The value of $\{x \in R \mid \log_{10} ((1.6)^{1-x^2} - (0.625)^{6(1+x)}) \in R\}$ is

What is the maximum value of the force $F$ such that the block shown in the arrangement does not move (in $N$)? (Given: $m = \sqrt{3} \ kg$,$\mu = \frac{1}{2\sqrt{3}}$,$\theta = 60^\circ$)

$A$ disc of radius $R$ and thickness $\frac{R}{6}$ has a moment of inertia $I$ about an axis passing through its centre and perpendicular to its plane. The disc is melted and recast into a solid sphere. The moment of inertia of the sphere about its diameter is

If $a_k$ is the coefficient of $x^k$ in the expansion of $(1+x+x^2)^n$ for $k=0, 1, 2, \ldots, 2n$,then $a_1+2a_2+3a_3+\ldots+2na_{2n}$ is equal to

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{\cos 3x - \cos x}{x^2}, & \text{for } x \neq 0 \\ \lambda, & \text{for } x = 0 \end{cases}$ and if $f$ is continuous at $x = 0$,then $\lambda$ is equal to

If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^2+x, (x \neq 0)$, then