The number of solutions of the equation \cos \left( {x + {\pi \over 3}} \right)\cos \left( {{\pi \over 3} - x} \right) = {1 \over 4}{\cos ^2}2x, $$x\in [-3\pi ,3\pi ]$$ is :

Let S = \left\{ {\theta \in [ - \pi ,\pi ] - \left\{ { \pm \,\,{\pi \over 2}} \right\}:\sin \theta \tan \theta + \tan \theta = \sin 2\theta } \right\}. If $$T=\sum _{\theta \in S}\cos 2\theta$$, then T + n(S) is equal to :

The number of solutions of $$|\cos x|=\sin x$$, such that $$-4\pi \le x\le 4\pi$$ is :

Let $$S=\left\{\theta \in [0,2\pi ]:8^{2{\sin }^2\theta }+8^{2{\cos }^2\theta }=16\right\}.$$ Then n(s) + \sum\limits_{\theta \in S}^{} {\left( {\sec \left( {{\pi \over 4} + 2\theta } \right)\cos ec\left( {{\pi \over 4} + 2\theta } \right)} \right)} is equal to:

Let $$S=\left\{\theta \in \left(0,\frac{\pi }{2}\right):\sum _{m=1}^9\sec \left(\theta +(m-1)\frac{\pi }{6}\right)\sec \left(\theta +\frac{m\pi }{6}\right)=-\frac{8}{\sqrt{3}}\right\}$$. Then

The number of elements in the set $$S=\left\{x\in \mathbb{R}:2\cos \left(\frac{x^2+x}{6}\right)=4^x+4^{-x}\right\}$$ is :

The number of ways to distribute 30 identical candies among four children C1, C2, C3 and C4 so that C2 receives at least 4 and at most 7 candies, C3 receives at least 2 and at most 6 candies, is equal to :

The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is :

Let $$\alpha$$ be a root of the equation 1 + x2 + x4 = 0. Then, the value of $$\alpha$$1011 + $$\alpha$$2022 $$-$$ $$\alpha$$3033 is equal to :

Let f(x) be a quadratic polynomial such that f($$-$$2) + f(3) = 0. If one of the roots of f(x) = 0 is $$-$$1, then the sum of the roots of f(x) = 0 is equal to :

The number of distinct real roots of x4 $$-$$ 4x + 1 = 0 is :

Let $$A = \{ x \in R:|x + 1|

Let a, b $$\in$$ R be such that the equation $$a{x}^2-2bx+15=0$$ has a repeated root $$\alpha$$. If $$\alpha$$ and $$\beta$$ are the roots of the equation $$x^2-2bx+21=0$$, then $${\alpha }^2+{\beta }^2$$ is equal to :

The sum of all the real roots of the equation $$(e^{2x}-4)(6e^{2x}-5e^x+1)=0$$ is

The number of distinct real roots of the equation x7 $$-$$ 7x $$-$$ 2 = 0 is

If the sum of the squares of the reciprocals of the roots $$\alpha$$ and $$\beta$$ of the equation 3x2 + $$\lambda$$x $$-$$ 1 = 0 is 15, then 6($$\alpha$$3 + $$\beta$$3)2 is equal to :

Let {S_1} = \left\{ {x \in R - \{ 1,2\} :{{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0} \right\} and $$S_2=\left\{x\in R:3^{2x}-3^{x+1}-3^{x+2}+27\le 0\right\}$$. Then, $$S_1\cup S_2$$ is equal to :

Let S be the set of all integral values of $$\alpha$$ for which the sum of squares of two real roots of the quadratic equation $$3x^2+(\alpha -6)x+(\alpha +3)=0$$ is minimum. Then S :

If $$\alpha ,\beta ,\gamma ,\delta$$ are the roots of the equation $$x^4+x^3+x^2+x+1=0$$, then $${\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{\delta }^{2021}$$ is equal to :

The minimum value of the sum of the squares of the roots of $$x^2+(3-a)x+1=2a$$ is: