Minimum integral value of $\alpha$ for which graph of $f(x) = ||x -2| -\alpha|-5$ has exactly four $x-$intercepts-
$6$
$4$
$7$
$5$
Let $f(\theta ) = \sin \theta (\sin \theta + \sin 3\theta )$, then $f(\theta )$
The graph of function $f$ contains the point $P (1, 2)$ and $Q(s, r)$. The equation of the secant line through $P$ and $Q$ is $y = \left( {\frac{{{s^2} + 2s - 3}}{{s - 1}}} \right)$ $x - 1 - s$. The value of $f ‘ (1)$, is
Let function $f(x) = {x^2} + x + \sin x - \cos x + \log (1 + |x|)$ be defined over the interval $[0, 1]$. The odd extensions of $f(x)$ to interval $[-1, 1]$ is
Let $A$ denote the set of all real numbers $x$ such that $x^3-[x]^3=\left(x-[x]^3\right)$, where $[x]$ is the greatest integer less than or equal to $x$. Then,
Suppose $f$ is a function satisfying $f ( x + y )= f ( x )+ f ( y )$ for all $x , y \in N$ and $f (1)=\frac{1}{5}$. If $\sum \limits_{n=1}^m \frac{f(n)}{n(n+1)(n+2)}=\frac{1}{12}$, then $m$ is equal to $...............$.