If the domain of the function $f(x)=\cos ^{-1}\left(\frac{2 x-5}{11-3 x}\right)+\sin ^{-1}\left(2 x^2-3 x+1\right)$ is the interval $[\alpha, \beta]$, then $\alpha+2 \beta$ is equal to :

Let O be the vertex of the parabola $x^2=4 y$ and Q be any point on it. Let the locus of the point P , which divides the line segment OQ internally in the ratio $2: 3$ be the conic C . Then the equation of the chord of $C$, which is bisected at the point $(1,2)$, is :

Let $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$ be vectors such that $|\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|=\sqrt{29}$ and $\overrightarrow{\mathrm{c}} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \overrightarrow{\mathrm{d}}$. If $\lambda_1, \lambda_2\left(\lambda_1>\lambda_2\right)$ are the possible values of $(\vec{c}+\vec{d}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$, then the equation $\mathrm{K}^2 x^2+\left(\mathrm{K}^2-5 \mathrm{~K}+\lambda_1\right) x y+\left(3 \mathrm{~K}+\frac{\lambda_2}{2}\right) y^2-8 x+12 y+\lambda_2=0$ represents a circle, for K equal to :

Let $f: \mathbf{R} \rightarrow(0, \infty)$ be a twice differentiable function such that $f(3)=18, f^{\prime}(3)=0$ and $f^{\prime \prime}(3)=4$.

Then $\lim\limits _{x \rightarrow 1}\left(\log _e\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^2}}\right)$ is equal to :