The range of the function f(x)=sin- 1[x2 - (1/3)]-cos- 1[x2 + (2/3)] is (where, [x] represents the greatest integer value of x )

Question

The range of the function f(x)=sin- 1[x2 - (1/3)]-cos- 1[x2 + (2/3)] is (where, [x] represents the greatest integer value of x ) (A) [- π , 0] (B)  - π , 0  (C)  0 , π  (D)  0 , π , - π

Tardigrade Admin · Accepted Answer

[x2 - (1/3)]=[x2 + (2/3) - 1]=[x2 + (2/3)]-1=k(let) Now, the function is sin- 1k-cos- 1(k + 1) For the above function to be defined, -1≤ k≤ 1 &-1≤ (k + 1)≤ 1 ∴ k= - 1,0 ∴ Range = s i n- 1 (- 1) - c o s- 1 (0) , s i n- 1 (0) - c o s- 1 (1) = - π , 0

Q. The range of the function $f (x) = s i n^{- 1} [x^{2} - \frac{1}{3}] - co s^{- 1} [x^{2} + \frac{2}{3}]$ is (where, $[x]$ represents the greatest integer value of $x$ )

$[- π, 0]$

${- π, 0}$

${0, π}$

${0, π, - π}$

Q. The range of the function f(x)=sin−1[x2−31​]−cos−1[x2+32​] is (where, [x] represents the greatest integer value of x )

[−π,0]

{−π,0}

{0,π}

{0,π,−π}

[x2−31​]=[x2+32​−1]=[x2+32​]−1=k(let) Now, the function is sin−1k−cos−1(k+1) For the above function to be defined, −1≤k≤1&−1≤(k+1)≤1 ∴k={−1,0} ∴ Range ={sin−1(−1)−cos−1(0),sin−1(0)−cos−1(1)} ={−π,0}

Q. The range of the function $f (x) = s i n^{- 1} [x^{2} - \frac{1}{3}] - co s^{- 1} [x^{2} + \frac{2}{3}]$ is (where, $[x]$ represents the greatest integer value of $x$ )

$[- π, 0]$

${- π, 0}$

${0, π}$

${0, π, - π}$