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(D) Given the equation $\int_{0}^{x} f(t) dt = f^2(x) - 1$. Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus,w

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both $(A)$ and $(C)$

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(D) Given the equation $\int_{0}^{x} f(t) dt = f^2(x) - 1$. Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus,we get: $f(x) = 2f(x)f'(x)$. Assuming $f(x) 
eq 0$,we divide by $f(x)$ to obtain $f'(x) = \frac{1}{2}$. Integrating $f'(x) = \frac{1}{2}$ with respect to $x$,we get $f(x) = \frac{x}{2} + c$. Substituting $x = 0$ into the original equation: $\int_{0}^{0} f(t) dt = f^2(0) - 1 \Rightarrow 0 = f^2(0) - 1 \Rightarrow f(0) = \pm 1$. Thus,$c = \pm 1$,so $f(x) = \frac{x}{2} + 1$ or $f(x) = \frac{x}{2} - 1$. In both cases,$f'(x) = \frac{1}{2} > 0$,so $f$ is monotonically increasing for all $x \in R$. Also,the graph of $f(x) = \frac{x}{2} \pm 1$ is a straight line. Therefore,both $(A)$ and $(C)$ are correct.

List $I$	List $II$
$P.$ The number of polynomials $f(x)$ with non-negative integer coefficients of degree $\leq 2$,satisfying $f(0)=0$ and $\int_0^1 f(x) dx=1$,is	$1.$ $8$
$Q.$ The number of points in the interval $(-\sqrt{13}, \sqrt{13})$ at which $f(x)=\sin(x^2)+\cos(x^2)$ attains its maximum value,is	$2.$ $2$
$R.$ $\int_{-2}^2 \frac{3x^2}{1+e^x} dx$ equals	$3.$ $4$
$S.$ $\frac{\int_{-1/2}^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx}{\int_0^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx}$ equals	$4.$ $0$

$A$ continuous and differentiable function $f$ satisfies the condition $\int_{0}^{x} f(t) dt = f^2(x) - 1$ for all real $x$. Then:

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