Let \(f\) be a differentiable function satisfying \(f ( x )=\frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} f \left(\frac{\lambda^{2} x }{3}\right) d \lambda, x >0\) and \(f (1)=\sqrt{3}\). If \(y=f(x)\) passes through the point \((\alpha, 6)\), then \(\alpha\) is equal to \(.........\)

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three units vectors such that \(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0} .\) If \(\lambda=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}} \) and \(\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}},\) then the ordered pair \((\lambda, {\mathrm{\vec d}})\) is equal to 

Let \(p(x)\) be a quadratic polynomial such that \(p(0)= 1\) . If \(p(x)\) leaves remainder \(4\) when divided by \(x-1\) and it leaves remainder \(6\) when divided by \(x+ 1\) ; then

The product of all possible values of \(\alpha\), for which \(\displaystyle\lim_{x \to 0}\left(\dfrac{1 - \cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}\right) = 2\), is:

The probability distribution of a random variable X is given below:

X	\(4k\)	\(\frac{30}{7} k\)	\(\frac{32}{7} k\)	\(\frac{34}{7} k\)	\(\frac{36}{7} k\)	\(\frac{38}{7} k\)	\(\frac{40}{7} k\)	\(6k\)
P(X)	\(\frac{2}{15}\)	\(\frac{1}{15}\)	\(\frac{2}{15}\)	\(\frac{1}{5}\)	\(\frac{1}{15}\)	\(\frac{2}{15}\)	\(\frac{1}{5}\)	\(\frac{1}{15}\)

If \(E ( X )=\frac{263}{15}\), then \(P ( X <20)\) is equal to :

If the coefficients of the middle terms in the binomial expansions of \((1 + \alpha x)^{26}\) and \((1 - \alpha x)^{28}\), \(\alpha \neq 0\), are equal, then the value of \(\alpha\) is:

A value of \(\theta  \in  (0, \pi /3)\), for which \(\left| {\begin{array}{*{20}{c}}
  {1 + {{\cos }^2}\,\theta }&{{{\sin }^2}\,\theta }&{4\,\cos \,6\theta } \\ 
  {{{\cos }^2}\,\theta }&{1 + {{\sin }^2}\,\theta }&{4\,\cos \,6\theta } \\ 
  {{{\cos }^2}\,\theta }&{{{\sin }^2}\,\theta }&{1 + 4\,\cos \,6\theta } 
\end{array}} \right| = 0\), is

Let \(A\,(4, -4)\) and \(B\,(9,6)\) be points on the parabola \(y^2 = 4x\) . Let \(C\) be chosen on the arc \(AOB\) of the parabola, where \(O\) is the origin, such that the area of \(\Delta ACB\) is maximum. Then, the area (in sq. units) of \(\Delta ACB,\) is

The sum of the solutions of the equation \(\left| {\sqrt x  - 2} \right| + \sqrt x \left( {\sqrt x  - 4} \right) + 2 = 0\left( {x > 0} \right)\) is equal to

The number of solutions of the equation \(\cos 2 \theta \cos \frac{\theta}{2}+\cos \frac{5 \theta}{2}=2 \cos ^3 \frac{5 \theta}{2}\) in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) is :

In a tournament, a team plays \(10\) matches with probabilities of winning and losing each match as \(\frac{1}{3}\) and \(\frac{2}{3}\) respectively. Let \(x\) be the number of matches that the team wins, and \(y\) be the number of matches that team loses. If the probability \(\mathrm{P}(|\mathrm{x}-\mathrm{y}| \leq 2)\) is \(\mathrm{p}\), then \(3^9 \mathrm{p}\) equals ...........

If \(\frac{{dy}}{{dx}} + \frac{3}{{{{\cos }^2}\,x}}\,y = \frac{1}{{{{\cos }^2}\,x}},\) \(x \in \left( {\frac{{ - \pi }}{3},\frac{\pi }{3}} \right)\) and \(y\left( {\frac{\pi }{4}} \right) = \frac{4}{3}\), then \(y\left( { - \frac{\pi }{4}} \right)\) equals

Let \(z \in C\) with \(Im(z) = 10\) and it satisfies \(\frac{{2z - n}}{{2z + n}} = 2i - 1\) for some natural number \(n\). Then