The least positive integer n for which 1-i-1-in-2- -11-x-sin -1 x- x-0-1 - x - 1 is-A- 2B- 6C- 8D- 4

The correct option is D 4 sec^ 1 (1/x) + sin^ 1 x = cos^ 1 x + sin^ 1 x = π/2 ∴(1 + i/1 i)^n = 1 ⇒[(1 + i)^2/2]^n = i^n = 1 ⇒ least n = 4 ...

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Standard XI

Mathematics

Trigonometric Ratios Using Right Angled Triangle

The least pos...

Question

The least positive integer n for which ${(\frac{1 + i}{1 - i})}^{n} = \frac{2}{π} (s e c^{- 1} \frac{1}{x} + s i n^{- 1} x), x \neq 0, - 1 \leq x \leq 1$ is:

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Solution

The correct option is D
4

$s e c^{- 1} (\frac{1}{x}) + s i n^{- 1} x = c o s^{- 1} x + s i n^{- 1} x = \frac{π}{2}$

$∴ {(\frac{1 + i}{1 - i})}^{n} = 1 \Rightarrow {[\frac{(1 + i)^{2}}{2}]}^{n} = i^{n} = 1$

$\Rightarrow$ least n = 4

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Similar questions

Q. The value of x for which

s i n (c o t^{- 1} (1 + x)) = c o s (t a n^{- 1} x)

is:

Q. The least positive integer

n

for which

{(\frac{1 + i}{1 - i})}^{n} = \frac{2}{π} {sin}^{- 1} (\frac{1 + x^{2}}{2 x})

, where

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, is

Q. For any positive integer

n

, define

f_{n} : (0, \infty) \to R

f_{n} (x) = n \sum j = 1 {tan}^{- 1} (\frac{1}{1 + (x + j) (x + j - 1)})

for all

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(

Here, the inverse trigonometric function

{tan}^{- 1} x

assumes values in

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Then, which of the following statement(s) is (are) TRUE?

Q. For any positive integer

n

, define

f_{n} : (0, \infty) \to R

f_{n} (x) = n \sum j = 1 {tan}^{- 1} (\frac{1}{1 + (x + j) (x + j - 1)})

for all

x ϵ (0, \infty)

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(Here, the inverse triagonometric function

{tan}^{- 1} x

assume values in

(- \frac{π}{2}, \frac{π}{2})

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Then, which of the following statement(s) is (are) TRUE?

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
(a)

{s i n}^{- 1} (3 x / 5) + {s i n}^{- 1} (4 x / 5) = {s i n}^{- 1} x

(b)

{c o s}^{- 1} x + {s i n}^{- 1} (\frac{1}{2} x) = \frac{π}{6}

a \leq {t a n}^{- 1} (\frac{1 - x}{1 + x}) \leq b

where

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then

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(a)

(0, π)

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(0, π / 4)

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(- π / 4, π / 4)

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(π / 4, π / 2)

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a \leq {({s i n}^{- 1} x)}^{3} + {({c o s}^{- 1} x)}^{3} \leq b

then (a,b) is equal to

(\frac{π^{3}}{32}, \frac{7 π^{3}}{8})

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The least positive integer n for which (1+i1−i)n=2π(sec−11x+sin−1x),x≠0,−1≤x≤1 is:

The correct option is D 4 sec−1(1x)+sin−1x=cos−1x+sin−1x=π2 ∴(1+i1−i)n=1⇒[(1+i)22]n=in=1 ⇒ least n = 4

The least positive integer n for which ${(\frac{1 + i}{1 - i})}^{n} = \frac{2}{π} (s e c^{- 1} \frac{1}{x} + s i n^{- 1} x), x \neq 0, - 1 \leq x \leq 1$ is:

The correct option is D
4

$s e c^{- 1} (\frac{1}{x}) + s i n^{- 1} x = c o s^{- 1} x + s i n^{- 1} x = \frac{π}{2}$

$∴ {(\frac{1 + i}{1 - i})}^{n} = 1 \Rightarrow {[\frac{(1 + i)^{2}}{2}]}^{n} = i^{n} = 1$

$\Rightarrow$ least n = 4