If $$\mathrm{f}(x)=\int \frac{x^2 \mathrm{~d} x}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)}$$ and $$\mathrm{f}(0)=0$$, then $$\mathrm{f}(1)$$ is

A line $$\mathrm{L}_1$$ passes through the point, whose p. v. (position vector) $$3 \hat{i}$$, is parallel to the vector $$-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$$. Another line $$\mathrm{L}_2$$ passes through the point having p.v. $$\hat{i}+\hat{j}$$ is parallel to vector $$\hat{i}+\hat{k}$$, then the point of intersection of lines $$L_1$$ and $$L_2$$ has p.v.

If $$y=\tan ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right)$$, then $$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)$$ at $$x=0$$ is

The expression $$(p \wedge \sim q) \vee q \vee(\sim p \wedge q)$$ is equivalent to