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Arts (English Medium) कक्षा १२ - CBSE Question Bank Solutions for Mathematics

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Mathematics
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Let f: R → R defined by f(x) = x4. Choose the correct answer

[1] Relations and Functions
Chapter: [1] Relations and Functions
Concept: undefined >> undefined

Let f: R → R defined by f(x) = 3x. Choose the correct answer

[1] Relations and Functions
Chapter: [1] Relations and Functions
Concept: undefined >> undefined

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The value of `"tan"^-1 (1/2) + "tan"^-1(1/3) + "tan"^-1(7/8)` is ____________.

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Solve for x : `"sin"^-1  2"x" + "sin"^-1  3"x" = pi/3`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

The value of `"tan"^-1 (3/4) + "tan"^-1 (1/7)` is ____________.

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

If `"tan"^-1 2  "x + tan"^-1 3  "x" = pi/4`, then x is ____________.

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

`"tan" (pi/4 + 1/2 "cos"^-1 "x") + "tan" (pi/4 - 1/2 "cos"^-1 "x") =` ____________.

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

`"tan"^-1 1/3 + "tan"^-1 1/5 + "tan"^-1 1/7 + "tan"^-1 1/8 =` ____________.

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

`"cos"^-1["cos"(2"cot"^-1(sqrt2 - 1))]` = ____________.

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

`"cos" (2  "tan"^-1 1/7) - "sin" (4  "sin"^-1 1/3) =` ____________.

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

The value of `"cos"^-1 ("cos" ((33pi)/5))` is ____________.

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

`"sin"^-1 (1 - "x") - 2  "sin"^-1 "x" = pi/2`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

If `6"sin"^-1 ("x"^2 - 6"x" + 8.5) = pi,` then x is equal to ____________.

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

`"cos"^-1 1/2 + 2  "sin"^-1 1/2` is equal to ____________.

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

`"tan"^-1 1 + "cos"^-1 ((-1)/2) + "sin"^-1 ((-1)/2)`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

`"sin"^-1 (1/sqrt2)`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

`"tan"^-1 (sqrt3)`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

`"cos"^-1 (1/2)`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

`"sin"^-1 ((-1)/2)`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

If `"sin" {"sin"^-1 (1/2) + "cos"^-1 "x"} = 1`, then the value of x is ____________.

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined
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