Let \({f_k}\left( x \right) = \frac{1}{k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)\;,x \in R\) and \(k \ge 1\), then \({f_4}\left( x \right) - {f_6}\left( x \right)\) is equal to

The  the circle passing through the foci of the \(\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1\) and having centre at \((0,3) \) is

Two parabolas have the same focus \((4,3)\) and their directrices are the \(x\)-axis and the \(y\)-axis, respectively. If these parabolas intersects at the points \(A\) and \(B\), then \((A B)^2\) is equal to :

The image of the line \(\frac{{x - 1}}{3} = \frac{{y - 3}}{1} = \frac{{z - 4}}{{ - 5}}\) in the plane \(2x - y + z + 3 = 0\)  is the line :

The value of \(\int\limits_{0}^{2 \pi} \frac{x \sin ^{8} x}{\sin ^{8} x+\cos ^{8} x} d x\) is equal to

For a triangle ABC, let \( \vec{p}=\vec{BC}, \vec{q}=\vec{CA} \) and \( \vec{r}=\vec{BA} \). If \( |\vec{p}|=2\sqrt{3}, |\vec{q}|=2 \) and \( cos\hat{\theta}=\frac{1}{\sqrt{3}} \) where θ is the angle between \( \vec{P} \) and \( \vec{q} \) then \( |\vec{p}\times(\vec{q}-3\vec{r})|^{2}+3|\vec{r}|^{2} \) is equal to :

Let \(P\) be a variable point on the parabola \(y=4 x^{2}+1\). Then the locus of the mid-point of the point \(P\) and the foot of the perpendicular drawn from the point \(P\) to the line \(y=x\) is:

The distance of the point \((1,1,9)\) from the point of intersection of the line \(\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}\) and the plane \(x+y+z=17\) is

Let a conic \(\mathrm{C}\) pass through the point \((4,-2)\) and \(\mathrm{P}(\mathrm{x}, \mathrm{y}), \mathrm{x} \geq 3\), be any point on \(\mathrm{C}\). Let the slope of the line touching the conic \(\mathrm{C}\) only at a single point \(\mathrm{P}\) be half the slope of the line joining the points \(P\) and \((3,-5)\). If the focal distance of the point \((7,1)\) on \(C\) is \(d\), then \(12 \mathrm{~d}\) equals ...........

Statement \(-1\) : The system of linear equations \(x + \left( {\sin \,\alpha } \right)y + \left( {\cos \,\alpha } \right)z = 0\) \(x + \left( {\cos \,\alpha } \right)y + \left( {\sin \alpha } \right)z = 0\) \(x - \left( {\sin \,\alpha } \right)y - \left( {\cos \alpha } \right)z = 0\) has a non-trivial solution for only one value of \(\alpha \) lying in the interval \(\left( {0\,,\,\frac{\pi }{2}} \right)\)  Statement \(-2\) : The equation in \(\alpha \) \(\left| {\begin{array}{*{20}{c}}
  {\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\ 
  {\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\ 
  {\cos {\mkern 1mu} \alpha }&{ - \sin {\mkern 1mu} \alpha }&{ - \cos {\mkern 1mu} \alpha } 
\end{array}} \right| = 0\) has only one solution lying in the interval \(\left( {0\,,\,\frac{\pi }{2}} \right)\)

Let \(\overrightarrow{\mathrm{OA}}=2 \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=6 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{b}}\) and \(\overrightarrow{\mathrm{OC}}=3 \overrightarrow{\mathrm{b}}\), where \(O\) is the origin. If the area of the parallelogram with adjacent sides \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{OC}}\) is \(15\) sq. units, then the area (in sq. units) of the quadrilateral \(\mathrm{OABC}\) is equal to :

The locus of the centres of the circles, which touch the circle, \(x^2 + y^2 = 1\) externally, also touch the \(y-\) axis and lie in the first quadrant is

Let \(A=\left[a_{i j}\right], a_{i j} \in Z \cap[0,4], 1 \leq i, j \leq 2\). The number of matrices \(A\) such that the sum of all entries is a prime number \(p \in(2,13)\) is \(........\).