Left $$f$$ be an odd function defined on the set of real numbers such that for $$x\geq 0, f(x)=3\sin x+4\cos x$$. Then $$f(x)$$ at $$x= \displaystyle -\frac{11\pi}{6}$$ is equal to :
$$f(x) = 3 \sin x + 4 \cos x $$ $$x \geq 0$$
$$f(x) =5 \sin (x+\alpha)$$
$$\alpha = \sin^{-1}\left ( \dfrac{4}{5} \right )$$
$$f\left
( \dfrac{-11 \pi}{6}\right ) = -f \left ( \dfrac{11\pi}{6} \right ) =-[3
\sin \left ( \dfrac{11 \pi}{6}\right ) + 4 \cos \left ( \dfrac{11 \pi}{6}
\right )]$$
$$=-\left [ 3 \sin (2 \pi - \dfrac{\pi}{6})+4 \cos (2\pi - \dfrac{\pi}{6})\right ]$$
$$=-\left [ -\dfrac{3}{2}+2\sqrt{3} \right ]=\dfrac{3}{2} - 2 \sqrt{3}$$
Let $$f$$ be an injective function with domain $$\{x, y, z\}$$ and range $$\{1, 2, 3\}$$ such that exactly one of the following statements is correct and the remaining are false $$f(x) =1, f(y) \neq 1, f(z) \neq$$ 2. The value of $$f^{-1}(1)$$ is:
Let $$ f_{k}(x)=\displaystyle \frac{1}{k}(\sin ^{k}x+\cos ^{k}x)$$ where $$ x\in R$$ and $$ k\geq 1$$. Then $$ f_{4}(x)-f_{6}(x)$$ equals
Let $$f:\left( 1,3 \right) \rightarrow R$$ be a function defined by $$f(x)=\cfrac { x\left[ x \right] }{ 1+{ x }^{ 2 } } $$, where $$[x]$$ denotes the greatest integer $$\le x$$. Then the range of $$f$$ is:
For $$x\in R, x\neq 0, x\neq 1$$, let $$f_{0}(x) = \dfrac {1}{1 - x}$$ and $$f_{n + 1} (x) = f_{0} (f_{n}(x)), n = 0, 1, 2, ....$$ Then the value of $$f_{100}(3) + f_{1}\left (\dfrac {2}{3}\right ) + f_{2} \left (\dfrac {3}{2}\right )$$ is equal to:
If $$f(x) = \dfrac {2^{x} + 2^{-x}}{2}$$, then $$f (x + y)\cdot f (x - y)$$ is
Let $$A = \left \{x_{1}, x_{2}, x_{3}, ...., x_{7}\right \}, B = \left \{y_{1}, y_{2}, y_{3}\right \}$$. The total number of functions $$f : A\rightarrow B$$ that are on to and there are exactly three element $$x$$ in $$A$$ such that $$f(x) = y_{2}$$ is equal to