The $x$-coordinate changes on the curve $y=3 x^5+15 x-8$ at the rate of $\frac{1}{5}$ units/sec. $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ are the points on the curve at which the $y$-coordinate changes at the rate of 6 units/sec, then the slope of $A B=$
In $\triangle A B C, \angle B=90^{\circ}$ and $(b+a)$ is always a constant. In order that $\triangle A B C$ encloses the maximum area, $\angle C=$
If $\int \frac{(x-1) d x}{(x+1) \sqrt{x^3+x^2+x}}=A \cdot \tan ^{-1} \sqrt{f(x)}+$ constant, then the ordered pair $(A, f(-1))=$
If $f\left(\frac{2 x+3}{3 x+5}\right)=x+4, x \neq \frac{-5}{3}, \frac{2}{3}$ and $\int f(x) d x=A x+B \ln |3 x-2|+C$, then $3 B-A=$
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