If the area of the region bounded by the curves, y=x^2,y= 1 x and the lines y=0 and x=t(t>1) is 1 sq. unit, then t is equal to

Let $f(x)=\lim _{\theta \rightarrow 0}\left(\frac{\cos \pi x-x^{\left(\frac{2}{\theta}\right)} \sin (x-1)}{1+x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x \in \mathbf{R}$. Consider the following two statements : (I) $f(x)$ is discontinous at $x=1$. (II) $f(x)$ is continous at $x=-1$. Then,

The area of the region $\{(x, y) : 0 \leq y \leq 6 - x, y^2 \geq 4x - 3, x \geq 0\}$ is:

The value of ∫₀¹ x·eˣ dx is:

Let $[\cdot]$ denote the greatest integer function. Then the value of $\displaystyle\int_0^3 \left(\dfrac{e^x + e^{-x}}{[x]!}\right) dx$ is :

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f\left(\dfrac{x+y}{3}\right) = \dfrac{f(x) + f(y)}{3}$ for all $x, y \in \mathbb{R}$, and $f'(0) = 3$. Then the minimum value of the function $g(x) = 3 + e^x f(x)$, is: