If the Rolle's theorem holds for the function \(f(x) = 2x^3 + ax^2 + bx\) in the interval \([-1, 1 ]\) for the point \(c = \frac{1}{2}\) , then the value of \(2a + b\) is

The domain of \(f(x)=\frac{\log _{(x+1)}(x-2)}{e^{\log _e x}-(2 x+3)}, x \in R\) is

Let \(f : R \to R\) be a function such that \(f(2 - x)\, = f(2 + x)\) and \(f(4 -x)\, = f(4 + x)\), for all \(x \in  R\) and \(\int\limits_0^2 {f\left( x \right)\,dx = 5} \) . Then the value of \(\int\limits_{10}^{50} {f\left( x \right)\,\,dx} \) is

lf the mean deviation of the numbers \(1, 1 + d, . . . ,1 + 100d\) from their mean is \(255,\)  then a value of \(d\) is

Let a line pass through two distinct points \(P(-2,-1,3)\) and \(Q\), and be parallel to the vector \(3 \hat{i}+2 \hat{j}+2 \hat{k}\). If the distance of the point Q from the point \(\mathrm{R}(1,3,3)\) is 5 , then the square of the area of \(\triangle P Q R\) is equal to :

The solution curve, of the differential equation \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}+3=5 \frac{\mathrm{dy}}{\mathrm{dx}}\), passing through the point \((0,1)\) is a conic, whose vertex lies on the line :

If \(A\) denotes the sum of all the coefficients in the expansion of \(\left(1-3 x+10 x^2\right)^n\) and \(B\) denotes the sum of all the coefficients in the expansion of \(\left(1+x^2\right)^n\), then :

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 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 ...........

Let \(C_1\) be the circle in the third quadrant of radius 3 , that touches both coordinate axes. Let \(\mathrm{C}_2\) be the circle with centre \((1,3)\) that touches \(\mathrm{C}_1\) externally at the point \((\alpha, \beta)\). If \((\beta-\alpha)^2=\frac{m}{n}, \operatorname{gcd}(m, n)=1\), then \(m+n\) is equal to :

The equations of the sides \(AB\) and \(AC\) of a triangle \(ABC\) are \((\lambda+1) x +\lambda y =4 \text { and } \lambda x +(1-\lambda) y +\lambda=0\) respectively. Its vertex \(A\) is on the \(y\)-axis and its orthocentre is \((1,2)\). The length of the tangent from the point \(C\) to the part of the parabola \(y^2=6 x\) in the first quadrant is

For \(\alpha, \beta \in \mathrm{R}\) and a natural number \(\mathrm{n}\), let \(A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|\). Then \(2 A_{10}-A_8\)

Let \(a \in R\) and let \(\alpha, \beta\) be the roots of the equation \(x^2+60^{\frac{1}{4}} x+a=0\). If \(\alpha^4+\beta^4=-30\), then the product of all possible values of \(a\) is \(......\)