Let $f:[1, \infty) \rightarrow \mathbf{R}$ be a differentiable function defined as $f(x)=\int_1^x f(\mathrm{t}) \mathrm{dt}+(1-x)\left(\log _{\mathrm{e}} x-1\right)+\mathrm{e}$.

Then the value of $f(f(1))$ is :

Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f^{\prime \prime}(x)=g^{\prime \prime}(x)$ for all $x \in \mathbf{R}, f^{\prime}(1)=2 g^{\prime}(1)=4$ and $g(2)=3 f(2)=9$. Then $f(25)-g(25)$ is equal to :

Let $\mathrm{A}=\{1,4,7\}$ and $\mathrm{B}=\{2,3,8\}$. Then the number of elements, in the relation $R=\left\{\left(\left(a_1, b_1\right),\left(a_2, b_2\right)\right) \in((A \times B) \times(A \times B)): a_1+b_2\right.$ divides $\left.a_2+b_1\right\}$ is $\_\_\_\_$ .

From the point $(-1,-1)$, two rays are sent making angles of $45^{\circ}$ with the line $x+y=0$. These rays get reflected from the mirror $x+2 y=1$. If the equations of the reflected rays are $\mathrm{a} x+\mathrm{b} y=9$ and $c x+d y=7, a, b, c, d \in \mathbf{Z}$, then the value of $a d+b c$ is $\_\_\_\_$ .