("u", "v", "w")| = |("n", "f", "w"),(l, "e", "u"),("m", "d", "v")|`

Without expanding the determinants, show that `|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|` = 0

Evaluate the following limit:

`lim_(x -> 3) [sqrt(x + 6)/x]`

Evaluate the following limits: `lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`

Evaluate the following limit:

`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`

Evaluate the following limits: if `lim_(x -> 1)[(x^4 - 1)/(x - 1)] = lim_(x -> "a") [(x^3 - "a"^3)/(x - "a")]`, find all the value of a.

Evaluate the following limit:

`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`

Evaluate the following limits: if `lim_(x -> 5)[(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.

Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`

Evaluate the following limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`

Evaluate the following limits: `lim_(y -> 1) [(2y - 2)/(root(3)(7 + y) - 2)]`

Evaluate the following limits: `lim_(x -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`

Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`

if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.

Evaluate the following Limits: `lim_(x -> "a") ((x + 2)^(5/3) - ("a" + 2)^(5/3))/(x - "a")`

Evaluate the following Limit:

`lim_(x -> 0) ((1 + x)^"n" - 1)/x`

Examine whether the function is continuous at the points indicated against them :

f(x) `{:( = (x^2 + 18x - 19)/(x - 1)",",  "for"  x ≠ 1),(= 20",",  "for"  x = 1):}}` at x = 1