Match the items of List-I with those of the i | vedclass

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(A) $A)$ Range of $\cos ^2 x \in [0, 1]$.$\Rightarrow 1+\cos ^2 x \in [1, 2]$.$\Rightarrow [1+\cos ^2 x] \in \{1, 2\}$.$\Rightarrow \sec ^{-1}[1+\cos

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$A-V, B-IV, C-I, D-II$

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(A) $A)$ Range of $\cos ^2 x \in [0, 1]$.$\Rightarrow 1+\cos ^2 x \in [1, 2]$.$\Rightarrow [1+\cos ^2 x] \in \{1, 2\}$.$\Rightarrow \sec ^{-1}[1+\cos ^2 x] \in \{\sec ^{-1} 1, \sec ^{-1} 2\}$. Thus, $A-V$.$B)$ Given $f(x+\frac{1}{x}) = x^2+\frac{1}{x^2} = (x+\frac{1}{x})^2 - 2$.Let $t = x+\frac{1}{x}$. Since $|x+\frac{1}{x}| \ge 2$, the domain of $f(t)$ is $(-\infty, -2] \cup [2, \infty)$. However, if we consider the function definition $f(t) = t^2-2$, the domain is $R$. Given the context of such problems, $f(x) = x^2-2$ is defined for all $x \in R$. Thus, $B-IV$.$C)$ $f(x+y) = f(x)+f(y)$ is Cauchy's functional equation, so $f(x) = kx$.Given $f(1) = 5$, we have $k(1) = 5 \Rightarrow k = 5$.So $f(x) = 5x$, which is an odd function since $f(-x) = 5(-x) = -f(x)$. Thus, $C-I$.$D)$ $\sin ^{-1} x - \cos ^{-1} x + \sin ^{-1}(1-x) = 0$.For $x=0$: $\sin ^{-1}(0) - \cos ^{-1}(0) + \sin ^{-1}(1) = 0 - \frac{\pi}{2} + \frac{\pi}{2} = 0$. (Correct)For $x=\frac{1}{2}$: $\sin ^{-1}(\frac{1}{2}) - \cos ^{-1}(\frac{1}{2}) + \sin ^{-1}(\frac{1}{2}) = \frac{\pi}{6} - \frac{\pi}{3} + \frac{\pi}{6} = 0$. (Correct)Thus, $D-II$.

List-$I$	List-$II$
$A$. Range of $\sec ^{-1}\left[1+\cos ^2 x\right]$, where $[.]$ denotes the greatest integer function	$I$. Odd function
$B$. Domain of $f(x)$ where $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$	$II$. $\left\{0, \frac{1}{2}\right\}$
$C$. $f(x+y)=f(x)+f(y) ; f(1)=5$	$III$. $\left\{\sec ^{-1} 5, \sec ^{-1} 4\right\}$
$D$. $\sin ^{-1} x-\cos ^{-1} x+\sin ^{-1}(1-x)=0 \Rightarrow x \in$	$IV$. $R$
	$V$. $\left\{\sec ^{-1} 1, \sec ^{-1} 2\right\}$

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