The largest interval lying in left( -frac{pi} | vedclass

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(B) The function $f(x) = 4^{-x^2} + \cos^{-1}\left( \frac{x}{2} - 1 \right) + \log(\cos x)$ is defined if all its components are defined.$1$. For $\co

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$\left[ 0, \frac{\pi}{2} \right)$

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(B) The function $f(x) = 4^{-x^2} + \cos^{-1}\left( \frac{x}{2} - 1 \right) + \log(\cos x)$ is defined if all its components are defined.$1$. For $\cos^{-1}\left( \frac{x}{2} - 1 \right)$ to be defined,we must have $-1 \leq \frac{x}{2} - 1 \leq 1$.Adding $1$ to all parts: $0 \leq \frac{x}{2} \leq 2$.Multiplying by $2$: $0 \leq x \leq 4$.$2$. For $\log(\cos x)$ to be defined,we must have $\cos x > 0$.In the interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$,$\cos x > 0$ is always true.$3$. Combining these conditions with the given interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$:We need $x \in [0, 4]$ $AND$ $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$.Since $\frac{\pi}{2} \approx 1.57$,the intersection is $[0, \frac{\pi}{2})$.Thus,the largest interval is $\left[ 0, \frac{\pi}{2} \right)$.

The largest interval lying in $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ for which the function $f(x) = 4^{-x^2} + \cos^{-1}\left( \frac{x}{2} - 1 \right) + \log(\cos x)$ is defined is:

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