If the population grows at the rate of $5 \%$ per year,then the time taken for the population to become double is (Given $\log 2=0.6912$ ) (in $years$)

$A$ tangent is drawn at any point $P(x, y)$ on a curve,which passes through $(1, 1)$. The tangent cuts the $X$-axis and $Y$-axis at $A$ and $B$ respectively. If $AP:BP = 3:1$,then:

Let $f$ be a twice differentiable non-negative function such that $(f(x))^2 = 25 + \int_{0}^{x} ((f(t))^2 + (f'(t))^2) dt$. Then the mean of $f(\log_e(1)), f(\log_e(2)), \ldots, f(\log_e(625))$ is equal to:

If the population grows at the rate of $8 \%$ per year,then the time taken for the population to be doubled is $\quad$ (given $\log 2=0.6912$ ) (in $years$)

Given $\frac{d^2 y}{d x^2}+\cot x \frac{d y}{d x}+4 y \operatorname{cosec}^2 x=0$. Changing the independent variable $x$ to $z$ by the substitution $z=\log \tan \frac{x}{2}$,the equation is changed to

The bacteria increases at a rate proportional to the number of bacteria present. If the original number $N$ doubles in $4$ hours,then the number of bacteria in $12$ hours will be: (in $N$)