Let $$\mathrm{f}(x)$$ be positive for all real $$x$$. If $$\mathrm{I}_1=\int_\limits{1-\mathrm{h}}^{\mathrm{h}} x \mathrm{f}(x(1-x)) \mathrm{d} x$$ and $$\mathrm{I}_2=\int_\limits{1-\mathrm{h}}^{\mathrm{h}} \mathrm{f}(x(1-x)) \mathrm{d} x$$, where $$(2 h-1)>0$$, then $$\frac{I_1}{I_2}$$ is
The mirror image of the point $$(1,2,3)$$ in a plane is $$\left(-\frac{7}{3},-\frac{4}{3},-\frac{1}{3}\right)$$. Thus, the point _________ lies on this plane.
If $$\alpha=3 \sin ^{-1} \frac{6}{11}$$ and $$\beta=3 \cos ^{-1}\left(\frac{4}{9}\right)$$, where the inverse trigonometric functions take only the principal values, then the correct option is
$$\text { The value of } \sec ^2\left(\tan ^{-1} 2\right)+\operatorname{cosec}^2\left(\cot ^{-1} 3\right) \text { is }$$