Consider a function f defined by f(x)= sin -1((2 tan (x/2)/1+ tan 2 (x)2)) then

Question

Consider a function f defined by f(x)= sin -1((2 tan (x/2)/1+ tan 2 (x)2)) then (A) Range of f(x) is [(-π/2), (π/2)] (B) y=f(x) and y= sin -1( sin x) have same graphs. (C) y = f ( x ) is periodic with period 2 π. (D) Number of solutions of f ( x )=0 in [0,4 π] is 5 .

Tardigrade Admin · Accepted Answer

f ( x )= sin -1( sin x ), x ≠(2 n +1) π ⇒ f(x) and y= sin -1( sin x) are not identical <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/ebc548a9574fd706133dad1ecbc8fa4f-.png" /> Range of f ( x )=[(-π/2), (π/2)] and period =2 π

Q. Consider a function $f$ defined by $f (x) = sin^{- 1} (\frac{2 t a n \frac{x}{2}}{1 + t a n ^{2} \frac{x}{2}})$ then

Range of $f (x)$ is $[\frac{- π}{2}, \frac{π}{2}]$

$y = f (x)$ and $y = sin^{- 1} (sin x)$ have same graphs.

$y = f (x)$ is periodic with period $2 π$ .

Number of solutions of $f (x) = 0$ in $[0, 4 π]$ is 5 .

$f (x) = sin^{- 1} (sin x), x \neq = (2 n + 1) π$
$\Rightarrow f (x)$ and $y = sin^{- 1} (sin x)$ are not identical

Range of $f (x) = [\frac{- π}{2}, \frac{π}{2}]$ and period $= 2 π$

Q. Consider a function f defined by f(x)=sin−1(1+tan22x​2tan2x​​) then

Range of f(x) is [2−π​,2π​]

y=f(x) and y=sin−1(sinx) have same graphs.

y=f(x) is periodic with period 2π.

Number of solutions of f(x)=0 in [0,4π] is 5 .