If the circles \(x^2+y^2+5 k x+2 y+k=0\) and \(2 x^2+2 y^2+2 k x+3 y-1=0, k \in R\) intersect at points \(P\) and \(Q\), then the line \(4 x+5 y-k=0\) passes through \(P\) and \(Q\) for

If the plane \(x-y+z+4=0\) divides the line joining the points \(\mathrm{P}(2,3,-1)\) and \(\mathrm{Q}(1,4,-2)\) in the ratio \(l: m\), then \(l+m\) is

If \(\bar{a}=2 \bar{i}-3 \bar{j}+5 \bar{k}\) and \(\bar{b}=-\bar{i}+3 \bar{j}+3 \bar{k}\) are two vectors, then the vector of magnitude 28 units in the direction of the vector \(\bar{a}-\bar{b}\) is

When the origin is shifted to \((2,3)\) the transformed equation \(x^2+3 x y-2 y^2+17 x-7 y-11=0\), then the original equation of curve is

The real valued function \(f: \mathrm{R} \rightarrow\left[\frac{5}{2}, \infty\right)\) defined by \(f(x)=|2 x+1|+1|x-2|\) is

If \(x=\sec \theta-\cos \theta, y=\sec ^n \theta-\cos ^n \theta\) then \(\frac{d y}{d x}=\)

\(1+\sin x+\sin ^2 x+\sin ^3 x+\ldots+\infty=4+2 \sqrt{3}\) and \(0 \lt x \lt \pi\), \(x \neq \frac{\pi}{2}\) then \(x=\)

If for the reaction \(\mathrm{CCl}_4(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{Cl}(\mathrm{g})\) the following data is given
\(\begin{aligned} & \Delta_{\mathrm{vap}} \mathrm{H}^\theta \mathrm{CCl}_4(\mathrm{l})=30 \mathrm{~kJ} \mathrm{~mol}^{-1} \text {, vap = vaporization } \\ & \Delta_{\mathrm{f}} \mathrm{H}^\theta{ }_{\mathrm{CCl}_4}=-136.0 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{f}=\text { formation } \\ & \Delta_{\mathrm{a}} \mathrm{H}_{\mathrm{C}}^{\mathrm{q}}=714.0 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{a}=\text { atomization } \\ & \Delta_{\mathrm{a}} \mathrm{H}^\theta \mathrm{Cl}_2=242.0 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{a}=\text { atomization }\end{aligned}\)
the bond mean enthalpy of \(\mathrm{C}-\mathrm{Cl}\) in \(\mathrm{CCl}_4(\mathrm{l})\) is

The polydispersity index of a polymer containing 10 molecules with molecular mass \(1.0 \times 10^4\) and 10 molecules with molecular mass \(1.0 \times 10^5\) is approximately.

A cyclist comes to a skidding stop in \(10 \mathrm{~m}\). During this process, the force on the cycle due to the road is \(200 \mathrm{~N}\) and is directly opposite to the motion. How much work does the road do on the cycle?

If the electronic configuration of \(M^{3+}\) is [Xe] \(4 f^3\), then \(M^{3+}\) is

If \(x d y+\left(y+y^2 x\right) d x=0\) and \(y=1\) at \(x=1\). then

Define \(f: R \rightarrow R\) by \(f(x)=\max \{x+1,1-x, 2\}\). Then, \(f\) is