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 :

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that the quadratic equation $f(x) \mathrm{m}^2-2 f^{\prime}(x) \mathrm{m}+f^{\prime \prime}(x)=0$ in m , has two equal roots for every $x \in \mathbf{R}$. If $f(0)=1, f^{\prime}(0)=2$, and ( $\alpha, \beta$ ) is the largest interval in which the function $f\left(\log _{\mathrm{e}} x-x\right)$ is increasing, then $\alpha+\beta$ is equal to

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