Let $y=f(x)$ be a thrice differentiable function in $(-5,5)$. Let the tangents to the curve $y=f(x)$ at $(1, \mathrm{f}(1))$ and $(3, \mathrm{f}(3))$ make angles $\frac{\pi}{6}$ and $\frac{\pi}{4}$, respectively with positive $x$-axis. If $27 \int_1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3} \quad$ where $\alpha, \quad \beta$ are integers, then the value of $\alpha+\beta$ equals

Let $f$ be a positive function. Let

${I_1} = \int_{1 - k}^k {x\,f\left\{ {x(1 - x)} \right\}} \,dx$, ${I_2} = \int_{1 - k}^k {\,f\left\{ {x(1 - x)} \right\}} \,dx$

when $2k - 1 > 0.$ Then ${I_1}/{I_2}$ is

Let $a, b, c$ be non-zero real numbers such that ; $\int\limits_0^1 {} (1 + cos^8x) (ax^2 + bx + c) dx$ $= \int\limits_0^2 {} (1 + cos^8x) (ax^2 + bx + c) dx$ , then the quadratic equation $ax^2 + bx + c = 0$ has :

The minimum value of the function $f(x)=\int \limits_0^2 e^{|x-t|} d t$ is

The true solution set of the inequality,$\sqrt{5x-6-x^2}+\left( \frac{\pi }{2}\int\limits_{0}^{x}{dz} \right)>x\int\limits_{0}^{\pi }{{{\sin }^{2}}xdx}$ is:

Let $\mathrm{f}$ be a non-negative function in $[0,1]$ and twice differentiable in $(0,1) .$ If $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \,d t=\int \limits_{0}^{x} f(t) \,d t$ $0 \leq x \leq 1$ and $f(0)=0$, then $\lim \limits _{x \rightarrow 0} \frac{1}{x^{2}} \int \limits_{0}^{x} f(t)\, d t:$