Solution:
Solution
\( <p>f (x) = sin <sup>-1</sup> \(\left[ {{x}^{2}}+\cfrac{1}{2} \right]+\)cos <sup>-1</sup>\(\left[ {{x}^{2}}-\cfrac{1}{2} \right]\)<br /> </p> <p>= sin <sup>-1</sup> \(\left[ {{x}^{2}}+\cfrac{1}{2} \right]+\) cos <sup>-1</sup> \(\left[ {{x}^{2}}+\cfrac{1}{2}-1 \right]\)<br /> </p> <p>= sin <sup>-1</sup> \(\left[ {{x}^{2}}+\cfrac{1}{2} \right]+\) cos <sup>-1</sup> \(\left( \left[ {{x}^{2}}+\cfrac{1}{2} \right]-1 \right)\)<br /> </p> <p>Since \({{x}^{2}}+\cfrac{1}{2}\) \(\ge \cfrac{1}{2}\), \(\left[ {{x}^{2}}+\cfrac{1}{2} \right]\) = 0 or 1 as<br /> Sin <sup>-1</sup>\(\left[ {{x}^{2}}+\cfrac{1}{2} \right]\) is defined only for these two values<br /> </p> <p>Hence \(\left[ {{x}^{2}}+\cfrac{1}{2} \right]=\) 0<br /> </p> <p>⇒f (x) = sin <sup>-1</sup> 0 + cos <sup>-1</sup> ( - 1) = \(\pi \left[ {{x}^{2}}+\cfrac{1}{2} \right]=\) 1<br /> ⇒ f (x) = sin -1 (1) + cos -10 = π</p> <p>Therefore range of f (x) = {\(\pi \)}</p>\)
