Verified Solution • 2025–26 Session moderate

QJEEM07Ma5941: JEE Main 2007 Mathematics - Class 11 JEEM07Ma5941 Solution: The largest interval lying in for which the function , is…

The largest interval lying in (−π2,π2)\left( { - {\pi \over 2},{\pi \over 2}} \right) for which the function

f(x)=4−x2+cos⁡−1(x2−1)f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right) +log⁡(cos⁡x)+ \log \left( {\cos x} \right) ,

is defined, is

Multiple Choice Options

  • A [−π4,π2)\left[ { - {\pi \over 4},{\pi \over 2}} \right)
  • B [0,π2)\left[ {0,{\pi \over 2}} \right)
  • C [0,π]\left[ {0,\pi } \right]
  • D (−π2,π2)\left( { - {\pi \over 2},{\pi \over 2}} \right)

Step-by-Step Verified Solution

f(x)=4−x2+cos⁡−1(x2−1)+log⁡(cos⁡ x)f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right) + \log \left( {\cos \,x} \right)

f(x)f\left( x \right) is defined if −1≤(x2−1)≤1- 1 \le \left( {{x \over 2} - 1} \right) \le 1 and \cos \,x > 0

or     0≤x2≤2  \,\,\,\,0 \le {x \over 2} \le 2\,\, and \,\, - {\pi \over 2} < x < {\pi \over 2}

or    \,\,\, 0≤x≤40 \le x \le 4   \,\, and \,\, - {\pi \over 2} < x < {\pi \over 2}

∴\therefore      x ∈[0,π2)\,\,\,\,\,x\, \in \left[ {0,{\pi \over 2}} \right)

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