প্রমান কর যে, (a) $A × \left ( B \cap C \right ) = \left ( A × B \right ) \cap \left ( A × C \right )$ ,যেখানে A, B ও C তিনটি সেট। 

(b)  যদি x=f(t) এবং y= g(t) হলে প্রমাণ কর যে,   $\frac{d^{2} y}{dx^{2}} = \frac{x_{1} y_{2} - y_{1} x_{2}}{x_{1}^{3}}$  

ধরি, $(x, y) \in A \times(B \cap C)$

$\Rightarrow \mathrm{x} \in \mathrm{A}, \mathrm{y} \in(\mathrm{B} \cap \mathrm{C})$

$\Rightarrow \mathrm{x} \in \mathrm{A},(\mathrm{y} \in \mathrm{B}$ এবश $\mathrm{y} \in \mathrm{C})$

$\Rightarrow(x \in A, y \in B)$ এবং $(x \in A, y \in C)$

$\Rightarrow(x, y) \in(A \times B)$ এবং $(x, y) \in(A \times C)$

$\Rightarrow(\mathrm{x}, \mathrm{y}) \in(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})$

$\therefore \mathrm{A} \times(\mathrm{B} \cap \mathrm{C}) \subset(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})$

ধরি, $(x, y) \in(A \times B) \cap(A \times C)$

$\Rightarrow(\mathrm{x}, \mathrm{y}) \in(\mathrm{A} \times \mathrm{B})$ এবং $(\mathrm{x}, \mathrm{y}) \in(\mathrm{A} \times \mathrm{C})$

$\Rightarrow(x \in A, y \in B)$ এবং $(x \in A, y \in C)$

$\Rightarrow \mathrm{x} \in \mathrm{A},(\mathrm{y} \in \mathrm{B}$ এবং $\mathrm{y} \in \mathrm{C})$

$\Rightarrow \mathrm{x} \in \mathrm{A}, \mathrm{y} \in(\mathrm{B} \cap \mathrm{C})$

$\Rightarrow(\mathrm{x}, \mathrm{y}) \in \mathrm{A} \times(\mathrm{B} \cap \mathrm{C})$

$\therefore(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C}) \subset \mathrm{A} \times(\mathrm{B} \cap \mathrm{C})$

$\therefore \mathrm{A} \times(\mathrm{B} \cap \mathrm{C})=(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})$

(b) যদি $\mathrm{x}=\mathrm{f}(\mathrm{t})$ এবং $\mathrm{y}=\mathrm{g}(\mathrm{t})$ হলে প্রমাণ কর যে, $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{x}_{1} \mathrm{y}_{2}-\mathrm{y}_{1} \mathrm{x}_{2}}{\mathrm{x}_{1}{ }^{3}}$

সমাধান: ধরি, $\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{x}_{1} ; \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{y}_{1} ; \frac{\mathrm{dx}_{1}}{\mathrm{dt}}=\mathrm{x}_{2} \otimes \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{y}_{2} \quad \therefore \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}$

আবার, $\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}\right)=\frac{\mathrm{x}_{1} \mathrm{y}_{2}-\mathrm{y}_{1} \mathrm{x}_{2}}{\mathrm{x}_{1}^{2}}$

$\therefore \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{\frac{\mathrm{x}_{1} \mathrm{y}_{2}-\mathrm{y}_{1} \mathrm{x}_{2}}{\mathrm{x}_{1}{ }^{2}}}{\mathrm{x}_{1}}=\frac{\mathrm{x}_{1} \mathrm{y}_{2}-\mathrm{y}_{1} \mathrm{x}_{2}}{\mathrm{x}_{1}{ }^{3}}$

(Proved)