JEE MAIN - Mathematics (2007 - No. 4)
The largest interval lying in $$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$ for which the function
$$f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right)$$$$ + \log \left( {\cos x} \right)$$,
is defined, is
$$f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right)$$$$ + \log \left( {\cos x} \right)$$,
is defined, is
$$\left[ { - {\pi \over 4},{\pi \over 2}} \right)$$
$$\left[ {0,{\pi \over 2}} \right)$$
$$\left[ {0,\pi } \right]$$
$$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$
Explanation
$$f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right) + \log \left( {\cos \,x} \right)$$
$$f\left( x \right)$$ is defined if $$ - 1 \le \left( {{x \over 2} - 1} \right) \le 1$$ and $$\cos \,x > 0$$
or $$\,\,\,\,0 \le {x \over 2} \le 2\,\,$$ and $$\,\, - {\pi \over 2} < x < {\pi \over 2}$$
or $$\,\,\,$$ $$0 \le x \le 4$$ $$\,\,$$ and $$\,\, - {\pi \over 2} < x < {\pi \over 2}$$
$$\therefore$$ $$\,\,\,\,\,x\, \in \left[ {0,{\pi \over 2}} \right)$$
$$f\left( x \right)$$ is defined if $$ - 1 \le \left( {{x \over 2} - 1} \right) \le 1$$ and $$\cos \,x > 0$$
or $$\,\,\,\,0 \le {x \over 2} \le 2\,\,$$ and $$\,\, - {\pi \over 2} < x < {\pi \over 2}$$
or $$\,\,\,$$ $$0 \le x \le 4$$ $$\,\,$$ and $$\,\, - {\pi \over 2} < x < {\pi \over 2}$$
$$\therefore$$ $$\,\,\,\,\,x\, \in \left[ {0,{\pi \over 2}} \right)$$
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